Approximating the Square Function: Mathematical Tricks and Other Methods

[PeteSampras]

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?

 

[PeroK]

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 $0^{1/n} = 0$.

 

[nuuskur]

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.

 

[PeteSampras]

Thanks. I don't understand your idea, because $f'(x_0)$ does not exist for x_0=1

 

[PeteSampras]

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

 

[nuuskur]

Thanks. I don't understand your idea, because $f'(x_0)$ 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]

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 $x$.

 

[nuuskur]

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.

 

[PeteSampras]

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 don't 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]

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 $x = 1$.

 

[Fred Wright]

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.

 

[haruspex]

You might as well work in terms of $x^{\frac 1n}$, for x near zero,
It depends what you want from an approximation. Generally you want such as an easy way to generate numerical solutions. What form will x be in? If in scientific notation, $x=a\times 10^m$, you have $a^{\frac 1n}\times 10^{\frac mn}$. For large n you can ignore $a^{\frac 1n}$, and it’s easy to reduce m so that it is less than n.
This leaves you with $10^y$, 0<y<1. For that, you can use $e^{y\ln(10)}$ or $e^{y\ln(10)+\ln(a)}$.

But is the problem as stated in post #1 as it was given to you? If not, please post the entire original problem.