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Find the Maclaurin series generated by the function
f(x)=(x^(2))/(1+x). Does this series converge at
x=1 ?

$3$ . Find the Maclaurin series generated by the function $f(x)=\frac{x^{2}}{1+x}$ . Does this series converge at $x=1$ ?

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Find the vertex of the transformed function

Full solution

Q.

3

. Find the Maclaurin series generated by the function

f(x)=\frac{x^{2}}{1+x}

. Does this series converge at

x=1

?

Expand $f(x)$ : To find the Maclaurin series, we need to expand $f(x)$ as a power series around $x=0$ . $\newline$ $f(x) = x^2 \cdot \left(\frac{1}{1+x}\right)$ $\newline$ We can use the geometric series formula $\frac{1}{1-r} = 1 + r + r^2 + r^3 + \ldots$ , where $|r| < 1$ . $\newline$ Here, $r = -x$ .
Use Geometric Series Formula: So, $f(x) = x^2 \cdot (1 - x + x^2 - x^3 + \ldots)$ Now we distribute $x^2$ to each term in the series. $f(x) = x^2 - x^3 + x^4 - x^5 + \ldots$
Distribute $x^2$ : The Maclaurin series for $f(x)$ is therefore: $f(x) = x^2 - x^3 + x^4 - x^5 + \ldots$
Find Maclaurin Series: To check if the series converges at $x=1$ , we substitute $x=1$ into the series. $f(1) = 1^2 - 1^3 + 1^4 - 1^5 + ...$ $f(1) = 1 - 1 + 1 - 1 + ...$ This is an alternating series with terms that do not approach zero, so it does not converge.

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