BAM! Mr. Tarrou In this calculus lesson we're going to be looking at a four part
question that involves Part A writing a third-grade Taylor polynomial part be
working with and writing the first non three zero terms of Maclaurin series
gonna be doing a product of a couple of Maclaurin series basically we're going
to be looking at Part C which is going to be evaluating or estimating some
function at a value that function which is going to be based off and using the
Taylor polynomial that we come up with in Part A and then Part D we're going to
be finding the upper bound of the error or the alternating using the alternating
series error theorem basically so that's going to be our four parts in the
description you'll find time stamps to jump ahead to each of these four parts
you'll find time stamps or not time stamps but links to the AP Central
College Board website if you want to find some old AP tests that you can use
to study for and find the test by the marking schemes find the master graders
comments you can see where students earn and did not earn their points you can
find pie lists for all of my calculus lessons and if you want to find some
problems to help you look at some AP style problems to review you can find
that as well so our question here says a function f has derivatives of all orders
for all real numbers of x a portion of the graph of f that's this graph right
here and that is concave up is shown along with the line tangent to the graph
of F at X is equal to zero so we get this set up out of our way we're me
pulling some information off of this graph and this table as we go through
the four parts certainly the first part of this question as we get my fancy
picture at a picture-in-picture out of there we're gonna write the third degree
Taylor polynomial for F about X is equal to zero remember this Taylor polynomial
is something that we're coming up with that is going to be able to approximate
values of function f centered about about the points of X is equal to zero
okay so our third degree Taylor polynomial
centered about zero is going to be equal to F of zero plus the first derivative
of F at zero over one factorial times X minus C X minus the center at what
you're going to construct this Taylor polynomial which is zero so I'm just
writing that for notes purposes this F double prime of zero over two factorial
times X minus zero to the second plus the third derivative of F at 0 over 3
factorial times X minus zero to the third power so that's sort of the
framework we're going to be using you might very well be able to just simply
write the answer but I'm going to in case anyone struggling help you figure
this out so P sub 3 of X is equal to okay F of 0
remember this is F and that functions value at 0 is equal to 4 so f of 0 is 4
now the value of the derivative of f of 0 the slope of F at the instantaneous
rate of change at 0 is well we have the tangent line given to us in the graph
well yeah of course we can just look at a graph and say oh the slope of the line
is rise over run so we're gonna go down 3 and over 1 so the slope there is equal
to negative 3 over 1 factorial which is 1 times X to the first now we have plus
f double prime of 0 yeah second derivative of f at 0 so the second
derivative of f at 0 is equal to 1 so we're looking at 1 over 2 factorial is 2
times x squared and then plus the third derivative of F at 0 is equal to
negative 12 over 7
times 1 over 3 factorial which is 1 over 6x cubed so the answer to Part A here
our third degree Taylor polynomial is going to be 4 minus 3x plus x squared
over 2 of course the 6 and the 12 will cancel
giving us negative 2 so we have minus 2/7 X cubed I'm just making sure I
didn't write something silly we are good Part B coming up right now so Part B it
says for the first part of Part B we're going to write the first 3 nonzero terms
of the Maclaurin series for e to the X and of course the difference like
difference between a Maclaurin series and a Taylor series is that the center
is at 0 and we have that list of common Maclaurin series that we do generally
need to have memorized for the AP tests I'm not sure what your teachers and
college professors may possibly be doing but like we're looking for the sort of
elementary series such as 1 over X 1 over 1 plus X the natural log of X e to
the X sine of X cosine of X of course if you know the sine of X you can find the
Maclaurin series for the cosine of X by simply taking the derivative and then we
have arc tan arc sine and 1 plus X to the K power at least as is defined as
the common Maclaurin series in the book that I teach from lorrison calculus well
for e to the X we're looking for 1 plus X to the first over 1 factorial plus x
squared over 2 factorial plus X cubed over 3 factorial and that continues on
until you get to that general term of X to the N power over n factorial and then
because it's a series it just goes on forever plus dot dot dot
but the question simply asked for us to write the first three nonzero terms not
write out the entire sort of common Maclaurin series so
there's our answer to the first part here now we're going to write the second
degree Taylor polynomial for e to the x times f of X we're going to look at that
Taylor polynomial that we came up with just in Part A and combine that with the
first three nonzero terms of a Maclaurin series Taylor polynomial that Center
that Taylor polynomial was zero so we're all great all we have to do is take that
basically what we have here the first three nonzero terms of our Maclaurin
series for e to the X 1 plus X plus 1/2 x squared and multiply that with the
Taylor polynomial that we just found which was now normally I want to write
my Taylor polynomials or just my polynomials in general with the powers
going from high to low but here we want just a second degree Taylor polynomial
so I really want to work with those terms that are in these polynomials with
the degrees or powers of 0 and those small powers so I'm gonna leave it the
way it was written originally which is 4 minus 3 X plus 1/2 x squared minus 2/7 x
cubed and start just distributing this together but not all of it we only want
a second degree Taylor polynomial so we're going to do 1 times 4 which is 4
we have 1 times negative 3 X so negative 3x and then plus a 1/2 x squared I'm
going to go ahead and stop with that multiplication that distribute that
distribution with our first term because I've just gotten to a power of 2 no
reason to go to the negative 2 sevenths X cubed now we'll move on to X we have x
times 4 which is going to be 4x and we have x times negative 3x which is
going to be of course negative 3x squared and okay if we do the x times
the 1/2 x squared we're going to have a something with a third power third
degree so we're going to go to now our third term then our first factor we have
1/2 x squared times 4 which is going to give a small of course at the 4 is equal
to 2 combine like terms and we're done
when you glance at this might feel a little bit tricky but basically as long
as you have that that common Maclaurin series memorized for e to the X this is
really just a distribution problem so we have negative we have 2 minus 3 is
negative 1 negative 1 plus a 1/2 is negative 1/2 x squared we've got a 4
minus 3x which is going to give us a plus 1 X and plus 4 and in case I'd
unless I just made some kind of silly error we are good to go that is the
answer to Part B Part C coming up right now so for Part C we have let H be the
function defined by H of X is equal to the definite integral from 0 to X of F
of T DT so basically we have like an accumulation function here starting from
accumulating a rate of change for H of X off of 0 and we're going to use the
Taylor polynomial found in Part A to approximate for H of 1/2 C so H is
related to very closely to the function basically if f of X we just don't have
an X here because that's that independent variable is being used as an
upper limit here on our accumulating function and a Taylor polynomial is used
to approximate write this third degree Taylor polynomial that we had in Part A
I say that but I say it's this one like it's up on the board it's 2's is created
to be an estimate of f of X so it makes sense that we should
be able to accumulate or integrate with our Taylor polynomial to find to
estimate an area under that curve off of zero okay so we have here that H of one
half is equal to the definite integral from 0 to 1/2 of F of T DT but it said
use the Taylor polynomial that we found in part a let's try to go back and get
my notes so we're looking here at equals to the definite integral from 1 0 to 1/2
of we had negative 2/3 negative 2/7 X cubed plus 1/2 x squared minus 3x plus 4
DX okay so let's go ahead and double check here
make sure to make a copier okay here we go
so we're going to say that we've got we're gonna write an equal sign next to
each line of work case there's some kind of small error and we don't to get
marked off for linking error we're looking at raise that power by 1 and
divide so negative 2 over 28 times X to the fourth raised that power by 1 and
divide so we're looking at 1 6 X cubed minus 3 over 2x squared plus 4x that
upper limit of 1/2 lower limit of 0 all of these terms have a factor of 0 so
when we plug in 0 for X these are all going to be 0 so we're just going to
focus on the 1/2 so we have negative 2 over 28 times 1/2 to the fourth power
plus 1/6 1/2 to the third minus 3 over 2 1/2 squared plus 4 times 1/2 minus
again all of this is equal to zero when we plug in our lower limit of zero so we
have 2 4 8 16 and right off let's see here yeah we have negative that's going
to be 1 over 14 because that reduces times 1 over 1 4 8 16 plus 1 6 1 over 8
minus 3 over 2 1/4 plus 2 plus 0 so we have 14 times 16 10 times 140 is 10
times 14 is 140 we have 6 times 10 which is 60 we have 6 times 4 which is 24 so
we're looking at 224 so negative 1 over 2 24 plus 1 over 48 minus 3 over 8 plus
2 and this is what happens when you make up your own numbers this is going to
come out to a nice easy rounded off number not really of 11:03 over 6 72 so
to summarize all that up h of one half it is approximately equal to using the
third degree Taylor polynomial from Part A which estimates f of X of 1103 over
672 our last part coming up of course right now for Part D our last part
assume the Maclaurin series for H converges to H of X for all real numbers
X and the individual terms of the series H of 1/2 alternate sign and decrease in
absolute value to 0 basically just setting up the conditions here that are
going to allow us to safely answer the actual question what is the upper bound
the air for the approximation of H of 1/2 well they basically just told us
that what we're dealing with is an alternating series that the individual
terms of the series H of 1/2 alternate signs and they're decreasing and
decreasing in absolute value until they get to 0 ok so the upper bound for an
alternating series is less than or equal to the last negated or neglected excuse
me term well we found function H of
basically H of X by integrating our f of T function or FO function we and the
approximation came from the third degree Taylor polynomial well if we worked with
and integrated a third degree Taylor polynomial to estimate basically H of X
and then approximate H of 1/2 then the first negated term is going to come from
negated or neglected term is going to come from the fourth degree or that last
term not the not the last term of the third degree Taylor polynomial but that
then fourth term so what we're going to be doing is saying that we're going to
be integrating or finding the first neglected term by integrating from 0 to
1/2 0 to 1/2 of ok to find that fourth degree term of the Taylor polynomial we
need to work with the fourth derivative of F about 0 which is going to be that
value numerically is 26 over then 4 factorial so 4 times 3 times 2 times 1
is going to be 24 next to the fourth DX alright ok so our approximation for H of
1/2 was created and found by using the third degree Taylor polynomial we
integrated it yes we got up to sort of like a fourth power but that was after
integration so now we're going to go to the next term the first neglected term
which were to create by integrating the fourth
power term of that Taylor polynomial so we have the definite integral from 1/2 0
to 1/2 of what 13 over 12x to the fourth DX that is going to give us 13 over 12
times 5x to the fifth 1/2 to 0 so we're looking at 13 over 60 times 1/2 to the
fifth power then minus 0 so we're looking at 13 over 60 times 2 to the
fifth 2 4 8 16 32 and that comes out to a nice small number of 13 over 1920 so
the maximum amount we can be off from our estimation of H of 1/2 is 13 over 19
over 13 over 1900 and 20 I miss it true hey go to your homework
you