Mathematica Series expansion from the red book on special functions by Richard Ask

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The discussion centers on verifying a series expansion from Richard Ask's book using Mathematica. The user encountered a discrepancy in the expansion, specifically between the expected term and their calculated term. Suggestions include using Mathematica's Series command to expand both components of the expression to the appropriate order. It is emphasized that both expansions must be accurate to order j^{-2} to avoid errors in the final multiplication. The correct approach leads to the expected result of the series expansion without the incorrect term originally calculated.
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I want to check my calculations via mathematica.

In the book I am reading there's this expansion:
$$\frac{(1+\frac{1}{j})^x}{1+x/j}=1+\frac{x(x-1)}{2j^2}+\mathcal{O}(1/j^3)$$

though I get instead of the term ##\frac{x(x-1)}{2j^2}## in the rhs the term: ##-\frac{x(x+1)}{2j^2}##.

So I want to check by mathematica if my calculations are correct, how do you suggest me to implement it in mathematica?

Thanks!
 
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I don't have experience with Mathematica, but I suggest you try the example x=2 to see if your expansion is right.
 
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Define ##k \equiv 1/j## then use the command Series to get the series expansion about ##k=0##.
The book is correct.
 
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DrClaude said:
Define ##k \equiv 1/j## then use the command Series to get the series expansion about ##k=0##.
The book is correct.
Yeah I know.
I thought to myself to expand ##(1+1/j)^x = 1+x/j+x(x-1)/(2j^2)+\mathcal{O}(1/j^3)## and ##(1+x/j)^{-1}=1-x/j+\mathcal{O}(1/j^2)## and then multiply both factors; but it seems not to be the correct approach.
 
MathematicalPhysicist said:
Yeah I know.
I thought to myself to expand ##(1+1/j)^x = 1+x/j+x(x-1)/(2j^2)+\mathcal{O}(1/j^3)## and ##(1+x/j)^{-1}=1-x/j+\mathcal{O}(1/j^2)## and then multiply both factors; but it seems not to be the correct approach.

You need to expand both to order j^{-2}: <br /> \begin{align*}<br /> \left(1 + \frac 1j\right)^x &amp;= 1 + \frac xj + \frac{x(x-1)}{2j^2} + O(j^{-3}) \\<br /> \left(1 + \frac xj \right)^{-1} &amp;= 1 - \frac xj + \frac{x^2}{j^2} + O(j^{-3}) \end{align*}. Now multiply these together, ignoring any term of order j^{-3} or higher: \begin{split}<br /> \left(1 - \frac xj + \frac{x^2}{j^2} + O(j^{-3})\right) + \frac xj \left( 1 - \frac xj + O(j^{-2}) \right) + <br /> \frac{x(x-1)}{2j^2} \left(1 + O(j^{-1})\right) \\= 1 + \frac{x(x-1)}{2j^2} + O(j^{-3}).\end{split} Omitting +x^2/j^2 from the first bracket will give the incorrect <br /> 1 - \frac{x(x+1)}{2j^2} + O(j^{-3}) from your original post.
 
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