How to prove if f(x) is infinitely differentiable

[j-lee00]

Say I have the function

f(x) = x^5ln(x)

I can differentiate f(x) say 6 times and I'm left with just ln(x). But this is not a proof!

How do I go about proving it?

Could I use the Taylor expansion?

 

[zhermes]

In general, I'm not sure what the best way to prove infinitive differentiability is, but in this case: taylor expansion would work very well (because there is such a nice clean answer).

 

[Office_Shredder]

If f(x) and g(x) are differentiable n times, then f(x)g(x) is differentiable n times (proof by induction). So if f(x) and g(x) are differentiable infinitely many times, so is f(x)g(x).

 

[j-lee00]

If f(x) and g(x) are differentiable n times, then f(x)g(x) is differentiable n times (proof by induction). So if f(x) and g(x) are differentiable infinitely many times, so is f(x)g(x).

Say f(x)=x^5 and g(x)=ln(x) but f(x) is not infinitely differentiable ?

 

[Tedjn]

x^5 is infinitely differentiable. Why? Also, if I ask you for the nth derivative of some function h(x), for any value of n, and you can give it to me, would you agree that h(x) is infinitely differentiable?

 

[j-lee00]

x^5 is infinitely differentiable. Why? Also, if I ask you for the nth derivative of some function h(x), for any value of n, and you can give it to me, would you agree that h(x) is infinitely differentiable?

So if the derivative = 0, then it is still differentiable?

 

[Tedjn]

Yes. To see this, just apply the definition of derivative to f(x) = 0.

 

[icystrike]

I'm wondering if i can go about saying that the taylor expansion of x^5 * ln(x) is equal to the taylor expansion of x^5 multiply the taylor expansion of ln(x). Since ln(x) is infinitely differentiable, x^5 * ln(x) is also infinitely differentiable..

 

[Wizlem]

The taylor series is unique for a given function so if you find a series which fits the taylor series form, then it is the taylor series of that function. The problem with x⁵*ln(x) is that ln(x) doesn't have a taylor series expansion about 0.

 

[statdad]

Say I have the function

f(x) = x^5ln(x)

I can differentiate f(x) say 6 times and I'm left with just ln(x). But this is not a proof!

How do I go about proving it?

Could I use the Taylor expansion?

If you differentiate f 6 times you are not left with \ln(x).

 

[icystrike]

The taylor series is unique for a given function so if you find a series which fits the taylor series form, then it is the taylor series of that function. The problem with x⁵*ln(x) is that ln(x) doesn't have a taylor series expansion about 0.

Simply out of curousity.. Where can i find the proof for the uniqueness of taylor series of function?

 

[Landau]

Simply out of curousity.. Where can i find the proof for the uniqueness of taylor series of function?

Well, there's not really anything to prove: the Taylor series is defined as some special power series, namely whose coefficients a_n are explicitly prescribed: some constant times the n-th derivative. Otherwisre see here.

Interesting (in complex analysis): two convergent power series which coincide on an infinite set with 0 as accumulation point, must be equal. See e.g. Serge Lang's book, p.62.

 

[Office_Shredder]

Using power series is a weak way to do this. For example, you could expand ln(x) as a power series around x=1, and x⁵ as a power series around x=1, but when you multiply them you get a power series with a radius of convergence of 1 (because that's the radius of convergence of ln(x)). So how do I know that the function is infinitely differentiable at x=7?

 

[Jilvin]

Here's what I would do:

1. Every polynomial of degree n can be differentiated (n+1) times before vanishing (it is n+1 times differentiable).
2. Thus, an infinite order polynomial is infinitely differentiable.
3. The power series expansion of ln x is of infinite degree. This expansion absorbs the x^5 term, merely creating another infinite degree expansion with each term 5 degrees higher. This combined expansion is infinitely differentiable.

I don't know what rigor my proof lacks, but that's my take if it's at all valuable.

NVM: I ignored radius of convergence as pointed out by the previous poster. HOWEVER, can one use analytic continuation to "cover" the entire domain of the function?

 

[statdad]

Here's what I would do:

1. Every polynomial of degree n can be differentiated (n+1) times before vanishing (it is n+1 times differentiable).

A polynomial is infinitely differentiable - every derivative after the n+1st is zero, but that is immaterial.

2. Thus, an infinite order polynomial is infinitely differentiable.

There is no such thing as an infinite order polynomial.

3. The power series expansion of ln x is of infinite degree. This expansion absorbs the x^5 term, merely creating another infinite degree expansion with each term 5 degrees higher. This combined expansion is infinitely differentiable.

This is correct heuristically, and may be all the OP needs. But I would expect that the statements made in (1) and (2) would still run afoul of his/her professor's grading. The correct terms must be used. Even so, it is possible to show that the function is infinitely differentiable without resorting to power series.

 

[Office_Shredder]

Multiplying two infinitely differentiable functions gives an infinitely differentiable function (the details for why are sketched above). This doesn't need to be more complicated than it is