Hadrian Tang

>Have you verified the answer? The result you get is: A. Ugly B. Needlessly complex C. 100% correct A. Evaluate can't simplify it B. The length exceeds Wolfram Alpha's input...

I think that this can easily solved by implementing cube root for complex numbers. Ideas?

I think I saw where the rounding error comes from - using Math.sin and Math.cos inside Parser.pow. The conversion between Frac and Javascript Number is what causes the error.

The workaround earlier is not useful when fractional answers are there... ~solve(8x^3-26x^2+3x+9,x) should=> [3,3/4,-1/2] It currently produces `[(-1/24)*(-13984+15120*i*sqrt(3))^(1/3)*2^(-1/3)+(-151/6)*(-13984+15120*i*sqrt(3))^(-1/3)*2^(1/3)+13/12,(1/48)*(-13984+15120*i*sqrt(3))^(1/3)*(1+i*sqrt(3))*2^(-1/3)+(151/12)*((-13984+15120*i*sqrt(3))^(1/3)*2^(-1/3))^(-1)*(-i*sqrt(3)+1)+13/12,(1/48)*(-13984+15120*i*sqrt(3))^(1/3)*(-i*sqrt(3)+1)*2^(-1/3)+(151/12)*((-13984+15120*i*sqrt(3))^(1/3)*2^(-1/3))^(-1)*(1+i*sqrt(3))+13/12]`.~ Another bug is present here. When evaluate is checked, the vector does...

Also this. ~solve(x^3-1/2x^2-13/2x-3,x) => `[(-1/3)*((315/4)*i*sqrt(3)-221/2)^(1/3)*2^(-1/3)+(-79/12)*((315/4)*i*sqrt(3)-221/2)^(-1/3)*2^(1/3)+1/6,(1/6)*((315/4)*i*sqrt(3)-221/2)^(1/3)*(1+i*sqrt(3))*2^(-1/3)+(79/24)*(((315/4)*i*sqrt(3)-221/2)^(1/3)*2^(-1/3))^(-1)*(-i*sqrt(3)+1)+1/6,(1/6)*((315/4)*i*sqrt(3)-221/2)^(1/3)*(-i*sqrt(3)+1)*2^(-1/3)+(79/24)*(((315/4)*i*sqrt(3)-221/2)^(1/3)*2^(-1/3))^(-1)*(1+i*sqrt(3))+1/6]` should=> [-1/2,3,-2]~

1. The idea here is to follow https://github.com/jiggzson/nerdamer/pull/337#issuecomment-353820872 and only minimize clone()s when a notable performance difference is present. 2. This proposal attempts to implement BigNumbers (BigDecimals) in terms of...

While #134 and #148 are closed, this is still not implemented. sum(1/(2^x),x,1,Infinity) won't evaluate.

sum(1/(2^x),x,1,Infinity) should=> 1

Request reopen

True, but when we talk about infinite series/products/integrals, we do not write things like limit(sum(1/(2^n),n,1,x),x,Infinity), but we instead write sum(1/(2^n),n,1,Infinity). The implementation could use some limits though.