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I am trying to expand

$$1 - \left(1-\frac{0.511}{v}\right)^{0.5}\left(1+\frac{0.511}{v}\right)^{0.5} $$

for large $v$, however using a binomial expansion on the two terms I get the expression

$$\frac{0.511^2}{4v^2}$$

which is exactly half the correct answer.

Can anyone explain what is going wrong here?


Context from a Comment to an Answer:

My working is:

$$1 - \left(1-\frac{0.511}v\right)^{0.5}\left(1+\frac{0.511}v\right)^{0.5}$$ I use the first two terms of the expansion $= 1 - \left(1-\frac{0.511}{2v}\right)\left(1+\frac{0.511}{2v}\right) = \frac{0.511^2}{4v^2}$ (**)

However this is not right because if I take a different route to the answer and spot that what is square rooted is the difference of 2 squares I obtain:

$$1 - \left(1-\frac{0.511^2}{v^2}\right)^{0.5}$$ I use the first two terms of the expansion $= 1 - 1 + \frac{0.511^2}{2v^2} = \frac{0.511^2}{2v^2}$

Here lies the confusion, I am sure something is going wrong here (**) as before that I obtain the same result from both methods when I set v=100

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    $\begingroup$ Not if yo won't show us your work. $\endgroup$ – B. Goddard Jun 28 '17 at 16:01
  • $\begingroup$ Use the binomial approximation $\endgroup$ – samjoe Jun 28 '17 at 16:02
  • $\begingroup$ Incorporating the comment that the OP made on my answer, with improved formatting, I now think there is enough context to reopen the question. $\endgroup$ – robjohn Jun 29 '17 at 13:15
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Not enough terms have been included. An answer is being computed to the second order using information only to the first order. The second order terms need to be included in each factor to get the correct second order term in the result: $$ \begin{align} &\left(1-\frac{0.511}{v}\right)^{0.5}\\ &=1+\overbrace{\ \left(\frac{0.5}1\right)\ }^{\large\binom{0.5}{1}}\left(-\frac{0.511}{v}\right)+\overbrace{\left(\frac{0.5}1\right)\left(\frac{-0.5}2\right)}^{\large\binom{0.5}{2}}\left(-\frac{0.511}{v}\right)^2+O\!\left(\frac1{v^3}\right)\\[12pt] &=1-\frac{0.511}{2v}-\frac{0.511^2}{8v^2}+O\!\left(\frac1{v^3}\right)\tag{1} \end{align} $$ and $$ \begin{align} &\left(1+\frac{0.511}{v}\right)^{0.5}\\ &=1+\overbrace{\ \left(\frac{0.5}1\right)\ }^{\large\binom{0.5}{1}}\left(\frac{0.511}{v}\right)+\overbrace{\left(\frac{0.5}1\right)\left(\frac{-0.5}2\right)}^{\large\binom{0.5}{2}}\left(\frac{0.511}{v}\right)^2+O\!\left(\frac1{v^3}\right)\\[12pt] &=1+\frac{0.511}{2v}-\frac{0.511^2}{8v^2}+O\!\left(\frac1{v^3}\right)\tag{2} \end{align} $$ Multiply $(1)$ and $(2)$ together: $$ \begin{align} &\left(1-\frac{0.511}{2v}-\color{#C00}{\frac{0.511^2}{8v^2}}+O\!\left(\frac1{v^3}\right)\right)\left(1+\frac{0.511}{2v}-\color{#090}{\frac{0.511^2}{8v^2}}+O\!\left(\frac1{v^3}\right)\right)\\ &=1-\frac{0.511}{2v}+\frac{0.511}{2v}-\frac{0.511^2}{4v^2}-\color{#C00}{\frac{0.511^2}{8v^2}}-\color{#090}{\frac{0.511^2}{8v^2}}+O\!\left(\frac1{v^3}\right)\\ &=1-\frac{0.511^2}{2v^2}+O\!\left(\frac1{v^3}\right)\tag{3} \end{align} $$ The second order terms are significant. Their omission in the computation caused the error.

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  • $\begingroup$ My working is: 1 - (1-0.511/v)^0.5(1+0.511/v)^0.5 I use the first two terms of the expansion = 1 - (1-0.511/2v)(1+0.511/2v) ** = 0.511^2 / 4v^2 However this is not right because if I take a different route to the answer and spot that what is square rooted is the difference of 2 squares I obtain: 1 - (1-0.511^2/v^2)^0.5 I use the first two terms of the expansion = 1 - 1 + 0.511^2 / 2v^2 = 0.511^2 / 2v^2 Here lies the confusion, I am sure something is going wrong here (**) as before that I obtain the same result from both methods when I set v=100 $\endgroup$ – Will Jun 28 '17 at 19:28
  • $\begingroup$ @Will: You are looking for a second order answer using only first order approximations. As suggested in my answer above, you need to use the second order terms of the binomial expansions. The second order terms are significant, especially when making a second order approximation. $\endgroup$ – robjohn Jun 29 '17 at 0:09
  • $\begingroup$ @Will: The information you supply in your comment above should have been included in your question. If you add it to your question, your question may be reopened and some of those who downvoted may reconsider. $\endgroup$ – robjohn Jun 29 '17 at 8:04
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hint

$$(1+a)^\alpha(1-a)^\alpha=(1-a^2)^\alpha$$

$$(1-b)^\alpha=1-\alpha\cdot b $$ for very small b.

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    $\begingroup$ On second thought, I believe this is what lead the OP to the problem they had. The second order terms are significant. $\endgroup$ – robjohn Jun 28 '17 at 16:29

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