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Let \(\Large f(x) = log(1-x)\) c) Show that the formula for the power series holds analogously for \(\Large k=0\) (although the functions \(\Large f^{(0)}(x)=f(x)\) and \(\Large f^{(k)}(x)\) for \(\Large k > 0 \) are quite different).
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is there total solution or is it a just part of solution there ?
expand log(1-x) using taylor series at x=0 .. that must be equivalent to formula
is it the blok which starts with "Series expansion at x=0:" ?
\[ \frac 1 {1-x} =\sum_{n=0}^\infty x^n\\ \int_{0}^t \frac 1 {1-x} =\sum_{n=0}^\infty \int_{0}^t x^n\\ \ln(1-t) =\sum_{n=0}^\infty \frac{ t^{n+1}}{n+1} \]
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Of course |t|<1
:) thank you very much mr Elias, it was a suprise of you
@hash.nuke thank you too naturally
yw
np!!
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