# Extracting rational coefficients from an irrational sum [closed]

I have a function that returns results that always can be represented as $p+q\,\pi^2+r\ln2+s\ln^22+t\sqrt2$ where $p,q,r,s,t\in\mathbb Q$. But the actual Mathematica expression returned not always in exactly this form, e.g. it can be

1/Sqrt[2] + 1/54 (34 - 3 π^2 + (-8 + Log[8]) Log[64])


It's not difficult to extract rational coefficients from this expression manually, but I need to process hundreds of expressions, so I want to make this process automated. How can I do this?

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## closed as off-topic by Mr.Wizard♦Aug 15 '14 at 17:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

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Please include multiple examples on which to test our methods. –  Mr.Wizard Aug 9 '14 at 18:20
I recommend using PowerExpand see e.g. How can I simplify log(512) to 9log(2)?. It is faster than appropriate useage of FullSimplify or Simplify with ComplexityFunction. Try e.g. Expand @ PowerExpand[ 1/Sqrt[2] + 1/54 (34 - 3 \[Pi]^2 + (-8 + Log[8]) Log[64])]. It yields 17/27 + 1/Sqrt[2] - \[Pi]^2/18 - (8 Log[2])/9 + Log[2]^2/3. –  Artes Aug 10 '14 at 16:42

Probably you will have to adapt my basic idea to your special needs.

mem = 1/Sqrt[2] + 1/54 (34 - 3 π^2 + (-8 + Log[8]) Log[64]) //
ExpandAll // Apply[List, #] &
{mem[[1]], mem[[2]]/Sqrt[2], mem[[3]]/Pi^2, mem[[4]]/Log[2],
mem[[5]]/Log[2]^2} // FullSimplify


The key is: after ExpandAll the summands are sorted in the same order, containing the factors:

{1, Sqrt[2], Pi^2, Log[2], Log[2]^2}


here a function, which calculates the rational coefficients in your desired order:

pqrstCoeff =
Function[# // ExpandAll // Apply[List, #] & //
Divide[#, {1, Sqrt[2], Pi^2, Log[2], Log[2]^2}] & //
FullSimplify // Part[#, {1, 3, 4, 5, 2}] &];


and you may verify:

givenExpression =
1/Sqrt[2] + 1/54 (34 - 3 π^2 + (-8 + Log[8]) Log[64])
pqrst = {p, q, r, s, t} = givenExpression // pqrstCoeff

pqrst // Element[#, Rationals] &


with results:

{17/27, -(1/18), -(8/9), 1/3, 1/2}

True

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