Find Taylor Series from a function and its interval of convergence

[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?

 

[PeroK]

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?

How did you find the derivatives?

 

[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)}

 

[PeroK]

f⁰(x)=(1+x)^-4
f¹(x)= \frac{-4}{(1+x)⁵} [\itex]<br /> f²(x)= \frac{20}{(1+x)&lt;sup&gt;6&lt;/sup&gt;} [\itex]&lt;br /&gt; f&lt;sup&gt;3&lt;/sup&gt;(x)= \frac{-120}{(1+x)&amp;lt;sup&amp;gt;7&amp;lt;/sup&amp;gt;} [\itex]&amp;lt;br /&amp;gt; f&amp;lt;sup&amp;gt;4&amp;lt;/sup&amp;gt;(x)= \frac{840}{(1+x)&amp;amp;lt;sup&amp;amp;gt;8&amp;amp;lt;/sup&amp;amp;gt;} [\itex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; I think that I got the Taylor Series of f&amp;amp;amp;lt;sup&amp;amp;amp;gt;n&amp;amp;amp;lt;/sup&amp;amp;amp;gt;(x)=\frac{(-1)&amp;amp;amp;lt;sup&amp;amp;amp;gt;n+1&amp;amp;amp;lt;/sup&amp;amp;amp;gt;*(n-1)!}{2&amp;amp;amp;lt;sup&amp;amp;amp;gt;n+4&amp;amp;amp;lt;/sup&amp;amp;amp;gt;}

&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Can you fix that latex? In any case, it doesn&amp;amp;amp;#039;t look right. It looks like a mixture of parts of the solution.

 

[deagledoubleg]

Fixed

 

[PeroK]

Fixed

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

 

[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?

 

[PeroK]

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?

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

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

 

[deagledoubleg]

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

 

[PeroK]

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

 

[deagledoubleg]

Does the above look better?