Does that actually depend on that number being a power of 2?
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Replying to @sl2c
i'm quite sure, since if you make the substitution u= 2x you can see why
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Replying to @InertialObservr @sl2c
a power of 2 is not necessary in this integral
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Replying to @mtbatchelor @sl2c
in order for the result to be 2^n, surely it should be
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ohh or are you saying that a^n would equally as valid?
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Replying to @mtbatchelor @sl2c
oh yes, i agree with that!
3:56 PM - 8 Jul 2019
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