# Taylor series proof by induction

I am kind of stuck with this problem. I know the basics of induction proof and how to use it to prove stuff about basic arithmetic and geometric series and so on, but the problem below seems a bit new and unfamiliar to me.

The problem is about proving how a Taylor series develops using induction proof.

The problem goes like this:

Given the function $$f(x)=\frac{1}{x^2}$$

Prove by induction that $$f^{(k)}(x)=\frac{(-1)^{k}(k+1)!}{x^{k+2}}$$ for all $$k\geq 0$$.

Is there anyone out there who is willing to provide a solution to this problem and help me and expand my horizon/ help in my learning?

All help is welcomed and appreciated.

• Can you do the base step ($k=0$)? Can you do the inductive step (if $k=n$ works, so does $k=n+1$)? Edit in your efforts so we know where you need help.
– J.G.
Oct 17, 2021 at 17:38
• @J.G.: To be specific it is the last step $(k=n+1)$ where I'm unsure of how to make the expression the same on both sides of the equals sign and basically prove the problem Oct 17, 2021 at 17:47
• Then you should edit in your attempt to prove$$\frac{d}{dx}\frac{(-1)^n(n+1)!}{x^{n+2}}=\frac{(-1)^{n+1}(n+2)!}{x^{n+3}}.$$
– J.G.
Oct 17, 2021 at 17:48

Let $$g(k)=f^{(k)}(x)=\frac{(-1)^k(k+1)!}{x^{k+2}}$$

For $$g(0)$$, $$f(x)=\frac{1×1!}{x^2}$$

$$g(1)=\frac{\partial(x^{-2})}{\partial{x}}=\frac{-2}{x^3}=\frac{(-1)(2)!}{x^{3}}$$

Therefore $$g(k)$$ is true for $$k=0$$ and $$1$$.

Let $$g(k)=f^{(k)}(x)=\frac{(-1)^k(k+1)!}{x^{k+2}}$$ be true for all $$k≥2$$.

Therefore,$$g(p)=f^{(p)}(x)=\frac{(-1)^p(p+1)!}{x^{p+2}}$$

And

$$g(p+1)=f^{(p+1)}(x)=\frac{(-1)^{p+1}(p+2)!}{x^{p+3}}$$$$\Rightarrow\frac{\partial g(p)}{\partial x}=\frac{(-1)^{p+1}(p+2)!}{x^{p+3}}\Rightarrow\frac{\partial{\frac{(-1)^{p}(p+1)!}{x^{p+2}}}}{\partial x}=\frac{(-1)^{p+1}(p+2)!}{x^{p+3}}$$

Now we need to calculate $$\frac{\partial{\frac{(-1)^{p}(p+1)!}{x^{p+2}}}}{\partial x}$$ Taking the constants out, we are left with $$(-1)^p(1+p)!\frac{\partial{(x^{-p-2}})}{\partial x}$$ This gives us $$(-1)^p(1+p)!(-2-p)(x^{-p-3})$$$$\Rightarrow (-1)^p(1+p)!(-1)(p+2)\frac{1}{x^{p+3}}\Rightarrow\frac{(-1)^{p+1}(p+2)!}{x^{p+3}}$$

Therefore for every $$p$$, $$g(p)$$ as well as $$g(p+1)$$ is true.

Hence proved.

I am not very sure about what I have done is correct or not.... Maybe there is some mistake. For that pardon me.

• Thank you for providing a clear and well-formulated solution! So far I haven't been able to spot any flaws in your solution. If I come across any, I will let you know. Either way, I still appreciate your help :-) Oct 17, 2021 at 18:05
• Everything looks pretty solid from what I can tell. The only thing I thought was a bit unclear was in your last line. How did you make $(1+p)!\cdot(p+2)=(p+2)!$? Oct 17, 2021 at 19:08
• $$(1+p)!.(p+2)=(p+2).(p+1)!=(p+2).(p+1).p.(p-1).(p-2)......3.2.1=(p+2)!$$ I think now it's clear :) Oct 18, 2021 at 1:03
• You can take it as a property $$n(n-1)!=n!$$ and put $n=p+2$ If you still have any doubt you can refer to "Defination" article of en.m.wikipedia.org/wiki/Factorial Oct 18, 2021 at 1:08
• Thank you so much! Thanks to you I now have a couple of more tools in my mathematical toolbelt :-) Oct 18, 2021 at 6:22