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In this question, I have to write Taylor's series expansion of the function $f(x) = ln(x+n)$ about x = 0, where n ≠ 0 is a known constant.

I have done the following: enter image description here

But my professor handed me back the answer as shown in red above.

Could somebody give me a hand and tell me what did I do wrong here?

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Your first two lines look fine, but you then use the expression for $f'(x)$ where you want $f'(x_0) = f'(0)$. – Dylan Moreland Jul 9 '11 at 15:28
It might be helpful to write $\log(x +n)$ as $\log n + \log( 1 + \frac{x}{n})$. – Geoff Robinson Jul 9 '11 at 15:38
up vote 1 down vote accepted

The notation $$ \left.\frac{d^rf}{dx^r}\right|_{x=x_0} $$ means that you have to plug in the Taylor expansion the $n$-th derivative evaluated at $x=x_0$ (in your case $x_0=0$).

Thus $$ \left.\frac{df}{dx}\right|_{x=0}=\left.\frac1{x+n}\right|_{x=0}=\frac1n $$ and so on.

Mind that in the Taylor expansion $x$ needs to appear only through its powers $x$, $x^2$, $x^3$, ....

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You did not evaluate the derivatives at $x=0$!


Another approach to the problem would be as follows. Clearly $n>0$, else we have trouble at $0$.

Since $x+n=n(1+x/n)$, we have $$\ln(x+n)=\ln n + \ln\left(1+\frac{x}{n}\right).$$

Now you can just quote the power series for $\ln(1+y)$, which has undoubtedly already been done.

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This all comes down to $$\left.\frac{df}{dx}\right|_{x=x_0}$$. You need to actually plug in the value of $x_0$ for $x$ at those points. So when your teacher writes $x/n - x/2n^2 + ...$ it's because she has already substituted 0 in for $\dfrac{x}{x_0 + n} - \dfrac{x}{2(n + x_0)^2} + ...$

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$\displaystyle \frac{1}{n+x} = \frac{1}{n}\left(\frac{1}{1+\frac{x}{n}}\right) = \frac{1}{n}\sum_{k \ge 0}\frac{(-1)^kx^k}{n^k} = \sum_{k \ge 0}\frac{(-1)^kx^k}{n^{k+1}}$, integrating both sides, we have $\displaystyle \ln(x+n) = \sum_{k \ge 0}\frac{(-1)^kx^{k+1}}{n^{k+1}(k+1)}+\mathcal{C} $. If we put $x = 0$, we get $\mathcal{C} = \ln(n)$, and we are done.

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