How to prove the following successive differential by means of mathematical induction.

Question

In my previous question, (asked here: Prove the following successive differential)

I discovered how $f^n(x)$ of the given function can be $\displaystyle \frac{n!}{2a}\left(\frac{1}{(a-x)^{n+1}}+\frac{(-1)^n}{(a+x)^{n+1}}\right)$, and was then told that I can prove that this statement is true by means of induction on $n$.

For this to be done, I had to calculate $f^{n+1}(x)$ or $\displaystyle\frac{d}{dx}\left(\frac{n!}{2a}\left(\frac{1}{(a-x)^{n+1}}+\frac{(-1)^n}{(a+x)^{n+1}}\right)\right)$

The following is a rigorous proof

Base step:

$$f^1(x)=\displaystyle \frac{1!}{2a}\left(\frac{1}{(a-x)^{1+1}}+\frac{(-1)^1}{(a+x)^{1+1}}\right)$$

$$=\frac{1}{2a}\left(\frac{1}{(a-x)^2}+\frac{(-1)}{(a+x)^2}\right)$$

which is true. Hence we assume it is true for $f^n(x)$ and proceed to prove it for $f^{n+1}(x)$

Inductive step:

Firstly, I applied the product rule of differentiation to attain the following equation: $$f^{n+1}(x)=\displaystyle \frac{d}{dx}\left(\frac{n!}{2a}\right)\cdot\left(\frac{1}{(a-x)^{n+1}}+\frac{(-1)^n}{(a+x)^{n+1}}\right)+\frac{n!}{2a}\cdot\frac{d}{dx}\left(\frac{1}{(a-x)^{n+1}}+\frac{(-1)^n}{(a+x)^{n+1}}\right)$$

$$= \displaystyle \frac{n!}{2a}\cdot\left(\frac{d}{dx}\left(\frac{1}{(a-x)^{n+1}}\right)+\frac{d}{dx}\left(\frac{(-1)^n}{(a+x)^{n+1}}\right)\right)$$

from here, to avoid the use of the quotient rule, I decided to rewrite the fractions as $(a-x)^{-n-1}$ and $(-1)^n(a+x)^{-n-1}$ and used the chain rule, making the substitution $u=a-x$ and $n=a+x$

$$\displaystyle =\frac{n!}{2a}\cdot\left((-n-1)u^{-n-2}\frac{du}{dx}+(-1)^n(-n-1)n^{-n-2}\frac{dn}{dx}\right)$$

and then substituted the original expression for $u$ and $n$ back into the equation

$$= \displaystyle \frac{n!}{2a}\cdot\left((-n-1)(a-x)^{-n-2}(-1)+(-1)^n(-n-1)(a+x)^{-n-2}\right)$$

$$\displaystyle =\frac{n!}{2a}\cdot\left(\frac{(-n-1)(-1)}{(a-x)^{n+2}}+\frac{(-1)^n(-n-1)}{(a+x)^{n+2}}\right)$$

$$\displaystyle = \frac{n!}{2a}\cdot\left(\frac{n+1}{(a-x)^{n+2}}+\frac{(-1)^{n+1}(n+1)}{(a+x)^{n+2}}\right)$$

note; to attain $(-1)^{n+1}$, I factored out $(-1)$ from $(-n-1)$, and then multiplied it by $(-1)^n$

$$=\frac{(n+1)(n!)}{2a}\cdot\displaystyle \left(\frac{1}{(a-x)^{n+2}}+\frac{(-1)^{n+1}}{(a+x)^{n+2}}\right)$$

$$=\displaystyle\frac{(n+1)!}{2a}\left(\frac{1}{(a-x)^{n+2}}+\frac{(-1)^{n+1}}{(a+x)^{n+2}}\right)$$

which completes the proof.

Long question short, are there any errors in this proof? (In addition, if I made any mistakes while typing up the proof, such as sign errors or missing terms, please either edit it or notify me. Thank you)

Answer 1

First of all (not a mistake, but it can be done better):

$$f^{n+1}(x)=\displaystyle \frac{d}{dx}\left(\frac{n!}{2a}\right)\cdot\left(\frac{1}{(a-x)^{n+1}}+\frac{(-1)^n}{(a+x)^{n+1}}\right)+\frac{n!}{2a}\cdot\frac{d}{dx}\left(\frac{1}{(a-x)^{n+1}}+\frac{(-1)^n}{(a+x)^{n+1}}\right)$$

In this case you can use the property, that $\frac{d}{dx}(\alpha g(x))= \alpha(\frac{d}{dx} g(x)) $, so $$f^{n+1}(x)=\displaystyle \frac{n!}{2a}\cdot\frac{d}{dx}\left(\frac{1}{(a-x)^{n+1}}+\frac{(-1)^n}{(a+x)^{n+1}}\right)$$ Less writing mean less opportunities to make a mistake $\ddot\smile$

making the substitution (...) $n=a+x$

You are already using $n$, so You should use another sign, for example $v$.

The rest of your proof looks OK.