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4 years ago

6

What is the relationship between the place-value position of

1 answer:

o-na [289]4 years ago

3 0

Answer:7

Step-by-step explanation:

Because it’s the place value of 5

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

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

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