+9665 I think of \(e^{i\theta} = \cos \theta + i\sin \theta\)......
Let me try :) (Im just a kid don't blame me if I did anything wrong....... Just point out the mistake......)
\(4\sin x = e^{x}\\ 4\sin x = e^{(i^4 \cdot x)}\quad\boxed{i^4 = 1}\\ 4\sin x = e^{i\cdot(xi^3)}\\ 4\sin x = e^{i\cdot(-ix)}\quad\boxed{i^3 = -i}\\ 4\sin x = \cos(-ix)+i\sin(-ix)\)
We are going to introduce 4 formulae here:
1) \(\sin(-x) = -\sin x\)
2)\(\cos(-x) = \cos x\)
3)\(\cos(ix) = \cosh x = \dfrac{e^x + e^{-x}}{2}\)
4)\(\sin(ix) = i\sinh x = \dfrac{e^{-x} - e^{x}}{2i}\)
Now continue.....
Last step done was:
\(4\sin x = \cos(-ix) + i\sin(-ix)\\ 4\sin x = \cos(ix) -i\sin(ix)\quad\boxed{\text{Formulae 1 and 2}}\\ 4\sin x = \dfrac{e^x + e^{-x}}{2}+\dfrac{e^x - e^{-x}}{2}=e^x\quad\boxed{\text{Formulae 3 and 4}}\)
Nope it just goes back to 4 sin x = e^x....... Sob sob :(
Another method!!
\(4\sin x = e^x\\ \)
We are going to introduce the definition of sin x this time....... (If it doesn't work Im going to use Taylor series :( )
\(4\left(\dfrac{e^{ix}-e^{-ix}}{2i}\right) = e^x\\ \dfrac{2e^{ix}-2e^{-ix}}{i}=e^x\\ 2ie^{-ix} - 2ie^{ix} = e^x\quad\boxed{\dfrac{1}{i} = -i}\\ 2i\left(\dfrac{1}{e^{ix}}-e^{ix}\right)=e^x\\ 2i\left(\dfrac{1-e^{2ix}}{e^{ix}}\right)=e^x\)
Np.
I never give up!!!
Another method!!
\(4\left(x-\dfrac{x^3}{3!}-\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{x^9}{9!}\cdot \cdot\; \cdot\right)=e^x\)
Still np......
Sigh maybe I am just too dull and can't see the method......
PS I am 14 :)
