For a postive integers

we have

for a fraction

we also have

I can complete the above proofs but don't know to prove for irrational numbers index ? Is ?

Moreover , are and ?

Thank you !

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- September 3rd 2009, 02:44 AMsimplependulumDifferentiation proof
For a postive integers

we have

for a fraction

we also have

I can complete the above proofs but don't know to prove for irrational numbers index ? Is ?

Moreover , are and ?

Thank you ! - September 3rd 2009, 03:13 AMProve It
- September 3rd 2009, 03:38 AMynj
First we have: (a is the indepedent variable).

Let be a irrational number. Let be a rational sequence with .

We get a sequence of functions .

Now we are to prove that , so that .

There is a theorem saying that:

If 1: ,

2: uniformly converges at ,

3: converges at at least one point ,

then .

we know that for every x, ,so 3 is true.

1 is clearly true.

Now we only have to show 2.Let

let , it suffices to show that

Let , then

so

We have .

so ,the second inequality is true since is uniformly continous on .

So ,we are done..

So tough a proof...(Wink)(Evilgrin) - September 3rd 2009, 05:56 AMHallsofIvy
Or, if you first prove that the derivative of ln(x) is 1/x, starting from , take the ln of both sides: ln(y)= r ln(x).

Now, differentiating both sides with respect to x, . Multiply both sides by y: .

That's easier than ynj's proof but requires first proving that the derivative of ln(x) is 1/x as well as the chain rule. - September 3rd 2009, 09:05 AMKrizalid