# Find Taylor Series from a function and its interval of convergence

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1. Apr 8, 2016

### deagledoubleg

Let f(x) = (1+x)-4
Find the Taylor Series of f centered at x=1 and its interval of convergence.

$$\sum_{n=0}^\infty f^n(c)\frac{(x-c)^n}{n!}$$ is general Taylor series form

My attempt

I found the first 4 derivatives of f(x) and their values at fn(1). Yet from here I do not know how to find the taylor series, from there I should be able to finish it myself. Any ideas on how to find the Taylor series here?

Last edited: Apr 8, 2016
2. Apr 8, 2016

### PeroK

How did you find the derivatives?

3. Apr 8, 2016

### deagledoubleg

f0(x)= $$\frac{(1)}{(1+x)^4}$$
f1(x)= $$\frac{(-4)}{(1+x)^5}$$
f2(x)= $$\frac{(20)}{(1+x)^6}$$
f3(x)= $$\frac{(-120)}{(1+x)^7}$$
f4(x)= $$\frac{(840)}{(1+x)^8}$$

I think that I got the Taylor Series of $$f^n(x)=\frac{(-1)^(n+1)*(n-1)!}{2^(n+4)}$$

Last edited: Apr 8, 2016
4. Apr 8, 2016

### PeroK

Can you fix that latex? In any case, it doesn't look right. It looks like a mixture of parts of the solution.

5. Apr 8, 2016

Fixed

6. Apr 8, 2016

### PeroK

You can see that your general formula doesn't match the first four derivatives.

7. Apr 8, 2016

### deagledoubleg

True, I can see that now. I think the issue is the (n-1)! in the numerator. I can't find something that works because I want it to be n! but to stop at 4. like 7*6*5*4, but not include *3*2*1. To do this would I just divide that by 6?

8. Apr 8, 2016

### PeroK

For the time being, just calculate $f^{(n)}(x)$. You're not aiming at anything.

And, yes, $7*6*5*4 = 7!/6$

9. Apr 8, 2016

### deagledoubleg

I think that I have $$f^n(x)=\frac{(-1)^(n)*\frac{(n-1)!}{6}}{(1+x)^(n+4)}$$

10. Apr 8, 2016

### PeroK

You need to check that $(n-1)!$.

11. Apr 8, 2016

### deagledoubleg

Does the above look better?