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I am reading a textbook and try to understand the following $$\frac{(x+\Delta x)^{1-\gamma}}{1-\gamma}=\frac{x^{1-\gamma}}{1-\gamma}+x^{-\gamma}\Delta x+o(\Delta x)$$ It says, that they use the Taylor expansion with respect to $\Delta x$. If you have a chance to check the book: It is "Mathematical Modeling in Economics, Ecology and the Environment" by Natali Hritonenko and Yuri Yatsenko. You can find the equation on page 127. Thanks

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    $\begingroup$ Can you explain what you do not understand? $\endgroup$
    – amWhy
    Oct 10 '17 at 16:17
  • $\begingroup$ Check the page because I dont see that in the book $\endgroup$ Oct 10 '17 at 16:17
  • $\begingroup$ Also, could you include a bit more textual context for those of us who don’t have the book? $\endgroup$ Oct 10 '17 at 16:22
  • $\begingroup$ Thank you very much for your comments. It is about optimal control in vintage capital models. A part of the model is approximated by using a Taylor expansion. I tried to follow the steps by using mathematica, but my calculations were different to those of the textbook. As I can see, John Doe (second answer below), has answered my question and cannot find a mistake. So I have to check my input in mathematica. Thanks for your comments $\endgroup$ Oct 11 '17 at 8:39
  • $\begingroup$ Dear @Isham, I found that passage in the 2nd edition of the book (should be the newest). It is in chapter 5.4.3 (Method of Lagrange Multipliers) after equation (5.60). It is said, that it is page 127. Thank you for your comment. $\endgroup$ Oct 11 '17 at 9:59
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Let $$f(t)=\frac{(x+t)^{1-\gamma}}{1-\gamma}$$ then its Taylor expansion at $0$ is $$f(t)=f(0)+f'(0)t+o(t).$$ Are you able to obtain the textbook's formula?

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  • $\begingroup$ Thank you for editing my post. John Doe answered my question in the meantime. $\endgroup$ Oct 11 '17 at 8:40
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Let $f(\Delta x)=(x+\Delta x)^{1-\gamma}$. Then $f'(\Delta x)=(1-\gamma)(x+\Delta x)^{-\gamma}$. Now expand about $0$: $$\begin{align}f(\Delta x)&=f(0)+\Delta x \cdot f'(0)+o(\Delta x)\\&=x^{1-\gamma}+\Delta x\cdot (1-\gamma)x^{-\gamma}+o(\Delta x)\\\implies \frac{f(\Delta x)}{1-\gamma}&=\frac{x^{1-\gamma}}{1-\gamma}+\Delta x\cdot x^{-\gamma}+o(\Delta x)\end{align}$$ I don't think there is a mistake

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  • $\begingroup$ Thank you very much. I can follow your explanation. Thanks for that. I tried to understand the explanations of the textbook by typing the formula into Mathematica. Obviously it is wrong. Can you help me with that? My input is like Series[(x + [Delta]x)^(1 -[Gamma])/(1 -[Gamma]), {[Delta]x, 0, 2}] $\endgroup$ Oct 11 '17 at 9:34
  • $\begingroup$ And there is something else. I used in my Mathematica input n=2. Because it is said in the textbook, that "...Using the Taylor's expansion $$\frac{(x+\Delta x)^{1-\gamma}}{1-\gamma}=\frac{x^{1-\gamma}}{1-\gamma}+x^{-\gamma}\Delta x+o(\Delta x)$$ for $\frac{x^{1-\gamma}}{1-\gamma}$ up to the second order with respect to $o(\Delta x)$, we obtain that..." Where is the second order included? Or in other words, it seems, that you do not have that in your formula (@John Doe), what is not a problem at all with respect to your equation. Thank you! $\endgroup$ Oct 11 '17 at 9:50
  • $\begingroup$ @KerimBenHamida You're welcome :) Unfortunately I am not familiar with Mathematica so I cannot help with that. As for the other bit, my Taylor expansion is up to first order in $\Delta x$ - you can expand further and get an additional term $$\frac{(\Delta x)^2}{2}f''(0)$$where $f''(0)=-\gamma(1-\gamma)x^{-\gamma -1}$Then the full Taylor expansion would be $$\frac{f(\Delta x)}{1-\gamma}=\frac{x^{1-\gamma}}{1-\gamma}+\Delta x\cdot x^{-\gamma}-\frac{(\Delta x)^2}{2}\gamma x^{-\gamma -1}+o((\Delta x)^2)$$which is to second order. Note $o(\Delta x)=O((\Delta x)^2)$, so maybe this is what was meant. $\endgroup$
    – John Doe
    Oct 11 '17 at 13:38

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