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I came across an interesting approximation today in an indian olympiad paper (page 6). It essentially boils down to:

$$V(x)=\frac{k}{\sqrt{x^2+4/3}}+\frac{k}{\sqrt{(x+1)^2+1/3}}+\frac{k}{\sqrt{(x-1)^2+1/3}}$$ $$\approx k\sqrt{\frac 34}(3+\frac {9}{16}x^2)$$

Even before I got down to attempt to solve this step, I realized a problem. expression 1 decreases with $x$, since $x$ is in the denominator of all three terms in it. But expression 2 increases with $x$, since $x$ is in the numerator. But both are supposed to be approximately equal. How do we justify this?

Gaurang Tandon
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2 Answers2

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While it is true that for large $x$, the left side tends to $0$ and the right side tends to $\infty$, these are approximations for small $x$. As Daniel Fischer notes, for small $x$, we need to apply a bit more care.

The first three terms in the Taylor Series for $\frac1{\sqrt{1+x}}$ are $$ (1+x)^{-1/2}=1\color{#090}{-\tfrac12x}\color{#C00}{+\tfrac38x^2}\color{#00F}{-\tfrac5{16}x^3}+O\!\left(x^4\right)\tag1 $$ Thus, $$ \begin{align} \left(\tfrac43+x^2\right)^{-1/2} &=\tfrac{\sqrt3}2\left(1+\tfrac34x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1\color{#090}{-\tfrac38x^2}\color{#C00}{+O\!\left(x^4\right)}\right)\tag2 \end{align} $$ and $$ \begin{align} \left(\tfrac13+(x+1)^2\right)^{-1/2} &=\left(\tfrac43+2x+x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1+\tfrac32x+\tfrac34x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1\color{#090}{-\tfrac34x-\tfrac38x^2}\color{#C00}{+\tfrac{27}{32}x^2+\tfrac{27}{32}x^3}\color{#00F}{-\tfrac{135}{128}z^3}+O\!\left(x^4\right)\right)\\ &=\tfrac{\sqrt3}2\left(1-\tfrac34x+\tfrac{15}{32}x^2-\tfrac{27}{128}x^3+O\!\left(x^4\right)\right)\tag3 \end{align} $$ and $$ \begin{align} \left(\tfrac13+(x-1)^2\right)^{-1/2} &=\left(\tfrac43-2x+x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1-\tfrac32x+\tfrac34x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1\color{#090}{+\tfrac34x-\tfrac38x^2}\color{#C00}{+\tfrac{27}{32}x^2-\tfrac{27}{32}x^3}\color{#00F}{+\tfrac{135}{128}z^3}+O\!\left(x^4\right)\right)\\ &=\tfrac{\sqrt3}2\left(1+\tfrac34x+\tfrac{15}{32}x^2+\tfrac{27}{128}x^3+O\!\left(x^4\right)\right)\tag3 \end{align} $$ Adding, we get $$ \begin{align} &\left(\tfrac43+x^2\right)^{-1/2}+\left(\tfrac13+(x+1)^2\right)^{-1/2}+\left(\tfrac13+(x-1)^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(3+\tfrac9{16}x^2+O\!\left(x^4\right)\right)\tag4 \end{align} $$ Therefore, we get an approximation accurate to $O\!\left(x^4\right)$, which is why the plots match so closely, as noted by Claude Leibovici.

Because this is an even function, there will only be terms in the expansion with even exponents of $x$. This is why the terms with $x$ and $x^3$ vanish.

robjohn
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  • Hello Rob do you have any idea here ?https://math.stackexchange.com/questions/2587694/convergence-of-seriess-of-the-generalized-mean-sum-n-1-infty-left-frac – Guy Fsone Jan 18 '18 at 15:10
  • Being almost blind, there are a lot of things I do not percieve. So, my wife (she does not know anything in science) tells that, with colors, your answer makes mathemeatics beautiful to her. I wanted you to know that your are also fascinating old ladies ! This does not look to be a very serious site. Thanks and cheers. – Claude Leibovici Jan 18 '18 at 15:17
  • Wow, I'm amazed at how you used the result of taylor expansion of linear in denominator of $x$ in that of $x^2$ before (3). Great! – Gaurang Tandon Jan 18 '18 at 15:19
  • @ClaudeLeibovici: It is my opinion that the simpler things appear, the more enticing they are. What is done with color to group related terms could be done in monochrome, but perhaps with added verbiage or formulas. I think the color helps to simplify, yet to convey added information. Unfortunately, color does not work for everyone, and for that I apologize. Tell your wife thanks, and that I will continue to use color when useful. – robjohn Jan 18 '18 at 15:53
  • @GaurangTandon: Thanks! I have included an extra order in the expansion to show that the $x^3$ terms cancel giving us an approximation to $O!\left(x^4\right)$ instead of just $O!\left(x^3\right)$. This explains why the plots are so close, as Claude Leibovici notes. – robjohn Jan 18 '18 at 16:13
  • Please, don't apologize for anything. I fully agree that, for such problems, colors are really showing many things. What I was amazed by is my wife's comments about your post I was reading, she standing next to me. – Claude Leibovici Jan 18 '18 at 16:17
  • @robjohn Excellent! :D – Gaurang Tandon Jan 19 '18 at 02:10
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Consider $$y(a,b)=\frac{1}{\sqrt{(x+a)^2+b}}$$ Using Taylor series around $x=0$, you get $$y(a,b)=\frac{1}{\sqrt{a^2+b}}-\frac{a x}{\left(a^2+b\right)^{3/2}}+\frac{x^2 \left(2 a^2-b\right)}{2 \left(a^2+b\right)^{5/2}}+\frac{x^3 \left(3 a b-2 a^3\right)}{2 \left(a^2+b\right)^{7/2}}+O\left(x^4\right)$$ Using this for the complete expression with the proper values of $a,b$ $$V(x)=\frac{ 3\sqrt 3}2k\left(1+\frac{3 x^2}{16}+O\left(x^4\right) \right)$$

Claude Leibovici
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  • I have only been taught Taylor's series for sin, cos, tan, and ln :( Can you point a link where I can read about taylor expansion of polynomial under square root under denominator? Thanks a ton! – Gaurang Tandon Jan 18 '18 at 13:57
  • @GaurangTandon. Forget the $k$ in both expressions and plot the two functions for $-\frac 12<x<\frac 12$. You will be amazed to see how close they are. For $x=\pm 0.5$, they differ by $0.00634$. For Taylor series, just go to https://en.wikipedia.org/wiki/Taylor_series. May I confess that I felt in love with them more than 60 years ago ? Cheers. – Claude Leibovici Jan 18 '18 at 14:05
  • Wow! You're right, the plots seem to be superimposed on each other :-O But, the WIkipedia page doesn't have any info on such a particular expansion as in this question, or maybe I am unable to identify it :( – Gaurang Tandon Jan 18 '18 at 14:17
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    @GaurangTandon. Just look at the very first equation and ... apply ! Impressive isn't it ? And we can do much much better if we want. – Claude Leibovici Jan 18 '18 at 14:21
  • @ClaudeLeibovici Wonderful! Thanks! :D – Gaurang Tandon Jan 18 '18 at 14:26
  • @Semiclassical. Thanks for pointing the fact that I wrote $y(a)$ instead of $y(a,b)$ – Claude Leibovici Jan 18 '18 at 14:27