Something a bit trippy I saw in a book

I was reading a book today (Game, Set, and Math, by Ian Stewart) which had the following in it:

\int e^x = e^x

e^x - \int e^x = 0

(1 - \int) e^x = 0

e^x = \frac{1}{1 - \int} \times 0

\frac{1}{1 - \int}

is the sum of a geometric series with first term 1 and common ratio

\int

, so:

e^x = (1 + \int + \int ^2 + \int ^3 + \int^4 + \cdots) \times 0

e^x = (1 + \int + \int \int + \int \int \int + \int \int \int \int + \cdots) \times 0

e^x = 0 + \int 0 + \int\int 0 + \int\int\int 0 + \int\int\int\int 0 + \cdots

Which could conceivably be written as follows, since you could say that

\int 0 = 1

and

\int \int 0 = \int 1 = x

...:

e^x = 0 + 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots

Which is the familiar Maclaurin series.

Is this nonsense? Apparently it is possible to treat integral signs in this way using 'Banach spaces'... I dunno. It certainly caught me off-guard.

Thoughts?

Reply 1

BJack

19

Well the result is certainly correct.

How cool is that?

I'm used to seeing differential operators treated like that, that is pretty nice though - and presumably rigorous since Stewart included it, I know nothing of Banach spaces though.

Reply 4

Adjective

OP

12

When you say you're 'used to seeing differential operators treated like that', what do you mean? Something like:

(x + x^2 + x^3 + x^4) \, dx = 1 + 2x + 3x^2 + 4x^3

? Or something completely different? (Is dx even what you mean by a differential operator, or is it d/dx?)

Reply 5

BJack

19

Adje

When you say you're 'used to seeing differential operators treated like that', what do you mean?

I'd imagine it's more like when making substitutions for integrations, e.g.:

Let

u=3x

Then

\frac{du}{dx} = 3 \Rightarrow dx = du/3

Reply 6

username167718

4

Adje

Apparently it is possible to treat integral signs in this way using 'Banach spaces'...

Are you saying this "proof" is actually meaningful, in some way?! Because it clearly doesn't look like it :s-smilie:

Pretty cool trick, though.

Reply 7

DeanK2

I imagin it is more like this:

y + (d2y/dx2) = x^5

Let D = d/dx, D^2 = d2y/dx2, etc.

(1+D*D)y = x^5

y = (1- D^2 + D^4 - D^6 ...)x^5 (expanding as a power series).

y = x^5 - 20x^3 + 120x.

Of course, my exxample does not do justice - it is quite a powerul method.

Reply 8

Lster

I think it's probably an abuse of notation or something "similar".

Reply 9

qgujxj39

17

That looks really cool, actually.

What about the missed out constant from the initial integration though, where's that gone? You could equally do it like this:

\int e^x = e^x - 69

(because

e^x

and

e^x - 69

are both equally valid antiderivatives of

e^x

)

e^x - \int e^x = 69

(1 - \int) e^x = 69

e^x = \frac{1}{1 - \int} \times 69

\frac{1}{1 - \int}

is the sum of a geometric series with first term 1 and common ratio

\int

, so:

e^x = (1 + \int + \int ^2 + \int ^3 + \int^4 + \cdots) \times 69

e^x = (1 + \int + \int \int + \int \int \int + \int \int \int \int + \cdots) \times 69

e^x = 69 + \int 69 + \int\int 69 + \int\int\int 69 + \int\int\int\int 69 + \cdots

Then you could say that

\int 69 = 69x

and

\int \int 69 = \int 69x = \frac{69x^2}{2}

...:

e^x = 69 + 69x + \frac{69x^2}{2} + \frac{69x^3}{6} + \frac{69x^4}{24} + \cdots

Which isn't correct.
But still, I obviously have no clue what I'm talking about, I've seen nothing like this before. We need DFranklin or someone to come and explain it.

Reply 10

DFranklin

18

tommm

That looks really cool, actually.

What about the missed out constant from the initial integration though, where's that gone?Yeah, that bothered me too (there's actually a potential constant on each of the integrations, as it stands).

Simplest is to consider

\int_0^x

instead of

\int

; it's not as "neat looking" but it doesn't involve arbitrary constants.

I'll put this down under "abuse of notation" as well, at least as presented. There are times when it works, but a large part of the theory with Banach spaces is actually pinning down when it works and proving it. You don't just grab an operator and say "oh well, we can divide by it using a Maclaurin series" without checking the result actually converges.

Reply 11

DeanK2

I imagine using integration in this way spring from the use of the Heaveside operator, shown with an example in my earlier post.

However, you have to be careful to take certain steps to make the operations rigorous - for example, the operator can be trated like a real number, but it obviously does not commute.

However, I think Tom has shown why people need to be careful using this kind of shortcut / rule in which operators can be treated as real numbers (even though it can be made rigorous).

Reply 12

Zhen Lin

12

I'm still puzzled by the Heaviside operator notation - what exactly is the benefit of being able to "factorise"

\displaystyle \frac{d^2 y}{dx^2} - 5 \frac{dy}{dx} + 6y = 0

into

\displaystyle \left( \frac{d}{dx} - 2 \right) \left( \frac{d}{dx} - 3 \right)y = 0

? A friend asked the DE lecturer about this one day and apparently he said something about the annihilation of the operator...? :confused:

Reply 13

DFranklin

18

Zhen: Here's one thing. When you "solve" D^2y-5Dy+6 = 0 by forming t^2-5t+6 = 0, saying t=2 or t = 3 and so y=Ae^2t + Be^3t, you're really just using a "magic formula".

And in particular, if you had D^2y-4Dy+4, you'd get a somewhat different y = (A+Bt)e^2t, for no obvious reason.

But if instead you look at (D-2)(D-3)y = 0, then we can write z=(D-3)y and solve (D-2)z=0, and then solve (D-3)y = z without any "magic". So we've reduced the problem to the solving of two easier first order DEs.

Moreover, if we have (D-2)(D-2)y = 0, you'll find the same approach gets you the (A+Bt)e^2t solution.

Reply 14

Barny

17

DFranklin

Yeah, that bothered me too (there's actually a potential constant on each of the integrations, as it stands).

Simplest is to consider

instead of

; it's not as "neat looking" but it doesn't involve arbitrary constants.

I'll put this down under "abuse of notation" as well, at least as presented. There are times when it works, but a large part of the theory with Banach spaces is actually pinning down when it works and proving it. You don't just grab an operator and say "oh well, we can divide by it using a Maclaurin series" without checking the result actually converges.

But isn't the integral of zero over any interval zero?

Reply 15

generalebriety

16

Barny

But isn't the integral of zero over any interval zero?

It is.

Something seems rather fishy about this to me. I suspect something clever is going on underneath the surface, but also that it's nowhere near as simple as the author has made out...

Reply 16

Swayum

18

Zhen Lin

I'm still puzzled by the Heaviside operator notation - what exactly is the benefit of being able to "factorise"

\displaystyle \frac{d^2 y}{dx^2} - 5 \frac{dy}{dx} + 6y = 0

into

\displaystyle \left( \frac{d}{dx} - 2 \right) \left( \frac{d}{dx} - 3 \right)y = 0

? A friend asked the DE lecturer about this one day and apparently he said something about the annihilation of the operator...? :confused:

Are you asking what's the point of writing it as (D^2 - 5D +7)y? If you've got spare time, I thought it was explained fairly well in these two lectures at MIT:

http://www.archive.org/download/MITMITOCW18.03DifferentialEquationsVideoLecturesSpring2004/mitocw18.03lec1310mar2003220k.mp4 (98 MB)

http://www.archive.org/download/MITMITOCW18.03DifferentialEquationsVideoLecturesSpring2004/mitocw18.03lec1412mar2003220k.mp4 (107 MB)

(they follow on from each other but are conducted over 2 days but you can probably skip the second)

Reply 17

Zhen Lin

12

DFranklin

Moreover, if we have (D-2)(D-2)y = 0, you'll find the same approach gets you the (A+Bt)e^2t solution.

Ah. Very interesting. I didn't think of it that way. (In hindsight, I should have recognised it as composition of operators rather than some strange kind of product.) What about the commutativity of these operators? Are there any important rules?

Reply 18

DFranklin

18

Barny

But isn't the integral of zero over any interval zero?

Good point - I should have looked more carefully. I don't think you can sensibly make it rigourous without a whole bunch of extraneous assumptions then. But I might be wrong.

Reply 19

generalebriety

16

Zhen Lin

Ah. Very interesting. I didn't think of it that way. (In hindsight, I should have recognised it as composition of operators rather than some strange kind of product.) What about the commutativity of these operators? Are there any important rules?

Well, they're obviously commutative, because: (D-a)(D-b)f = D(D-b)f - a(D-b)f = D^2 f - Dbf - aDf + abf = (D-b)(D-a)f. However, if a and b are functions of x, it's a different story. Consider, for instance, (D-x)(D-1)f and (D-1)(D-x)f - the obvious problem is that you get -xDf on the left and -Dxf on the right, which aren't the same (use product rule).

Something a bit trippy I saw in a book

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