Further Maths Question! D:

I have to prove by induction that
f(x)=(e^x)*sin(x) implies (f(x))^n) = (2^(n/2))*(e^x)*sin(x+(n*pi)/2)

So far I tried to say: f(x)^n , when n = 1 is equal to (e^x)*sin(x) on one side, and on the other side is....(here's the problem)

f(x)^1 = 2^(1/2)*(e^x)*sin(x+ pi/4) = 2^(1/2)*(e^x)*(cos(x)*sin(pi/4)+sin(x)*cos(pi/4)) = 2^(1/2)*(e^x)*(2^(1/2)/2)*(cosx+sinx) which leads me to (e^x)*(sinx+cosx)

This can't be right, cause sinx+cosx cant be the same as sin(x).

Where did I go wrong?

Thanks

Question

Further Maths Question! D:

I have to prove by induction that 
f(x)=(e^x)*sin(x) implies (f(x))^n) = (2^(n/2))*(e^x)*sin(x+(n*pi)/2)

So far I tried to say: f(x)^n , when n = 1 is equal to (e^x)*sin(x) on one side, and on the other side is....(here's the problem)

f(x)^1 = 2^(1/2)*(e^x)*sin(x+ pi/4) = 2^(1/2)*(e^x)*(cos(x)*sin(pi/4)+sin(x)*cos(pi/4)) = 2^(1/2)*(e^x)*(2^(1/2)/2)*(cosx+sinx) which leads me to (e^x)*(sinx+cosx)

This can't be right, cause sinx+cosx cant be the same as sin(x).

Where did I go wrong?

Thanks

anonymous · Answer

sorry, logs kick my butt.

anonymous · Answer

this is what calculators are made for.   i agree with ehuman

anonymous · Answer

I see where I did go wrong, $$f^n (x) \neq [f(x)]^n$$ it means the nth differential of $$f(x)$$.
eg: $$f^1(x) = f^\prime (x)$$