my friend said that he got (sqrt3)/2 for the very last convergence problem because apparently -sqrt3 gives an imaginary number when plugging it back into the series, so sqrt(3) is the diameter of convergence. Idk if this is right at alll but i thought id share it lmaoo
@revisnow
12 күн бұрын
Considering imaginary solutions is definitely important! However, I don't think that here it's a concern since we don't have x^(n/2). This can be problematic because if n is odd, we are taking the square root of x to some odd power, which would necessitate that x >= 0. However, here, since we have x^(2n), whose domain in this context is the set of all real numbers, it wouldn't be a problem. Also, even if -sqrt(3) gave an imaginary number, I don't think that sqrt(3) / 2 would solve that problem. But thanks for sharing your thought on this!
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