Limit with taylor

Nov 2018
32
3
Italy
I've tried studying the limit with taylor,(untill o(x^6)) but it result to me that the limit is equal to-5/6 and not 17/12... Any tips about this limit? I mean where do i get wrong?
Thanks


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Nov 2018
32
3
Italy
I've tried studying the limit with taylor,(untill o(x^6)) but it result to me that the limit is equal to-5/6 and not 17/12... Any tips about this limit? I mean where do i get wrong?
Thanks


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Here it is how i did it...
Where did i get wrong? I'm getting crazy.. :(


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Jun 2013
1,110
590
Lebanon
for the numerator you should get

\(\displaystyle \left(x^2-x^3+\frac{11 x^4}{12}\right)-\left(x^2-x^3-\frac{x^4}{2}\right)=\frac{17}{12}x^4\)

denominator is OK
 
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Nov 2018
32
3
Italy
for the numerator you should get

\(\displaystyle \left(x^2-x^3+\frac{11 x^4}{12}\right)-\left(x^2-x^3-\frac{x^4}{2}\right)=\frac{17}{12}x^4\)

denominator is OK
Mmm i'm not able to get that 11x^4/12
Where do i get wrong?
Thank you so much for your time...

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Dec 2013
2,000
757
Colombia
Your coefficients of $x^4$ and $x^5$ in the series for $\ln^2{(1+x)}$ look wrong to me. I think you forgot that there are two terms for each.

Also, $$\frac1{1-f(x)} \approx 1 + f(x) + f^2(x) + f^3(x) + \ldots$$ so

\begin{align}
\frac1{1-x} &= x + x^2 + x^3 + x^4 + o(x^4) \\
\log{(1+x)} &= x - \tfrac12 x^2 + \tfrac13x^3 - \tfrac14x^4 + o(x^3) \\[12pt]
\log^2{(1+x)} &= x^2 - x^3 + (\tfrac14 + \tfrac23) x^4 + o(x^4) \\
&= x^2 - x^3 + \tfrac{11}{12} x^4 + o(x^4) \\
&= x^2\big(1 - x + \tfrac{11}{12} x^2 + o(x^2)\big) \\[8pt]
\log{(1+x^2)} &= x^2 - \tfrac12 x^4 + o(x^4) \\
\log{(1+x^2)}-x^3 &= x^2 - x^3 - \tfrac12 x^4 + o(x^4) \\
&= x^2\big(1 - x - \tfrac12 x^2 + o(x^2)\big) \\[8pt]
\frac1{\log^2{(1+x)}} &= \frac1{x^2\big(1 - x + \tfrac{11}{12} x^2 + o(x^2)\big)} \\
&= \frac1{x^2}\cdot\frac1{1 - x + \tfrac{11}{12} x^2 + o(x^2)} \\
&= \frac1{x^2} \cdot \frac1{1 - \big(x - \tfrac{11}{12} x^2 + o(x^2)\big)} \\
&= \frac1{x^2} \bigg(1 + \big(x - \tfrac{11}{12} x^2 + o(x^2)\big) + \big(x - \tfrac{11}{12} x^2 + o(x^2)\big)^2 + o(x^2)\bigg) \\
&= \frac1{x^2} \bigg(1 + \big(x - \tfrac{11}{12} x^2\big) + \big(x^2\big) + o(x^2)\bigg) \\[8pt]
\frac1{\log{(1+x^2)}-x^3} &= \frac1{x^2\big(1 - x - \tfrac12 x^2 + o(x^2)\big)} \\
&= \frac1{x^2} \cdot \frac1{1 - x - \tfrac12 x^2 + o(x^2)} \\
&= \frac1{x^2} \cdot \frac1{1 - \big(x + \tfrac12 x^2 + o(x^2)\big)} \\
&= \frac1{x^2} \bigg(1 + \big(x + \tfrac12 x^2 + o(x^2)\big) + \big(x + \tfrac12 x^2 + o(x^2)\big)^2 + o(x^2)\bigg) \\
&= \frac1{x^2} \bigg(1 + \big(x + \tfrac12 x^2\big) + \big(x^2\big) + o(x^2)\bigg) \\[12pt]
\frac1{\log{(1+x^2)}-x^3} - \frac1{\log^2{(1+x)}} &= \frac1{x^2} \bigg(1 + \big(x + \tfrac12 x^2\big) + \big(x^2\big) + o(x^2)\bigg) - \frac1{x^2} \bigg(1 + \big(x - \tfrac{11}{12} x^2\big) + \big(x^2\big) + o(x^2)\bigg) \\
&= \frac1{x^2}\bigg[\bigg(1 + \big(x + \tfrac12 x^2\big) + \big(x^2\big) + o(x^2)\bigg) - \bigg(1 + \big(x - \tfrac{11}{12} x^2\big) + \big(x^2\big) + o(x^2)\bigg)\bigg] \\
&= \frac1{x^2}\big(\tfrac12 x^2 - (-\tfrac{11}{12}x^2) + o(x^2)\big) \\
&= \frac{x^2}{x^2}\big(\tfrac12 + \tfrac{11}{12} + o(1)\big) \\
&= \frac{17}{12} + o(1)
\end{align}
 
Last edited:
Nov 2018
32
3
Italy
Your coefficients of $x^4$ and $x^5$ in the series for $\ln^2{(1+x)}$ look wrong to me. I think you forgot that there are two terms for each.
I controlled it 10 times... The coefficients got from the taylor series are correct for the Numerator... :(

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Nov 2018
32
3
Italy
Your coefficients of $x^4$ and $x^5$ in the series for $\ln^2{(1+x)}$ look wrong to me. I think you forgot that there are two terms for each.
Omg... I thought it was factorial the denominator of the log Taylor series... What a shame omg. Thx mate

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