Understanding Iterative Derivatives of Log Functions

[Kolika28]

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?

 

[BvU]

I don't know where to start

A little googling perhaps ?

 

[PeroK]

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?

 

[FactChecker]

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.

 

[Kolika28]

A little googling perhaps ?

I tried to google it, but I still did not understand.

 

[Kolika28]

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.

 

[Kolika28]

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?

 

[BvU]

general formula for the derivative of lnx?

For the n^th derivative

 

[Kolika28]

For the n^th 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.

 

[FactChecker]

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.

 

[HallsofIvy]

The first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, 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 $$f^{(n)}(x)= 0=(-1)^{n+1}(n-``1)! x^{-n}$$. It remains to use induction to prove that.
When n= 1, the derivative is $$\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.
Now assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$

 

[PeroK]

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.

 

[Kolika28]

The first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, 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 $$f^{(n)}(x)= 0=(-1)^{n+1}(n-``1)! x^{-n}$$. It remains to use induction to prove that.
When n= 1, the derivative is $$\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.
Now assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$

Thank you so much! I finally understand it. I really appreciate your help! :)