Iterative derivatives of log

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Homework Statement:

Let f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =
87 178 291 200.

Relevant Equations:

n=k
n=k+1
Derivative of lnx is 1/x.
I have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?
 

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BvU
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PeroK
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Homework Statement: Let f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =
87 178 291 200.
Homework Equations: n=k
n=k+1
Derivative of lnx is 1/x.

I have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?
You can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?
 
  • #4
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The general outline of a basic inductive proof is in two steps:
1) Prove it is true for N=1
2) Prove that, assuming it is true for N=n, prove it is true for N=n+1.

There is another variation where in step #2 you assume that it is true for N##\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.
 
  • #5
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A little googling perhaps ?
I tried to google it, but I still did not understand.
 
  • #6
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You can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?
Induction is one of the things I struggle with in math, that why it was not obvious to me. Thanks for the tips by the way.
 
  • #7
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The general outline of a basic inductive proof is in two steps:
1) Prove it is true for N=1
2) Prove that, assuming it is true for N=n, prove it is true for N=n+1.

There is another variation where in step #2 you assume that it is true for N##\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.
Thank you for the tips and tricks. But how do I prove for n when I don't have a general formula? Do I have to make a general formula for the derivative of lnx?
 
  • #8
BvU
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general formula for the derivative of lnx?
For the nth derivative
 
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For the nth derivative
Yes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.
 
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Do you know what the first derivative of ##ln(x)## and of ##x^{-n}## are? That should be all you need.

PS. When you are stuck by an intimidating problem, do what you can do. You might find out that you can do a lot more than you anticipated.
 
  • #11
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The first thing I would do is calculate a few derivatives. With f(x)= ln(x), [tex]f'= 1/x= x^{-1}[/tex], [tex]f''(x)= -x^{-2}[/tex], [tex]f'''= 2x^{-3}[/tex], [tex]f^{IV}= -6x^{-4}[/tex], [tex]f^{V}= 24x^{-5}[/tex], etc.
Did you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is [tex]f^{(n)}(x)= 0=(-1)^{n+1}(n-``1)! x^{-n}[/tex]. It remains to use induction to prove that.
When n= 1, the derivative is [tex]\frac{1}{x}= x^{-1}[/tex]. The formula gives [tex](-1)^2(0!)x^{-1}= x^{-1}[/tex] so is true for x= 1.
Now assume that for some n= k, [tex]f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}[/tex]. Then [tex]f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}[/tex]
 
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PeroK
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Yes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.
If you want to find a derivative, you differentiate the function! If you want to find the second derivative, you differentiate the first derivative. And so on.
 
  • #13
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The first thing I would do is calculate a few derivatives. With f(x)= ln(x), [tex]f'= 1/x= x^{-1}[/tex], [tex]f''(x)= -x^{-2}[/tex], [tex]f'''= 2x^{-3}[/tex], [tex]f^{IV}= -6x^{-4}[/tex], [tex]f^{V}= 24x^{-5}[/tex], etc.
Did you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is [tex]f^{(n)}(x)= 0=(-1)^{n+1}(n-``1)! x^{-n}[/tex]. It remains to use induction to prove that.
When n= 1, the derivative is [tex]\frac{1}{x}= x^{-1}[/tex]. The formula gives [tex](-1)^2(0!)x^{-1}= x^{-1}[/tex] so is true for x= 1.
Now assume that for some n= k, [tex]f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}[/tex]. Then [tex]f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}[/tex]
Thank you so much!!! I finally understand it. I really appreciate your help! :)
 

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