# Taylor series

1. Apr 8, 2009

### nns91

1. The problem statement, all variables and given/known data

Let f be a function with derivatives of all orders and for which f(2)=7. When n is odd, the nth derivative of f at x=2 is 0. When n is even and n=>2, the nth derivative of f at x=2 is given by f(n) (2)= (n-1)!/3n

a. Write the sixth-degree Taylor polynomial for f about x=2.
b. In the Taylor series for f about x=2, what is the coefficient of (x-2)(2n) for n =>1 ?

2. Relevant equations

Taylor series

3. The attempt at a solution

a. I got the sixth-degree series.
b. Will it just be (n-1)!/3n / (2n)! ??
c. So will the Taylor series be $$\sum$$ (n-1)/3n * (x-2)^2n from n=1 to infinity ??

2. Apr 8, 2009

### e(ho0n3

No. Try again.

No. See part b.

3. Apr 8, 2009

### nns91

Can you give me some hints ?

4. Apr 8, 2009

### e(ho0n3

You made a very simple mistake with the n's. Looking at some examples should reveal them. Try 2n = 2 and 2n = 4.

5. Apr 8, 2009

### nns91

For 2n=2, the coefficient will be (n-1)!/6 right ?

So should it be (n-1)!/3^n /n! ??

6. Apr 8, 2009

7. Apr 8, 2009

### nns91

So for n> or =1, 2n is always even so for the coeffecient we have to use the formula that is given: (n-1)!/3^n but also the denominator contains an even factorial so it will be 2n

so will the coefficent be (n-1)!/3^n / 2n!

8. Apr 8, 2009

### e(ho0n3

Still wrong. If n is even, the coefficient of the (x-2)n in the Taylor series is f(n)(2)/n! = (n-1)!/(3nn!), so when you substitute n for 2k (I will use k instead of n because I believe this is what is confusing you), what do you get?

9. Apr 9, 2009

### nns91

So the coefficient will be (n-1)!/(3^n*n!) ??

10. Apr 9, 2009