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.

2 Answers
Jan 24, 2018

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.

Jan 24, 2018

# (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