Find the taylor series of the function $$f(x) = 8(x-12)ln(x-12)$$ about $x = 13$. Give the taylor polynomial of degree 3 as your answer. Then find the interval of convergence for the series.

I found the taylor polynomial of degree 3 to be the following: $$8(x-13) + \frac82(x-13)^2 -\frac43(x-13)^3$$ not sure if this is right though. I haven't been able to write the series in sigma notation, and therefore haven't been able to find the interval of convergence. The problem doesen't require you to write out the series in sigma form though, so if there's any other way to find the intervall of convergence, that would work as well.


1 Answer 1


The Taylor expansion of a function $f$ about $x=13$ to get a third degree polynomial is (ignoring the rest, infinite terms :

$$f(x) = f(13) + f'(13) + \frac{f''(13)}{2!}(x-13)^2 + \frac{f'''(13)}{3!}(x-13)^3 + \dots$$

substituting for your given function $f$ should give you an expression.

You can calculate the first few derivatives and check to see if you can recognize a pattern.

The expansion around $x=13$ of your given function, which you should make out after some calculations, is :

$$\sum_{n \geq 2}\frac{8(-1)^n(x-13)^n}{n(n-1)} + 8(x-13)$$

After you've figured that stuff out, can you figure out the convergence ?

  • $\begingroup$ Yes, i was able to figure out the interval of convergence from the series you wrote out. Do you think that generally, for any taylor series centered around $x = a$ you could find the interval of convergence from the absolute value of $$(x-a)< 1 ?$$ $\endgroup$
    – Pame
    Nov 18, 2017 at 12:17

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