Taylor expansion

In summary: O(x^2))/(x-x^2/2+x^3/6-x^4/24+x^5/120 ...) = 1+O(x) = 1+O(exp(-1))= 1+O(0.367879) = 1.367879...In summary, the conversation discusses how to expand the expression (1-exp(-1))^-1 as a Taylor series. The suggested approach is to first use long division to get a Taylor series for 1/(1-x), and then substitute x=exp(-1) into that series. The final result is an exact Taylor expansion with only one term, unless the problem was meant to find an approximate value for
  • #1
Callisto
41
0
Hi

How do you expand

(1-exp(-1))^-1

as Taylor series

Callisto
 
Last edited:
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  • #2
Try long division to get a Taylor (or McClaurin) series for f(x) = 1/(1-x) for a start.

Then put x=exp(-1) into that series, and there you go!
 
  • #3
Taylor series for

1/(1-x) = (1-x)^-1 = 1 + x^2 + x^3 + x^4 +...= SUMi x^i

Then sub in exp(-1) for x
That's it?
 
  • #4
Yes. That's it.
 
  • #5
Cheers James R!
 
  • #6
a constant has a trivial Taylor expansion

The expression you submitted is a constant, it doesn't depend on x.
All derivatives of this function with respect to x (the assumed variable) are zero.

Therefore the Taylor expansion contains only one term (and is exact):

f(x) = 1/(1-exp(-1)) + 0 + 0 + 0 + 0 + 0 + 0 + ... = 1/(1-exp(-1))

(unless you did not formulate your problem correctly)
 
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  • #7
Technically, you're right of course, lalbatros, but your solution doesn't seem to fit the spirit of the question which was asked. Maybe the question was a little unclear, though...
 
  • #8
You are probably right James.
The question probably was: "how can I get an approximate value for 1/(1-exp(-1)) using Taylor expansions".
Then, there are numerous ways to choose how to expand.
For example:

1/(1-exp(-x)) = 1 /(1-sumi((-x)^i / i!))

or explicitely:

1/(1-exp(-x)) = 1/(x-x^2/2+x^3/6-x^4/24+x^5/120 ...)
 

What is Taylor expansion?

Taylor expansion is a mathematical concept that allows us to approximate a complex function using a simpler polynomial function. It is named after the mathematician Brook Taylor.

Why do we use Taylor expansion?

We use Taylor expansion to simplify complex functions and make them easier to work with. It also allows us to approximate values of a function at a certain point without having to know the exact function.

What is the formula for Taylor expansion?

The formula for Taylor expansion is given by:
f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... + f^n(a)(x-a)^n/n!
where f(x) is the function we want to approximate, a is the point around which we are approximating, and f'(a), f''(a), f'''(a), ..., f^n(a) are the derivatives of the function at point a.

What is the difference between Taylor series and Taylor expansion?

Taylor series is the infinite expansion of a function, while Taylor expansion is a finite approximation of a function using a certain number of terms. Taylor series is used to represent an entire function, while Taylor expansion is used to approximate a function at a specific point.

What are the applications of Taylor expansion?

Taylor expansion has various applications in fields such as physics, engineering, and economics. It is used to approximate solutions to differential equations, calculate derivatives and integrals, and make predictions in various scientific and mathematical models.

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