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

In summary: The question that was asked is this: what is the coefficient of (x-2)^(2n) when n is at least 1?By the way, what is 99!/100!?In summary, the function f has derivatives of all orders and f(2)=7. The nth derivative of f at x=2 is 0 when n is odd, and when n is even and n>=2, the nth derivative is given by f(n)(2)=(n-1)!/3n. The sixth-degree Taylor polynomial for f about x=2 was obtained and the coefficient of (x-2)^(2n) for n>=1 in the Taylor series for f about x=2 was determined
  • #1
nns91
301
1

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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  • #2
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.
 
  • #3
Can you give me some hints ?
 
  • #4
You made a very simple mistake with the n's. Looking at some examples should reveal them. Try 2n = 2 and 2n = 4.
 
  • #5
For 2n=2, the coefficient will be (n-1)!/6 right ?

So should it be (n-1)!/3^n /n! ??
 
  • #6
nns91 said:
For 2n=2, the coefficient will be (n-1)!/6 right ?
No. Try again and please show your work.
 
  • #7
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
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
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! ?
 

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

What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms. It is named after the mathematician Brook Taylor.

What is the purpose of a Taylor series?

The purpose of a Taylor series is to approximate a complicated function with a simpler one, making it easier to perform calculations and analyze the behavior of the function.

How do you calculate a Taylor series?

A Taylor series is calculated by taking the derivatives of a function at a specific point and plugging in those values into the formula for the Taylor series. The number of terms used in the series determines the accuracy of the approximation.

What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is centered at a specific point, while a Maclaurin series is centered at 0. This means that a Maclaurin series is a special case of a Taylor series where the center point is 0.

What are the practical applications of Taylor series?

Taylor series are used in a variety of fields including engineering, physics, and economics. They are used to approximate solutions to differential equations, analyze the behavior of functions, and make predictions in various systems.

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