Taylor Series for f(x) with nth Derivatives and Coefficients | Homework Help

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Homework Help Overview

The problem involves finding the Taylor series for a function f at x=2, given specific conditions about its derivatives. The original poster states that f(2)=7, and provides information about the nth derivatives of f at x=2, particularly noting that odd derivatives are zero and even derivatives follow a specific formula.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of the sixth-degree Taylor polynomial and the coefficients for even powers in the series. There are attempts to clarify the expressions for coefficients and the series representation, with some participants questioning the correctness of previous responses.

Discussion Status

The discussion is ongoing, with participants providing hints and corrections to each other's reasoning. There is a focus on clarifying the coefficients for the Taylor series, particularly for even indices, and some participants are encouraged to show their work to facilitate understanding.

Contextual Notes

Participants are working under the constraints of the problem statement, which specifies the behavior of the derivatives and the need to derive the Taylor series without providing complete solutions. There is a noted confusion regarding the indexing and factorial expressions involved.

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 \sum (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 \sum (n-1)/3n * (x-2)^2n from n=1 to infinity ??
No. See part b.
 
Can you give me some hints ?
 
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?
 
So the coefficient will be (n-1)!/(3^n*n!) ??
 
  • #10
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! ?
 

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