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    Download
    00:00:01
    to look at examples of finding the
    00:00:03
    derivative and integral of a power
    00:00:05
    series let's get to it
    00:00:09
    all right so here's our first example we
    00:00:12
    want to find the radius and interval of
    00:00:14
    convergence for f of x f Prime of X and
    00:00:17
    the integral of f of x DX okay and we
    00:00:20
    are given f of x f of x is equal to the
    00:00:23
    sum from n equals 1 to Infinity of x
    00:00:26
    divided by 3 raised to the power of n
    00:00:28
    and so this right here is our power
    00:00:31
    series that we are going to want to find
    00:00:33
    the derivative and the integral of all
    00:00:36
    right so if f of x equals this power
    00:00:38
    series then we need to find F Prime of X
    00:00:41
    the derivative of this power series and
    00:00:43
    the integral of f of x DX which is the
    00:00:46
    integral of this power series now
    00:00:48
    finding the derivative and integral of a
    00:00:50
    power series is fairly simple if you
    00:00:52
    have a power series of this form this is
    00:00:55
    the general form of a power Series right
    00:00:57
    we have the sum from n equals 0 to
    00:00:59
    Infinity of a sequence a sub n times the
    00:01:02
    quantity x minus C to the power of n
    00:01:04
    what you can do if you have a power
    00:01:06
    Series in this form is find the
    00:01:08
    derivative and insert role by using the
    00:01:11
    power rule for derivatives and the power
    00:01:13
    rule for integration right so for the
    00:01:16
    derivative what we will do is multiply
    00:01:18
    the exponent down for the quantity that
    00:01:21
    involves X and then subtract 1 from that
    00:01:23
    exponent right so we'll multiply n down
    00:01:25
    so we have n times a sub n times the
    00:01:28
    quantity x minus C to the power of n
    00:01:30
    minus 1. we subtract 1 from that
    00:01:33
    exponent and after you do that after you
    00:01:35
    multiply down n and subtract one from
    00:01:37
    the exponent you have the derivative of
    00:01:40
    your power series you will leave
    00:01:42
    everything else alone everything else in
    00:01:44
    the power series meaning a sub n
    00:01:46
    whatever is not part of the quantity
    00:01:49
    Where you have x minus C to the power of
    00:01:51
    n that is going to remain unchanged the
    00:01:54
    only part that is affected by taking the
    00:01:56
    derivative with respect to X of the
    00:01:59
    power series is the quantity involving x
    00:02:02
    minus C all right now to find the
    00:02:04
    integral of a power series we will do a
    00:02:06
    similar process except instead of using
    00:02:09
    the power rule for derivatives we will
    00:02:11
    use the power rule for integration so
    00:02:13
    instead of subtracting one from the
    00:02:15
    exponent we are going to add one to the
    00:02:16
    exponent so x minus C to the power of n
    00:02:19
    will become x minus C to the power of n
    00:02:22
    plus one and then we will divide by that
    00:02:25
    new exponent of n plus one okay and then
    00:02:29
    everything else will remain the same
    00:02:31
    just like when we take a derivative of a
    00:02:33
    power series a sub n is not going to be
    00:02:36
    changed in any way the only part of the
    00:02:39
    power series that is affected by the
    00:02:40
    integration process is the quantity
    00:02:42
    Where you have x minus C the exponent of
    00:02:45
    that quantity increases by one and then
    00:02:47
    of course you need to divide everything
    00:02:49
    in your power series by that new
    00:02:51
    exponent of n plus 1. okay nothing else
    00:02:55
    needs to change and then one more note
    00:02:57
    about integration is that because we are
    00:03:00
    integrating a function you do need to
    00:03:02
    include a plus c that constant of
    00:03:05
    integration and so while it's important
    00:03:07
    to remember that constant it actually
    00:03:09
    doesn't not affect the interval or
    00:03:12
    radius of convergence of the integral of
    00:03:14
    the power Series in any way right
    00:03:16
    because that plus C that constant is on
    00:03:19
    the outside of the power series and so
    00:03:22
    it has no real effect on the radius or
    00:03:24
    interval of convergence okay and so
    00:03:26
    that's how you find the derivative and
    00:03:28
    integral of a power Series so if we go
    00:03:30
    back to our example let's start by
    00:03:33
    finding the derivative F Prime of X of
    00:03:35
    our function and then we'll find the
    00:03:37
    integral of our function and then we'll
    00:03:39
    work on finding the radius and interval
    00:03:41
    of convergence for all three of those
    00:03:44
    functions all right and now before we
    00:03:46
    find F Prime of X I want to rewrite this
    00:03:48
    power series a little bit we can rewrite
    00:03:51
    it to have both the numerator and
    00:03:53
    denominator to the power of n that's
    00:03:55
    going to make it a little bit easier for
    00:03:57
    us to see how to take the derivative and
    00:03:59
    how to take the integral so what I'm
    00:04:01
    going to do is rewrite this to be x to
    00:04:03
    the power of n divided by 3 to the power
    00:04:05
    of n all right so let me do that we will
    00:04:08
    have x to the power of n divided by 3 to
    00:04:11
    the power of n and now let's take the
    00:04:13
    first derivative of that function we
    00:04:16
    will have the sum from n equals 1 to
    00:04:19
    Infinity of the derivative of x to the
    00:04:21
    power of n divided by 3 to the power of
    00:04:24
    n and remember we're taking this
    00:04:26
    derivative with respect to X so we're
    00:04:28
    going to multiply the exponent of X down
    00:04:30
    and then subtract 1 from that exponent
    00:04:33
    and then everything else is going to
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    remain the same we will not change this
    00:04:38
    3 to the power of n in the denominator
    00:04:40
    that is going to stay there and not
    00:04:42
    change in any way so we will have n
    00:04:45
    times x to the power of n minus 1
    00:04:48
    divided by 3n and that is going to be
    00:04:50
    the derivative of our function okay and
    00:04:54
    so now we can work on finding the
    00:04:55
    integral the integral of f of x DX will
    00:04:59
    be equal to the sum from n equals 1 to
    00:05:02
    Infinity of the integral of x to the
    00:05:05
    power of n divided by 3 to the power of
    00:05:07
    n now once again 3 to the power of n is
    00:05:09
    not going to change the only part that
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    is affected by the integration process
    00:05:14
    is this x to the power of n and so if we
    00:05:17
    integrate that we're going to add 1 to
    00:05:19
    the exponent and then divide it by that
    00:05:22
    new exponent which will be n plus 1. so
    00:05:24
    the integral will be x to the power of n
    00:05:27
    plus 1 divided by 3 to the power of n
    00:05:30
    times n plus 1. all right because we're
    00:05:34
    dividing by that new exponent 3 to the
    00:05:37
    power of n will be multiplied by n plus
    00:05:39
    1 because that needs to be in the
    00:05:41
    denominator of this series all right and
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    then one more thing don't forget that
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    when you are integrating a function you
    00:05:48
    need to add that constant of plus C so
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    in your own work make sure to include C
    00:05:54
    plus that integral because that would be
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    part of the integral of the original
    00:05:59
    Power series however I'm going to leave
    00:06:01
    that out because as I explained earlier
    00:06:03
    this constant of integration is not
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    going to have any effect on the radius
    00:06:08
    and interval of convergence for this
    00:06:10
    power series all right so do remember to
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    include the plus C or if you want to
    00:06:15
    write C plus before this series for your
    00:06:17
    integral but I'm going to ignore it for
    00:06:20
    now so that we can focus on the radius
    00:06:22
    and interval of convergence which are
    00:06:24
    not affected by that constant okay and
    00:06:26
    so now we have successfully found the
    00:06:28
    derivative and the integral of our
    00:06:31
    original function and so now what we
    00:06:33
    want to do is determine the radius and
    00:06:36
    interval of convergence for each of
    00:06:38
    these three functions now something
    00:06:40
    that's important to know about a power
    00:06:42
    series and its derivative and its
    00:06:44
    integral is that all three of those
    00:06:46
    functions or all three of those power
    00:06:48
    series are going to have the same radius
    00:06:51
    of convergence and they're also going to
    00:06:54
    have a very similar interval of
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    convergence the only difference is going
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    to be whether the endpoints for that
    00:07:01
    interval are included or Not Included
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    the derivative or the integral could
    00:07:06
    gain or lose an end point for the that
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    interval of convergence and so what's
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    going to happen here is we're only going
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    to need to find the radius of
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    convergence one time we're going to find
    00:07:18
    it for the original Power series and
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    then we're going to use that information
    00:07:22
    to find the interval of convergence for
    00:07:25
    all three of them all right so what we
    00:07:28
    want to do here is use the ratio test
    00:07:30
    for our original Power series find our
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    radius of convergence and then our
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    interval of convergence as well all
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    right and so remember how the ratio test
    00:07:39
    Works we're going to be taking the limit
    00:07:41
    as n approaches Infinity of the absolute
    00:07:44
    value of a sub n plus 1 divided by a sub
    00:07:47
    n and that will be for our original
    00:07:50
    Power series okay now because we have a
    00:07:53
    rational expression for a sub n what I'm
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    going to do instead of writing out a
    00:07:57
    very complex fraction of a fraction
    00:07:59
    divided by a fraction is I'm going to
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    write out a sub n plus 1 and then I'm
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    going to multiply it by the reciprocal
    00:08:05
    of a sub n and that will be the same as
    00:08:08
    dividing by that fraction action all
    00:08:10
    right and so let me show you what I mean
    00:08:11
    this will be equal to the Limit as n
    00:08:14
    approaches Infinity all the absolute
    00:08:16
    value of a sub n plus 1 which is going
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    to be this expression right here but
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    each of these ends will become n plus
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    one so we'll have x to the power of n
    00:08:25
    plus one divided by three to the power
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    of n plus one and that will be
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    multiplied by the reciprocal of a sub n
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    which is the same as dividing by a sub n
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    so we're going to flip the numerator and
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    denominator and have 3 to the power of n
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    divided by x to the power of n okay and
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    so now we can simplify and try to
    00:08:44
    evaluate this limit by first splitting
    00:08:47
    up these powers for x and three x to the
    00:08:51
    power of n plus 1 will be x to the power
    00:08:53
    of n times x to the first Power and 3 to
    00:08:56
    the power of n Plus One will be 3 to the
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    power of n times three to the power of
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    one so we'll have that this is equal to
    00:09:02
    the Limit as n approaches Infinity of
    00:09:06
    the absolute value of x to the power of
    00:09:08
    n times x to the first Power times three
    00:09:11
    to the power of n divided by three to
    00:09:13
    the power of n times three to the first
    00:09:15
    Power Times x to the power of n all
    00:09:18
    right now in the numerator we have x to
    00:09:19
    the power of n and we also have that in
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    the denominator and so those are going
    00:09:23
    to cancel out and then we also have 3 to
    00:09:26
    the power of n in the numerator and
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    denominator as well and so those are
    00:09:30
    going to cancel out and what that's
    00:09:31
    going to leave us with is the limit as n
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    approaches Infinity of the absolute
    00:09:37
    value of x divided by 3. okay and now
    00:09:40
    remember we're looking at the limit as n
    00:09:42
    approaches infinity and we no longer
    00:09:45
    have any n's in this expression so
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    really what we have here is a limit of a
    00:09:50
    constant now X is a variable but in
    00:09:53
    terms of n this limit is the limit of a
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    constant because there's no ends within
    00:09:59
    this expression so this limit will just
    00:10:01
    be equal to the absolute value of x
    00:10:03
    divided by 3. all right and we could
    00:10:05
    rewrite that to be one-third times the
    00:10:08
    absolute value of x because because 1 3
    00:10:10
    is always going to be positive so we
    00:10:12
    don't need the absolute value bars
    00:10:14
    around that and now this is the value of
    00:10:17
    this limit but from the ratio test we
    00:10:19
    know that when the value of this limit
    00:10:21
    is less than one that is when our series
    00:10:25
    converges and so to find the interval of
    00:10:27
    convergence we need to set this result
    00:10:30
    to be less than one all right now if we
    00:10:33
    multiply both sides by 3 we'll have that
    00:10:35
    the absolute value of x is less than 3
    00:10:38
    and now from this statement we can
    00:10:41
    determine the radius of convergence for
    00:10:44
    all three power series or all three
    00:10:47
    functions all right whenever you have
    00:10:49
    the absolute value of x or the absolute
    00:10:51
    value of x minus C in this case C would
    00:10:54
    be zero since that is less than 3 we
    00:10:57
    know that the radius of convergence is
    00:11:00
    three okay so from that statement right
    00:11:02
    there we know that the radius is equal
    00:11:04
    to 3 and that will be the radius for the
    00:11:07
    original Power series the derivative of
    00:11:09
    the power series and the integral of the
    00:11:11
    power series okay and then for the
    00:11:14
    interval of convergence remember that
    00:11:15
    the absolute value of x Less Than 3
    00:11:18
    would be the same as saying this that X
    00:11:21
    is less than 3 but greater than negative
    00:11:24
    three all right that's what the absolute
    00:11:26
    value of x less than three is implying
    00:11:29
    the values of X for which our series is
    00:11:32
    convergent lies between negative three
    00:11:34
    and three and that is going to be the
    00:11:37
    case for all three of these functions or
    00:11:39
    all three of these power series what's
    00:11:41
    going to differ between the three series
    00:11:44
    or between the three functions is
    00:11:46
    whether or not these endpoints are
    00:11:48
    included within that interval of
    00:11:51
    convergence and so now if I clean up my
    00:11:53
    work we know that the radius of
    00:11:54
    convergence is three for all three of
    00:11:56
    those series or all three of those
    00:11:58
    functions but our interval of
    00:12:00
    convergence will be some form of
    00:12:02
    negative three to three right we're
    00:12:04
    going to have to test these endpoints
    00:12:06
    for all three of the series to see
    00:12:09
    whether their interval of convergence
    00:12:11
    should either include or not include
    00:12:13
    negative 3 and 3. okay and so let's
    00:12:16
    start with our original series let's
    00:12:19
    start by checking the endpoints for f of
    00:12:21
    x so when x equals negative 3 our series
    00:12:25
    will look like this we will have the sum
    00:12:27
    from n equals 1 to Infinity of negative
    00:12:30
    three to the power of n divided by 3 to
    00:12:33
    the power of n right we just replaced x
    00:12:36
    with negative three and now we could
    00:12:38
    rewrite this to be the sum from n equals
    00:12:40
    1 to Infinity of negative three divided
    00:12:43
    by three to the power of n because both
    00:12:45
    the numerator and denominator are raised
    00:12:48
    to that power of n so we could just take
    00:12:49
    that whole fraction to that power and
    00:12:52
    negative 3 divided by three is equal to
    00:12:54
    negative one so if we rewrite that we'll
    00:12:57
    have the sum from n equals 1 to Infinity
    00:12:59
    of negative 1 to the power of n and this
    00:13:03
    right here is a Divergent series right
    00:13:06
    we just have the sum from n equals 1 to
    00:13:09
    Infinity of negative 1 to the power of n
    00:13:11
    which will alternate between being
    00:13:13
    negative 1 and positive one the terms of
    00:13:15
    this series are going to be 1 negative 1
    00:13:18
    1 negative 1 1 and so the sum will never
    00:13:21
    converge to one specific value which
    00:13:24
    means that this series diverges
    00:13:27
    all right and so because our power
    00:13:28
    series diverges when x equals negative
    00:13:31
    three we will not include that in our
    00:13:34
    interval of convergence we will leave
    00:13:36
    that bracket open and we will not change
    00:13:38
    it to a closed bracket all right we're
    00:13:41
    not going to include negative 3 in the
    00:13:43
    interval of convergence okay but what
    00:13:45
    about positive 3 what about when x
    00:13:48
    equals positive 3 well when x equals
    00:13:51
    three our series will look like this
    00:13:53
    we'll have the sum from n equals one to
    00:13:55
    Infinity of 3 to the power of n divided
    00:13:58
    by 3 to the power of n right we just
    00:14:00
    replaced x with 3 and so we have 3 to
    00:14:03
    the power of n divided by itself which
    00:14:05
    is actually equal to one and so we have
    00:14:08
    the sum from n equals 1 to Infinity of
    00:14:11
    one and this series is just a series
    00:14:13
    where every single term is positive one
    00:14:16
    so we have One Plus One Plus One Plus
    00:14:18
    One Forever And so the sum of that
    00:14:21
    series will not converge to a specific
    00:14:22
    value either in fact you could check
    00:14:25
    this one with the Divergence test by
    00:14:27
    taking the limit as n approaches
    00:14:29
    Infinity of the sequence from your
    00:14:32
    series which in this case is 1 and the
    00:14:34
    limit as n approaches Infinity of 1 will
    00:14:37
    just be one because the limit of a
    00:14:39
    constant is just that constant and one
    00:14:41
    is not equal to zero and so from the
    00:14:43
    Divergence test we know that this series
    00:14:46
    diverges and so our power series also
    00:14:50
    diverges when x equals positive three
    00:14:53
    and so we also will not include positive
    00:14:56
    three for our interval of convergence
    00:14:58
    this interval right here negative three
    00:15:00
    to three not including the endpoints is
    00:15:03
    the interval of convergence for this
    00:15:06
    power series neither of the endpoints
    00:15:08
    are going to be included okay so that's
    00:15:11
    it for the original series but now we
    00:15:13
    need to check the derivative and then we
    00:15:15
    will check the integral so let's start
    00:15:17
    by checking x equals negative three our
    00:15:20
    series for the derivative will look like
    00:15:22
    this we'll have the sum from n equals 1
    00:15:24
    to Infinity of n times negative 3 to the
    00:15:28
    power of n minus 1 divided by 3 to the
    00:15:31
    power of n right we just replaced x with
    00:15:34
    negative 3. okay and so if we simplify
    00:15:37
    we can split up this power we'd have
    00:15:40
    negative three to the power of n times a
    00:15:42
    negative three to the power of negative
    00:15:43
    one so I'm just going to rewrite that
    00:15:46
    here we'll have negative 3 to the power
    00:15:48
    of negative one all right and then
    00:15:50
    notice that we have negative 3 to the
    00:15:52
    power of n and three to the power of n
    00:15:54
    and so we can combine that to be one
    00:15:57
    fraction to the power of n so this will
    00:15:59
    be equal to the sum from n equals one to
    00:16:02
    Infinity of n times negative three to
    00:16:05
    the power of negative 1 times negative 3
    00:16:08
    divided by three to the power of n okay
    00:16:11
    so we just separated negative 3 to the
    00:16:13
    power of n and three to the power of n
    00:16:14
    to be their own fraction but they're
    00:16:17
    still multiplied by the rest of the
    00:16:19
    parts of our series and by the way
    00:16:20
    negative 3 to the power of negative 1 is
    00:16:23
    the same as negative one-third if we
    00:16:26
    move this negative 3 to the denominator
    00:16:28
    this exponent would become positive and
    00:16:31
    so if we rewrite that this would be
    00:16:32
    equal to the sum from n equals 1 to
    00:16:35
    Infinity of n divided by negative three
    00:16:37
    and then this part negative 3 divided by
    00:16:40
    3 is negative one so we're multiplying
    00:16:43
    by negative one to the power of n all
    00:16:45
    right and so this is our series that
    00:16:48
    results when x equals negative 3 for the
    00:16:50
    derivative and this is an alternating
    00:16:52
    series so if we want to know whether it
    00:16:54
    converges or diverges we can use the
    00:16:57
    alternating series test and the
    00:16:59
    alternating series test says that R
    00:17:01
    series is convergent if it meets two
    00:17:03
    requirements the first one being that
    00:17:06
    the limit as n approaches Infinity of
    00:17:08
    our sequence is zero and the second one
    00:17:10
    being that our sequence needs to be
    00:17:12
    decreasing but in this case we only have
    00:17:14
    to check the first requirement because
    00:17:16
    watch what's going to happen if we take
    00:17:19
    the limit as n approaches Infinity of
    00:17:21
    our sequence in this case if we ignore
    00:17:24
    the part that makes it alternating and
    00:17:25
    just look at n divided by negative
    00:17:27
    negative 3 what will be the limit as n
    00:17:29
    approaches Infinity of n divided by
    00:17:31
    negative three well as n approaches
    00:17:33
    Infinity this numerator is going to
    00:17:36
    increasingly get larger and larger and
    00:17:38
    larger and so we have Infinity divided
    00:17:40
    by negative 3 and infinity divided by a
    00:17:43
    negative fixed value is just going to
    00:17:45
    make infinity negative infinity and so
    00:17:49
    what we find here is that this limit is
    00:17:52
    not equal to zero which means that from
    00:17:54
    the alternating series test or more
    00:17:56
    specifically we are kind of using the
    00:17:59
    Divergence test here but either way from
    00:18:01
    the Divergence test or the alternating
    00:18:03
    series test we can conclude that this
    00:18:06
    series diverges all right and so since
    00:18:09
    it diverges
    00:18:10
    the interval of convergence for our
    00:18:13
    derivative will not include negative 3.
    00:18:16
    so we're going to leave that Open
    00:18:18
    Bracket open we're not going to close it
    00:18:21
    because we're not including that value
    00:18:22
    our series does not converge when x
    00:18:25
    equals negative 3. okay so we checked
    00:18:27
    negative 3 but now let's check positive
    00:18:30
    3 for our derivative when x equals
    00:18:32
    positive 3 we will have the sum from n
    00:18:35
    equals 1 to Infinity of n times 3 to the
    00:18:38
    power of n minus 1 divided by 3 to the
    00:18:40
    power of n all right now if we simplify
    00:18:42
    it 3 to the power of n minus 1 can be
    00:18:45
    split into two parts we'll have 3 to the
    00:18:47
    power of n and then three to the power
    00:18:49
    of negative one so this will be equal to
    00:18:51
    the sum from n equals 1 to Infinity of n
    00:18:55
    times three to the power of n times 3 to
    00:18:58
    the power of negative 1 divided by 3 to
    00:19:00
    the power of n but now we have 3 to the
    00:19:02
    power of n in the numerator and
    00:19:04
    denominator so those are going to cancel
    00:19:06
    out and then 3 to the power of negative
    00:19:08
    1 we can just move that 3 to the
    00:19:10
    denominator later to make that exponent
    00:19:12
    positive one and so this will be equal
    00:19:14
    to the sum from n equals 1 to Infinity
    00:19:16
    of n divided by 3. and now if we use the
    00:19:20
    Divergence test on this series just take
    00:19:22
    the limit as n approaches Infinity of
    00:19:25
    this sequence we will find that this
    00:19:27
    series will diverge right the limit as n
    00:19:30
    approaches Infinity of n divided by 3 is
    00:19:34
    equal to Infinity n is approaching
    00:19:36
    infinity and infinity divided by 3 will
    00:19:39
    still be positive infinity and that's
    00:19:41
    not equal to zero which means that our
    00:19:45
    series diverges all right that's from
    00:19:48
    the Divergence test if the limit of your
    00:19:51
    sequence from your series does not equal
    00:19:53
    zero then it diverges okay and so since
    00:19:56
    the derivative diverges when x equals
    00:19:59
    positive 3 that bracket will also remain
    00:20:01
    open and this is the complete interval
    00:20:04
    of convergence for the derivative all
    00:20:07
    right it actually has the same interval
    00:20:09
    of convergence as the original series
    00:20:11
    but we still had to check to make sure
    00:20:14
    that that was true there is a
    00:20:16
    possibility that it could be different
    00:20:18
    all right now you are going to see with
    00:20:20
    our integral that it will be slightly
    00:20:23
    different and so let's work on that next
    00:20:25
    let's check the endpoints for our
    00:20:27
    integral series let's start by checking
    00:20:30
    x equals negative 3 when x equals
    00:20:32
    negative 3 we'll have the sum from n
    00:20:35
    equals 1 to Infinity of negative 3 to
    00:20:38
    the power of n plus 1 divided by 3 to
    00:20:40
    the power of n times n plus one okay and
    00:20:43
    then we can simplify this numerator by
    00:20:45
    breaking up this exponent we'll have
    00:20:47
    negative 3 to the power of n times
    00:20:49
    negative three to the power of one or
    00:20:52
    negative 3 to the first power so I'm
    00:20:54
    just going to rewrite that immediately
    00:20:55
    we'll have negative 3 to the power of n
    00:20:57
    times negative 3 to the first Power and
    00:21:00
    then notice that we have negative 3 to
    00:21:01
    the power of n divided by 3 to the power
    00:21:04
    of n and so we can combine them to be
    00:21:06
    one fraction to the power of n so this
    00:21:09
    will be equal to the sum for many equals
    00:21:11
    1 to Infinity of negative 3 divided by 3
    00:21:14
    to the power of n times negative 3
    00:21:17
    divided by n plus 1. all right we just
    00:21:20
    combined those two parts to be their own
    00:21:23
    fraction and that's multiplied by
    00:21:25
    everything else in that series all right
    00:21:28
    now negative 3 divided by 3 is negative
    00:21:30
    one so what we really have here
    00:21:33
    is negative 1 to the power of n times
    00:21:35
    negative 3 divided by n plus one and so
    00:21:38
    we have an alternating series here that
    00:21:40
    we can test the convergence of by using
    00:21:43
    the alternating series test and so from
    00:21:45
    the alternating series test we can check
    00:21:47
    our two requirements to see if the
    00:21:49
    series converges the first one is the
    00:21:51
    limit as n approaches Infinity of the
    00:21:55
    sequence from our series and that's
    00:21:57
    going to exclude the part that makes it
    00:21:59
    alternating so we're just going to look
    00:22:00
    at negative 3 divided by n plus one so
    00:22:03
    the limit as n approaches Infinity of
    00:22:05
    negative three divided by n plus one
    00:22:07
    needs to be equal to zero in this case
    00:22:10
    it will be because we have a fixed value
    00:22:13
    divided by n plus one where n is
    00:22:16
    approaching Infinity so this denominator
    00:22:18
    is increasing it is approaching infinity
    00:22:20
    and a fixed value divided by Infinity
    00:22:23
    will always be zero so the first
    00:22:25
    requirement is met and the second one is
    00:22:28
    that our sequence needs to be decreasing
    00:22:30
    in other words a sub n plus 1 needs to
    00:22:32
    be less than or equal to a sub n the
    00:22:35
    next term of the sequence needs to be
    00:22:36
    less than or equal to the previous term
    00:22:38
    all right and so if we look at that over
    00:22:40
    here a sub n plus 1 will be negative 3
    00:22:43
    divided by n plus two right we have
    00:22:46
    negative 3 divided by n plus 1 plus 1
    00:22:48
    which is n plus two and then a sub n is
    00:22:52
    just going to be negative 3 divided by n
    00:22:53
    plus 1. so if we compare those two
    00:22:56
    fractions which one is less than or
    00:22:58
    equal to the other well the numerators
    00:23:01
    are the same so the denominator is the
    00:23:03
    only difference here and this
    00:23:04
    denominator is always one bigger than
    00:23:07
    this denominator right we have n plus 2
    00:23:09
    compared to n plus one and so since this
    00:23:13
    fraction will always have a bigger
    00:23:14
    denominator it will always have a
    00:23:16
    smaller value so a sub n plus 1 is less
    00:23:20
    than or equal to a sub n and so our
    00:23:22
    second requirement is met which means
    00:23:24
    that from the alternating series test we
    00:23:26
    can conclude that our series converges
    00:23:29
    and so because this series converges we
    00:23:32
    can say that this integral this function
    00:23:35
    will be defined for x equals negative 3.
    00:23:39
    negative 3 will be included in the
    00:23:42
    interval of convergence all right so now
    00:23:45
    we check negative three let's check
    00:23:46
    positive 3 and then that will be it for
    00:23:49
    this example all right and so when x
    00:23:51
    equals positive 3 will our integral
    00:23:55
    series converge or diverge let's take a
    00:23:57
    look when x equals positive three we
    00:23:59
    will have the sum from n equals 1 to
    00:24:01
    Infinity of 3 to the power of n plus one
    00:24:04
    divided by 3 to the power of n times M
    00:24:07
    plus 1. okay we just replaced X in our
    00:24:10
    series with positive 3. now we can split
    00:24:14
    up this numerator once again 3 to the
    00:24:16
    power of n plus 1 is the same as 3 to
    00:24:18
    the power of n times three to the first
    00:24:20
    power so we'll have three to the power
    00:24:22
    of n times three to the first Power and
    00:24:24
    now we have 3 to the power of n in the
    00:24:26
    numerator and denominator and so they're
    00:24:28
    going to cancel out and our series will
    00:24:31
    look like this we'll have the sum for
    00:24:32
    from n equals 1 to Infinity of 3 divided
    00:24:36
    by n plus one all right and so now we
    00:24:38
    have a fairly simple series that we can
    00:24:40
    determine the convergence or Divergence
    00:24:42
    of in this case we have a series that
    00:24:44
    looks similar to a p series specifically
    00:24:47
    the harmonic Series where p is equal to
    00:24:49
    one so I'm going to use a comparison
    00:24:51
    test here let's compare this using the
    00:24:55
    limit comparison test that's what I'm
    00:24:57
    going to choose to use here and we're
    00:24:59
    going to compare it to the sum from n
    00:25:02
    equals 1 to Infinity of 1 divided by n
    00:25:04
    which like I said is the harmonic series
    00:25:07
    or a p Series where p is equal to 1
    00:25:10
    which means that it diverges all right
    00:25:13
    and so if we can show that the limit of
    00:25:17
    this sequence divided by the sequence
    00:25:19
    from our comparison series is equal to a
    00:25:22
    finite value then this series will share
    00:25:25
    the same Divergence as this comparison
    00:25:28
    Series so we're going to look at the
    00:25:30
    limit as n approaches Infinity of 3 3
    00:25:33
    divided by n plus 1 divided by 1 divided
    00:25:36
    by n but that would be the same as
    00:25:38
    multiplying by the reciprocal which is
    00:25:40
    just n divided by one so we'll multiply
    00:25:42
    by n divided by one and this will be
    00:25:45
    equal to the Limit as n approaches
    00:25:48
    Infinity of 3n divided by n plus one all
    00:25:51
    right now for this limit as n approaches
    00:25:54
    Infinity we have a rational expression
    00:25:57
    where the numerator and denominator are
    00:25:59
    both algebraic and so what that means is
    00:26:01
    that we can look at the highest power of
    00:26:03
    n in the numerator and denominator and
    00:26:06
    that will help us determine what that
    00:26:07
    limit is equal to in this case we have
    00:26:10
    three times n to the first Power in the
    00:26:12
    numerator and N to the first Power in
    00:26:14
    the denominator so since those highest
    00:26:17
    powers of n are equal this limit as n
    00:26:19
    approaches Infinity will be equal to the
    00:26:22
    ratio of the coefficients between those
    00:26:24
    two terms with n to the highest power of
    00:26:26
    one so in this case we'll have 3 which
    00:26:28
    is the coefficient of this n divided by
    00:26:31
    one the coefficient of the this n so
    00:26:33
    this limit is equal to 3 divided by 1
    00:26:36
    which is equal to 3 which is a finite
    00:26:39
    value meaning it's not infinity or
    00:26:41
    negative infinity and so our series will
    00:26:43
    also be Divergent just like our
    00:26:46
    comparison series all right and so since
    00:26:48
    this series diverges when x equals 3 3
    00:26:52
    will not be included in the interval of
    00:26:55
    convergence for our power series which
    00:26:57
    is the integral of our original function
    00:27:00
    or our original series all right so
    00:27:03
    negative 3 to 3 including negative 3 is
    00:27:07
    the interval of convergence for our
    00:27:09
    integral series okay and so with that we
    00:27:12
    have found the radius of convergence
    00:27:14
    which is the same for all three
    00:27:16
    functions or all three series and the
    00:27:18
    interval of convergence for all three
    00:27:20
    series as well okay so you can see that
    00:27:23
    sometimes they will be the same but
    00:27:25
    other times they will be slightly
    00:27:27
    different and so with that let's look at
    00:27:29
    one more example for this video where
    00:27:31
    you will really see how the intervals of
    00:27:34
    convergence can be different for the
    00:27:36
    original compared to the derivative and
    00:27:38
    the integral all right so here's our
    00:27:40
    second and final example once again we
    00:27:42
    want to find the radius and interval of
    00:27:44
    convergence for f of x f Prime of X and
    00:27:47
    the integral of f of x DX and here's our
    00:27:50
    function this time f of x is equal to
    00:27:52
    this power series the sum from n equals
    00:27:55
    1 to Infinity of negative 1 to the power
    00:27:57
    of n times x minus 5 quantity to the
    00:28:00
    power of n divided by n times 5 to the
    00:28:03
    power of n and so let's start things off
    00:28:05
    by finding the derivative F Prime of X
    00:28:07
    of this series so F Prime of X will be
    00:28:11
    equal to the sum from n equals 1 to
    00:28:13
    Infinity of negative 1 to the power of n
    00:28:16
    times the derivative of x minus 5 to the
    00:28:19
    power of n right so remember anything
    00:28:22
    that does not have X within it we can
    00:28:24
    leave alone we're not going to be taking
    00:28:26
    the derivative of negative 1 to the
    00:28:27
    power of n or anything in the
    00:28:29
    denominator n or 5 to the power of n
    00:28:32
    those are all going to stay the same
    00:28:33
    because they do not have X within them
    00:28:36
    we're taking a derivative with respect
    00:28:38
    to X and so the only part we are
    00:28:40
    concerned with actually taking the
    00:28:41
    derivative of is x minus 5 to the power
    00:28:44
    of n right this whole quantity which has
    00:28:47
    X in it is raised to a power of n so
    00:28:49
    we're going to multiply that power down
    00:28:51
    and then subtract 1 from that power now
    00:28:54
    normally you would need to use the chain
    00:28:56
    rule here and then multiply by the
    00:28:58
    derivative of the inside function but in
    00:29:00
    this case the derivative is just going
    00:29:02
    to be 1 right the derivative of x is 1
    00:29:04
    and the derivative of negative 5 is 0.
    00:29:07
    so we don't need to worry about that we
    00:29:08
    just need to multiply n down and then
    00:29:10
    subtract one from that exponent all
    00:29:13
    right so we'll have n times x minus 5 to
    00:29:15
    the power of n minus 1 and that will be
    00:29:18
    divided by n times 5 to the power of n
    00:29:22
    okay and now notice that we have an n in
    00:29:25
    the numerator and denominator and so
    00:29:27
    those are going to cancel out and so F
    00:29:30
    Prime of X is really equal to the sum
    00:29:32
    from an n equals 1 to Infinity of
    00:29:34
    negative 1 to the power of n times x
    00:29:37
    minus 5 to the power of n minus 1
    00:29:39
    divided by 5 to the power of n okay so
    00:29:43
    if I clean up my work here there's our
    00:29:45
    actual derivative for this function we
    00:29:48
    no longer have that n in the denominator
    00:29:50
    or in the numerator and so now that we
    00:29:52
    have our derivative we can find the
    00:29:54
    integral of our series as well so the
    00:29:57
    integral of f of x DX will be equal to
    00:30:02
    the sum from n equals 1 to Infinity of
    00:30:05
    negative 1 to the power of n and then
    00:30:07
    we're going to integrate x minus 5 to
    00:30:10
    the power of n using the power rule of
    00:30:12
    integration all right and so in this
    00:30:14
    case we just need to add 1 to this
    00:30:15
    exponent so we'll have n plus 1 and then
    00:30:17
    we're going to divide by that new
    00:30:19
    exponent of n plus 1. so we're going to
    00:30:22
    have x minus 5 to the power of n plus 1
    00:30:25
    divided by n times 5 to the power of n
    00:30:28
    times M plus one and that's how we will
    00:30:31
    divide by that new power hour all right
    00:30:33
    and now unlike our derivative nothing is
    00:30:35
    going to cancel out here and so this
    00:30:37
    just is our integral okay and then just
    00:30:40
    like with our previous example don't
    00:30:42
    forget that when you integrate a
    00:30:43
    function you need to include that
    00:30:45
    constant of integration right somewhere
    00:30:48
    in here we would need to add that extra
    00:30:50
    constant of C you could have C plus this
    00:30:53
    power series but I'm not going to write
    00:30:55
    that explicitly just to save some space
    00:30:57
    with my work here but you should have
    00:30:59
    that in your own work when you write out
    00:31:02
    the integral I'm going to remove it
    00:31:04
    because the plus C doesn't have any
    00:31:06
    effect on the radius or interval of
    00:31:07
    convergence and so to simplify things as
    00:31:10
    we work on finding the radius and
    00:31:12
    interval of convergence for that power
    00:31:13
    series I'm just going to remove that
    00:31:16
    plus C okay but do remember that that is
    00:31:19
    a part of the integral for your power
    00:31:22
    series all right and so now we have the
    00:31:24
    derivative and the integral of our power
    00:31:26
    Series so now let's work on finding the
    00:31:28
    radius and interval of convergence for
    00:31:31
    all three functions or all three Series
    00:31:33
    so what we'll do is use the ratio test
    00:31:36
    on the original Power series to find the
    00:31:39
    radius of convergence and what our
    00:31:41
    endpoints are going to be for the
    00:31:43
    interval of convergence that we're going
    00:31:45
    to need to check for all three functions
    00:31:47
    all right so we'll take the limit as n
    00:31:50
    approaches Infinity of the absolute
    00:31:52
    value of a sub n plus 1 divided by a sub
    00:31:54
    n this is the limit for the ratio test
    00:31:57
    and so this will be equal to the Limit
    00:31:59
    as n approaches Infinity of the absolute
    00:32:02
    value of a sub n plus 1 which will be
    00:32:04
    negative 1 to the power of n plus 1
    00:32:06
    times x minus 5 to the power of n plus 1
    00:32:09
    divided by n plus 1 times 5 to the power
    00:32:13
    of n plus 1. all right so we had a lot
    00:32:15
    of n's to change to be n plus 1 but that
    00:32:18
    is a sub n plus 1 and then we need to
    00:32:21
    multiply that by the reciprocal of a sub
    00:32:23
    n which would be the same as dividing by
    00:32:25
    a sub n so we're going to flip the
    00:32:27
    numerator and denominator of this
    00:32:29
    expression and we will multiply by n
    00:32:32
    times times 5 to the power of n divided
    00:32:34
    by negative 1 to the power of n times x
    00:32:37
    minus 5 to the power of n all right and
    00:32:40
    so that is the limit that we need to
    00:32:42
    evaluate to find our radius and interval
    00:32:44
    of convergence now what we want to do is
    00:32:47
    simplify this a little bit and the first
    00:32:49
    thing that I notice is that these
    00:32:50
    absolute value bars are going to
    00:32:52
    completely eliminate negative 1 to the
    00:32:54
    power of n plus 1 and negative 1 to the
    00:32:56
    power of n because both of those
    00:32:58
    expressions can only be one of two
    00:33:00
    values they'll either be negative one or
    00:33:02
    positive one so if we're taking the
    00:33:04
    absolute value of that it's always going
    00:33:06
    to be positive one and so those two
    00:33:08
    parts are just completely eliminated the
    00:33:10
    absolute value bars take care of them
    00:33:12
    and then other than that I see some
    00:33:14
    powers that we can split up so we'll
    00:33:17
    have that this is equal to the Limit as
    00:33:20
    n approaches Infinity of the absolute
    00:33:22
    value of x minus 5 to the power of n
    00:33:25
    times x minus 5 all right that is this
    00:33:28
    power split up and then that's
    00:33:30
    multiplied by n times 5 to the the power
    00:33:32
    of n and that's all divided by M plus 1
    00:33:35
    times 5 to the power of n times 5 that
    00:33:39
    is this power split up and then we have
    00:33:41
    x minus 5 to the power of n okay now we
    00:33:45
    have 5 to the power of n in the
    00:33:46
    numerator and denominator and so those
    00:33:48
    are going to cancel out and we also have
    00:33:50
    x minus 5 to the power of n in the
    00:33:52
    numerator and denominator as well so
    00:33:55
    those are also going to cancel out so
    00:33:57
    what we're left with is the limit as n
    00:33:59
    approaches Infinity of the absolute
    00:34:01
    value of x minus 5 times n divided by n
    00:34:06
    plus 1 times 5. okay so those were all
    00:34:10
    the parts left over in our limit after
    00:34:12
    all of that cancellation and so now if I
    00:34:14
    clean up my work what I want to do now
    00:34:16
    is separate the ends from the X right we
    00:34:19
    can move those ends outside of the
    00:34:21
    absolute value bars since our values of
    00:34:24
    n are always going to be positive right
    00:34:26
    n is only defined for values from 1 to
    00:34:28
    Infinity which are all positive and so
    00:34:30
    they don't need to be within those
    00:34:32
    absolute value bars what we will have is
    00:34:34
    that this is equal to the Limit as n
    00:34:36
    approaches Infinity of n divided by n
    00:34:39
    plus 1 times the absolute value of x
    00:34:41
    minus 5 divided by 5. okay and now we
    00:34:45
    can take the limit as n approaches
    00:34:46
    Infinity of this expression fairly
    00:34:49
    easily and in this case we have a
    00:34:52
    rational expression where the numerator
    00:34:53
    and denominator are both algebraic we
    00:34:57
    have n to the first power divided by n
    00:34:59
    to the first power plus one so if we
    00:35:01
    compare those highest powers of n they
    00:35:03
    are equal right n to the first Power in
    00:35:06
    the numerator and N to the first Power
    00:35:08
    in the denominator and since those
    00:35:10
    highest powers of n are equal the limit
    00:35:12
    as n approaches Infinity of this
    00:35:14
    expression will be the ratio of the
    00:35:16
    coefficients for those two terms in this
    00:35:19
    case the limit will be one the
    00:35:21
    coefficient of this n divided by one the
    00:35:24
    coefficient of this n and so this limit
    00:35:26
    is equal to one divided by one which is
    00:35:28
    just one and so this is equal to one
    00:35:31
    times the app absolute value of x minus
    00:35:33
    five divided by five all right I just
    00:35:36
    removed the absolute value bars from the
    00:35:38
    denominator because 5 is always positive
    00:35:40
    and so this limit is equal to the
    00:35:43
    absolute value of x minus 5 divided by
    00:35:45
    5. right anything times 1 is just going
    00:35:48
    to be itself so we don't even need to
    00:35:49
    write that one this is what our limit is
    00:35:52
    equal to okay and now remember for the
    00:35:54
    ratio test the value of our limit needs
    00:35:57
    to be less than one in order for the
    00:35:59
    series to converge and so we need to set
    00:36:02
    this to be less than one and then solve
    00:36:04
    for x now if we multiply both sides by 5
    00:36:08
    that will give us that the absolute
    00:36:09
    value of x minus 5 is less than 5 and at
    00:36:13
    this point right here we can determine
    00:36:15
    the radius of convergence once you have
    00:36:18
    the absolute value of whatever
    00:36:20
    expression dealing with X that was in
    00:36:23
    your original Power Series right we had
    00:36:25
    x minus 5 to the power of n once you
    00:36:27
    have that x minus 5 in absolute value
    00:36:30
    bars less than some value that value is
    00:36:34
    the radius of convergence so we will say
    00:36:37
    that R is equal to 5. that is the radius
    00:36:41
    of convergence for all three of our
    00:36:43
    functions or all three of our power
    00:36:45
    series that is not going to change what
    00:36:48
    is going to change is the interval of
    00:36:51
    convergence in regards to the endpoints
    00:36:53
    and so if we solve for x here remember
    00:36:55
    that the absolute value bars just mean
    00:36:57
    that this expression is less than 5 but
    00:37:00
    greater than negative 5 so let's write
    00:37:02
    that out negative 5 is less than x minus
    00:37:05
    5 Which is less than 5 and so if we add
    00:37:07
    5 to all of these values what we'll have
    00:37:10
    is that zero is less than x Which is
    00:37:14
    less than 10 right negative 5 plus 5 is
    00:37:17
    0 x minus 5 plus 5 is just X and 5 plus
    00:37:21
    5 is 10. all right and so the interval
    00:37:24
    of convergence that we're going to be
    00:37:25
    looking at for all three of our series
    00:37:27
    is 0 to 10. and the only way that it's
    00:37:30
    going to change is if we need to to
    00:37:32
    include one of those two endpoints or
    00:37:35
    possibly even both okay and so if I
    00:37:37
    clean up my work here we now want to go
    00:37:39
    through each of our series and check to
    00:37:42
    see if the interval of convergence needs
    00:37:45
    to include the end points or not and so
    00:37:47
    let's start with our original Power
    00:37:49
    series we will Begin by looking at zero
    00:37:52
    as our first endpoint to check when x
    00:37:55
    equals zero our power series will look
    00:37:57
    like this well the sum from n equals 1
    00:37:59
    to Infinity of negative 1 to the power
    00:38:02
    of n times 0 minus five to the power of
    00:38:05
    n divided by n times 5 to the power of n
    00:38:08
    all right now zero minus five is
    00:38:10
    negative 5 and so we have negative 5 to
    00:38:12
    the power of n which means that we have
    00:38:15
    negative 5 to the power of n divided by
    00:38:17
    5 to the power of n so we can rewrite
    00:38:19
    this series to be the sum from n equals
    00:38:22
    one to Infinity of negative 1 to the
    00:38:24
    power of n divided by n times negative 5
    00:38:28
    divided by 5 to the power of n all right
    00:38:30
    and now negative 5 divided by 5 is equal
    00:38:33
    to negative one and so this is equal to
    00:38:36
    the sum from n equals one to Infinity of
    00:38:39
    negative one to the power of n divided
    00:38:41
    by n times negative one to the power of
    00:38:43
    n all right but now we can actually
    00:38:45
    simplify this further because we have
    00:38:47
    two parts that are negative one to the
    00:38:49
    power of n and they are actually going
    00:38:51
    to cancel each other out whenever this
    00:38:54
    is a negative one this will be negative
    00:38:56
    one negative 1 times negative 1 is
    00:38:58
    positive one and when this is positive
    00:39:00
    one and this is positive one one times
    00:39:02
    one is one and so this numerator is
    00:39:04
    always going to be positive one no
    00:39:06
    matter the value of N and so this series
    00:39:09
    is equal to the sum from n equals one to
    00:39:11
    Infinity of one divided by n which is
    00:39:15
    the harmonic Series right it is a p
    00:39:17
    series
    00:39:18
    where p is equal to 1 which means it
    00:39:21
    diverges okay and so since this series
    00:39:25
    diverges when x equals zero our power
    00:39:27
    series is a Divergent series zero will
    00:39:31
    not be included in this interval of
    00:39:33
    convergence and so we will leave that as
    00:39:35
    an Open Bracket we will not close it and
    00:39:37
    now we will move on to checking the
    00:39:39
    other endpoint when x equals 10. and so
    00:39:42
    let's check to see what happens when x
    00:39:44
    equals 10 when x equals 10 our series
    00:39:47
    will look like this well the sum from n
    00:39:49
    equals 1 to Infinity of negative 1 to
    00:39:52
    the power of n times 10 minus 5 to the
    00:39:55
    power of n divided by n times 5 to the
    00:39:57
    power of n all right so we replaced x
    00:39:59
    with 10 and 10 minus 5 is 5. so we have
    00:40:03
    5 to the power of n in the numerator and
    00:40:06
    5 to the power of n in the denominator
    00:40:08
    and so those are actually going to
    00:40:11
    cancel out which leaves us with the sum
    00:40:14
    from n equals 1 to Infinity of negative
    00:40:16
    1 to the power of n divided by n and now
    00:40:19
    we have the alternating harmonic series
    00:40:21
    which is a convergent series and if you
    00:40:24
    didn't know that you could use the
    00:40:26
    alternating series test here if you were
    00:40:28
    to use the alternating series test and
    00:40:30
    check the requirements the first one the
    00:40:33
    limit as n approaches Infinity of our
    00:40:35
    sequence if we remove the part that
    00:40:38
    makes it alternating we'll have one
    00:40:39
    divided by n and the limit as n
    00:40:41
    approaches Infinity of 1 divided by n is
    00:40:44
    zero so that first requirement is met
    00:40:47
    this limit of our sequence needed to be
    00:40:49
    zero and then if we check the second
    00:40:51
    requirement our sequence needs to be
    00:40:53
    decreasing so a sub n plus 1 needs to be
    00:40:56
    less than a sub n and so if we check
    00:40:58
    that over here a sub n Plus One will be
    00:41:00
    1 divided by M plus 1 and a sub n is 1
    00:41:03
    divided by n and 1 divided by n plus 1
    00:41:06
    is always less than or equal to 1
    00:41:08
    divided by n because this denominator is
    00:41:11
    always one bigger than this denominator
    00:41:13
    and when your denominator is bigger for
    00:41:16
    a fraction that fraction is always
    00:41:18
    smaller so a sub n plus 1 is less than
    00:41:21
    or equal to a sub n and so that second
    00:41:23
    requirement is met which means that this
    00:41:25
    alternating series is a convergent
    00:41:27
    alternating series and so since it
    00:41:30
    converges
    00:41:31
    we can conclude that x equals 10 should
    00:41:35
    be included in the interval of
    00:41:37
    convergence for our original Power
    00:41:39
    Series so we will change this Open
    00:41:41
    Bracket to be a closed bracket and now
    00:41:44
    we have the interval of convergence for
    00:41:46
    our original Power series f of x okay
    00:41:50
    and now we can move on to determining
    00:41:52
    the interval of convergence for the
    00:41:54
    derivative F Prime of X we're now going
    00:41:56
    to do that whole process over again and
    00:41:59
    check each of these endpoints for this
    00:42:01
    series which is different than the
    00:42:03
    original all right so let's work on our
    00:42:06
    endpoints again we will start by
    00:42:08
    checking x equals zero for our
    00:42:11
    derivative Series so we'll have the sum
    00:42:13
    from n equals 1 to Infinity of negative
    00:42:16
    1 to the power of n times 0 minus five
    00:42:19
    to the power of n minus 1 divided by 5
    00:42:21
    to the power of n okay we just replaced
    00:42:24
    this x with zero and so zero minus five
    00:42:27
    is negative five so we have negative 5
    00:42:30
    to the power of n minus 1 so this is
    00:42:32
    equal to the sum from n equals one to
    00:42:34
    Infinity of negative one to the power of
    00:42:37
    n times negative five to the power of n
    00:42:39
    minus one but we can split up that
    00:42:42
    exponent so we'll have negative 5 to the
    00:42:44
    power of n times negative 5 to the power
    00:42:46
    of negative one all right so we can
    00:42:49
    split up that power and then in a
    00:42:51
    denominator we have five to the power of
    00:42:53
    n all right and now we have negative
    00:42:55
    five to the power of n and 5 to the
    00:42:57
    power of n that we can combine to be
    00:42:59
    their own fraction to the power of n and
    00:43:02
    then we can move negative 5 to the
    00:43:03
    negative first power to the denominator
    00:43:05
    to make that power positive so we're
    00:43:07
    just going to have negative 5 in the
    00:43:09
    denominator and so this is equal to the
    00:43:11
    sum from n equals one to Infinity of
    00:43:14
    negative one to the power of n divided
    00:43:16
    by negative five times negative 5
    00:43:18
    divided by 5 to the power of n okay so
    00:43:21
    we just combine negative 5 to the power
    00:43:23
    of n and 5 to the power of n to be this
    00:43:25
    right here negative 5 divided by 5 to
    00:43:28
    the power of n and then we have negative
    00:43:29
    1 to the power of n divided by negative
    00:43:31
    five okay now negative 5 divided by 5 is
    00:43:34
    negative 1. so this will just become
    00:43:37
    negative 1 to the power of n and
    00:43:39
    negative 1 to the power of n times
    00:43:41
    negative 1 to the power of n is just
    00:43:43
    positive one as I explained for one of
    00:43:45
    our previous series all right and so
    00:43:48
    this series is equal to the sum from n
    00:43:50
    equals 1 to Infinity of positive one
    00:43:52
    divided by negative five and whenever
    00:43:55
    you have a series of just a constant
    00:43:57
    that series is going to diverge right
    00:44:00
    because if we use the Divergence test
    00:44:02
    we'll take the limit as n approaches
    00:44:04
    Infinity of this sequence the limit as n
    00:44:07
    approaches Infinity of negative
    00:44:09
    one-fifth is equal to negative one-fifth
    00:44:12
    which doesn't equal zero which means
    00:44:14
    that our series diverges from the
    00:44:16
    Divergence test all right so since it
    00:44:19
    diverges we can conclude that x equals
    00:44:22
    zero will not be included in our
    00:44:24
    interval of convergence for the
    00:44:26
    derivative all right so we're going to
    00:44:27
    leave that Open Bracket alone and we're
    00:44:30
    not going to make it closed we're not
    00:44:31
    include including x equals zero okay so
    00:44:34
    now let's move on to x equals ten let's
    00:44:37
    check the other endpoint and so when x
    00:44:39
    equals 10 our series is going to look a
    00:44:41
    little bit different and so let's take a
    00:44:43
    look when x equals 10 we will have the
    00:44:46
    sum from n equals 1 to Infinity of
    00:44:49
    negative 1 to the power of n times 10
    00:44:52
    minus 5 to the power of n minus 1
    00:44:54
    divided by 5 to the power of n okay so
    00:44:57
    we just replaced x with positive 10 and
    00:45:00
    10 minus 5 is 5 so we have 5 to the
    00:45:03
    power of n minus one so this will be
    00:45:05
    equal to the sum from n equals 1 to
    00:45:07
    Infinity of negative 1 to the power of n
    00:45:10
    times 5 to the power of n minus 1
    00:45:13
    divided by 5 to the power of n now once
    00:45:16
    again we can split up this exponent
    00:45:18
    we'll have 5 to the power of n times 5
    00:45:20
    to the power of negative one so I'm
    00:45:22
    going to rewrite this we will have 5 to
    00:45:25
    the power of n times 5 to the power of
    00:45:27
    negative one and now we have 5 to the
    00:45:30
    power of n in the numerator and
    00:45:31
    denominator here and so those are going
    00:45:34
    to cancel out and we can move 5 to the
    00:45:36
    power of negative 1 to the denominator
    00:45:38
    to make that exponent positive and so
    00:45:40
    what we'll have is that this is equal to
    00:45:42
    the sum from n equals 1 to Infinity of
    00:45:45
    negative 1 to the power of n divided by
    00:45:47
    5. okay and so we have an alternating
    00:45:50
    series that we can test the convergence
    00:45:52
    of by using the alternating series test
    00:45:54
    the part that makes it alternating is
    00:45:56
    negative 1 to the power of n and so if
    00:45:59
    we use the alternating series test and
    00:46:01
    check the first requirement we will be
    00:46:04
    looking at the limit as n approaches
    00:46:06
    Infinity of our sequence excluding the
    00:46:09
    part that makes it alternating so remove
    00:46:10
    negative 1 to the power of n and we're
    00:46:13
    taking the limit of one-fifth and the
    00:46:15
    limit of a constant is equal to that
    00:46:17
    constant and so one-fifth is the result
    00:46:20
    of the limit which is not equal to zero
    00:46:22
    which means that this alternating series
    00:46:25
    does not pass the alternating series
    00:46:27
    test and it will diverge all right and
    00:46:31
    so our series diverges which means that
    00:46:34
    positive 10 or x equals 10 is also not
    00:46:38
    included in the interval of convergence
    00:46:40
    for this series all right so the
    00:46:43
    interval of convergence will just be
    00:46:44
    from 0 to 10 not including either of
    00:46:48
    those endpoints this series diverges
    00:46:50
    when x equals 0 and when x equals 10.
    00:46:53
    but it will converge for all the values
    00:46:55
    between those two values of X okay so
    00:46:58
    now we have checked the original Power
    00:47:00
    series and the derivative of that power
    00:47:02
    series now we just need to check the
    00:47:04
    interval of convergence for the integral
    00:47:07
    of that power series and so let's start
    00:47:09
    once again with that first endpoint
    00:47:11
    let's check x equals zero when x equals
    00:47:14
    zero our integral series will look like
    00:47:17
    this we have the sum from n equals 1 to
    00:47:19
    Infinity of negative 1 to the power of n
    00:47:22
    times 0 minus 5 to the power of n plus 1
    00:47:26
    divided by n times 5 to the power of n
    00:47:29
    times n plus 1. okay so we just placed x
    00:47:32
    with zero and now we can simplify our
    00:47:35
    series and so 0 minus five is the same
    00:47:38
    as a negative five so I'm going to erase
    00:47:40
    this and we'll have negative 5 to the
    00:47:43
    power of n plus one and so let's split
    00:47:45
    up that power we'll have negative 5 to
    00:47:48
    the power of n times negative 5 to the
    00:47:50
    first Power and I'm just going to
    00:47:52
    rewrite that right here we have negative
    00:47:54
    5 to the power of n times negative 5 to
    00:47:57
    the first Power okay and now we can
    00:47:59
    combine negative 5 to the power of n and
    00:48:02
    5 to the power of n to be their own
    00:48:04
    fraction that we are going to simplify
    00:48:06
    to be negative 1 to the power of n right
    00:48:09
    this is like the third or fourth time
    00:48:10
    we've done this now but I'm still going
    00:48:12
    to write out the steps here this will be
    00:48:14
    equal to the sum from n equals 1 to
    00:48:16
    Infinity of negative 1 to the power of n
    00:48:19
    times negative 5 divided by n times n
    00:48:23
    plus 1 times negative 5 divided by 5 to
    00:48:27
    the power of n okay so 5 to the power of
    00:48:31
    n and negative 5 the Power Band have
    00:48:33
    become this fraction right here which is
    00:48:35
    raised to the power of n since the
    00:48:37
    numerator and denominator both had that
    00:48:39
    power of n now negative 5 divided by 5
    00:48:41
    is negative one and so we're going to be
    00:48:44
    multiplying negative 1 to the power of n
    00:48:46
    into that numerator and once again we
    00:48:49
    have negative 1 to the power of n times
    00:48:51
    the negative 1 to the power of n which
    00:48:53
    is always going to be positive no matter
    00:48:55
    what the value of n is when this is
    00:48:58
    negative this is negative negative times
    00:49:00
    negative is positive and when this is
    00:49:03
    positive and this is positive positive
    00:49:05
    times positive is positive and so those
    00:49:08
    parts are going to cancel out with each
    00:49:10
    other and we are going to be left with
    00:49:12
    the sum from n equals 1 to Infinity of
    00:49:15
    negative 5 divided by n times n plus 1.
    00:49:18
    now I'm going to distribute that n
    00:49:20
    through this quantity so we will have N
    00:49:22
    squared plus n okay and now we have a
    00:49:26
    series that we can determine the
    00:49:28
    convergence of and this one's going to
    00:49:30
    be a little bit trickier than some of
    00:49:31
    the the others if we were to use the
    00:49:33
    Divergence test for this series The
    00:49:35
    Limit would be equal to zero right if
    00:49:37
    you were to take the limit as n
    00:49:39
    approaches Infinity of this sequence the
    00:49:41
    denominator would become infinity and so
    00:49:43
    you'd have negative 5 divided by
    00:49:45
    Infinity which would be zero and so we
    00:49:48
    can't use a Divergence test it's
    00:49:49
    inconclusive instead I'm thinking we can
    00:49:52
    use a comparison test because this
    00:49:54
    series looks pretty close to a p series
    00:49:57
    or at least we can compare it to one so
    00:49:59
    if I clean up my work a little bit we
    00:50:01
    can compare this series to a p series of
    00:50:04
    the sum from n equals 1 to Infinity of
    00:50:06
    one divided by N squared I'm just
    00:50:09
    choosing that highest power of n from
    00:50:11
    the denominator okay and this is a p
    00:50:13
    Series where p is equal to two right two
    00:50:16
    is the power of n and 2 is greater than
    00:50:19
    1 which means that this P series
    00:50:21
    converges
    00:50:23
    and so what we want to do is use a
    00:50:25
    comparison test to compare this
    00:50:27
    comparison series to our Series right
    00:50:30
    here now I'm going to use the limit
    00:50:32
    comparison test I find that that is
    00:50:34
    easier to use when you are comparing to
    00:50:36
    a p series at least that's true in most
    00:50:38
    cases so that's what I'm going to use
    00:50:40
    here so I'm just going to write that
    00:50:41
    down here we are using the limit
    00:50:44
    comparison test okay and so what we will
    00:50:48
    do is take the limit as n approaches
    00:50:51
    Infinity of the sequence from the series
    00:50:53
    that we want to test divided by the
    00:50:55
    sequence from our comparison series now
    00:50:58
    we're going to have a fraction divided
    00:50:59
    by a fraction so instead of doing that
    00:51:01
    we're going to take this fraction and
    00:51:03
    multiply it by the reciprocal of this
    00:51:05
    one which is the same as dividing by 1
    00:51:08
    divided by N squared okay so we're going
    00:51:10
    to have is the limit as n approaches
    00:51:12
    Infinity of negative 5 divided by N
    00:51:15
    squared plus n times N squared divided
    00:51:18
    by 1. and that is the same as dividing
    00:51:21
    by 1 divided by N squared all right now
    00:51:23
    if we simplify we'll have the limit as n
    00:51:25
    approaches Infinity of negative 5 times
    00:51:28
    N squared divided by N squared plus n
    00:51:31
    okay now what we want to do is evaluate
    00:51:34
    this limit and if it's equal to any
    00:51:36
    finite value meaning it's not infinity
    00:51:38
    or negative Infinity we can conclude
    00:51:41
    that our series shares the same
    00:51:43
    convergence as as our comparison series
    00:51:45
    now in this case for this limit as n
    00:51:48
    approaches Infinity we have a rational
    00:51:50
    expression where the numerator and
    00:51:51
    denominator are algebraic so we can
    00:51:54
    compare the highest powers of n to
    00:51:56
    determine the value of the limit in this
    00:51:58
    case our highest power in the numerator
    00:52:00
    is N squared and the highest power in
    00:52:03
    the denominator is also N squared so
    00:52:05
    this limit will be equal to the ratio of
    00:52:08
    the coefficients for those terms so
    00:52:10
    we'll have negative 5 the coefficient of
    00:52:12
    this N squared divided by one the
    00:52:15
    coefficient of this N squared so this
    00:52:17
    limit is equal to negative 5 divided by
    00:52:19
    1 which is equal to negative 5 which is
    00:52:21
    a finite value and so that means that
    00:52:24
    our series is convergent just like our
    00:52:27
    comparison series okay so since our
    00:52:30
    series converges when x equals zero we
    00:52:32
    will include zero in our interval of
    00:52:35
    convergence for this integral series all
    00:52:39
    right so I'm going to erase that open
    00:52:40
    bracket and I'm going to make it a
    00:52:42
    closed bracket and now we can move on to
    00:52:45
    testing our final endpoint which is x
    00:52:47
    equals 10 and then we will have
    00:52:49
    officially found the interval of
    00:52:51
    convergence for all three of our series
    00:52:53
    or all three of our functions so if we
    00:52:56
    back it up here and check our last
    00:52:58
    endpoint of x equals 10. let's see what
    00:53:00
    our series will look like we'll have the
    00:53:02
    sum from n equals 1 to Infinity of
    00:53:05
    negative 1 to the power of n times 10
    00:53:07
    minus 5 to the power of n plus 1 divided
    00:53:10
    by n times 5 to the power of n times n
    00:53:14
    plus 1. okay once again we just replaced
    00:53:17
    x with 10 because that's what we're
    00:53:19
    setting x equal to all right now 10
    00:53:21
    minus 5 is 5. so I'll just erase that we
    00:53:25
    will have 5 to the power of n plus one
    00:53:28
    and if we split that power we'll have 5
    00:53:30
    to the power of n times 5 to the first
    00:53:32
    power so I'm just going to rewrite that
    00:53:34
    right away we will have 5 to the power
    00:53:36
    of n times 5 to the first Power and now
    00:53:39
    we have 5 to the power of n in the
    00:53:41
    numerator and denominator and so we can
    00:53:43
    cancel those out so this series will be
    00:53:45
    the sum from n equals 1 to Infinity of
    00:53:48
    negative one to the power of n times
    00:53:50
    five divided by n times n plus one all
    00:53:54
    right now if we distribute this n
    00:53:55
    through this quantity we'll have N
    00:53:57
    squared plus n so let's just do that
    00:53:59
    make this denominator a little nicer to
    00:54:01
    look at we'll have N squared plus n and
    00:54:04
    now to determine the convergence of this
    00:54:07
    series I think we should use the
    00:54:08
    alternating series test because this is
    00:54:10
    an alternating series due to this
    00:54:12
    negative 1 to the power of n that will
    00:54:15
    make the terms alternate between being
    00:54:16
    positive and negative and so if we use
    00:54:19
    the alternating series test we need to
    00:54:21
    check two requirements to see if our
    00:54:22
    series converges the first one will be
    00:54:25
    the limit as n approaches Infinity of
    00:54:28
    our sequence excluding the part that
    00:54:30
    makes it alternating so we'll remove
    00:54:31
    this negative 1 to the power of n and
    00:54:34
    just have five divided by N squared plus
    00:54:36
    n and this limit needs to be equal to
    00:54:39
    zero and so let's see if that's true as
    00:54:41
    n approaches Infinity we have 5 divided
    00:54:43
    by N squared plus n and so this
    00:54:46
    denominator is going to increasingly get
    00:54:48
    larger so we have a fixed value divided
    00:54:50
    by infinity or a fixed value divided by
    00:54:53
    an increasing denominator which will
    00:54:55
    always be zero all right so the first
    00:54:57
    requirement is met and now we need to
    00:54:58
    check the second requirement which is
    00:55:00
    that our sequence needs to be decreasing
    00:55:02
    in other words a sub n plus 1 needs to
    00:55:05
    be less than or equal to a sub n the
    00:55:07
    next term of the sequence needs to be
    00:55:09
    less than or equal to the previous term
    00:55:11
    so if we check that over here a sub n
    00:55:14
    plus 1 will be 5 divided by M plus 1
    00:55:17
    squared plus M plus 1 and a sub n is
    00:55:21
    just 5 divided by N squared plus n and
    00:55:24
    so if we compare these two fractions the
    00:55:27
    numerators are the same so the
    00:55:28
    denominators are going to decide which
    00:55:31
    one is smaller and so you can see that
    00:55:33
    the denominator for a sub n plus 1 is
    00:55:36
    always going to be bigger than the
    00:55:38
    denominator for a sub n right we have n
    00:55:41
    plus 1 squared plus n plus one that
    00:55:43
    that's always going to be bigger than
    00:55:46
    just N squared plus n this denominator
    00:55:49
    is squaring a bigger value and adding it
    00:55:52
    to a bigger value than this denominator
    00:55:54
    and so because the denominator is always
    00:55:56
    going to be bigger for this fraction the
    00:55:59
    value of this fraction will always be
    00:56:01
    smaller so we can conclude that a sub n
    00:56:04
    plus 1 is less than or equal to a sub n
    00:56:07
    which means that the second requirement
    00:56:09
    for the alternating series test is met
    00:56:11
    and so now both requirements for the
    00:56:14
    alternating series test are met which
    00:56:16
    means that this series
    00:56:18
    converges
    00:56:19
    okay so since this series converges the
    00:56:22
    power series that is the integral of the
    00:56:25
    original series converges when x equals
    00:56:27
    10 and so 10 needs to be included in the
    00:56:31
    interval of convergence okay so I'll
    00:56:33
    erase the Open Bracket and make it a
    00:56:36
    closed bracket and now we have our last
    00:56:39
    interval of convergence for this problem
    00:56:41
    we have now found f Prime of X and the
    00:56:45
    integral of f of x and we found their
    00:56:47
    respective intervals of convergence as
    00:56:49
    well as the radius of convergence which
    00:56:51
    is the same for the original series the
    00:56:53
    derivative and the integral okay and so
    00:56:56
    with that this was the last example for
    00:56:58
    this video so if you have any questions
    00:57:00
    feel free to leave those in the comments
    00:57:02
    but if you don't have any questions this
    00:57:04
    is all I had for now so I will see you
    00:57:06
    next time
    00:57:11
    foreign
    00:57:16
    [Music]

    Description:

    Example Problems For How to Find the Derivative & Integral of a Power Series (Calculus 2) In this video we look at practice problems of finding the derivative and integral of a power series, as well as their respective radius and interval of convergence. 🖥️ Join My Membership Site: https://www.jkmathematics.com//plus This video series is designed to help students understand the concepts of Calculus 2 at a grounded level. No long, boring, and unnecessary explanations, just what you need to know at a reasonable and digestible pace, with the goal of each video being shorter than the average school lecture! Calculus 2 requires a solid understanding of calculus 1, precalculus, and algebra concepts and techniques. This includes limits, differentiation, basic integration, factoring, equation manipulation, trigonometric functions, logarithms, graphing, and much more. If you are not familiar with these prerequisite topics, be sure to learn them first! Video Chapters: 0:00 Example 1 - (x/3)^n 27:39 Example 2 - ((-1)^n*(x-5)^n)/(n*5^n) 56:56 Outro 📝 Lesson Video For These Examples: https://www.youtube.com/watch?v=jloA_OtzG-I ⏩ Next Lesson: https://www.youtube.com/watch?v=ZZcWPxgoirE 📺 Calculus 2 Playlist: https://www.youtube.com/playlist?list=PLHdRLeAbIZE7KQ922piyRIDAj30XYcOIu 📺 Calculus 1 Playlist: https://www.youtube.com/playlist?list=PLHdRLeAbIZE6krGDQthaQRpLZFLD4ob54 🌐 Visit My Website: https://www.jkmathematics.com ⚡️Math Products I Recommend⚡️ Graphing Calculator: https://www.amazon.com/Texas-Instruments-TI-Nspire-Graphing-Calculator/dp/B07RL7HM5V?language=en_US Non-Graphing Calculator: https://www.amazon.com/Texas-Instruments-TI-30X-Scientific-Calculator/dp/B00SWWKEAY?language=en_US Financial Calculator: https://www.amazon.com/s?k=ba+2+plus&language=en_US Graphing Paper: https://www.amazon.com/KAISA-Punched-Loose-Leaf-Binders-150sheet/dp/B096V2RYR2?sp_csd=d2lkZ2V0TmFtZT1zcF9hdGY&language=en_US College Rule Paper: https://www.amazon.com/Mead-Filler-College-Punched-15326/dp/B000068UWB?language=en_US My Favorite Pencil: https://www.amazon.com/Quick-Mechanical-Pencil-Assorted-Barrel/dp/B07FC5NGBX?language=en_US My Favorite Erasers: https://www.amazon.com/Prismacolor-Premier-Magic-Erasers-3-Count/dp/B005ENNEJW?language=en_US Calculus Workbook: https://www.amazon.com/Essential-Calculus-Practice-Workbook-Solutions/dp/1941691242?sp_csd=d2lkZ2V0TmFtZT1zcF9hdGY&language=en_US ⚡️Textbooks I Use⚡️ Calculus 1 & 2: https://www.amazon.com/Calculus-Ron-Larson/dp/0547167024?language=en_US Calculus 3: https://www.amazon.com/Multivariable-Calculus-Ron-Larson/dp/1285060296?language=en_US Financial Mathematics: https://www.amazon.com/Mathematics-Investment-Credit-ACTEX-Academic/dp/1566987679?language=en_US ⚡️My Recording Equipment⚡️ iPad Air: https://www.amazon.com/Apple-iPad-Air-5th-Generation/dp/B09V3HJBN3?sp_csd=d2lkZ2V0TmFtZT1zcF9hdGY&th=1&language=en_US Apple Pencil (Gen 2): https://www.amazon.com/Apple-MU8F2AM-A-Pencil-Generation/dp/B07K1WWBJK?sp_csd=d2lkZ2V0TmFtZT1zcF9hdGY&language=en_US Tablet Desk Stand: https://www.amazon.com/OMOTON-Adjustable-Multi-Angle-Aluminum-Convenient/dp/B015FH0XKE?sp_csd=d2lkZ2V0TmFtZT1zcF9hdGY&th=1&language=en_US External Hardrive: https://www.amazon.com/Toshiba-HDTB420XK3AA-Canvio-Portable-External/dp/B079D3D8NR?language=en_US (Commissions earned on qualifying purchases) Find me on social media: Facebook: https://www.facebook.com/unsupportedbrowser Twitter: https://twitter.com/jk_mathematics Instagram: @jk_mathematics Found this video to be helpful? Consider giving this video a like and subscribing to the channel! Thanks for watching! Any questions? Feedback? Leave a comment! -Josh from JK Math #calculus Disclaimer: Please note that some of the links associated with the videos on my channel may generate affiliate commissions on my behalf. As an amazon associate, I earn from qualifying purchases that you may make through such affiliate links.

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