10

In the question: Geometrical interpretation of $(\sum_{k=1}^n k)^2=\sum_{k=1}^n k^3$ the mysterious last answer involves concepts from electrical engineering and signal processing:

Consider the time-signal [1,2,...,k] (A linear function or "triangle wave"). The LHS is the square of the DC component of the signal in the temporal/spatial domain which is straight forward to calculate. The RHS is the iterated convolution of the fourier transform of [1,2,3,...,k] three times and then DC component of that.

Can anyone supply details here? What does he mean triangle wave and DC component? Sounds like tuning a radio then proving a theorem, but that is what makes math great.

Community
  • 1
cactus314
  • 24,438
  • I'd rather guess "DC component" is the time average, the direct current component of a voltage signal. To be precise it should be $\frac 1 n \sum_{k=1}^n k$, but the factor $\frac 1 n$ probably doesn't matter that much. – Christoph Oct 14 '15 at 21:25

1 Answers1

5

Yes Christoph is on the right idea and there is some technical details involved here. Some FFTs normalize the sum to be $1/n$ and some don't and some normalize in a third way and so on, so I need to check which of these my local FFT actually does to be able to check myself. Anyway first of all we need a function to work with, and we choose the linear function below, basically $f(x) = x$ :

enter image description here

The theory behind this all is based on applying the convolution theorem of Fourier transforms:

$$\mathcal{F}(f^3) = \mathcal{F}(f)*\mathcal{F}(f)*\mathcal{F}(f)$$ together with the fact that for the frequency 0, the exponential factor becomes 1 and the fourier transform is a sum (or in the continous case an integral): $$\mathcal{F}(g)[0] = \sum_{k=1}^N g(k)$$ possibly times a factor, depending on normalization. This is what the left hand side amounts to. The right hand side is with the same reasoning $(\mathcal{F}(f)[0])^2$. Note the square is last so it is on the result of the sum.

mathreadler
  • 25,824
  • You basically restated that $(\mathcal F(f)[0])^2=(\mathcal{F}(f)\mathcal{F}(f)\mathcal{F}(f))[0]$ for $f(j)=j$. How is it shown without grunt work given that you need two convolutions? – A.S. Oct 15 '15 at 02:03
  • OK it's reasonable that $\mathcal{F}(f^3)[0] = \sum f(k)^3$ but can you prove the other side? $$ \big[ \mathcal{F}(f)\ast \mathcal{F}(f)\ast\mathcal{F}(f)\big][0] = \big[\sum f(k)\big]^2$$ – cactus314 Oct 18 '15 at 23:30
  • I think I had it in my head at some point, but sadly it seems to have run and hid. – mathreadler Oct 20 '15 at 08:51
  • @mathreadler this is a key point -- I cannot award the bounty without it. we know this formula is more general since the divisor function also does it $ \sum \tau(k)^3 = \big(\sum \tau(k)\big)^2$ Possibly this is not true for all $f$ - then just some short explanation of when this works. – cactus314 Oct 20 '15 at 11:30
  • Yes I agree you should not award the bounty unless I figure out an explanation why it is true for this particular $f$. – mathreadler Oct 20 '15 at 12:49