The Proof is Trivial!

Where is this coming from? I don't see how you know this unless you know how to evaluate the integral with

a=1

in the usual fashion.

That's exactly how you do know it.

Reply 2861

atsruser

11

Original post by james22

That's exactly how you do know it.

In that case, the dimensional analysis approach seems to be less useful than the standard approach, doesn't it? I'm not sure I see the point, apart from the fact that it's a slightly unusual application of the technique.

Reply 2862

james22

16

Original post by atsruser

In that case, the dimensional analysis approach seems to be less useful than the standard approach, doesn't it? I'm not sure I see the point, apart from the fact that it's a slightly unusual application of the technique.

I'm not sure if the standard approach works in the more general case though.

Reply 2863

atsruser

11

Original post by james22

I'm not sure if the standard approach works in the more general case though.

You mean for

I= \int_{-\infty}^\infty e^{-ax^2}dx

? If so, yes. You can calculate

I^2

as a double integral and convert to polars. It's only a change of variable from

x \rightarrow \sqrt{a}x

, after all.

Reply 2864

Tobybeta

2

Hey guys,
new to this thread, but love maths. Hopefully I'll contribute a bit to discussion!

Reply 2865

_Dreamville_

16

*!

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1402268286862.jpg30.6KB

Reply 2866

arrow900

12

Original post by ZSHNZ

*!

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Are you serious ?

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Reply 2867

Arieisit

3

Original post by ZSHNZ

*!

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Is that gcse?

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Reply 2868

user2020user

16

Continuing on from Arithmeticae's 469, Tarquin's one is 470 and jjpneed1's problems are 471 and 472 respectively.

Problem 473 **

Show that:

\displaystyle \int_{0}^{1} x^4 \left( 1 - x \right)^{20} \text{ d}x = \dfrac{1}{25 \binom{24}{4}}

Spoiler

(edited 9 years ago)

Reply 2869

james22

16

Original post by Khallil

Continuing on from Arithmeticae's 469, Tarquin's one is 470 and jjpneed1's problems are 471 and 472 respectively.

Problem 473 **

Show that:

\displaystyle \int_{0}^{1} x^4 \left( 1 - x \right)^{20} \text{ d}x = \dfrac{1}{25 \binom{24}{4}}

Spoiler

Solution 473**

No need for any complicated functions. Note that, when integrating by parts, the extra terms are all 0 and teh 2 minus terms cancel so we get

\displaystyle \int_{0}^{1} x^4 \left( 1 - x \right)^{20} \text{ d}x=\frac{4!}{21 \times 22 \times 23 \times 24}\displaystyle \int_{0}^{1} \left( 1 - x \right)^{24} \text{ d}x=\frac{4!}{21 \times 22 \times 23 \times 24 \times 25}=\dfrac{1}{25 \binom{24}{4}}

Reply 2870

user2020user

16

Original post by james22

...

I didn't think of integrating by parts but I think the solution to do with the Beta and Gamma functions is more elegant.

Edit: Somebody, use the

\Gamma

hammer!

(edited 9 years ago)

Reply 2871

Tarquin Digby

9

Original post by khallil

continuing on from arithmeticae's 469, tarquin's one is 470 and jjpneed1's problems are 471 and 472 respectively.

problem 473 **

show that:

\displaystyle \int_{0}^{1} x^4 \left( 1 - x \right)^{20} \text{ d}x = \dfrac{1}{25 \binom{24}{4}}

Spoiler

\displaystyle \int_0^1 x^4 (1-x)^{20} \, \mathrm d x = \mathrm B (5,21) = \frac {\Gamma (5) \Gamma (21)} {\Gamma (26)} = \frac {4! \, 20!} {25!} = \frac 1 {25 \binom {24} 4}

Reply 2872

james22

16

Original post by Khallil

I didn't think of integrating by parts but I think the solution to do with the Beta and Gamma functions is more elegant.

Edit: Somebody, use the

\Gamma

hammer!

Why use complicated functions when there is a perfectly good (and not even messy or long) elementary solution.

Reply 2873

user2020user

16

Original post by james22

Why use complicated functions when there is a perfectly good (and not even messy or long) elementary solution.

Because I said so.

Reply 2874

james22

16

Original post by Khallil

Why don't you add fractions, fam?

I can't do maths when tired.

Reply 2875

jjpneed1

2

That's pretty harsh Khallil I'm sure you've made those kinds of mistakes at some point :tongue:

To be fair your problem isn't very interesting, in fact if the beta function was taught at A-level I'm sure I'd see it in exercise 1A

Fun one but fairly simple:

Problem 474 *

I can write any (positive) number as a sum of arbitrarily many (positive) real numbers, e.g. 16 = 10 + 6 = 3.4 + 6.6 + 6, etc. When is the product of all elements in this sum maximised?

(edited 9 years ago)

Reply 2876

Zen-Ali

8

Original post by Khallil

when ur anoos quivers

I arrived at a similar answer. :—)

(Mine also quivered. :—))

Reply 2877

Smaug123

15

Original post by jjpneed1

That's pretty harsh Khallil I'm sure you've made those kinds of mistakes at some point :tongue:

To be fair your problem isn't very interesting, in fact if the beta function was taught at A-level I'm sure I'd see it in exercise 1A

Fun one but fairly simple:

Problem 474 *

I can write any (positive) number as a sum of arbitrarily many (positive) real numbers, e.g. 16 = 10 + 6 = 3.4 + 6.6 + 6, etc. When is the product of all elements in this sum maximised?

I can do this by Lagrange multipliers (assuming a finite sum, of course). That wouldn't be a (*) kind of question, though, so presumably there is a nicer way?

Reply 2878

jjpneed1

2

Original post by Smaug123

I can do this by Lagrange multipliers (assuming a finite sum, of course). That wouldn't be a (*) kind of question, though, so presumably there is a nicer way?

A very nice way. Though I would be interested to see how you do it with Lagrange multipliers as I initially tried that??

This was on a trinity admissions test by the way

Reply 2879

Smaug123

15

Original post by jjpneed1

A very nice way. Though I would be interested to see how you do it with Lagrange multipliers as I initially tried that??

This was on a trinity admissions test by the way

Hmm. I can Lagrange-multipliers it down to "Maximise

a^i

subject to a>0, i a natural number,

a i = 16

", anyway, because Lagrange multipliers on

a b c

subject to

a+b+c=16

yields

a=b=c

. Then the only problem is to find how many terms are in the sum.

That is, maximise

(\frac{16}{i})^i

for integer i. Let's move to the more general

x

as our sum (so x=16 above). Now, if i>x then we start raising something less than 1 to a high power, so that can't be a maximum. Our answer must therefore be less than x. Using i=1 gives us x straight off.

Suppose we'd found i. What would we know? We'd have that

\frac{x}{i}^i > \frac{x}{i+1}^{i+1}

, or

\frac{1}{i}^i > \frac{1}{i+1}^{i+1}x

, or that

x < \frac{i+1}{i}^i (i+1)

. That fraction term tends to

e

from below as i increases; in particular, we need

i+1 > \frac{x}{e}

, or

i > \frac{x}{e}-1

. That suggests using i = roundup(x/e - 1).

How tight is this bound? Not sure. It is conceivable that the "fraction term tends to…" line makes the bound slacker than it needs to be.

The Proof is Trivial!

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