Functions to power series Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series. f(x)=ln⁡4−x

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Functions to power series Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series.  f(x)=ln⁡4−x

l1koV · Accepted Answer

Given, f(x)=ln⁡4−x Find power series representation, the radius of converges, the interval of converges. Use the power series formula,  f(x)=ln⁡(1−xn)=−∑k=1∞xkk,−1≤x&amp;lt;1  Consider f(x)=ln⁡4−x2 f(x)=ln⁡(4−x2)12 Use the power rule of logarithmic, ln⁡m2=nln⁡m =12ln⁡(4−x2) Now multiple and divide &quot;4&quot; on the logarithmic function we get, =12ln⁡4(4−x24) =12ln⁡4(44−x24) =12ln⁡4(1−x24) Now use the product rule of logarithmic we get, ln⁡(mn)=ln⁡(m)+ln⁡(n) =12ln⁡4+ln⁡(1−x24) Now again use the power rule of logarithmic we get, =ln⁡412+ln⁡(1−x24) =ln⁡222+ln⁡(1−x24) =ln⁡2+ln⁡(1−x24) Now use the power series formula we get, =ln⁡2−∑k=1∞(x24)kk =ln⁡2−∑k=1∞(x2k4k)k For this we get interval, −2≤x≤2 Use Ratio test to check the series converges or diverges: Ratio test definition: Suppose we have series ∑an defined, L=limn→∞|an+1an| If L&amp;gt;1 then the series diverges, If L&amp;lt;1 then the series converges, If L=1 then the test fails. Consider, ak=x2k4kk ak+1=x2(k+1)4(k+1)(k+1) limk→∞|x2kx24k4(k+1)x2k4kk|=limk→∞|x2kx24k4(k+1)×4kkx2k| =limk→∞|x24|kk+1| =|x24|limk→∞|kk(1+1k)| =|x24|limk→∞|11+1k| =|x24

Jeffrey Jordon · Accepted Answer

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Functions to power series Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series. f(x)=lnsqrt{4-x}

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