Taylor series centered at c = 1

In summary, the conversation discusses finding the Taylor Series of 1/x centered at c=1. The attempt at a solution involves taking derivatives and recognizing a pattern of fn(c)=(-1)^n(n!) and simplifying to \sum_{n=0}^{\infty} (-1)^n (x-1)^n. However, the book's answer is incorrect and not related to the problem at hand.
  • #1
DrummingAtom
659
2

Homework Statement



Find the Taylor Series of 1/x centered at c = 1.

Homework Equations



[tex] \sum_{n=0}^{\infty} f^n (c) \frac{(x-c)^n}{n!}[/tex]

The Attempt at a Solution



I made a list of the derivatives:

f(x) = 1/x
f'(x) = -1/x2
f''(x) = 2/x3
f'''(x) = -6/x4

f(1) = 1
f'(1) = -1
f''(1) = 2
f'''(1) = -6

From this I see the pattern fn(c) = (-1)n(n!)

[tex] \sum_{n=0}^{\infty} (-1)^n(n!) \frac{(x-1)^n}{n!}[/tex]

Then I canceled the factorials and I'm left with

[tex] \sum_{n=0}^{\infty} (-1)^n (x-1)^n[/tex]

Checked my answer and it's way off.. Thanks for any help.
 
Physics news on Phys.org
  • #2
DrummingAtom said:

Homework Statement



Find the Taylor Series of 1/x centered at c = 1.

Homework Equations



[tex] \sum_{n=0}^{\infty} f^n (c) \frac{(x-c)^n}{n!}[/tex]

The Attempt at a Solution



I made a list of the derivatives:

f(x) = 1/x
f'(x) = -1/x2
f''(x) = 2/x3
f'''(x) = -6/x4

f(1) = 1
f'(1) = -1
f''(1) = 2
f'''(1) = -6

From this I see the pattern fn(c) = (-1)n(n!)

[tex] \sum_{n=0}^{\infty} (-1)^n(n!) \frac{(x-1)^n}{n!}[/tex]

Then I canceled the factorials and I'm left with

[tex] \sum_{n=0}^{\infty} (-1)^n (x-1)^n[/tex]

Checked my answer and it's way off.. Thanks for any help.

How did you check your answer to conclude "it's way off"? I agree with your answer. I'll probably disagree with your "check".
 
  • #3
Hmm well that's reassuring. We've had a couple problems with the answers in this book being off.

The back of the book said the answer is:

[tex]
\sum_{n=0}^{\infty} \frac {(-1)^{n+1}(x-5)^n} {4^{n+1}}
[/tex]

I'll just go ahead and ignore that answer for now until next class. Thanks for your help.
 
  • #4
DrummingAtom said:
Hmm well that's reassuring. We've had a couple problems with the answers in this book being off.

The back of the book said the answer is:

[tex]
\sum_{n=0}^{\infty} \frac {(-1)^{n+1}(x-5)^n} {4^{n+1}}
[/tex]

I'll just go ahead and ignore that answer for now until next class. Thanks for your help.

The book's answer doesn't even converge at x=1 and it's not expressed in powers of (x-1). It looks like the answer to some completely different exercise. Yes, ignore it.
 

What is a Taylor series centered at c = 1?

A Taylor series is a mathematical representation of a function that allows us to approximate the value of the function at a specific point by using a sum of terms involving the function's derivatives at that point. When a Taylor series is centered at c = 1, it means that the series is expanded around the point x = 1.

Why is a Taylor series centered at c = 1 useful?

A Taylor series centered at c = 1 is useful because it allows us to approximate the value of a function at a point close to 1 by using a finite number of terms. This can be especially helpful when the function is too complex to evaluate directly at that point.

How is a Taylor series centered at c = 1 calculated?

To calculate a Taylor series centered at c = 1, we use the formula: f(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)^2/2! + f'''(1)(x-1)^3/3! + ... where f'(1), f''(1), f'''(1), etc. are the first, second, third, etc. derivatives of the function evaluated at x = 1.

What are some applications of a Taylor series centered at c = 1?

A Taylor series centered at c = 1 has many applications in mathematics and physics. It is commonly used in engineering to approximate the behavior of a system, in economics to model financial data, and in computer science to develop algorithms for numerical analysis.

Can a Taylor series centered at c = 1 be used to approximate any function?

No, a Taylor series centered at c = 1 can only be used to approximate functions that are infinitely differentiable at x = 1. If a function is not differentiable at x = 1, the Taylor series will not converge and will not provide an accurate approximation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
417
  • Calculus and Beyond Homework Help
Replies
2
Views
711
  • Calculus and Beyond Homework Help
Replies
1
Views
259
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
17
Views
619
  • Calculus and Beyond Homework Help
Replies
2
Views
188
Replies
8
Views
993
  • Calculus and Beyond Homework Help
Replies
6
Views
391
  • Calculus and Beyond Homework Help
Replies
4
Views
310
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top