Question

Prove by induction: d^2n/dx^2n (xsinsx)=(-1)^n (xsins-2ncosx)?

I have proved this for n=1 and assumed true for n=k. However, I am running into some challenges when I try proving this for n=k+1. So far I have `d^(2(k+1))/dx^(2(k+1)) (xsinx)=(d^(2k+2))/d^(2k+2)(xsinx)`
From here on I am not quite sure how to expand it further, specifically how to expand the exponents.

Answer 1

For `n=1` the proposition is true as, using the product rule:

`d/dx (xsinx) = xd/dx sinx + sinx d/dx x = sinx +xcosx`

`d^2/dx^2 (xsinx) = d/dx( sinx +xcosx) = cosx -xsinx+cosx = (-1)(xsinx -2cosx)`

Suppose now that the proposition is true for `n=k`:

`(1) " " d^(2k)/dx^(2k) = (-1)^k(xsinx-2kcosx)`

Differentiating this equality twice we have:

`d^(2k+1)/dx^(2k+1) = (-1)^k(sinx+xcosx+2k sinx) = (-1)^k(xcosx+(2k+1) sinx)`

`d^(2k+2)/dx^(2k+2) = (-1)^k(-xsinx+cosx+(2k+1) cosx)`

`d^(2k+2)/dx^(2k+2) = (-1)^k(-xsinx+(2k+2) cosx)`

`d^(2(k+1))/dx^(2(k+1)) = (-1)^(k+1)(xsinx-2(k+1)cosx)`

so we can conclude that if `(1)` is true for `n=k` then it is also true for `n= k+1` and the proof by induction is complete.

Answer 2

`(d^(2n))/(dx^(2n)) xsinx = (-1)^n(xsinx-2ncosx) AA n in NN`

Explanation:

Induction Proof - Hypothesis

We seek to prove that the the expression:

`(d^(2n))/(dx^(2n)) xsinx = (-1)^n(xsinx-2ncosx) AA n in NN` ..... [A]

So let us test this assertion using Mathematical Induction:

Induction Proof - Base case:

We will show that the given result, [A], holds for `n=1`

When `n=1` the given expression gives:

`d^2/(dx^2) xsinx = (-1)(xsinx-2cosx)`
`" " = 2cosx - xsinx`

By direct differentiation of the given expression using the product rule, we have:

`d/dx xsinx = (x)(cosx)+(1)(sinx)`

`:. d^2/(dx^2) xsinx = (x)(-sinx)+(1)(cosx)+cosx`
`" " = 2cosx - xsinx`

Which is the same as the previous result, So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`(d^(2m))/(dx^(2m)) xsinx = (-1)^m(xsinx-2mcosx)` ..... [B]

Knowing that the above result is true, we can differentiate the expression using the product rule, giving:

`d/dx (d^(2m))/(dx^(2m)) xsinx = d/dx (-1)^m(xsinx-2mcosx)`

`:. (d^(2m+1))/(dx^(2m+1)) xsinx = (-1)^m{(x)(cosx)+(1)(sinx)+2msinx}`

`" " = (-1)^m{xcosx+sinx+2msinx}`

And if we differentiate again, we get:

`(d^(2m+2))/(dx^(2m+2)) xsinx = (-1)^m{(x)(-sinx) + (1)(cosx) +cosx+2mcosx}`

`" " = (-1)^m{-xsinx + 2cosx+2mcosx}`
`" " = (-1)^m{-xsinx + 2(m+1)cosx}`
`" " = (-1)^m(-1)(xsinx - 2(m+1)cosx)`

And so finally, we have:

`(d^(2(m+1)))/(dx^(2(m+1))) xsinx = (-1)^(m+1)(xsinx - 2(m+1)cosx)`

Which is the given expression [A] with `n=m+1`

Induction Proof - Summary

So, we have shown that if the given result [A] is true for `n=m`, then it is also true for `n=m+1` where `m gt 1`. But we initially showed that the given result was true for `n=1` so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for `n in NN`

Hence we have:

`(d^(2n))/(dx^(2n)) xsinx = (-1)^n(xsinx-2ncosx) AA n in NN` QED