i am giving $e^{\sqrt{x^2+1}}$ and asked to find the Maclaurin series for this term.

Here is my solution:

let $u=\sqrt{x^2+1}$, and given that we know that Maclaurin series for $e^x= 1+x+\frac{x^2}{2!} ...$ then: $$e^u= 1+u+\frac{u^2}{2}+...$$ hence: $$e^{\sqrt{x^2+1}}=1+\sqrt{x^2+1}+\frac{x^2+1}{2}+...$$

Am I doing it right? please help

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    $\begingroup$ Presumably, mean $u=\sqrt{x^2+1}$? $\endgroup$ – Thomas Andrews Aug 19 '17 at 20:28
  • $\begingroup$ No, you can only have powers of $x$, not $x$ inside a square-root. Next: You will use $e^u = 1 + u + ...$ only when $u$ is close to zero. $\endgroup$ – GEdgar Aug 19 '17 at 20:30
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    $\begingroup$ @ThomasAndrews you are correct, i mean i . dont see why it would work when you have $ln(x+1)$ but not when you have a bit more complex expression $\endgroup$ – Reddevil Aug 19 '17 at 20:33
  • $\begingroup$ Maybe you were expected to directly differentiate the first few times... $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 20:51
  • $\begingroup$ @GEdgar $e^u=1+u+\dots$ for any $u\in\Bbb C$, as $e^z$ is an entire function? $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 21:39

You can get the first terms of the series computing the derivatives of $f(x)=e^{\sqrt{x+1}}$. You get:$$\begin{align}f(x)&=e^{\sqrt{x+1}}\\f'(x)&=\frac{e^{\sqrt{x+1}}}{2\sqrt{x+1}}\\f''(x)&=\frac{e^{\sqrt{x+1}} \left(\sqrt{x+1}-1\right)}{4(x+1)^{3/2}}\\f^{(3)}(x)&=\frac{e^{\sqrt{x+1}} \left(x-3 \sqrt{x+1}+4\right)}{8(x+1)^{5/2}}\\f^{(4)}(x)&=\frac{e^{\sqrt{x+1}} \left(x \left(\sqrt{x+1}-6\right)+16\sqrt{x+1}-21\right)}{16(x+1)^{7/2}}\\f^{(5)}(x)&=\frac{e^{\sqrt{x+1}} \left(x^2+\left(47-10 \sqrt{x+1}\right) x-115 \sqrt{x+1}+151\right)}{32(x+1)^{9/2}}\end{align}$$Therefore$$\begin{align*}f(0)&=e\\f'(0)&=\frac e2\\\frac{f'(0)}{2!}&=0\\\frac{f^{(3)}(0)}{3!}&=\frac e{48}\\\frac{f^{(4)}(0)}{4!}&=-\frac{5e}{384}\\\frac{f^{(5)}(0)}{5!}&=\frac{3e}{320}\end{align*}$$So, the first terms of the Maclaurin series of $f(x)$ are$$e+\frac e2x+\frac e{48}x^3-\frac{5e}{384}x^4+\frac{3e}{320}x^5$$and therefore the first terms of the Maclaurin series of $e^{\sqrt{x^2+1}}$ are$$e+\frac e2x^2+\frac e{48}x^6-\frac{5e}{384}x^8+\frac{3e}{320}x^{10}.$$

  • $\begingroup$ Basically a copy of Raffaele's answer, which was posted almost 40 minutes ago. Suggesting deletion of this answer. $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 21:42
  • $\begingroup$ No, it is not. My computations are far simpler. If you don't believe me, try for yourself to compute the second derivative of $e^{\sqrt{x^2+1}}$ and compare it with the computation of the second derivative of $e^{\sqrt{x+1}}$. $\endgroup$ – José Carlos Santos Aug 19 '17 at 21:44
  • $\begingroup$ Ah, I missed that. I suppose it is fine then :-) $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 21:46

You have to compute the derivatives


$f^1(x)=\dfrac{e^{\sqrt{x^2+1}} x}{\sqrt{x^2+1}},\\f^2(x)=\dfrac{e^{\sqrt{x^2+1}} \left(\sqrt{x^2+1} x^2+1\right)}{\left(x^2+1\right)^{3/2}},\\f^3(x)=\dfrac{e^{\sqrt{x^2+1}} x \left(x^4+x^2+3 \sqrt{x^2+1}-3\right)}{\left(x^2+1\right)^{5/2}},\\f^4(x)=\dfrac{e^{\sqrt{x^2+1}} \left(6 \left(3-2 \sqrt{x^2+1}\right) x^2+3 \left(\sqrt{x^2+1}-1\right)+\sqrt{x^2+1} x^6+\left(\sqrt{x^2+1}+6\right) x^4\right)}{\left(x^2+1\right)^{7/2}},\ldots$

And then evaluate them at $x=0$

$f^0(0)=e,\;f^1(0)=0,\;f^2(0)=e,\;f^3(0)=0,\;f^4(0)=0,\;f^5(0)=0,\;f^6(0)=15 e,\;f^7(0)=0,\;f^8(0)=-525 e,\;f^9(0)=0,\;f^{10}(0)=34020 e$

And use the formula $$f(x)=\sum _{n=0}^{\infty } \frac{f^n(0) x^n}{n!}$$

and get $$f(x)=e\left(1+\frac{x^2}{2} +\frac{x^6}{48}-\frac{5 x^8}{384}+\frac{3 x^{10}}{320}+O(x^{11})\right)$$ Hope this helps

  • $\begingroup$ compute derivatives is never a good idea, especially in this case... $\endgroup$ – Surb Aug 19 '17 at 21:03
  • $\begingroup$ @Surb Let's hear your alternative way! $\endgroup$ – Jyrki Lahtonen Aug 19 '17 at 21:08
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    $\begingroup$ @Surb Its good if the original problem only asks for the first few terms. $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 21:19

That's the right path. So far, we have


However, this is certainly not a Maclaurin expansion for the reason that $\sqrt{x^2+1}$ is not a polynomial. This can be taken care of using the generalized binomial expansion, which has


And thus,


Likely, you do not like your terms like this and would like to collect like terms. This may be done:

$$e^{\sqrt{x^2+1}}=\sum_{n=0}^\infty a_nx^n$$



Likely, $a_n=eq_n$ for rational $q$. Since the above method does not provide good forms for $a_n$, it may be better to differentiate and equate terms:

$$f(x)=e^{\sqrt{x^2+1}}\\f'(x)=\frac x{\sqrt{x^2+1}}f(x)$$

One can see that

$$f(x)=\sum_{k=0}^\infty a_kx^k$$

$$f'(x)=\sum_{k=1}^\infty a_kkx^{k-1}=\sum_{k=0}^\infty a_{k+1}(k+1)x^k$$

$$\frac x{\sqrt{x^2+1}}=x(1+x)^{-1/2}=x\sum_{k=0}^\infty\binom{-1/2}kx^k$$

Thus, by Cauchy products,

$$\frac x{\sqrt{x^2+1}}f(x)=x\sum_{k=0}^\infty\sum_{n=0}^k\binom{-1/2}na_{k-n}x^k=\sum_{k=0}^\infty a_{k+1}(k+1)x^k=f'(x)$$

Equating parts, we find that


  • $\begingroup$ could you have done it any more complicated than so? i needed the maclaurin polynomial for the expression, sorry but this looks too sophisticated for me, im just a high school student. $\endgroup$ – Reddevil Aug 19 '17 at 20:41
  • $\begingroup$ If anyone wants the closed form for $a_n$, it will likely come out much nicer to apply Faà di Bruno's formula twice and directly use Taylor expanding with derivatives. Trying to solve $a_n$, WolframAlpha gives nasty hypergeomtric series and modified Bessel integrals. $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 20:42
  • $\begingroup$ @Reddevil Uh, well, you see... the short version is that$$(a+b)^c=a^c+ca^{c-1}b+\frac{c(c-1)}2a^{c-2}b^2+\frac{c(c-1)(c-2)}{2\times3}a^{c-3}b^3+\dots$$Apply this to $(1+x^2)^{1/2}$ to get expansions such as$$\sqrt{x^2+1}=1+\frac12x^2-\frac18x^4+\dots$$And then you would probably like to collect the terms together from there. $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 20:44
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    $\begingroup$ @JyrkiLahtonen It does appear though that these result in integer multiples of $e$. As to why, I am unsure. If you don't happen to know why this curious fact is occurring, it may prove to be an interesting question? $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 21:43
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    $\begingroup$ Oh, never mind. Directly differentiating will always give $e^{\sqrt{x^2+1}}$ plus some stuff, where the denominators are always of the form $(x^2+1)^{k/2}$, which always goes to $1$, hence the result is always an integer multiple of $e$. Sorry to bother. $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 21:45

One approach is to find a differential equation satisfied by the function and use this to find a recursive relation for the Taylor series coefficients.

Let's simplify the problem a bit by considering $f(x) = e^\sqrt{x+1}$. If we can find a Taylor series for $f$, say $f(x)=\sum_{n=0}^{\infty} a_n x^n$, then $e^\sqrt{x^2+1} = \sum_{n=0}^{\infty} a_n x^{2n}$.

$f$ satisfies the differential equation $$2 \; \sqrt{x+1} \; f'(x) = f(x)$$ with initial condition $f(0) = e$. With the Taylor series for $f$ as defined above, the differential equation translates to $$2 \; \sqrt{x+1} \; \sum_{j=0}^{\infty}(j+1) a_{j+1} x^j = \sum_{n=0}^{\infty} a_n x^n$$ Applying the binomial theorem to expand $\sqrt{x+1}$ as a power series, we have $$2 \; \sum_{i=0}^{\infty} \binom{1/2}{i}x^i \cdot \sum_{j=0}^{\infty}(j+1) a_{j+1} x^j = \sum_{n=0}^{\infty} a_n x^n$$

Extracting the coefficient of $x^n$ on each side of the equation, $$2\;\sum_{j=0}^n \binom{1/2}{n-j} (j+1)a_{j+1} = a_n$$ We can solve this equation for $a_{n+1}$, with result $$a_{n+1} = \left( \frac{1}{2}a_n - \sum_{j=0}^{n-1} \binom{1/2}{n-j} (j+1)a_{j+1} \right) / (n+1)$$ for $n \ge 1$. Together with $a_0 = e$ and $a_1 = f'(0) = e/2$, this relation allows us to compute as many coefficients $a_n$ as we want.


Composing Taylor series is always a fascinating task to me.

For a truncated series, in order to avoid quite nasty derivatives, what I should have done is first to replace $x^2$ by $t$ and then consider $$y=e^{\sqrt{x^2+1}}=e^{\sqrt{t+1}}\implies \log(y)={\sqrt{t+1}}=\sum_{n=0}^\infty\binom{\frac{1}{2}}{n} t^n$$ Truncating to $O(t^8)$, this would give $$\log(y)=1+\frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}-\frac{5 t^4}{128}+\frac{7 t^5}{256}-\frac{21 t^6}{1024}+\frac{33 t^7}{2048}+O\left(t^8\right)$$ and now rewrite $$y=e^{\log(y)}=e \,e^{\log(y)-1}$$ Defining $u={\log(y)-1}$ $$y=e e^u=e \sum_{n=0}^\infty \frac {u^n}{n!}\tag 1$$ So, $$u=\frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}-\frac{5 t^4}{128}+\frac{7 t^5}{256}-\frac{21 t^6}{1024}+\frac{33 t^7}{2048}+O\left(t^8\right)$$ $$u^2=\frac{t^2}{4}-\frac{t^3}{8}+\frac{5 t^4}{64}-\frac{7 t^5}{128}+\frac{21 t^6}{512}-\frac{33 t^7}{1024}+O\left(t^8\right)$$ $$u^3=u^2 u=\frac{t^3}{8}-\frac{3 t^4}{32}+\frac{9 t^5}{128}-\frac{7 t^6}{128}+\frac{45 t^7}{1024}+O\left(t^{8}\right)$$ $$u^4=u^3 u=\frac{t^4}{16}-\frac{t^5}{16}+\frac{7 t^6}{128}-\frac{3 t^7}{64}+O\left(t^{8}\right)$$ $$u^5=u^4 u=\frac{t^5}{32}-\frac{5 t^6}{128}+\frac{5 t^7}{128}+O\left(t^{8}\right)$$ $$u^6=u^5 u=\frac{t^6}{64}-\frac{3 t^7}{128}+O\left(t^{8}\right)$$ $$u^7=u^6u=\frac{t^7}{128}+O\left(t^{8}\right)$$ $$u^8=u^7 u=O\left(t^{8}\right)$$ Replacing in $(1)$, this leads to $$y=e\left(1+\frac{t}{2}+\frac{t^3}{48}-\frac{5 t^4}{384}+\frac{3 t^5}{320}-\frac{329 t^6}{46080}+\frac{731 t^7}{129024}+O\left(t^{8}\right) \right)$$ Now, replace $t$ by $x^2$.


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