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I need to find the infinite taylor series of $ \sqrt{x+1} $.

I tried to just find the deriviatives of $ \sqrt{x+1} $ and search for a pattern, and then prove by induction that indeed the pattern I found was correct, and then prove the Lagrange reminder strive to $ 0 $ as $ n\to \infty $ but it got very complicated and seems not effiecient.

Is there a reasonable way to find this taylor series ? This is a question from an exam, so I guess it shouldnt take too much work (as what I tried) because the time is limited.

Thanks in advance.

Here's what I found:

For any $ 1\leq k $ :

$ f^{\left(k\right)}\left(x\right)=\prod_{n=1}^{k}\left(\frac{1}{2}-n+1\right)\left(x+1\right)^{\left(\frac{1}{2}-k\right)} $

Thus, if we'll name the taylor expansion as $ T_{f,0} $ (taylor expansion of $ f=\sqrt{x+1} $ around $ 0 $ ) we'll get:

$ T_{f,0}=1+\sum_{k=1}^{m}\frac{f^{\left(k\right)}\left(0\right)}{m!}x^{m}=\sum_{k=1}^{m}\frac{1}{k!}\left(\prod_{n=1}^{k}\left(\frac{1}{2}-n+1\right)\right)x^{k} $

Thus, the reminder should be:

$ R_{m}\left(x\right)=\frac{f^{\left(m+1\right)}\left(x_{0}\right)}{\left(m+1\right)!}x^{m+1}=\frac{1}{\left(m+1\right)!}\prod_{n=1}^{m+1}\left(\frac{1}{2}-n+1\right)\left(x_{0}+1\right)^{\frac{1}{2}-\left(m+1\right)}x^{m+1} $

FreeZe
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2 Answers2

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The derivatives of $(1+x)^\alpha$ are relatively easy to find:

$$(1+x)^\alpha\to\alpha(1+x)^{\alpha-1}\to\alpha(\alpha-1)(1+x)^{\alpha-1}\to\alpha(\alpha-1)(\alpha-2)(1+x)^{\alpha-3}\to\cdots$$

and evaluate at $x=0$ as the falling factorials $(\alpha)_k$.

Then the Lagrange remainder reads

$$\frac{(\alpha)_{n+1}}{(n+1)!}(1+x^*)^{\alpha-n-1}x^{n+1}=\frac{(\alpha)_{n+1}}{(n+1)!}(1+x^*)^{\alpha}\left(\frac x{1+x^*}\right)^{n+1},$$ where $0\le|x^*|<|x|.$ Then for $-\frac12<x<1$, the last factor ensures an exponential decay (the others are bounded).

  • I wrote explicitly what you called $ \left(\alpha\right){n+1} $. thats the expression: $ |R{m}\left(x\right)|=\frac{1}{\left(m+1\right)!}|\prod_{n=1}^{m+1}\left(\frac{1}{2}-n+1\right)|\left(x_{0}+1\right)^{\frac{1}{2}}|\frac{x}{x_{0}+1}|^{m+1} $. How do you prove that $ \underset{m\to\infty}{\lim}|R_{m}\left(x\right)|=0 $ ? \ – FreeZe Aug 17 '20 at 09:28
  • @Waizman: said in my answer. –  Aug 17 '20 at 09:44
  • "We can see exponential decay" isnt mathematical proof, isnt it? – FreeZe Aug 17 '20 at 09:45
  • @Waizman: depends on the requested level of detail. –  Aug 17 '20 at 09:46
  • @YvesDaoust I am intrigued why the proof that the remainder converges to 0 requires $x>-\frac{1}{2}$ while the radius of convergence is 1. – Miguel Aug 17 '20 at 10:16
  • @Miguel for X>-0.5 this quotient will not strive to 0. I guess the limit calculation would be harder, by the qay I still couldnt prove that the limit ia actually 0, I tried to use sterlong approximantion for k! And it didnt lead my anywhere. If someone can prove it Ill be thankful – FreeZe Aug 17 '20 at 11:37
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    @Miguel: the Lagrange remainder leaves an uncertainty on $x^*$, so it is not "tight". –  Aug 17 '20 at 18:34
  • @YvesDaoust Then if we take for granted that the function is analytic finding the interval of convergence is stronger and easier than messing with the limit of the remainder? – Miguel Aug 17 '20 at 20:06
  • @Miguel: that's possible. –  Aug 17 '20 at 20:12
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I understand that you mean the Taylor series at 0. If you differentiate repeatedly, you see the pattern: $$ \begin{aligned} f(x)&=\sqrt{x+1} \\\\ f'(x)&=\frac{1}{2}\frac{1}{\sqrt{x+1}} \\\\ f''(x)&=-\frac{1}{2\cdot 2}\frac{1}{(x+1)^{\frac{3}{2}}} \\\\ f'''(x)&=\frac{1\cdot 3}{2\cdot 2\cdot 2}\frac{1}{(x+1)^{\frac{5}{2}}} \\\\ f^{(4)}(x)&=-\frac{1\cdot 3\cdot 5}{2\cdot 2\cdot 2\cdot 2}\frac{1}{(x+1)^{\frac{7}{2}}} \\\\ f^{(n)}(x)&=\dots \end{aligned} $$

Observe that:

  • The signs are alternating.
  • The numerators have the products of odd numbers.
  • The denominators have powers of two.
  • The powers of $\sqrt{x+1}$ are decreasing.

So we can guess the general form: $$ f^{(n)}(x)=(-1)^{n+1}\,\frac{1\cdot 3 \cdots (2n-3)}{2^n}\frac{1}{(x+1)^{\frac{2n+1}{2}}} $$ for $n\geq1$ whereas $f(0)=1$.

The proof by induction is simply taking another derivative. If you want a nice formula for the numerator, you can use this: $$ f^{(n)}(x)=(-1)^{n+1}\,\frac{(2(n-1))!}{2^{n-1}\,(n-1)!\,2^n}\frac{1}{(x+1)^{\frac{2n+1}{2}}} $$ where again the sign for the case $n=0$ is exceptional, since $f(0)=1$. Then the series is: $$ S(x)=1+\sum_{n=1}^\infty (-1)^{n+1}\,\frac{(2(n-1))!}{2^{n-1}\,(n-1)!\,2^n}\,\frac{x^n}{n!} $$

Finally, you could prove that the remainder converges to zero, but I think that it is easier (yet a bit less rigourous) to argue that the function is analytic in a neighbourhood of 0 (assuming known that $\sqrt{x}$ is analytic at $x=1$) then the series converges to $f(x)$ in the interval of convergence (which can be easily proved to be $(-1,1]$).

As for the requirement of being reasonable, I would put this question in a first-year Calculus exam and give half an hour.

Miguel
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  • "argue that the function is analytic in a neghbourhood of 0 " dont we need to prove that the reminder converges to zero in order to argue this claim ? – FreeZe Aug 17 '20 at 09:32
  • @Waizman It depends on what you take for granted. You may assume that $x^n$ is analytic in a neighbourhood of any positive point but, yes, there is some handwaving here. – Miguel Aug 17 '20 at 09:35
  • I think the form you suggested for the $ kth $ derivitative is not trivial, what I found was $ f^{\left(k\right)}\left(x\right)=\prod_{n=1}^{k}\left(\frac{1}{2}-n+1\right)\left(x+1\right)^{\left(\frac{1}{2}-k\right)} $, which makes things more complicated when you want to show the limit is zero in order to argue that the function is analytic at $ 0 $, also the form I reached dosent hold for $k=0 $ – FreeZe Aug 17 '20 at 09:39
  • I just noticed. This wasnt the function I asked about. my function is $ f\left(x\right)=\sqrt{x+1} $. you calculated taylor series for another function. (This is different because the pattern in your function is easier to find). – FreeZe Aug 17 '20 at 09:48
  • The pattern is exactly the same, just one step ahead. – Miguel Aug 17 '20 at 09:58
  • Except that the original function and the first deriviative is positive, so the series isnt alternating signs and its more xomplicated finding a closed formula for the $kth$ deriviative that will hold for any k – FreeZe Aug 17 '20 at 10:06
  • @Waizman The series is alternating except for the case $n=0$. Some exceptional cases at the beginning are not important to find the pattern. It happens with the logarithm too. – Miguel Aug 17 '20 at 10:07
  • But as you presented the series : 1+(sum), this is not rhe regular form of a power series as we define it, how do you calculate the convergence radius with Cauchy-Hadamard theorem in that case? What would be the sequence of the seriea which we should calculate the limsup of? – FreeZe Aug 17 '20 at 10:15
  • @Waizman In this case the quotient form would be much easier to compute the limit. Since you take the limit $n\to\infty$ the value $n=0$ is irrelevant. – Miguel Aug 17 '20 at 10:19
  • Im sorry to bother once again, but how come the first formula you presented : "so we can guess the general form:" dosent hold for n=1 ( we get the right result but multyplied by (-1)). But the second formula works. What am I missing here? – FreeZe Aug 17 '20 at 10:49