Taylor series for f about x=2 with even derivatives

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nns91
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Homework Statement



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 ?
c. Find the interval of convergence of the Taylor series for f about x=2. Show the work that leads to your answer


Homework Equations



Taylor series

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 [tex]\sum[/tex] (n-1)/3n * (x-2)^2n from n=1 to infinity ??
 
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nns91 said:
b. Will it just be (n-1)!/3n / (2n)! ??
No. Try again.

c. So will the Taylor series be [tex]\sum[/tex] (n-1)/3n * (x-2)^2n from n=1 to infinity ??
No. See part b.
 
You made a very simple mistake with the n's. Looking at some examples should reveal them. Try 2n = 2 and 2n = 4.
 
For 2n=2, the coefficient will be (n-1)!/6 right ?

So should it be (n-1)!/3^n /n! ??
 
nns91 said:
For 2n=2, the coefficient will be (n-1)!/6 right ?
No. Try again and please show your work.
 
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 coefficient be (n-1)!/3^n / 2n!
 
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?
 
You keep answering the question (b) that could have been asked, instead of the question (b) that actually was asked.

The question that could have been asked is this: what is the coefficient of (x-2)^n when n is even?

By the way, what is 99!/100! ?