Prove by induction on $n\geq1$ that $f^{(n)}(x)=n!a_n(1+x)^{\frac{1}{2}-n}$ for all $x\in(-1,\infty)$

Question

Prove by induction on $n\geq1$ that $f^{(n)}(x)=n!a_n(1+x)^{\frac{1}{2}-n}$ for all $x\in(-1,\infty)$ for $f(x)=\sqrt{(1+x)}$

I'm not sure how to prove this usually when I am asked to prove something by induction there are statements on both sides of the inequality that I can input values into but in this case I don't see how to do the proof.

Base case when n=1

$f^{(1)}(x)=1!a_1(1+x)^{\frac{1}{2}-1}$ $=(1+x)^{-\frac{1}{2}}$?

Induction hypothesis assume the equality above holds for n=k.

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

Inductive step show the inequality holds for $n=k+1$

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

Any advice would be greatly appreciated.

Thank you,

Answer 1

You must differentiate $$f^{(n)}(x)=n!a_n(1+x)^{\frac{1}{2}-n}$$ one more time with reapect to $x$! $$(f^{(n)}(x))'=n!a_n\left(\frac{1}{2}-n\right)(1+x)^{\frac{1}{2}-n-1}=...$$ chanching the variables from $n$ to $k$ we have $$(f^{(k)}(x))'=k!a_k\left(\frac{1}{2}-k\right)(1+x)^{\frac{1}{2}-k-1}$$ now we consider $$k!a_k\left( \frac{1}{2}-k\right)=k!\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\cdot ...\cdot \left(\frac{1}{2}-k+\right)\left(\frac{1}{2}-k\right)}{k!}=(k+1)!\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\cdot ...\cdot \left(\frac{1}{2}-k+1\right)\left(\frac{1}{2}-k\right)}{(k+1)!}$$ and this is $$(k+1)!a_{k+1}$$ so we have $$f^{(k+1})(x)=(k+1)!a_{k+1}(1+x)^{\frac{1}{2}-k-1}$$ and the proof ends here