Differential sum

Note that because $\dfrac{\mathrm{d^i}}{\mathrm{d}x^i} \sin (x) \in \{\pm \sin(x), \pm \cos(x)\}$ , the sum just transforms to (using chain rule for the $\frac{x}{2}$ ) $\displaystyle \sum_{i=0}^{\infty} (-1)^i (2^{-2i} \sin (x) + 2^{-2i-1} \cos(x) )$ This is just the sum of two geometric progressions, the sum of each is $\displaystyle \sum_{i=0}^{\infty} (-1)^i 2^{-2i} \sin (x) = \dfrac{\sin(x)}{1+\frac{1}{4}} = \frac{4}{5} \sin(x)$ $\displaystyle \sum_{i=0}^{\infty} (-1)^i 2^{-2i-1} \cos(x) = \dfrac{\frac{\cos(x)}{2}}{1+\frac{1}{4}} = \frac{2}{5} \cos(x)$ Thus, by subsituting $x=\frac{\pi}{6}$ , and using half-angle formula to determine $\sin \left ( \frac{\pi}{12} \right ) = \frac{\sqrt{3}-1}{2\sqrt{2}}$ and $\cos \left ( \frac{\pi}{12} \right )=\frac{\sqrt{3}+1}{2\sqrt{2}}$ , we get $\displaystyle \sum_{i=0}^{\infty} (-1)^i (2^{-2i} \sin (x) + 2^{-2i-1} \cos(x) )=\frac{3\sqrt{3}-1}{5\sqrt{2}}$ Thus, the product $ABCDE=\boxed{90}$ .

$\begin{aligned} \sum_{n=0}^\infty \frac {d^n}{dx^n}\sin \frac x2 & = \sin \frac x2 + \frac 12 \cos \frac x2 - \frac 1{2^2} \sin \frac x2 - \frac 1{2^3} \cos \frac x2 + \frac 1{2^4} \sin \frac x2 + \frac 1{2^5} \cos \frac x2 - \cdots \\ & = \sin \frac x2 \sum_{k=0}^\infty \frac {(-1)^k}{2^{2k}} + \frac 12 \cos \frac x2 \sum_{k=0}^\infty \frac {(-1)^k}{2^{2k}} \\ & = \left(\sin \frac x2 + \frac 12 \cos \frac x2 \right) \sum_{k=0}^\infty \left(-\frac 14\right)^k & \small \color{#3D99F6} \text{Putting }x = \frac \pi 6 \\ & = \left(\sin \frac \pi{12} + \frac 12 \cos \frac \pi{12} \right) \left(\frac 1{1+\frac 14}\right) \\ & = \left(\frac {\sqrt 3-1}{2\sqrt 2} + \frac 12 \cdot \frac {\sqrt 3+1}{2\sqrt 2} \right) \left(\frac 45 \right) \\ & = \frac {3\sqrt 3-1}{5\sqrt 2} \end{aligned}$

Therefore, $ABDE = 3\cdot 3 \cdot 5 \cdot 2 = \boxed{90}$ .

We know that $\frac{1}{1-x} = \sum_{r=0}^{\infty} x^r$

So what we need to find is $\displaystyle \frac{1}{1-D} \sin(\frac{x}{2})$ [here D denotes the operator $\frac{d}{dx}$ ]

Note :- Here $\frac{1}{1-D}$ is a symbolic notation and not a fraction. It represents the inverse of the operator $1-\frac{d}{dx}$ .

as we know that $\displaystyle D^{2n} (\sin(ax)) = -a^{2n}\sin(ax)$ for $n\geq0$

We have for a polynomial $\phi$ in $D^{2}$ .

$\displaystyle \phi(D^{2})(\sin(ax)) = \phi(-a^{2})(\sin(ax))$

Operating both sides with $\displaystyle \frac{1}{\phi(D^{2})}$ and then multiplying by $\large \frac{1}{\phi(-a^{2})}$ we have:-

$\frac{1}{\phi(D^{2})}(\sin(ax)) = \frac{1}{\phi(-a^{2})}(\sin(ax))$

Again it is needed to mention here that $\frac{1}{\phi(D^{2})}$ is nothing but the inverse of the operator $\phi(D^{2})$ .

So $\displaystyle \frac{1}{1-D} \sin(\frac{x}{2}) = \frac{1+D}{1-D^{2}} \sin(\frac{x}{2}) = \frac{1+D}{1+\frac{1}{4}}$ [Replacing $D^{2}$ by $-a^{2}$ . here $a=\frac{1}{2}$ ]

$\displaystyle = \frac{4}{5}(D+1)(\sin(\frac{x}{2}))$

$\displaystyle (D+1)(\sin(\frac{x}{2})) = (1+\frac{\text{d}}{\text{d}x})(\sin(\frac{x}{2})) = \sin(\frac{x}{2}) + \frac{1}{2}\cos(\frac{x}{2})$

So the value of the summation for any x(within valid range of convergence) = $\displaystyle \frac{4}{5}(\sin(\frac{x}{2}) + \frac{1}{2}\cos(\frac{x}{2}))$

Putting $x=\frac{\pi}{6}$ in it we get $\displaystyle \frac{3\sqrt{3}-1}{5\sqrt{2}}$ as our answer.

We could have also done it using the symbolic operator method for finding particular integrals of $e^{ax}$ . There we would have had to find out the imaginary parts and we could have arrived at the same answer. But I feel this method is easier as the polynomial in this case was very small.

These methods of problem solving using the symbolic operator $D$ is not uncommon and can be found in any elementary level Differential Equation book. These become really handy when we evaluate particular integrals for function in which we do not need to apply the method of variation of parameters.

Sharky gives a amazing approach to solve this question!

I will try to elaborate a different one:

Let $\displaystyle S=\sum_{n=0}^\infty{\frac{d^n}{dx^n}\sin{\frac x2}}$ , we can expand a little to see that $S=\sin{\frac x2}+\frac{dS}{dx}$

Then continuing calculate it:

$\begin{aligned} \frac{dS}{dx}-S&=-\sin{\frac x2}\\ \frac{d}{dx}e^{-x}S&=-e^{-x}\sin{\frac x2}\\ e^{-x}S&=-\int e^{-x}\sin{\frac x2}dx\\ &=2e^{-x}\cos{\frac x2}+4e^{-x}\sin{\frac x2}-4\bigg[-\int e^-x\sin{\frac x2}dx\bigg]\\ &=2e^{-x}\cos{\frac x2}+4e^{-x}\sin{\frac x2}-4e^{-x}S+C\\ 5e^{-x}S&=2e^{-x}\cos{\frac x2}+4e^{-x}\sin{\frac x2}+C\\ S&=\frac25\cos{\frac x2}+\frac45\sin{\frac x2}+C'e^x \{C=5C'\}\\ \end{aligned}$

To determine the value of $C$ , we need to go back to the original expression, so $\displaystyle S=\sin\frac x2+\frac12\cos\frac x2-\frac14\sin\frac x2-\frac18\cos\frac x2+\dots$ , we can predict the pattern of $S(0)$ , that is $\begin{aligned}S(0)&=\frac12-\frac1{2^3}+\frac1{2^5}-\frac1{2^7}+\dots\\&=\frac{1/2}{1+1/2^2}\\&=\frac25\end{aligned}$

Substitute when $x=0,S=\frac25$ into the result of integral above, leads to $C'=0$ , so $\displaystyle S(x)=\frac25\cos\frac x2+\frac45\sin\frac x2$ .

With the identity $\displaystyle \cos\frac\pi{12}=\frac{\sqrt6+\sqrt2}{4},\sin\frac\pi{12}=\frac{\sqrt6-\sqrt2}{4}$ , we can calculate the value of $S$ when $x=\frac\pi6$ : $S\Bigg|_{x=\frac\pi6}=\frac{3\sqrt6-\sqrt2}{10}=\frac{3\sqrt3-1}{5\sqrt2}$ Thus, $A=3,B=3,C=1,D=5,E=2, A\times B\times C\times D\times E=\boxed{90}$ .

$\text{let A}=\sum_{n=0}^{\infty} \dfrac{d^n}{dx^n} \sin{\left( \dfrac{x}{2} \right)}=\sum_{n=0}^{\infty} \dfrac{d^n}{dx^n} \Im{\left( e^{i\dfrac{x}{2}} \right)}=\Im{\left(\sum_{n=0}^{\infty} \dfrac{d^n}{dx^n} e^{i\dfrac{x}{2}} \right)}$

$A=\Im{\left(\sum_{n=0}^{\infty} \left(\dfrac{i}{2} \right)^n e^{i\dfrac{x}{2}} \right)}=\Im{\left(e^{i\dfrac{x}{2}} \sum_{n=0}^{\infty} \left(\dfrac{i}{2} \right)^n \right)}$

$A=\Im{\left(\left(\cos{\left( \dfrac{x}{2} \right)}+i\sin{\left( \dfrac{x}{2} \right)} \right)\left(\dfrac{1}{1-\dfrac{i}{2}}\right)\right)}=\Im{\left(\left(\cos{\left( \dfrac{x}{2} \right)}+i\sin{\left( \dfrac{x}{2} \right)} \right)\left(\dfrac{4+2i}{5}\right)\right)}$

$A=\Im{\left(\dfrac{4}{5}\cos{\left( \dfrac{x}{2}\right)}-\dfrac{2}{5}\sin{\left( \dfrac{x}{2}\right)}+i\left(\dfrac{4}{5}\sin{\left( \dfrac{x}{2}\right)}+\dfrac{2}{5}\cos{\left( \dfrac{x}{2}\right)}\right)\right)}$

$A=\dfrac{4}{5}\sin{\left( \dfrac{x}{2}\right)}+\dfrac{2}{5}\cos{\left( \dfrac{x}{2}\right)}$

$A=\dfrac{2}{5}\left( 2\dfrac{\sqrt{3}-1}{2\sqrt{2}}+\dfrac{\sqrt{3}+1}{2\sqrt{2}}\right)=\dfrac{3\sqrt{3}-1}{5\sqrt{2}}$

$\text{ so ABCD }=3\times 3\times 5\times 2=90$