evaluate:
d ^15/dx ^15 exp [(1 + sqrt(3)i) x]
i dont know how to approach this problem with the drivative to the 15th power.
firstly 1 + sqrt(3)i must be converted to polar form.. then im stuck
pls help me.
evaluate:
d ^15/dx ^15 exp [(1 + sqrt(3)i) x]
i dont know how to approach this problem with the drivative to the 15th power.
firstly 1 + sqrt(3)i must be converted to polar form.. then im stuck
pls help me.
if you think about it $\displaystyle \frac{D[e^{u(x)}]}{dx}=u'(x)\cdot{e^u(x)}$ right? Now if $\displaystyle u'(x)$ is a constant then we dont need to use chain rule we jsut have keep multiplying it so for example...remember $\displaystyle u(x)$ is a constant then we would have $\displaystyle f(x)=e^{u(x)}$...then $\displaystyle f'(x)=u'(x)\cdot{e^{u(x)}}$...now since it is a constant we just yank it out so $\displaystyle f''(x)=u'(x)\cdot{\frac{D[e^{u(x)}]}{dx}}$...so using this same concept of yanking out the constant we get $\displaystyle f^{(n)}(x)=u'(x)^{n}\cdot{e^{u(x)}}$...and in this case $\displaystyle u'(x)$ is a consant...so yeah
after obtaining an expression for the derivative...i got a different answer to ures... since the constant would be of high power.. (1 + sqrt(3) i ) ^ 15... expanding it would be silly.
so wouldnt u need to convert it into polar form and use De Moivre's theorem to simplify it?
if that is so i got something different to ures... i had -2^15 ????
hrmm... i just noticed u only included the imaginary part in the constant....y is that?
Hello,
Indeed, mathstud got wrong because he forgot 1.
This is correct.(1 + sqrt(3) i ) ^ 15
But don't bother with De Moivre's theorem
$\displaystyle (1+\sqrt{3} i)^2=(1-3+2\sqrt{3} i)=2(-1+\sqrt{3} i)$
So $\displaystyle (1+\sqrt{3} i)^3=(1+\sqrt{3} i)^2(1+\sqrt{3} i)=2\underbrace{(-1+\sqrt{3} i)(1+\sqrt{3} i)}_{(a-b)(a+b)=a^2-b^2}=2(-3-1)=-2^3$
So $\displaystyle (1+\sqrt{3} i)^{15}=((1+\sqrt{3} i)^3)^5=(-2^3)^5=-2^{15}$