Question

Okay so here's a link http://apcentral.collegeboard.com/apc/public/repository/ap10_calculus_bc_scoring_guidelines.pdf I get the first part of part a, but not the second part where it asks to write terms for the "f" function... can someone explain?

Answer 1

wut question?

Answer 2

second part of part a

Answer 3

there's so many part A's!!

Answer 4

http://openstudy.com/study#/updates/5325e73ae4b087b3d40fcd3e

Answer 5

im sorry hahah i didnt realize it had so many, question 6

Answer 6

could you go back to ur earlier question ?

Answer 7

the link i have posted...
u can find ur old quesitons at ur profile page also :
http://openstudy.com/users/auditi5#/users/auditi5/asked

Answer 8

The first part of 6 (a) asks for the taylor series expansion of cos(x) about x=0

the series expansion about x=0 is
\[f(x)= \sum_{n=0}^{\infty} \frac{ 1 }{ n! } f ^{(n)}\left( 0 \right) \ x^n\]

the derivatives of cos(x), evaluated at x=0 give
n d^n (cos(x)) at x=0
0 cos(x) 1
1 -sin(x) 0
2 -cos(x) -1
3 sin(x) 0
4 cos(x) 1

putting those values into the series definition you get
\[ \cos(x) \approx \frac{1}{0!}\cdot 1\cdot x^0+ \frac{1}{1!} \cdot 0 \cdot x^1
+\frac{1}{2!} \cdot -1 \cdot x^2 + \frac{1}{3!} \cdot 0 \cdot x^3+\frac{1}{4!} \cdot 1 \cdot x^4+...
\]
which simplifies to
\[ \cos(x) \approx 1 - \frac{2^2}{2!} + \frac{x^4}{4!} + ...+ (-1)^n \frac{x^{2n}}{(2n)!} \]

Answer 9

For the second part of 6(a), they want you to replace cos(x) with this series expansion in the definition of f(x)
\[ f(x)= \frac{\cos(x)}{x^2} - \frac{1}{x^2} \]
you get
\[ f(x) = \frac{1 - \frac{x^2}{2!} + \frac{x^4}{4!} + ...+ (-1)^n \frac{x^{2n}}{(2n)!}}{x^2} - \frac{1}{x^2} =\\
\frac{1}{x^2} - \frac{1}{2!} + \frac{x^2}{4!} + ... - \frac{1}{x^2}
\]
which simplifies to
\[ - \frac{1}{2!} + \frac{x^2}{4!} - \frac{x^4}{6!} +...\]

the general term (starting with n=0) looks like
\[ (-1)^{n+1} \frac{x^{2n}}{(2n+2)!} \]

Answer 10

hello, thank you! can you or anyone else explain part c?

Answer 11

For Q 6 ( c ), they tell you
\[ g(x)= 1 + \int_0^x f(t)\ dt\]
and they want the taylor series expansion

we already have the series expansion for f(x). so we can replace f(t) ( It does not matter if we call the variable "x" or "t" ) like this:
\[ g(x)= 1 + \int_0^x - \frac{1}{2!} + \frac{t^2}{4!} - \frac{t^4}{6!} +... \ dt\]

the integral, integrated term by term becomes
\[ - \frac{t}{2!} + \frac{t^3}{3\cdot 4!} - \frac{t^5}{5\cdot 6!} ... \bigg|_0^x= \\
- \frac{x}{2!} + \frac{x^3}{3\cdot 4!} - \frac{x^5}{5\cdot 6!} ...
\]
and using this result in the definition of g(x), we have
\[ g(1)= 1 - \frac{x}{2!} + \frac{x^3}{3\cdot 4!} - \frac{x^5}{5\cdot 6!} ... \]

Answer 12

Thanks!! One more question, sorry, for part B why is f prime of zero zero?

Answer 13

***part B why is f prime of zero zero?***

It takes some "out of the box" thinking... or, at least, slightly clever thinking

First, write down the the definition of the first few terms of the taylor series expansion of f(x) about x=0
\[ f(x)= \sum_{n=0}^{\infty} \frac{ 1 }{ n! } f ^{(n)}\left( 0 \right) \ x^n \\
f(x)= f(0) + f'(0) x + \frac{f''(0)}{2} x^2 + ...
\]
notice that the coefficient of the x term is f'(0)

Answer 14

we already figured out that the series expansion of f(x) expanded about x=0 is
\[ f9x) = - \frac{1}{2!} +0 \cdot x+ \frac{x^2}{4!} - \frac{x^4}{6!} +... \]
to make it more obvious, I inserted 0*x (we ordinarily would not write that term)
the point is, the coefficient of the x term is 0
that matches up with f'(0), meaning f'(0) = 0

Answer 15

That's what I thought but isnt't f double prime zero as well?

Answer 16

in the taylor series expansion, we have
\[ f(x)= f(0) + f'(0) x + \frac{f''(0)}{2} x^2 + ... \]

match \[ \frac{f''(0)}{2} \]with the coefficient of the x^2 term \( \frac{x^2}{4!}\)
in other words
\[ \frac{f''(0)}{2} = \frac{1}{4!} \]
and

\[ f''(0)= \frac{2}{4!} \]

Answer 17

Oh, I see. And by that method f'''(0)=0 as well. Thank you!

Answer 18

yes f''(0) = 1/12 and f'''(0) is 0