Integration

[andyandy]

maths question time again

Here we are given the following....

[latex] I_{n,p} = $$\int_{0}^{1} {x^p}{(1-x)^n} dx $$

p\ge 0, n\ge 0
[/latex]


and told to

1) Show that [latex] I_{n,p} = I_{p,n} [/latex]

2) prove that, for [latex] p\ge 1, {(n+1)}I_{p,n} = {p}I_{p-1,n+1} [/latex]

2II) and also that [latex] p\ge 1, {(p+n+1)}I_{p,n} = {p}I_{p-1,n} [/latex]

3) Hence proving that, if p and n are positive inegers,

[latex] I_{n,p} = \frac{{p!}{n!}}{{(p+n+1)}!} [/latex]

now, the answers can be found on page 15 here

which, for the most part I follow - i do however have (at least) a couple of things to clarify....

when you are given such an equation as -

[latex] I_{n,p} = $$\int_{0}^{1} {x^p}{(1-x)^n} dx $$

p\ge 0, n\ge 0
[/latex]

What is [latex] I_{n,p}
[/latex]
actually telling you? That you have a function (in this case of x) with variables n and p? Or that p and n are unknown constants?

Secondly, for the proof to part 1) with the sub u=1-x you get

[latex] I_{n,p} = $$\int_{0}^{1} {u^n}{(1-u)^p} du $$
[/latex]

which is given as showing that [latex] I_{n,p} = $$\int_{0}^{1} {x^n}{(1-x)^p} dx $$
[/latex]

But whilst these are in the same form i don't understand why this constitutes a proof - as x =! u



many thanks

 

[Thing]

That's not a question.

 

[andyandy]

That's not a question.

That's not an answer

 

[Anacoluthon64]

I was about to furnish an overview of the approach to each of the four questions (substituting y = x-1, integration by parts, etc.) in the OP when I refreshed the page to find that the focus of the question had changed quite a lot.

What I_p,n is telling you is that you have an expression involving two parameters (i.e. p and n) that determine the value of I. Generally, such subscripts are integers but they need not be. All it says is that we're interested in the more general case of I (being the integration) than just a specific case of, say, p = 3 and n = 2.

In the integration, u and x are so-called "dummy" variables. The variable of integration is irrelevant to a definite integral. ∫f(x)dx = ∫f(t)dt in the case of a definite integral over the same interval whenever f(x) is the same as f(t). Swapping each x for a t does not mean that x = t; we've just changed symbols, that's all.

As for the substitutions u = x^p and dv = (x-1)ⁿdx, they work out just fine. You may have made an error in your derivation.

What has this to do with S, M, M, and T? Shouldn't this be in "Puzzles?"

ETA: It would, for the sake of your respondents' good spirits, be most considerate of you if you refrained from repeatedly changing your post.

'Luthon64

 

[Unnamed]

What is [latex] I_{n,p}
[/latex]
actually telling you? That you have a function (in this case of x) with variables n and p? Or that p and n are unknown constants?

But whilst these are in the same form i don't understand why this constitutes a proof - as x =! u

I_n,p is a function of n and p, but not of x. The integration variable in a definite integral is a dummy variable, which disappears after the integration. As a dummy variable, it may be called anything you want, including x and u.

ETA: damn, beaten in both counts.

 

[andyandy]

I was about to furnish an overview of the approach to each of the four questions (substituting y = x-1, integration by parts, etc.) in the OP when I refreshed the page to find that the focus of the question had changed quite a lot.

What I_p,n is telling you is that you have an expression involving two parameters (i.e. p and n) that determine the value of I. Generally, such subscripts are integers but they need not be. All it says is that we're interested in the more general case of I (being the integration) than just a specific case of, say, p = 3 and n = 2.

In the integration, u and x are so-called "dummy" variables. The variable of integration is irrelevant to a definite integral. ∫f(x)dx = ∫f(t)dt in the case of a definite integral over the same interval whenever f(x) is the same as f(t). Swapping each x for a t does not mean that x = t; we've just changed symbols, that's all.

thanks - i'll reflect upon this - i think that makes sense to me.....

As for the substitutions u = x^p and dv = (x-1)ⁿdx, they work out just fine. You may have made an error in your derivation.

yes, i edited that part out upon reflection.....

What has this to do with S, M, M, and T? Shouldn't this be in "Puzzles?"

ETA: It would, for the sake of your respondents' good spirits, be most considerate of you if you refrained from repeatedly changing your post.

well, SMMT is science, maths, medicine and technology - it does tend to be largely science - but i think maths questions are best here (where there is a good level of expertise)

with regards to the changing nature of the OP - i did leave a note to say it was under edit - writing in latex is rather time consuming - and it requires rather a few edits for me to get it right.....
but i do apologise - i think i will first off write in latex in the test section before posting it as a thread in future....

 

[Yllanes]

with regards to the changing nature of the OP - i did leave a note to say it was under edit - writing in latex is rather time consuming - and it requires rather a few edits for me to get it right.....
but i do apologise - i think i will first off write in latex in the test section before posting it as a thread in future....

You can use the preview function. I always do (and I have considerable experience writing in LaTeX).

Anyway, I think that posting doubts about mathematics is well within the domain of this board. This is an educational foundation after all.

Well, just to bring the question more into the 'Science' domain, I will say that this function
[latex]
$$
B(n,m) = \int_0^1\mathrm{d}x\ x^{m-1}(1-x)^{n-1}=\frac{\Gamma\Gamma(m)}{\Gamma(m+n)}
$$
[/latex]
(almost your integral) is called Euler's beta function. You can find a lot of information about it in Mathworld and many useful identities at functions.mathworld.com. It appears sometimes in physics. (I have an example from statistical physics in mind).

The beta function is analytical and defined on all C² except for countably many points. The points were B(a,b) doesn't work are those for which a=-k, with k in N or b = -l, with l in N (because the Gamma function is not defined there).

 

[Anacoluthon64]

Anyway, I think that posting doubts about mathematics is well within the domain of this board. This is an educational foundation after all.

Okay, fair enough. It's just highly unusual to encounter such a specialised topic in this particular sub-forum, and the OP raised little in the way of doubt; only what seemed like a calculus or statistics homework question. And I do understand now that andyandy was rather more seeking genuinely to understand than prove some obscure point.

'Luthon64

 

[andyandy]

Okay, fair enough. It's just highly unusual to encounter such a specialised topic in this particular sub-forum, and the OP raised little in the way of doubt; only what seemed like a calculus or statistics homework question. And I do understand now that andyandy was rather more seeking genuinely to understand than prove some obscure point.

'Luthon64

as an explanation...

i'm currently working through the Open university Maths MSc diagnostic quiz for the calculus module, i'm thinking about giving it a go - although there's a good deal of reading up that i need to do before i could hope to do so.....

hence i'm grateful to have JREF as a place where certain questions i have can be answered

for those interested, the quiz (and answers) are available here

 

[andyandy]

Well, just to bring the question more into the 'Science' domain, I will say that this function
[latex]
$$
B(n,m) = \int_0^1\mathrm{d}x\ x^{m-1}(1-x)^{n-1}=\frac{\Gamma\Gamma(m)}{\Gamma(m+n)}
$$
[/latex]
(almost your integral) is called Euler's beta function. You can find a lot of information about it in Mathworld and many useful identities at functions.mathworld.com. It appears sometimes in physics. (I have an example from statistical physics in mind).

cool - the wolfram link is excellent.

 

[Iamme]

maths question time again

Here we are given the following....

[latex] I_{n,p} = $$\int_{0}^{1} {x^p}{(1-x)^n} dx $$

p\ge 0, n\ge 0
[/latex]


and told to

1) Show that [latex] I_{n,p} = I_{p,n} [/latex]

2) prove that, for [latex] p\ge 1, {(n+1)}I_{p,n} = {p}I_{p-1,n+1} [/latex]

2II) and also that [latex] p\ge 1, {(p+n+1)}I_{p,n} = {p}I_{p-1,n} [/latex]

3) Hence proving that, if p and n are positive inegers,

[latex] I_{n,p} = \frac{{p!}{n!}}{{(p+n+1)}!} [/latex]

now, the answers can be found on page 15 here

which, for the most part I follow - i do however have (at least) a couple of things to clarify....

when you are given such an equation as -

[latex] I_{n,p} = $$\int_{0}^{1} {x^p}{(1-x)^n} dx $$

p\ge 0, n\ge 0
[/latex]

What is [latex] I_{n,p}
[/latex]
actually telling you? That you have a function (in this case of x) with variables n and p? Or that p and n are unknown constants?

Secondly, for the proof to part 1) with the sub u=1-x you get

[latex] I_{n,p} = $$\int_{0}^{1} {u^n}{(1-u)^p} du $$
[/latex]

which is given as showing that [latex] I_{n,p} = $$\int_{0}^{1} {x^n}{(1-x)^p} dx $$
[/latex]

But whilst these are in the same form i don't understand why this constitutes a proof - as x =! u



many thanks

Oh my! This looks like something Mozart composed. No kidding!

 

[Zygar]

Why are we helping you with your homework?

 

[69dodge]

[latex]
$$
B(n,m) = \int_0^1\mathrm{d}x\ x^{m-1}(1-x)^{n-1}=\frac{\Gamma\Gamma(m)}{\Gamma(m+n)}
$$
[/latex]

Can I ask you an unrelated question?

In elementary calculus books (and historically, I think?), I've always seen the dx appear at the end. But sometimes, I've seen it come before the rest of the integrand, like you just wrote it. Where is that convention common, and how did it arise? I figure, putting the dx at end makes it easier to see what the integrand is, without extra parentheses:[latex]$\int f + g\,dx$[/latex] is clear, but what does [latex]$\int dx\ f + g$[/latex] mean?

 

[andyandy]

Why are we helping you with your homework?

It's not homework - it's for my own interest. I did already explain that.

Even if it were "homework" i don't see why you believe someone should be precluded from asking questions about it - study work is meant to stimulate personal research and ultimately lead to a greater understanding of the set topic - and asking for help in doing so is not something that should be derided....

 

[andyandy]

Can I ask you an unrelated question?

In elementary calculus books (and historically, I think?), I've always seen the dx appear at the end. But sometimes, I've seen it come before the rest of the integrand, like you just wrote it. Where is that convention common, and how did it arise? I figure, putting the dx at end makes it easier to see what the integrand is, without extra parentheses:[latex]$\int f + g\,dx$[/latex] is clear, but what does mean?

I would suppose it's to lessen any chance for ambiguity -

[latex]$\int dx\ f + g$[/latex]

could be mistaken for [latex]$\int dx\ [/latex] multiplied by f and plus g.....

having the dx at the end gives a helpful bound for the integral's end....

that would be my guess....

 

[Yllanes]

In elementary calculus books (and historically, I think?), I've always seen the dx appear at the end. But sometimes, I've seen it come before the rest of the integrand, like you just wrote it. Where is that convention common, and how did it arise? I figure, putting the dx at end makes it easier to see what the integrand is, without extra parentheses:[latex]$\int f + g\,dx$[/latex] is clear, but what does [latex]$\int dx\ f + g$[/latex] mean?

Short answer: It doesn't mean anything special, is just more convenient to write when dealing with very long integrals. Also, it makes long formulas easier to 'parse' and it also requires a little less space when typeset properly, which is also very important. For short expressions it's worse, in your example it would need parentheses.

Long answer: It's very common in physics, but uncalled for in that example, I just wrote it like that in automatic. The reason is that many times we have to integrate over many different variables or have very long and ugly integrands. Example:
[latex]
$$
\begin{align*}
-\dot\phi^\dagger(x) \vec \nabla\phi(x)=\int \mathrm{d}^3 \vec x\int \frac{\mathrm{d}^3 \vec k}{(2\pi)^3
2E(\vec k)}\int \frac{\mathrm{d}^3 \vec q}{(2\pi)^3 2E(\vec q)} (\vec k \cdot \vec q)&\left[\ad k \ak q e^{i (k^0-q^0)x^0} e^{i(\vec q -\vec k)\cdot\vec x}\right.\\
&- \ad k \bd q e^{i (k^0+q^0)x^0} e^{-i(\vec q +\vec k)\cdot\vec x}\\
&- \bb k \ak q e^{-i (k^0+q^0)x^0} e^{i(\vec q +\vec k)\cdot\vec x}\\
& + \left.\bb k \bd q e^{i (q^0-k^0)x^0} e^{i(\vec k -\vec q)\cdot\vec x}\right]
\end{align*}
$$
[/latex]
Writing the dx (or whatever) at the beginning makes the formula much clearer in those cases. It's one of those things (like setting c=1) that seems confusing when first seen, but that becomes much more convenient with use. In pure mathematics it is not common to have such ugly formulas.

Also, sometimes it makes sense conceptually to write it like that. For example, in the integral I just wrote, the fractions with the d^3k are all part of the measure, the definition of the integral and have a physical meaning. This is an integral from QFT. We are decomposing a field (or the product of two fields) which is a function of x = (t,x,y,z) in terms of a superposition of plane waves. This is basically a Fourier transform. In principle, we should integrate over the four components of the momentum k. But we have a constraint, because m² = p₀² - (p₁² + p₂² + p₃² )so only three variables are independent. If we do this calculations properly, we see that the way to integrate over momentum is to use as our measure:
[latex]
$$
\int \frac{\mathrm{d}^3 \vec q}{(2\pi)^3 2E(\vec q)} = \int \mathrm{d}\tilde q
$$
[/latex]
All this belongs to the definition of our integral, so it is reasonable to write it next to the integral sign. We know that if we apply that object to a function of p we get a function of x and that this relation is Lorentz invariant. If we do it without the E in the denominator we don't get anything invariant. So if you want to give any meaning to this convention you can think of something like that. But the main reason why it's used is that it is much more convenient. You may think there is little difference, but if you have to spend all day writing these things, that small difference adds up. Same goes with setting c = 1. If you only have to write a few formulas you may think that putting the c is not too much work and makes things clearer. But when you have to write many of them every day you find that the c gets in the way, makes expressions uglier and more difficult to parse and is just unnecessary conceptually.

 

[Yllanes]

Also, it makes long formulas easier to 'parse' and it also requires a little less space when typeset properly, which is also very important.

Three 'alsos' in the same sentence... This is what happens when you don't write a post linearly.

Anyway, not that it matters, but my integral should be:

[latex]
\small
$$
\begin{align*}
- \int \mathrm{d}^3\vec{x} \ \dot\phi^\dagger(x) \vec \nabla\phi(x)=\int \mathrm{d}^3 \vec x\int \frac{\mathrm{d}^3 \vec k}{(2\pi)^3
2E(\vec k)}\int \frac{\mathrm{d}^3 \vec q}{(2\pi)^3 2E(\vec q)} (\vec k \cdot \vec q)&\left[a^\dagger(\vec k) a(\vec q) e^{i (k^0-q^0)x^0} e^{i(\vec q -\vec k)\cdot\vec x}\right.\\
&- a^\dagger(\vec k) b^\dagger(\vec q) e^{i (k^0+q^0)x^0} e^{-i(\vec q +\vec k)\cdot\vec x}\\
&- b(\vec k) a(\vec q) e^{-i (k^0+q^0)x^0} e^{i(\vec q +\vec k)\cdot\vec x}\\
& + \left. b(\vec k) b^\dagger(\vec q) e^{i (q^0-k^0)x^0} e^{i(\vec k -\vec q)\cdot\vec x}\right]
\end{align*}
$$
[/latex]
(I copied it from a document in my computer and forgot I had defined a few shorthands).

It represents one of the two terms of the spatial components of momentum for a complex scalar field, before simplifying of course. After adding the other term (whose lhs is the same as this one, with the dot over the second phi and the gradient with the first one) we are left with something like this: [latex]\scriptsize $\vec P = \int \mathrm{d}^3 \tilde k \ \vec k(a^\dagger a + b^\dagger b)$[/latex] Someone who knows QM can think of a⁺ and a as the creation and destruction operators from the harmonic oscillator (here we have an infinite array of oscillators). a is for particles and b for antiparticles

 

[Zygar]

It's not homework - it's for my own interest. I did already explain that.

Even if it were "homework" i don't see why you believe someone should be precluded from asking questions about it - study work is meant to stimulate personal research and ultimately lead to a greater understanding of the set topic - and asking for help in doing so is not something that should be derided....

I just felt like harassing you. Especially since I don't really feel like reading the math and trying to figure it out.

 

[andyandy]

ok, another quickie

I'm working through higher ODE .....
http://www.sosmath.com/diffeq/second/variation/variation.html

and following through a couple of different methods -

but i'm stuck with the method for solving the following simultaneous equation

[latex] acosx + b sinx = 0 [/latex]


[latex] -asinx +bcosx = 1+tanx

[/latex]

i'm sure it shouldn't be too difficult to work out but i can't figure out how to get the answer

[latex] a=sinx(1+tanx) b=cosx(1+tanx) [/latex]

ideas?

thanks

 

[Zygar]

ok, another quickie

I solved

for a. Then plugged it into

I ended up with a long convoluted solution that probably would have been a lot shorter if I could remember my trig identities, but I got these answers:

a=-sinx(1+tanx) b=cosx(1+tanx)

Seems like you were missing a "-".