1. ## Maclaurin Series Problem

Find the maclaurin polynomials of orders n=0,1,2,3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation.

cos(pi*x)

http://img382.imageshack.us/img382/5889/48183237ut4.png

This is the answer from the solution mannual

Can anybody tell me how it finds f(k) first? I mean not just for this question because all the solution about maclaurin and taylor, the solution finds f(k) first.

Another question, I do not understand the NB means? Why is it n/2 ?

2. Originally Posted by soleilion
Find the maclaurin polynomials of orders n=0,1,2,3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation.

cos(pi*x)

http://img382.imageshack.us/img382/5889/48183237ut4.png

This is the answer from the solution mannual

Can anybody tell me how it finds f(k) first? I mean not just for this question because all the solution about maclaurin and taylor, the solution finds f(k) first.

Another question, I do not understand the NB means? Why is it n/2 ?
I cant open your attachment...but if you mean $\sum_{x=0}^{\infty}cos(x\pi)=\sum_{x=0}^{\infty}(-1)^{x}$ which is divergent

3. Originally Posted by Mathstud28
I cant open your attachment...but if you mean $\sum_{x=0}^{\infty}cos(x\pi)=\sum_{x=0}^{\infty}(-1)^{x}$ which is divergent

Can you see this right now????

4. Mathstud28

are you still online?

5. Can anybody help me, please?

Thank you

6. Originally Posted by soleilion
Find the maclaurin polynomials of orders n=0,1,2,3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation.

cos(pi*x)

http://img382.imageshack.us/img382/5889/48183237ut4.png

This is the answer from the solution mannual

Can anybody tell me how it finds f(k) first? I mean not just for this question because all the solution about maclaurin and taylor, the solution finds f(k) first.

Another question, I do not understand the NB means? Why is it n/2 ?

the $f^k$ are the kth derivatives so lets get started

$f(x)=\cos(\pi x)$
$f'(x)=-\pi \sin(\pi x)$
$f''(x)=-\pi^2 \cos(\pi x)$
$f'''(x)=\pi^3 \sin(\pi x)$
$f^{(4)}(x)=\pi^4 \cos(\pi x)$

Now we need to evaluate each of these at the center for this series at x=0
so we get...

$f(0)=\cos(\pi 0)=1$
$f'(0)=-\pi \sin(\pi 0)=0$
$f''(0)=-\pi^2 \cos(\pi 0)-\pi^2$
$f'''(0)=\pi^3 \sin(\pi 0)=0$
$f^{(4)}(0)=\pi^4 \cos(\pi 0)=\pi^4$

Now we can use these in taylor theorem for a power series centered at a is

$f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(a)}{n!}(x-a)^n$

so we get

$1-\frac{\pi^2}{2!}x^2+\frac{\pi^4}{4!}x^4-...$

we get this because everyother term is zero.

I hope this helps

7. Originally Posted by TheEmptySet
the $f^k$ are the kth derivatives so lets get started

$f(x)=\cos(\pi x)$
$f'(x)=-\pi \sin(\pi x)$
$f''(x)=-\pi^2 \cos(\pi x)$
$f'''(x)=\pi^3 \sin(\pi x)$
$f^{(4)}(x)=\pi^4 \cos(\pi x)$

Now we need to evaluate each of these at the center for this series at x=0
so we get...

$f(0)=\cos(\pi 0)=1$
$f'(0)=-\pi \sin(\pi 0)=0$
$f''(0)=-\pi^2 \cos(\pi 0)-\pi^2$
$f'''(0)=\pi^3 \sin(\pi 0)=0$
$f^{(4)}(0)=\pi^4 \cos(\pi 0)=\pi^4$

Now we can use these in taylor theorem for a power series centered at a is

$f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(a)}{n!}(x-a)^n$

so we get

$1-\frac{\pi^2}{2!}x^2+\frac{\pi^4}{4!}x^4-...$

we get this because everyother term is zero.

I hope this helps

Do you see the answer? Do you understand what NB means?

8. Originally Posted by soleilion
Do you see the answer? Do you understand what NB means?
I have no idea what NB stands for. I "think" it is saying Note: that [x] is floor function.