Can someone check my work on an induction proof

  • Thread starter miglo
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  • #1
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Homework Statement


using mathematical induction, show that
d^n/dx^n(lnx)= (-1)^(n-1)*((n-1)!/x^n)


Homework Equations





The Attempt at a Solution



basis step:
for n=1 we get d/dx(lnx)=(-1)^0*((0)!/x^1)
which gives d/dx(lnx)=1/x

inductive step:
assuming it holds for all n, d^n/dx^n=(-1)^(n-1)*((n-1)!/x^n)
for n+1, d^(n+1)/dx^(n+1)=(-1)^n*(n!/x^(n+1))

d/dx[d^n/dx^n(lnx)]=d/dx((-1)^(n-1)*((n-1)!/x^n))
=-((n(n-1)!x^(n-1))/x^(2n))(-1)^(n-1)
=(-1)^n*(n!*x^(-n-1))
=(-1)^n*(n!/x^(n+1)

any help would be greatly appreciated, and sorry if its a bit hard to read, hopefully ill be posting using latex soon
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
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You have a 2n where there should be a n but apart from that, it looks good.
 
  • #3
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hunt mat is right about the 2n.

one other detail. in the induction step you say 'assume true for all n'; you don't want to say that because that is exactly what you are trying to prove. you want to assume it is true for some particular, but not specified, value of n, and show that then it must hold for the n+1 case.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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956

Homework Statement


using mathematical induction, show that
d^n/dx^n(lnx)= (-1)^(n-1)*((n-1)!/x^n)


Homework Equations





The Attempt at a Solution



basis step:
for n=1 we get d/dx(lnx)=(-1)^0*((0)!/x^1)
which gives d/dx(lnx)=1/x

inductive step:
assuming it holds for all n, d^n/dx^n=(-1)^(n-1)*((n-1)!/x^n)
for n+1, d^(n+1)/dx^(n+1)=(-1)^n*(n!/x^(n+1))
As eczeno says, this is what you want to prove. Instead say, "Assume it holds for some k" (I prefer to use "k" rather than "n" again, specifically to make it clear that we are NOT assuming what we want to prove.

d/dx[d^n/dx^n(lnx)]=d/dx((-1)^(n-1)*((n-1)!/x^n))
=-((n(n-1)!x^(n-1))/x^(2n))(-1)^(n-1)
As Hunt Mat says, the "x^(2n)" should be x^(n-1+1)= x^n

=(-1)^n*(n!*x^(-n-1))
=(-1)^n*(n!/x^(n+1)

any help would be greatly appreciated, and sorry if its a bit hard to read, hopefully ill be posting using latex soon
 
  • #5
97
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thanks a lot for the replies
now i see why i shouldnt have said assuming for all n, i havent had much practice with induction, so ill make sure i wont start assuming what i want to prove

however i dont see how i went wrong with the x^2n?

i got the x^2n from using the quotient rule on (n-1)!/x^n
so i would get

(d/dx(n-1)!*x^n-d/dx(x^n)*(n-1)!)/(x^n)^2
= (0-nx^(n-1)*(n-1)!)/x^2n
=-(nx^(n-1)*(n-1)!)/x^2n

?, did i do something wrong here? or should i have brought the x^n to the numerator making it x^-n and then using the power rule?
 
  • #6
242
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using the power rule is much simpler and leaves less room for error, though now that i know you were using the quotient rule, your work is probably right (it is hard to tell because it is very hard to read, try putting some spaces in).

-(( n (n-1)! x^(n-1) ) / x^(2n) ) (-1)^(n-1) <-- this is correct! sorry!

now that i look even more carefully, your work is correct. i think we all missed the x^(n-1) in the numerator (it is unusual to use the quotient rule in this situation).
 
  • #7
97
0
thanks!
yeah it does look very messy so i understand why you guys thought i had made an error
even i had a bit of trouble reading it again looking for a flaw haha
 

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