# Taylor Expansion for very small and very big arguments

• deuteron
deuteron
Homework Statement
Expand the given functions for very small and very big arguments ##x## giving ##3## terms
Relevant Equations
##T_nf[x;a]= \displaystyle\sum_k \frac 1 {k!}\ f^{(k)}|_a\ (x-a)^k##
The function is
$$f(x)=\sqrt{1-x}$$

and we are expected to expand it using Taylor expansion for very small ##(x<<1)## and very big ##(x>>1)## arguments

My thought process was the following:

$$T_2f[x;x_0]=\sqrt{1-x_0} -\frac 12 \frac 1{\sqrt{1-x_0}}(x-x_0) -\frac 14 \frac 1 {\sqrt{1-x_0}^3}(x-x_0)^2$$

and then assuming small arguments, namely assuming ##x_0<<1##, I did the following approximation:

$$T_2f[x;x_0]\approx \sqrt{1} -\frac 12 \frac 1 {\sqrt{1}} (x) -\frac 1 4 \frac 1 {\sqrt{1}^3} (x)^2 = 1-\frac 12 x -\frac 14 x^2$$

For big arguments, I assumed ##x_0>>1## and made the following approximation:

$$T_2f[x;x_0]\approx i\sqrt{a} +\frac 12 \frac 1 {i\sqrt{a}} (a) +\frac 14 \frac 1 {i\sqrt{a}^3} a^2$$

However, the real answers are, for small arguments:

$$f(x<<1)= 1-\frac 12 x - \frac 18 x^2$$

and for big arguments:

$$f(x>>1)= i\sqrt{a} \sqrt{1-\frac 1x} = i\sqrt{x} \ (1-\frac 1 {2x} -\frac 1{8x^2})$$

I see that my mistake is to assume the expansion point to be very big instead of assuming big ##x##, but I don't know how to do that, does anyone understand the in between steps for the solution?

Use the binomial expansion: $$(1 + x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \dots + \frac{\alpha(\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + \dots,\qquad |x| < 1.$$ For small $|x|$, $$f(x) = (1 - x)^{1/2}$$ can be expanded as is. A Taylor series of $x^{1/2}$ about $1$ should also yield the same result, bearing in mind that we are looking at $(1 - x)^{1/2}$ not $(1 + x)^{1/2}$. For large negative $x$, $$(1 - x)^{1/2} = (1 + |x|)^{1/2} = |x|^{1/2}(1 + |x|^{-1})^{1/2}$$ can be expanded. For large positive $x$, $$f(x) = ix^{1/2}(1 - x^{-1})^{1/2}$$ can be expanded.

deuteron

## What is a Taylor Expansion?

A Taylor Expansion is a series representation of a function as a sum of terms calculated from the values of its derivatives at a single point. It provides an approximation of the function near that point.

## How does the Taylor Expansion behave for very small arguments?

For very small arguments, the Taylor Expansion often converges quickly, and only a few terms are needed to achieve a good approximation. This is because the higher-order terms, which involve higher powers of the small argument, become negligible.

## What challenges arise when using Taylor Expansion for very large arguments?

For very large arguments, the Taylor Expansion may converge slowly or not at all. The higher-order terms can become very large, leading to inaccuracies and potential divergence. Alternative series expansions or asymptotic approximations are often used in these cases.

## Can Taylor Expansion be used for all functions?

Not all functions can be represented by a Taylor Expansion. A function must be infinitely differentiable at the point of expansion, and the series must converge to the function value. Functions with singularities or discontinuities at the expansion point are not suitable for Taylor Expansion.

## What are practical applications of Taylor Expansion for small and large arguments?

For small arguments, Taylor Expansion is often used in physics and engineering to simplify complex functions around equilibrium points. For large arguments, it is used in asymptotic analysis to approximate functions in limits, though often other methods like asymptotic series or Padé approximants are preferred.

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