# Analysis: prove that ln(x) is a smooth function (i.e. infinitely differentiable)

1. Dec 1, 2011

### yossup

1. The problem statement, all variables and given/known data

Prove that f(x) is a smooth function (i.e. infinitely differentiable)

2. Relevant equations

ln(x) = $\int^{x}_{1}$ 1/t dt

f(x) = ln(x)

3. The attempt at a solution

I was thinking about using taylor series to prove ln(x) is smooth but i'm strictly told to NOT assume f(x) = ln(x) beforehand.

Last edited: Dec 1, 2011
2. Dec 1, 2011

### Office_Shredder

Staff Emeritus
I'm confused. When you say you can't assume f(x)=ln(x), then how is f(x) defined?

3. Dec 1, 2011

### yossup

i guess i meant - we're supposed to derive that f(x) is smooth from f(x) = $\int^{x}_{1}$ 1/t dt, rather than f(x)=ln(x)

4. Dec 2, 2011

### Office_Shredder

Staff Emeritus
Ok. Can you calculate f'(x) from that definition?

5. Dec 2, 2011

### yossup

f'(x) is just 1/t, so i just go about proving that 1/t is infinitely differentiable?

6. Dec 2, 2011

### Office_Shredder

Staff Emeritus
That's correct, except it's 1/x not 1/t

7. Dec 2, 2011

### yossup

ah right. i know it's intuitive that 1/x is infinitely differentiable but what do we use to rigorously prove it? taylor series?

8. Dec 2, 2011

### Office_Shredder

Staff Emeritus
Well, can you calculate the derivative of 1/x? How about the second derivative of 1/x? The third derivative?

Once you see a pattern, think about a proof by induction

9. Dec 2, 2011

### yossup

thank you.

10. Dec 2, 2011

### Deveno

one caveat, which i feel i ought to mention:

the domain of definition *matters*.

you can define f on (0,∞), but not on any interval containing 0 (because f is undefined).

a similar restriction holds for the derivatives.

so it's not enough just to say f is smooth....you have to say WHERE f is smooth.