Higher order derivatives calculation

[4445]

Does anyone know why this is true?
View attachment 7432

 

[S.G. Janssens]

Is $M$ a non-negative integer?
Do the numbers in parentheses all indicate derivatives, while $(-4)^M$ is an ordinary power?
I cannot see why what you wrote would be true, then. For example, it fails for $F(x) = e^x$.

 

[4445]

Thank you for your answer. F(x) = (e^x)*sin(x). The numbers in parenthesis indicate derivatives

 

[MarkFL]

I think I would begin by looking at the $n$th order derivative of a product formula:

\(\displaystyle \frac{d^n}{dx^n}\left(f(x)\cdot g(x)\right)=\sum_{k=0}^n\left({n \choose k}f^{(n-k)}(x)\cdot g^{(k)}(x)\right)\)

And use this formula in an inductive proof, where the formula in question is your induction hypothesis $P_M$, where $M\in\mathbb{N_0}$.

Can you proceed?

 

[MarkFL]

As a follow up, let's first compute:

\(\displaystyle \frac{d^4}{dx^4}\left(e^x\sin(x)\right)=e^x\sin(x)+4e^x\cos(x)-6e^x\sin(x)-4e^x\cos(x)+e^x\sin(x)=-4e^x\sin(x)\)

And so our induction hypothesis is:

\(\displaystyle \frac{d^{4(M+1)}}{dx^{4(M+1)}}\left(e^x\sin(x)\right)=(-4)^{M+1}\left(e^x\sin(x)\right)\)

It is easy to see now, that taking the 4th derivative of both sides, we get:

\(\displaystyle \frac{d^{4((M+1)+1)}}{dx^{4((M+1)+1)}}\left(e^x\sin(x)\right)=(-4)^{(M+1)+1}\left(e^x\sin(x)\right)\)

Since we have obtained $P_{M+2}$ from $P_{M+1}$, this completes our proof by induction.