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I've worked out the Taylor series for $y(x)=\ln (1+2x)$ "up to and including terms of 0(x5)": $$2x - 2x^2 + {8x^3\over 3} - 4x^4 + {32x^5\over 5}$$ Now I have to use the series to estimate the value of the definite integral $$\int_0^\frac 14 \frac 1x\ln (1+2x)dx$$ Every method I've found has involved a series with a formula involving only 'x' and 'n' as variables, while the $\ln (1+2x)$ series is represented by $$\sum_{n=0}^\infty {f^n (0)x^n\over n!}$$ I am unsure both how to incorporate the new $\frac 1x$ into the Taylor series I'm using, and how to estimate the definite integral with it.

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    $\begingroup$ Just divide the series by $x$? $\endgroup$ – Kenny Lau Sep 14 '17 at 8:25
  • $\begingroup$ $$\frac1x\sum_{n=0}^\infty \frac{f^n(0)x^n}{n!}=\sum_{n=0}^\infty \frac{f^n(0)x^{n-1}}{n!}$$ $\endgroup$ – Faiq Irfan Sep 14 '17 at 8:30
  • $\begingroup$ Why stop the expansion of $x \mapsto \ln(1+2x)$ at $x^5$ ? The power series expansion of this function is given for all $x \in \displaystyle \Big] -\frac{1}{2}, \frac{1}{2} \Big[$ by: $$ \ln(1+2x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1} 2^n x^n}{n}. $$ And then you can integrate this power series on $[0, 1/4]$. $\endgroup$ – jibounet Sep 14 '17 at 8:31
  • $\begingroup$ He wants to approximate $\endgroup$ – neonpokharkar Sep 14 '17 at 8:40
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    $\begingroup$ @neonpokharkar : I understand but my point is that the expansion $$ \ln(1+2x) = 2x - 2x^2 + \frac{8x^3}{3} - 4x^4 + \frac{32x^5}{5} + o(x^5) $$ is only valid in a "small" neighborhood of $0$ because $o(x^5)$ is a function of $x$ which goes to $0$ as $x$ goes to $0$. It is unclear to me whether this polynomial approximation of the function is valid on the interval $[0, 1/4]$ (because the $o(x^5)$ term might not be neglectable anymore). $\endgroup$ – jibounet Sep 14 '17 at 8:48
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$$\ln (1+2x)\approx 2x - 2x^2 + {8x^3\over 3} - 4x^4 + {32x^5\over 5}$$ Multiplying by $\frac{1}{x}$ $$\frac{1}{x}\ln (1+2x)\approx 2- 2x + {8x^2\over 3} - 4x^3 + {32x^4\over 5}$$$$$$

$$\int_0^{\frac{1}{4}} \frac{1}{x} \ln (1+2x) \approx \int_0^{\frac{1}{4}} ( 2-2x+\frac{8x^2}{3}-4x^3+\frac{32x^4}{5})$$$$\int_0^{\frac{1}{4}} \frac{1}{x} \ln (1+2x) \approx \Biggr[2x-x^2+\frac{8x^3}{9}-x^4+\frac{32x^5}{25}\Biggr]_0^{\frac{1}{4}}$$$$\int_0^{\frac{1}{4}} \frac{1}{x} \ln (1+2x) \approx 0.448733$$

It is a good estimate,

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  • $\begingroup$ You should improve/modify your answer because these are not equalities. You are missing the $o(x^5)$ [or $o(x^4)$] term in each line. $\endgroup$ – jibounet Sep 14 '17 at 8:42
  • $\begingroup$ Done and done and done $\endgroup$ – neonpokharkar Sep 14 '17 at 8:47

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