Question

Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x) = x2/(x4 + 16)

Answer 1

Answer:

Given the function:  $f(x) =\frac{x^2}{x^4+16}$

A geometric series is of the form of :

$\sum_{n=0}^{\infty} ar^n$

Now, rewrite the given function in the form of $\frac{a}{1-r}$ so that we can express the representation as a geometric series.

$\frac{x^2}{x^4+16}$

Now, divide numerator and denominator by $x^4$ we get;

$\frac{\frac{1}{x^2}}{1+\frac{16}{x^4}}$ = $\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2}$

Therefore, we now depend on the geometric series which is;

$\frac{1}{1+x} =\sum_{n=0}^{\infty} (-1)^n x^n$

let $x \rightarrow x^2$ then,

$\frac{1}{1+x^2} =\sum_{n=0}^{\infty} (-1)^n x^{2n}$

to get the power series let $x \rightarrow \frac{4}{x^2}$

so,

$\frac{1}{1+(\frac{4}{x^2})^2} =\sum_{n=0}^{\infty} (-1)^n (\frac{4}{x^2})^{2n}$

Multiply both side by $\frac{1}{x^2}$ we get;

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =\frac{1}{x^2} \cdot \sum_{n=0}^{\infty} (-1)^n (\frac{4}{x^2})^{2n}$

or

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =x^{-2} \cdot \sum_{n=0}^{\infty} (-1)^n (16)^n (x^{-2})^{2n}$

or

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =\sum_{n=0}^{\infty} (-1)^n (16)^n x^{-4n} \cdot x^{-2}$

Using $x^n \cdot x^m = x^{n+m}$

we have,

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =\sum_{n=0}^{\infty} (-1)^n (16)^n x^{-4n-2}$

therefore, the power series representation centered at x =0 for the given function is: $\sum_{n=0}^{\infty} (-1)^n (16)^n x^{-4n-2}$

Answer 2

This mathematical problem indicated above relates to the use of Power Series Representation as expressed in the given equation.

What is Power Series in Mathematics?

Power Series is a method in mathematics that is used to derive the solution to some types of differential equations.

What are the steps of the solution to the question above?

The function given is: f(x) = x²/(x⁴ + 16)

$\sum _{n=0} ^{\infty }$ $ar^{n}$

If this is rewritten, in the form of a geometric series, we have:

x²/(x⁴ +16); Next we divide by x⁴

Hence we have:

$X^{1/2}$/(1+ 16/X⁴)

= X$^{1/2}$/ (1+(4/x²))²

At this point, we refer to the geometric series:

1/(1+x) = $\sum _{n=0} ^{\infty }$ (-1)ⁿXⁿ

Given that x → x² then,

1/1+x² = $\sum _{n=0} ^{\infty }$ (-1)ⁿX²ⁿ

Because we need to arrive at the power series, let x → 4/x²

Therefore,

1/(1+(4/x²)² = $\sum _{n=0} ^{\infty }$ (-1)ⁿ(4/x²)²ⁿ; we proceed to multiply both sides by 1/x² and this results to:

(1/x²)/(1+(4/x²)² = (1/x²) * $\sum _{n=0} ^{\infty }$(-1)ⁿ (4/x²)²ⁿ

Applying xⁿ * $x^{m}$ = $x^{n+m}$; from this, we can arrive at:

(1/xⁿ)/(1+(4/x²) = $\sum _{n=0} ^{\infty }$ (-1)ⁿ(16)ⁿ x⁻⁴ⁿ⁻².

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