# Approximate square function

## Homework Statement:

Is there some way to approximate this function, for x=1?

## Relevant Equations:

The function is ##(x-1)^{1/n}##, with ##n## integer and n>1 and n=odd.
Obviously, a priori it is not possible tu use the Taylor series because the derivative ##\sim (x-1)^{1/n-1}## is not well defined in x=1.

Is there any mathematical trick? or, other approximation?

nuuskur

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PeroK
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Homework Statement:: Is there some way to approximate this function, for x=1?
Relevant Equations:: The function is ##(x-1)^{1/n}##, with ##n## integer and n>1 and n=odd.

Obviously, a priori it is not possible tu use the Taylor series because the derivative ##\sim (x-1)^{1/n-1}## is not well defined in x=1.

Is there any mathematical trick? or, other approximation?
Do you mean for ##x \approx 1##?

At ##x = 1## you have ##0^{1/n} = 0##.

$$f'(x_0) = \lim _{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0} \Rightarrow f'(x_0)(x-x_0) +f(x_0) = f(x) +o(|x-x_0|),\ x\to x_0 .$$
Take $x_0 =1 + \delta$. If $|x-x_0|$ is small, then it's a pretty good linear approximation.

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Thanks. I dont understand your idea, beacause ##f'(x_0)## does not exist for x_0=1

Do you mean for ##x \approx 1##?

At ##x = 1## you have ##0^{1/n} = 0##.
Thanks. This is my doubt.

As for example, if is there any approximation of ##x^{1/3}## near x=0, where Taylor is not possible because the function is not differentiable

Thanks. I dont understand your idea, beacause ##f'(x_0)## does not exist for x_0=1
Indeed, but if you can calculate value of derivative exactly just a bit further from $1$, then you could approximate linearly around a very small neighborhood of $x_0>1$.

PeroK
Homework Helper
Gold Member
Thanks. This is my doubt.

As for example, if is there any approximation of ##x^{1/3}## near x=0, where Taylor is not possible because the function is not differentiable
If ##x## is small, then powers of ##x## are smaller. But, ##x^{1/3} > x##, so you are not going to get a power series approximation to ##x## in powers of ##x##.

For the record, non-smooth approximation is hard. Most of the theory I've come across always assumes differentiability or absolute continuity or something "nice". Non-smooth continuous functions need not be nice at all.

Indeed, but if you can calculate value of derivative exactly just a bit further from $1$, then you could approximate linearly around a very small neighborhood of $x_0>1$.
I dont understand :(
If ##x## is small, then powers of ##x## are smaller. But, ##x^{1/3} > x##, so you are not going to get a power series approximation to ##x## in powers of ##x##.
This mean that, in my problem ##(1-x)^{1/n}## is not possible to approximate the function for ##x_0 \approx 1##?

PeroK
Homework Helper
Gold Member
This mean that, in my problem ##(1-x)^{1/n}## is not possible to approximate the function for ##x_0 \approx 1##?
I don't see how you can do it with a Taylor series. One problem is that the derivatives are unbounded near ##x = 1##.

For an approximation, I suggest you expand using the binomial theorem and take the limit as x approaches 1. For ##x\gt 1## $$(x-1)^{\frac{1}{n}}=(x)^{\frac{1}{n}}(1-\frac{1}{x})^{\frac{1}{n}}=(x)^{\frac{1}{n}}\sum_{k=0}^{\infty} \begin{pmatrix} \frac{1}{n} \\ k \end{pmatrix} (\frac{1}{x})^k$$Use the identity$$\begin{pmatrix} \frac{1}{n} \\ k \end{pmatrix}=\frac{(-1)^k}{k!} \prod_{j=0}^{k-1} (j-\frac{1}{n})$$ for each term in the series.

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haruspex