Verify this statement if f(x) is infinitely differentiable

[utkarshakash]

Homework Statement

f(mx)=f(x) + (m-1)xf'(x)+\dfrac{(m-1)^2}{2!} x^2 f''(x)...

Homework Equations

Taylor's Series

The Attempt at a Solution

If I approximate the LHS of the eqn using Taylor's polynomial,
f(mx)=f(mx)+mxf'(mx)+\dfrac{(mx)^2f''(mx)}{2!}+...

But, I'm lost from here. It doesn't resemble RHS at all!

 

[ehild]

Let be y=m the variable and k=x a constant. Do the Taylor expansion of f(ky) about y=1.

ehild

 

[utkarshakash]

Let be y=m the variable and k=x a constant. Do the Taylor expansion of f(ky) about y=1.

ehild

Doing so gives me f(k)+(ky-k)f'(k)+\dfrac{(ky-k)^2f"(k)}{2!}+...

which indeed is equal to RHS! Brilliant! But why did you assume x as a constant and m as a variable?

 

[haruspex]

To get f(x) on the right, you need something of the form f(x+t) on the left. What does that make t equal to?

 

[utkarshakash]

To get f(x) on the right, you need something of the form f(x+t) on the left. What does that make t equal to?

t=(m-1)x

 

[ehild]

Doing so gives me f(k)+(ky-k)f'(k)+\dfrac{(ky-k)^2f"(k)}{2!}+...

which indeed is equal to RHS! Brilliant! But why did you assume x as a constant and m as a variable?

because that solves the problem The right side is very much like a Taylor expansion only you have m-1 instead of x-xo and f' is multiplied by x and f" multiplied by x², as if x was the coefficient of the independent variable, so it must be m and the expansion is about m-1.

 

[utkarshakash]

because that solves the problem The right side is very much like a Taylor expansion only you have m-1 instead of x-xo and f' is multiplied by x and f" multiplied by x², as if x was the coefficient of the independent variable, so it must be m and the expansion is about m-1.

Nice! What about this question
$$f(\dfrac{x^2}{x+1}) = f(x) - \dfrac{x}{x+1}f'(x)+\left( \dfrac{x}{x+1} \right) ^2 f"(x)+... $$

This doesn't look as simple as the previous one. There are alternating + and - signs along with x/x+1 as the coefficient in each term.

 

[Ray Vickson]

Nice! What about this question
$$f(\dfrac{x^2}{x+1}) = f(x) - \dfrac{x}{x+1}f'(x)+\left( \dfrac{x}{x+1} \right) ^2 f"(x)+... $$

This doesn't look as simple as the previous one. There are alternating + and - signs along with x/x+1 as the coefficient in each term.

Write $x^2/(x+1) = x + t$ and figure out the value of $t$.

BTW: in both these questions, just having $f$ infinitely differentiable may not be not enough: you might also need $f$ to be analytic. There are infinitely-differentiable functions which do not equal their Maclauren or Taylor series, but that is for functions expanded about a fixed point $x_0$; this problem is different, so I am not really sure whether or not there are counterexamples in this case. However, if $f$ is known to be analytic you are basically done.

 

[utkarshakash]

Write $x^2/(x+1) = x + t$ and figure out the value of $t$.

BTW: in both these questions, just having $f$ infinitely differentiable may not be not enough: you might also need $f$ to be analytic. There are infinitely-differentiable functions which do not equal their Maclauren or Taylor series, but that is for functions expanded about a fixed point $x_0$; this problem is different, so I am not really sure whether or not there are counterexamples in this case. However, if $f$ is known to be analytic you are basically done.

But what about the factorials that I get in the denominator in the expansion of f(x+t)?

 

[Ray Vickson]

But what about the factorials that I get in the denominator in the expansion of f(x+t)?

What has that got to do with anything? Of course the Taylor expansion has factorials in the denominators---so what?

 

[haruspex]

But what about the factorials that I get in the denominator in the expansion of f(x+t)?

Looks to me like the question has left out the factorials by mistake.

 

[utkarshakash]

Looks to me like the question has left out the factorials by mistake.

Hmm, that might be true. Here's another expression which I'm finding difficult to solve.

f(x)=f(0)+xf'(x)-\frac{x^2}{2!}f"(x)+.....

Here t=0. Now, if I expand it as Taylor's series, I get
$$f(0)+xf'(0)+x^2/2! f''(0)...$$
which does not contain alternating + and - signs as in the original question. Also, the argument of f is 0 for all terms in my case.

 

[haruspex]

Hmm, that might be true. Here's another expression which I'm finding difficult to solve.

f(x)=f(0)+xf'(x)-\frac{x^2}{2!}f"(x)+.....

Here t=0. Now, if I expand it as Taylor's series, I get
$$f(0)+xf'(0)+x^2/2! f''(0)...$$
which does not contain alternating + and - signs as in the original question. Also, the argument of f is 0 for all terms in my case.

Try rearranging the given equation

Spoiler

with f(0) one side and all else the other

 

[utkarshakash]

Try rearranging the given equation

Spoiler

with f(0) one side and all else the other

$$f(0) = f(x) -xf'(x)+\frac{x^2}{2!} f''(x) - \frac{x^3}{3!}f'''(x)+...$$

I still can't figure out what to do next.

 

[haruspex]

$$f(0) = f(x) -xf'(x)+\frac{x^2}{2!} f''(x) - \frac{x^3}{3!}f'''(x)+...$$

I still can't figure out what to do next.

As before, the standard form is f(x+t) = f(x) + tf'(x) + ... So you need to find substitutions for x and t in there to turn f(x+t) into f(0) etc.

 

[utkarshakash]

As before, the standard form is f(x+t) = f(x) + tf'(x) + ... So you need to find substitutions for x and t in there to turn f(x+t) into f(0) etc.

Thanks!

 

[HallsofIvy]

By the way, is the only condition here that f be infinitely differentiable? If that is the case then it does not necessarily follow that the Taylor's series for f itself converges to f. For that you need that f is "analytic".