If the Taylor Series of ln⁡(x) is known: ln⁡(x)=(x−1)−12(x−1)2+13(x−1)2−14(x−1)4+15(x−1)5−… Can one find the Taylor series of f(x)=x1−x2 by manipulating the Taylor series of ln⁡(x) ?

Question

If the Taylor Series of ln⁡(x) is known:  ln⁡(x)=(x−1)−12(x−1)2+13(x−1)2−14(x−1)4+15(x−1)5−…  Can one find the Taylor series of  f(x)=x1−x2  by manipulating the Taylor series of ln⁡(x) ?

Joseph Lewis · Accepted Answer

Make the substitution x=1−u2 in order to obatin:  ln⁡(1−u2)=−u2−u42−u63−…=−∑k=1∞u2kk  Based on the ratio test, the corresponding function is analytic for |u|&amp;lt;1. Prescisely,  limk→∞|ak+1ak|=limn→∞(kk+1)|u2|=|u2|&amp;lt;1→|u|&amp;lt;1  Hence we can differentiate the obtained expression in order to get:  −2u1−u2=−2∑k=1∞u2k−1→f(x)=x1−x2=∑k=1∞x2k−1=x+x3+x5+…  whenever |x|&amp;lt;1, and we are done.  Hopefully this helps!

Paul Mitchell · Accepted Answer

This isnt

user_27qwe · Accepted Answer

Notice that: ∫f(x)dx=∫x1−x2dx=−12∫duu=−12ln⁡|1−x2|+C (via the u-substitution u=1−x2) From your note: ln⁡(x)=(x−1)−12(x−1)2+13(x−1)2+... So substitute 1−x2 into x in the natural log taylor expansion, and take the derivative to obtain a series representation of f(x).

If the Taylor Series of \ln(x) is known: \ln(x)=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^2-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-... Can one find

Answered question

Answer & Explanation

New Questions in Calculus 2