Proof by induction

[ken74]

I just dont understand the k+1 part. The powers for (-1)^k+1, doesn't make sense to me. I thought it should be (-1)^k+1+1....Please help

 

[MarkFL]

Hello, and welcome to FMH!

Let's look at the given induction hypothesis:

[MATH]\frac{d^{2n}y}{dx^{2n}}=(-1)^n2^{2n}\cos(1-2x)[/MATH]
As our induction step, we need to differentiate this twice. What do you get when carrying out this operation?

 

[ken74]

that is all I could do and I'm stuck beyond this point

 

[MarkFL]

This is what I meant:

Begin with the induction hypothesis:

[MATH]\frac{d^{2n}y}{dx^{2n}}=(-1)^n2^{2n}\cos(1-2x)[/MATH]
Differentiate once:

[MATH]\frac{d^{2n+1}y}{dx^{2n+1}}=(-1)^n2^{2n+1}\sin(1-2x)[/MATH]
Differentiate again:

[MATH]\frac{d^{2n+2}y}{dx^{2n+2}}=-(-1)^n2^{2n+2}\cos(1-2x)[/MATH]
[MATH]\frac{d^{2(n+1)}y}{dx^{2(n+1)}}=(-1)^{n+1}2^{2(n+1)}\cos(1-2x)[/MATH]
We have derived \(P_{n+1}\) from \(P_n\), thereby completing the proof by induction.

 

[Steven G]

MarkFL wrote out the EQUATION nicely.

What you should for n=1 is not correct.

Like I said you have an equation. Compute the lhs for n=1, compute the rhs for n=1 and look at this I got the same results!

Assume for n=k that : lhs with n=k equals the rhs with n=k.

Then show true, possible using the assumption, for n=k+1. Show that the lhs really does equal the rhs when n=k+1.

Please try again. The 1st two steps are simple.

 

[ken74]

Thank you so so much to Jomo and MarkFL. Much appreciated

 

[Steven G]

Thank you so so much to Jomo and MarkFL. Much appreciated

Please reply back with your work.

 

[ken74]

below is my work. let me know i I did correctly. thank you

 

[Steven G]

When n=k, you had 2k+1
When n=k+1, had 2k+2 which is not correct. Please try again.