Maclurin series for $e^{\sqrt{x^2+1}}$

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: $$e^{\sqrt{x^2+1}}=1+\sqrt{x^2+1}+\frac{x^2+1}{2}+...$$

• Presumably, mean $u=\sqrt{x^2+1}$? – Thomas Andrews Aug 19 '17 at 20:28
• 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. – GEdgar Aug 19 '17 at 20:30
• @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 – Reddevil Aug 19 '17 at 20:33
• Maybe you were expected to directly differentiate the first few times... – Simply Beautiful Art Aug 19 '17 at 20:51
• @GEdgar $e^u=1+u+\dots$ for any $u\in\Bbb C$, as $e^z$ is an entire function? – 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$$e+\frac e2x^2+\frac e{48}x^6-\frac{5e}{384}x^8+\frac{3e}{320}x^{10}.$$

• Basically a copy of Raffaele's answer, which was posted almost 40 minutes ago. Suggesting deletion of this answer. – Simply Beautiful Art Aug 19 '17 at 21:42
• 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 $e^{\sqrt{x+1}}$. – José Carlos Santos Aug 19 '17 at 21:44
• Ah, I missed that. I suppose it is fine then :-) – Simply Beautiful Art Aug 19 '17 at 21:46

You have to compute the derivatives

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

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

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

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

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

$$e^{\sqrt{x^2+1}}=\sum_{k=0}^\infty\frac{(x^2+1)^{k/2}}{k!}$$

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

$$(x^2+1)^{k/2}=\sum_{n=0}^\infty\binom{k/2}nx^{2n}$$

And thus,

$$e^{\sqrt{x^2+1}}=\sum_{k=0}^\infty\frac1{k!}\sum_{n=0}^\infty\binom{k/2}nx^{2n}$$

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

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

where

$$a_{2n+1}=0\\a_{2n}=\sum_{k=0}^\infty\binom{k/2}n\frac1{k!}$$

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

$$a_k=\begin{cases}e,&k=0\\0,&k=1\\\frac1k\sum_{n=0}^{k-2}\binom{-1/2}na_{k-2-n},&k>1\end{cases}$$

• 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. – Reddevil Aug 19 '17 at 20:41
• 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. – Simply Beautiful Art Aug 19 '17 at 20:42
• @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. – Simply Beautiful Art Aug 19 '17 at 20:44
• @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? – Simply Beautiful Art Aug 19 '17 at 21:43
• 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. – 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 $x^2$.