How can I find the $n$ th derivatve of this function

Question

Let $f(x)=\sqrt{1+\sqrt{1-x}}\,\,\,\,\,\ \forall x\in[0,1].$

It is not difficult to find first few derivatives of this function as

$$f'(x)=-\dfrac1{2^2}(1+\sqrt{1-x})^{-1/2}(1-x)^{-1/2}$$

$$f''(x)=-\frac1{2^4}(1+\sqrt{1-x})^{-3/2}(1-x)^{-1}-\frac1{2^3}(1+\sqrt{1-x})^{-1/2}(1-x)^{-3/2}$$

$$f'''(x)=-\frac3{2^6}(1+\sqrt{1-x})^{-5/2}(1-x)^{-3/2}-\frac3{2^5}(1+\sqrt{1-x})^{-3/2}(1-x)^{-2}-\frac3{2^4}(1+\sqrt{1-x})^{-1/2}(1-x)^{-5/2}$$

However it is difficult to observe a pattern between these derivatives.
How can I find the $n$th derivative of $f$?

Did you mean this one? I just thought to use a subtitution like $x=\sin^2\theta.$ But not sure. — Bumblebee, Dec 28 '16 at 18:19
@Nil: that is a clever idea. You may also notice that $$ f(x)=\frac{1}{\sqrt{2}}\left(\sqrt{1-\sqrt{x}}+\sqrt{1+\sqrt{x}}\right)$$ — Jack D'Aurizio, Dec 28 '16 at 18:22
Are you trying to get the nth derivative as a function of x or do you want it to be evaluated at some point? — David Quinn, Dec 28 '16 at 18:22
You may also differentiate multiple times both sides of $$ f(x),f'(x) = -\frac{1}{4\sqrt{1-x}}.$$ — Jack D'Aurizio, Dec 28 '16 at 18:25
If you are just interested in $f^{(n)}(0)$, through the Taylor series of $\sqrt{1\pm x}$ it is straightforward to deduce the Taylor series of $$ f(x)=\frac{1}{\sqrt{2}}\left(\sqrt{1-\sqrt{x}}+\sqrt{1+\sqrt{x}}\right).$$ — Jack D'Aurizio, Dec 28 '16 at 18:28
@JackD'Aurizio: You give me a nice idea. I will try it. Thank you. — Bumblebee, Dec 28 '16 at 18:33

Olivier Oloa · Accepted Answer · 2016-12-29T02:25:06.230

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Edit. One may use the following result.

Hint. Let $f$ be $n$-times differentiable over some interval $(0,b).$ Then

$$ \left(\frac{d}{dx}\right)^n\left[f(\sqrt{x})\frac{}{}\right]=\sum_{k=0}^{n-1}(-1)^k\frac{(n+k-1)!}{k!(n-k-1)!}\frac{1}{(2\sqrt{x})^{n+k}}f^{(n-k)}(\sqrt{x}).\tag1 $$

Since $$ \sqrt{1+\sqrt{1-x}}=\frac1{\sqrt{2}}\left(\sqrt{1+\sqrt{x}}+\sqrt{1-\sqrt{x}}\right) $$then one may apply $(1)$ to $$ f(x)=\frac1{\sqrt{2}}\left(\sqrt{1+x}+\sqrt{1-x}\right),\qquad 0<x<1,\tag2 $$ observing that $$ f^{(k)}(x)=\frac1{\sqrt{2}}\left(\frac{(-1)^{k-1}(2k)!}{(2k-1)2^{2k}k!}\frac1{(1+x)^{k-\frac12}}-\frac{(2k)!}{(2k-1)2^{2k}k!}\frac1{(1-x)^{k-\frac12}}\right). \tag3 $$

edited Dec 29 '16 at 02:25

answered Dec 28 '16 at 19:07

Olivier Oloa

123,429

I differentiate few times and observe that your $(1)$ is absolutely correct. How did you obtain it? – Bumblebee Dec 30 '16 at 18:06
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@Nil I had this in my notes, but I've found a useful source (pp. 157-158): https://archive.org/details/Comtet_Louis_-_Advanced_Coatorics – Olivier Oloa Dec 30 '16 at 18:41
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Thanks for your helpful link. It explains everything even more than I expect :) – Bumblebee Dec 30 '16 at 19:20
@Nil You are very welcome. – Olivier Oloa Dec 30 '16 at 19:21

Jack D'Aurizio · Answer 2 · 2016-12-28T21:46:52.893

Just a partial answer. Since $$ f(x)=\frac{1}{\sqrt{2}}\left(\sqrt{1+\sqrt{x}}+\sqrt{1-\sqrt{x}}\right)\tag{1}$$ and $$ \sqrt{1-x} = -\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n-1)4^n}x^n \tag{2}$$ we have that: $$ \sqrt{1-x}+\sqrt{1+x} = -2\sum_{n\geq 0}\frac{\binom{4n}{2n}}{(4n-1)16^n} x^{2n} $$ and: $$ f(x) = -\sqrt{2}\sum_{n\geq 0}\frac{\binom{4n}{2n}}{(4n-1)16^n}\,x^{n} \tag{3} $$ so that: $$ f^{(n)}(0) = -\sqrt{2}\,\frac{\binom{4n-2}{2n-1}(n-1)!}{16^n}\tag{4} $$ for any $n\geq 1$. About the general expression of $f^{(n)}(x)$, there is no shame in exploiting our pattern matching abilities. It looks like: $$ f^{(n)}(x)=\sum_{k=1}^{n} c(n,k) \left(1+\sqrt{1-x}\right)^{-(2k-1)/2}(1-x)^{n-k/2} \tag{5}$$ and it is enough to differentiate both sides of $(5)$ wrt to $x$ to derive a recurrence relation for the coefficients $c(n,k)$.

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Nice derivation. (+1) – Olivier Oloa Dec 28 '16 at 21:16

How can I find the $n$ th derivatve of this function

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