calculus 2 question

Question

suppose f^n (x)= -36 sin(6x) and f'(0)=6 and f(0)= -5

f(π /4)= ?

Answer 1

There seems to be something amiss about this question.

Assuming fⁿ to mean the nth derivative with respect to x, where n=0⇒f(x) and integer n≥0:

fⁿ(x)=-36sin(6x)=-6²sin(6x)

fⁿ⁺¹(x)=-6³cos(6x)

fⁿ⁺²(x)=6⁴sin(6x)=-6²fⁿ(x)

fⁿ⁺³(x)=6⁵cos(6x)=-6²fⁿ⁺¹(x).

Therefore we can write, in general: fⁿ⁺²(x)=-6²fⁿ(x). This means that it doesn't matter whether n is even or odd.

If we let y=fⁿ(x), then y"=-6²y, that is, y"+6²y=0 which has the general solution:

y=Asin(6x)+Bcos(6x), implying fⁿ(x)=Asin(6x)+Bcos(6x) (A and B are constants).

We can also write fⁿ(x)=-6²f^n-2(x). This is an iterative DE, so sooner or later f^n-2(x) is the same as f⁰(x)=f(x).

[NOTE: If n=2, f"(x)=-36f(x) and f(x)=Asin(6x)+Bcos(6x).]

fⁿ(0)=-6²f^n-2(0). If we consider the initial conditions we have 2 consecutive derivatives (n=0 and n=1) and in each case we have a non-zero number.

When we consider fⁿ and fⁿ⁺¹ (two consecutive derivatives), we can see that the variable term is sin(6x) or cos(6x). Now, when x=0 we get either fⁿ(0)=0 and fⁿ⁺¹(0)≠0 or we get fⁿ(0)≠0 and fⁿ⁺¹(0)=0. So it's not possible to have f(0)≠0 and f'(0)≠0. We have also seen the iterative DE as the general solution, which involves both sin(6x) and cos(6x).

If f(x)=Asin(6x)+Bcos(6x), f(0)=B=-5; f'(0)=6A=6, so A=1 and f(x)=sin(6x)-5cos(6x).

If this reasoning is correct f(π/4)=-1, because sin(6x)=sin(3π/2)=-1 and cos(3π/2)=0.

This satisfies the initial conditions and, when n=2, also satisfies:

f²(x)=-36f(x); f"(x)=-36sin(6x)-180cos(6x).