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That is your formula: <math>|\frac{1}{2^n}\sum_{x=0}^{2^n-1}(-1)^{f(x)}|^2</math>. And how you can explain for example this <math>|\frac{1}{2^2}\sum_{x=0}^{2^2-1}(-1)^{f(x)}|^2=|\frac{1}{2^2}((-1)^{f(00)}+(-1)^{f(01)}+(-1)^{f(10)}+(-1)^{f(11)}|^2</math>?
Ah, but you say, x is 0, but 0's can't be so much ( <math>2^n</math> ).
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