induction...? (1 Viewer)

[aimhigh10]

hey,

does anyone feel like they can help with this:
<DIR>Use induction to show that the nth derivative of x^−1 is
(((−1)^n).n!) / (x^n+1 )

thanks

 

[addikaye03]

hey,

does anyone feel like they can help with this:
<DIR>Use induction to show that the nth derivative of x^−1 is
(((−1)^n).n!) / (x^n+1 )

thanks

For n=1, d/dx (x^-1)=-x^-2

n=2, d^2/dx^2 (x^-1)=2x^-3=2.1.x^-3

n=3, d^3/dx^3 (x^-1)=-6x^-4=-(3.2.1)x^-4 where d^3/dx^3 notates the 3rd derivative of (x^-1)

n=4, d^4/dx^4 (x^-1)=24x^-5=4.3.2.1x^-5

...

n=n, d^n/dx^n (x^-1)= [(-1)^n.n!]/(x^(n+1)]

Seems pretty obvious just via inspection. Is an inductive proof required for the question? or did you just assume it would require induction.

 

[aimhigh10]

yeah i thought it was pretty obvious via inspection too
but the question does say use induction.. so i dont know

 

[lyounamu]

blah blah blah blah blah

we assume that it is true for n=k
i.e. d^kx/dy^k = (-1)^k k!/x^(k+1)

then we now prove that it is also true for n=k+1
i.e. d^(k+1)x/dy^(k+1) = (-1)^(k+1)k!/x^(k+2)

LHS = d/dx (d^kx/dy^k) = d/dx ((-1)^k k!/x^(k+1))
= d/dx ((-1)^k k! x^-(k+1))
= ((-1)^(k+1) (k+1)! x^-(k+2)
= RHS

blah blah blah blah

 

[aimhigh10]

once again,
lyounamu you are the best : )
and thanks for the help addikaye03 too : )

 

[addikaye03]

blah blah blah blah blah

we assume that it is true for n=k
i.e. d^kx/dy^k = (-1)^k k!/x^(k+1)

then we now prove that it is also true for n=k+1
i.e. d^(k+1)x/dy^(k+1) = (-1)^(k+1)k!/x^(k+2)

LHS = d/dx (d^kx/dy^k) = d/dx ((-1)^k k!/x^(k+1))
= d/dx ((-1)^k k! x^-(k+1))
= ((-1)^(k+1) (k+1)! x^-(k+2)
= RHS

blah blah blah blah

Ahh yeah ofcourse, well done man.