Shortcut to taylor series of f, given taylor series of g

In summary, the conversation is about finding the series of f(x) = (e^{(x-1)^2}-1)/(x-1)^2, with a given Taylor series for g(x) = e^{(x-1)^2}. The person is struggling to find an easy way to solve it and is asking for help.
  • #1
54stickers
29
0
So, I have the series of [itex]g(x) = e^{(x-1)^{2}} = 1 + (x-1)^{2} + \frac{(x-1)^{4}}{2} + \frac{(x-1)^{6}}{6} + ... + \frac{(x-1)^{2n}}{n!} [/itex]

and I am asked to find the series of [itex]f(x) = \frac{e^{(x-1)^{2}}-1}{(x-1)^{2}}[/itex] for x [itex]\neq[/itex] 1 and f(1) = 1. The Taylor series is centered about x = 1

I am told that there is an easy way to do this, but I don't see it.

any help would be appreciated, thanks
 
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  • #2
54stickers said:
So, I have the series of [itex]g(x) = e^{(x-1)^{2}} = 1 + (x-1)^{2} + \frac{(x-1)^{4}}{2} + \frac{(x-1)^{6}}{6} + ... + \frac{(x-1)^{2n}}{n!} [/itex]

and I am asked to find the series of [itex]f(x) = \frac{e^{(x-1)^{2}}-1}{(x-1)^{2}}[/itex] for x [itex]\neq[/itex] 1 and f(1) = 1. The Taylor series is centered about x = 1

I am told that there is an easy way to do this, but I don't see it.

any help would be appreciated, thanks

1. What do you get for e(x - 1)2 - 1?
2. What do you get if you divide the result from #1 by (x - 1)2?

Is this a homework problem?
 
  • #3
I see it now, thanks

It's an old AP problem that I am doing to prepare for the test
 

1. How do you find the Taylor series of f from the Taylor series of g?

The Taylor series of f can be found by taking the derivatives of g and plugging them into the general formula for the Taylor series. This formula is f(x) = g(a) + g'(a)(x-a) + g''(a)(x-a)^2/2! + g'''(a)(x-a)^3/3! + ...

2. Is the Taylor series the same for all functions?

No, the Taylor series will vary depending on the function. Each function has its own unique series based on its derivatives and the point at which it is being expanded.

3. How do you know when to stop adding terms in the Taylor series?

The Taylor series is an infinite series, but we can approximate the function by using a finite number of terms. The error in this approximation decreases as we add more terms, so it is up to the individual to decide how many terms to use based on the desired level of accuracy.

4. Can the Taylor series be used to find the value of a function at any point?

Yes, the Taylor series can be used to find the value of a function at any point within its radius of convergence. This means that it can only be used for values of x that are close to the point at which the series is expanded.

5. What is the relationship between the Taylor series and the original function?

The Taylor series is an infinite polynomial that is used to approximate the original function. As we add more terms, the approximation becomes more accurate and will eventually converge to the original function within its radius of convergence. This allows us to use the Taylor series to study the behavior and properties of a function.

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