Integral help

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Reply 20

Gregorius

Original post by Zacken

Poor students. :wink:

<loud cackling> 😎

Reply 21

DFranklin

Not the same, but amusing:

\displaystyle \int_0^\infty \text{sinc}(x) \,dx = \dfrac{\pi}{2}

\displaystyle \int_0^\infty \text{sinc}(x) \text{sinc}(x/3) \,dx = \dfrac{\pi}{2}

\displaystyle \int_0^\infty \text{sinc}(x) \text{sinc}(x/3) \text{sinc}(x/5) \,dx = \dfrac{\pi}{2}

...

\displaystyle \int_0^\infty \text{sinc}(x) \text{sinc}(x/3) \text{sinc}(x/5) ... \text{sinc}(x/13) \,dx = \dfrac{\pi}{2}

\displaystyle \int_0^\infty \text{sinc}(x) \text{sinc}(x/3) \text{sinc}(x/5) ... \text{sinc}(x/15) \,dx =

\dfrac{467807924713440738696537864469 \pi}{935615849440640907310521750000}

(the fraction is approx 0.4999999999926 pi)

Reply 22

Zacken

Original post by DFranklin

Not the same, but amusing:

That is deviously cruel, I'd do the first few integrals manually, see that they come out to

\frac{\pi}{2}

and then spend an hour futilely trying to prove that it was so for all

I_n

Reply 23

DFranklin

Original post by Zacken

That is deviously cruel, I'd do the first few integrals manually, see that they come out to

\frac{\pi}{2}

and then spend an hour futilely trying to prove that it was so for all

I_n

Well, since you mention it...

FromMathOverflow

As a prank, Jonathan Borwein reported this to Maple, claiming there was a bug in the software. Maple computer scientist Jacques Carette spent 3 days trying to figure out the problem. Then he realized: There was no bug! That's what these integrals really equal!

Reply 24

Zacken

Original post by DFranklin

Well, since you mention it...

Hahahaha! Oh, that's hilarious. :rofl:

Are you a user on math.stackexchange/overflow? :-)

Reply 25

DFranklin

Original post by Zacken

Hahahaha! Oh, that's hilarious. :rofl:

Are you a user on math.stackexchange/overflow? :-)

No, it just comes up when googling about quirky math stuff quite often...

Reply 26

Zacken

Original post by DFranklin

No, it just comes up when googling about quirky math stuff quite often...

On that topic, the internet needs a more maths-oriented search engine, looking up maths stuff is horrible.

Reply 27

DFranklin

Oh, and apparently if you add a 2 cos x factor to the integral, then you get pi /2 right up until

\displaystyle \int_0^\infty 2 \cos(x)\, \text{sinc}(x) \text{sinc}(x/3) ... \text{sinc}(x/111) \,dx = \frac{\pi}{2}

\displaystyle \int_0^\infty 2 \cos(x)\, \text{sinc}(x) \text{sinc}(x/3) ... \text{sinc}(x/113) \,dx \neq \frac{\pi}{2}

(paper doesn't give the actual answer).

http://schmid-werren.ch/hanspeter/publications/2014elemath.pdf

Reply 28

Zacken

Original post by DFranklin

Oh, and apparently if you add a 2 cos x factor to the integral, then you get pi /2 right up until

\displaystyle \int_0^\infty 2 \cos(x)\, \text{sinc}(x) \text{sinc}(x/3) ... \text{sinc}(x/111) \,dx = \frac{\pi}{2}

\displaystyle \int_0^\infty 2 \cos(x)\, \text{sinc}(x) \text{sinc}(x/3) ... \text{sinc}(x/113) \,dx \neq \frac{\pi}{2}

(paper doesn't give the actual answer).

http://schmid-werren.ch/hanspeter/publications/2014elemath.pdf

Convolutions look really interesting, I can barely comprehend it, given that my knowledge barely scratches through Fourier Transform - what course and year do you tend to meet convolution products?

Reply 29

DFranklin

Original post by Zacken

Convolutions look really interesting, I can barely comprehend it, given that my knowledge barely scratches through Fourier Transform - what course and year do you tend to meet convolution products?

Was 1st year at Cambridge. Comes up a fair bit if you do image / audio processing too.

Reply 30

Zacken

Original post by DFranklin

Was 1st year at Cambridge. Comes up a fair bit if you do image / audio processing too.

Would that fall under an applied course? I can see it being part of some series topic in perhaps the differential equations course?

Reply 31

DFranklin

Original post by Zacken

Would that fall under an applied course? I can see it being part of some series topic in perhaps the differential equations course?

Yeah it was applied - I don't remember the name for sure, but Linear Systems is kind of ringing a bell (and would make sense given the content, which I remember better).

It was basically Wave + Heat Equation, Fourier Series, Fourier Transforms, 2nd Order ODEs (Sturm-Louiville).

Reply 32

Zacken

Original post by DFranklin

Ah, thanks for that.

Reply 33

atsruser

Original post by Zacken

Convolutions look really interesting, I can barely comprehend it, given that my knowledge barely scratches through Fourier Transform - what course and year do you tend to meet convolution products?

Suppose you have a linear system like a circuit composed of resistors, capacitors, and so on; a convolution can be used to calculate the output of the system when a signal is applied to it. So for example, you may have a "low pass filter"; this is a circuit that attenuates frequencies of signals higher than 1 KHz, say.

If you feed a 1V 500 Hz sine wave into it, it will come out as a 1V 500 Hz sine wave.
A 1V 2 KHz sine wave may come out as a 0.5 V 2 KHz sine wave. What happens when you feed a complex signal into, comprising a continuous range of frequencies, such as you get in human speech? You'll expect the higher frequencies to be quieter, and the lower ones to be left alone, but how would you calculate it exactly?

To find that out, you need to know the unit impulse response

h(t)

of the system; that is its output when you feed it a "impulse" of size 1 (a Dirac delta function); an impulse is an instantaneous spike. So you feed an impulse into the system and measure its output, which will generally oscillate for a bit, or spike and then die away gradually - that is

h(t)

You can then think of a real signal as being a continuous summation of weighted, time shifted impulses (i.e.

a\delta(t)

where

a

is the weight i.e. the amplitude); when you feed this signal into your system, it feels a continuous series of impulses, of varying strengths as time goes on, and to each impulse, it generates a weighted impulse response

ah(t)

as output, each delayed relative to the earlier outputs.

But to find the whole output signal at some time

t

, you have to add up all of those continuous, shifted, weighted

h(t)

s - that is what a convolution does. To do it, you need an integral, which is:

O(t) = \int_{-\infty}^\infty h(\tau) I(t-\tau) \ d\tau

That integral is the convolution of the input function

I(t)

and the impulse response

h(t)

. The presence of the

t-\tau

indicates that we are adding up the signals *before* time

t

, as measured by the time variable

\tau

(and in fact that we have to reverse the input function to get it to line up with the impulse response function properly). You can indicate this also as:

(h * I)(t) = \int_{-\infty}^\infty h(\tau) I(t-\tau) \ d\tau

That's a brief overview of the interpretation of a convolution in this particular application, but they have lots of mathematical aspects too that you can look up e.g. in Laplace and Fourier transform theory amongst others. (Laplace transforms play nicely with convolutions; you can think of a Laplace transform as mapping a product in one vector space to a product in another vector space)

[BTW, now I've written all that, I could probably just have pointed you at the Wikipedia article on convolution, which has nice pictures too - convolution is best understood with diagrams]

Reply 34

shamika

Original post by DFranklin

Not the same, but amusing:

\displaystyle \int_0^\infty \text{sinc}(x) \,dx = \dfrac{\pi}{2}

\displaystyle \int_0^\infty \text{sinc}(x) \text{sinc}(x/3) \,dx = \dfrac{\pi}{2}

\displaystyle \int_0^\infty \text{sinc}(x) \text{sinc}(x/3) \text{sinc}(x/5) \,dx = \dfrac{\pi}{2}

...

\displaystyle \int_0^\infty \text{sinc}(x) \text{sinc}(x/3) \text{sinc}(x/5) ... \text{sinc}(x/13) \,dx = \dfrac{\pi}{2}

\displaystyle \int_0^\infty \text{sinc}(x) \text{sinc}(x/3) \text{sinc}(x/5) ... \text{sinc}(x/15) \,dx =

\dfrac{467807924713440738696537864469 \pi}{935615849440640907310521750000}

(the fraction is approx 0.4999999999926 pi)

This is fairly well known; my integral knowledge is particularly crap but I did know of this. Cannot for the life of me figure out where I first saw it though

Reply 35

shamika

Original post by Zacken

Convolutions look really interesting, I can barely comprehend it, given that my knowledge barely scratches through Fourier Transform - what course and year do you tend to meet convolution products?

They're a basic concept for Fourier transforms so the very first course they are introduced. Currently I think they are second year in both Oxford and Cambridge (definitely in Cambridge, in Methods). Imperial was second year when I was there, assume it's the same now. No idea about Warwick

Reply 36

DFranklin

Original post by shamika

This is fairly well known; my integral knowledge is particularly crap but I did know of this. Cannot for the life of me figure out where I first saw it though

I'm sure I first saw it in the context of "patterns that break down unexpectedly"; it was fairly new to me (last 2-3 years at most).

I think I'm more impressed with a computer algebra system finding the exact solution (since it appears it didn't know the "trick" or the developer would not have spent days working out what was goiing on) than anything else.

Also, if I am understanding the proof correctly, any positive sequence

a_n

with a_1 = 1 and

\sum_1^{N-1} a_n \leq 2, \sum_1^N a_n > 2

will have

\displaystyle \int_0^\infty \prod_1^{N-1} \text{sinc}(a_n x)\,dx = \frac{\pi}{2}

\displaystyle \int_0^\infty \prod_1^{N} \text{sinc}(a_n x)\,dx \neq \frac{\pi}{2}

which allows for lots of fun if you pick a suitable slowly diverging sequence.

Reply 37

Zacken

Original post by shamika

This brings up an interesting point that I've been meaning to raise recently:

A-Level syllabi have been shrinking, in terms of content, for the past twenty years or so - has first year uni material been shifted/shrinked as well to allocate for the reduction in the syllabi?

Quite a few of the old timers post about how much more advanced their maths A-Level was back in their day and DFranklin mentioned that convolutions was in Cambridge first year, which is where my question comes from.

Reply 38

Zacken

Original post by atsruser

Thank you very much for this, I've read it and I'll try digesting it soon - re-reading and consulting diagrams at the same time.

I must say, your explanations are very nice. Every time I see you post on a thread, I open it up to see if it's one of these such explanations - there was one you did explaining Huygen's principle beautifully and another with you explaining how Jacobian's were just stretching/compressing the axes in three dimensions, etc... They're really nice! :smile:

Reply 39

Gregorius

Original post by Zacken

A-Level syllabi have been shrinking, in terms of content, for the past twenty years or so - has first year uni material been shifted/shrinked as well to allocate for the reduction in the syllabi?

Yes, very much so. Though I feel I must say that the brightest students these days are just as bright as the best in my day (back when dinosaurs roamed the earth).

Quite a few of the old timers post about how much more advanced their maths A-Level was back in their day and DFranklin mentioned that convolutions was in Cambridge first year, which is where my question comes from.

On the other hand, when I went up to Cambridge the first year was a huge jump over school mathematics - even taught to the level of the seventh term Oxbridge exams. One of the things that has changed is the size of that jump; I think it's finally sunk in with universities realizing that they really don't want to have good students drop out or fail so early through no great fault of their own.