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: 
$$e^{\sqrt{x^2+1}}=1+\sqrt{x^2+1}+\frac{x^2+1}{2}+...$$
Am I doing it right? please help
 A: 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$$e+\frac e2x^2+\frac e{48}x^6-\frac{5e}{384}x^8+\frac{3e}{320}x^{10}.$$
A: 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
$$a_k=\begin{cases}e,&k=0\\0,&k=1\\\frac1k\sum_{n=0}^{k-2}\binom{-1/2}na_{k-2-n},&k>1\end{cases}$$
A: 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
A: 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.
