BAM! Mr. Tarrou in this video we're gonna
be taking a look at number 6 from the
2018 AP Calculus BC test in the
description you'll find a link that you
jump ahead to any of these three parts
that we have you'll find a link to the
AP central website where you can
download this full PDF the solutions
marking schemes the master grader is
scoring commentary and where you can
learn where students made and did not
make their points and also we have about
650 other math lessons you'd like to
peruse it you know if you so choose I
would appreciate it okay so the
Maclaurin series for the natural log of
1 plus X is given by X minus x squared
over 2 plus X cubed over 3 minus x to
the 4th over 4 plus dot dot plus now
your general term negative 1 to the n
plus 1 power because we can see that our
signs are alternating so that's what
that's there 4 times X to the N over n
plus and then you'll another a little
plus dot dot dot because you don't want
to make this look like it is a Taylor
polynomial or Maclaurin polynomial but a
series it never stops now we do need to
remember some common Maclaurin series in
case we are being asked to recall those
on a calculus PSC exam they did last
year in this case they're actually
giving it to us so that's very nice on
its interval of convergence this series
converges to the natural log of 1 plus X
let F be the function defined by f of X
is equal to x times the natural log of 1
plus x over 3 so now that we you know
have this common Maclaurin series given
to us can we manipulate it a little bit
to fit this new function and we indeed
will be able to write the first four
nonzero terms and the general term of
the Maclaurin series for f of X ok
well what we're going to do is first of
all we're going to take an adjusted
version of this Maclaurin series and
multiply every term by X so for a we
have X and you might be able to simply
write the answer I'm going to show
the steps for teaching purposes so we
have x times okay so now the difference
between the natural log of 1 plus X in
the natural log of 1 plus x over 3 is
simply well just that one third constant
applied to each variable of X so I go so
as I go to write out the first four
nonzero terms and the general term for
the natural log of 1 plus x over 3 I'm
simply going to replace every X or
substitute in every place that I see an
X a 1/3 X now we're going to multiply
that X in as well as I'm going to go
ahead and do this 1/3 squared 1/3 cubed
1/3 fourth so we have x times X is x
squared we've got 3 squared is 9 and 9
times 2 is 18 we've got x squared now
being multiplied by X is going to be X
cubed and so on and so on just want to
go nice and slow to make sure I don't
make anything any silly errors so we
have your x times X is x squared X times
X cubed is x squared as X cubed x times
X cubed is X to the fourth so as we
continue this pattern I went ahead and
dead of course we've got 3 to the 4th
times 4 which is 3 24 and I'm looking
down because I left it as 4 times 3 to
the fourth but then we have over here
with our general term in the
construction of the general term this is
actually part of the reason why I want
to make sure that the denominator was
correct actually I could write this I
could write this as X to the n plus 1
over N times 3 to the N power one of the
reasons why I wrote that 1/3 out
separately to each of those powers has
just helped me to make sure that
wrote the general term correctly with at
3 to the N power in the denominator and
I made it really painfully obvious for
myself to make sure that where this was
X to the N is X to the n plus 1 at any
rate that is you know our answer to Part
A our first four nonzero terms in the
general term don't forget to put the dot
the plus dot dot dot at the end of that
answer Part B well we have a Maclaurin
series now in that general term so
whoops now we do we're gonna determine
the interval of convergence using that
general term which means that hopefully
as you know maybe watching this as
you're learning this concept I'm not
just reviewing we're gonna do the ratio
test which means we're gonna take the
limit as n approaches infinity of a sub
n plus 1 over a sub n because this is
basically sort of a while power series
and we're going to make sure then the
fat ratio is less than 1 so that power
series converges and then we're gonna
test the endpoints of this interval of
convergence because this is just going
to give us what kind of initially the
radius of convergence and then we'll
we'll know what the open interval is for
the interval of convergence so we have a
sub n plus 1 which means that we're
going to just look at this general term
and everywhere we see an n replace it
with an N so n plus 1 so we have
negative 1 to the n plus 1 plus 1 which
is n plus 2 we have times X to the n
plus 1 plus 1 which is n plus 2 over n
plus 1 as a factor times 3 to the n plus
1 now when I go ahead and and we have
that a sub n plus 1 term now divided by
a sub n I'm going to flip that up and
change that to a multiplication of the
reciprocal just to save some space on
the chalkboard
so we're going to multiply by not
divided by a sub n term but multiply by
the reciprocal of the a sub n term so
that means we're gonna multiply by n
times 3 to the N power over negative 1
to the n plus 1 times X to the n plus 1
and we're gonna make sure that that
limit stays less than 1 okay so is there
anything you know we can cancel out well
here we have n plus 1 basis of negative
1 and in the numerator over here we just
simply have one more well am i show this
cancellation but ultimately we have
absolute value symbols so that negative
one's gonna cancel out anyway in the
numerator of our first factor we have n
plus 2 bases of X and here we have n
plus 1 basis of X so here we have one
more base of X then we have in this
denominator so that's going to cancel
out we have n bases of 3 in the
numerator and we have n plus 1 basis of
3 in this denominator so these cancel
out we're gonna have one base of 3 one
factor of 3 left and that denominator
and nothing's gonna cancel with this n
over n plus 1 so we're looking at the
limit as n approaches infinity they did
tell us ashore work this might be a
slight overkill for some of us but I
think it's good
negative well negative what n times X
over 1 plus 1 times 3 again for the
ratio test we need to make sure the
limit as n approaches infinity of this
is equal to or less than 1 now the
absolute values are going to basically
just take care of this negative force
and remember we're letting n approach
infinity not X so you're looking at just
this n over n plus 1 and we don't need
to show like l'hopital's rule or some
kind of process to show that the limit
of basically n over n plus 1 is equal to
1 they're not expecting that much
to be shown we're just looking now at
and the absolute value of x over 3 is
less than 1 which means that if we
multiply both sides by 3 just as a
review we could look at this expression
and we could see the center of
convergence which we don't really need
to because we're told that it's a
Maclaurin series which is basically just
a special type of Taylor series where
the center of convergence is equal to 0
but we could see that this is the
absolute value of X minus 0 which means
that the center of convergence is 0 and
then that absolute value of this
binomial that has a coefficient of on
the X of 1 see I got rid of the
coefficient of 1/3 minus 0 is less than
3 when you have it in this X minus C
absolute value that is less than
basically you're looking at are your
radius of convergence so R I can't call
it an interval of convergence yet
because we haven't tested the endpoints
but we have so far a open interval of
negative 3 to 3 now that's not enable of
convergence but it's like we're almost
there now all we have to do is test the
endpoints we have a center of
convergence of 0 and a radius of
convergence of 3 so now that we have
this open interval that is the beginning
of us developing the final answer of
what our interval of convergence is we
need to test the endpoints ok so we're
going to do that by using this general
term and right not this whole tail
Maclaurin series but just the series in
terms of Sigma notation so with the
endpoint of negative x equals negative 3
we have the series where and starts at 1
and goes to infinity which I should be
including that correct yes okay we have
use and again this general term we've
got negative 1 to the n plus 1 power
times now not X because we're replacing
it with negative 3
so negative 3 to the n plus 1 power over
n times 3 to the N and all you have to
do is show the work as you start to
determine whether the series converges
or diverges so that you know whether to
include the endpoint of negative 3 or
not and I want to show a lot of work
just to make sure that I don't miss the
question and well you understand what
I'm doing so first thing I'm going to do
is bring out this negative 1 as its own
factor so we can see how it combines
with this negative 1 to the n plus 1
power when you multiply like bases of
course we add the exponents so rewrite
in that negative 3 to the n plus 1 power
we have with the expansion now negative
1 to the n plus 1 times negative 1 to
the n plus 1 times 3 to the n plus 1
multiplying these like bases the two
bases that have the value of negative 1
we have negative 1 n plus n is 2 n + 1
plus 1 is equal to 2 so we have this
series where negative we have negative 1
to the 2n plus 2 times 3 to the n plus 1
power over n times 3 to the N power this
negative 1 its exponent is 2 n if you
take a number multiply it by 2 it's even
if you add two it's to leave it and
negative 1 to an even power is positive
so the negative 1 goes away we have 3 to
the n plus 1 in the numerator and we
have 3 to the N in the denominator so
the numerator has one more base or
factor of 3 if you will so this is going
to the 3n and the denominators and a
cancel out with all but one of the
threes in the numerator and now in our
series notation all we have left is 3
over N and I'm going to go ahead and
just to make sure it's perfectly clear
we're going to pull this factor of 3 out
in front of the Sigma notation so that
we can see that we have 3 times the
summation where n starts at 1 and goes
to infinity
of one over n well that series but
basically the series of one over N is a
harmonic series which diverges so this
is harmonic give a reason please
harmonic series diverge so for the
purpose of this problem finding the
interval of convergence at least in
terms of the interval of convergence X
cannot be equal to negative 3 okay so
this point the season the left of
negative 3 is going to stay now let's
test X is equal to positive 3 and see if
this is going to be an open interval or
closed interval on our interval of
convergence one more time right out that
series notation that Sigma notation or
the series with Sigma notation foreign
series with Sigma notation so we have
summation where n starts 1 goes to
infinity one more time of negative 1 to
the n plus 1 power times 3 to the n plus
1 over N times 3 to the n problems like
this is really why in Chapter 9 with all
of your interval of convergence tests is
for this final step of finding your
Taylor and Maclaurin series to testing
the endpoints so we have 3 to the n plus
1 over 3 to the N so in the denominator
all of those factors of 3 are going to
cancel out leaving you with just one
factor of 3 in the numerator and there
is nothing to take care of that negative
1 to the n plus 1 so this is again
moving that 3 out front just to make
this abundantly clear we have 3 times
the summation where n starts at 1 and
goes to infinity of negative 1 to the n
plus 1 power over N and that is a
alternating harmonic series which
converges
thus three can be into our interval of
convergence and the interval of
convergence starts at with the open you
know it's open on the left side so from
negative three open two three closed or
you can say that negative 3 is less than
X which is less than or equal to three
is our interval convergence for our
series negative 1 to the n plus 1 power
x to the negative x to the n plus 1 over
3 times 3 to the N right that is Part B
Part C coming up right now and it's Part
C of the last question of the 2018
Calculus BC exam we're almost done let P
a 4 of X be the fourth degree Taylor
polynomial most as excited as if I was
really taking the test for F about X is
equal to 0 well I guess I technically
did but I didn't follow the you know
timed interval use the alternating
series error bound to find the upper
bound for the absolute value of P sub 4
of 2 minus F of 2 now we learned two
types of errors when dealing with
basically Taylor polynomial it's a
little like a legrasse the air and well
we have a nice fairly actually simple
process if the series happines happens
to be alternating the alternating series
remainder what the heck is this
expression about and what do they asking
us to do if a convergent alternating
series satisfies the condition where a
sub n plus 1 is less than or equal to a
sub n then the absolute value of the
remainder R sub n the absolute value
remainder R sub n is less than or equal
to the first neglected term so the
absolute value of the difference between
valued the series - the value that
you're getting from whatever degree
Taylor polynomial that you're working
with alternate well in this case
alternating series not any Taylor
polynomial is equal to the absolute
value of R sub n that is the absolute
value of your error and we're looking
for that upper bound of that error is
well less than or equal to a sub n plus
1 the first neglected term so what we
are going to do is we're going to simply
look at the fact that this is less than
or equal to what we're talking about the
fourth degree Taylor polynomial that's
alternating so we just need to look at
the fifth degree
basically term we're going to use this
expression and how it creates that well
that fifth degree and you kind of really
don't need to but I'm going to do that
for just teaching purposes anyway
glancing down to my notes to make sure I
don't mess this up so this is less than
or equal to the absolute value of
negative 1 to the n plus 1 well again we
have the second third and the fourth
degree terms here so we're looking at
the fifth degree term so we're looking
at negative 1 to the 4 plus 1 power
times X and we're looking at an x value
of 2 so 2 to the n plus 1 over N which
again is 4 right yeah cuz we're 4 n plus
1 is 4 times 3 to the fourth power now
that's going to be negative 1 out well
this absolute value so it's going to
cancel that out anyway
and so we're looking at 2 to the fifth
power over 4 times 3 to the fourth or
base we're looking at 32 over 3 24 and
yea so the upper bound is
equal to actually look I'm looking back
and forth now I've been better just to
do this if we look at this I got 2 to
the 5th is 2 4 8 16 32 4 times 3 to the
fourth so 8 so the upper bound of this
air is equal to 8 3 9 27 81 and that is
the end of our 2018 Calculus BC exam or
the end at least this is only video
you're watching of number 6
I'm mr. true hey go to your homework
you
