The Proof is Trivial!

Original post by ukdragon37

So you want a dag?

That's the one - I've never done any formal graph theory, and my only graph-theoretic knowledge is "that which can easily be done in Mathematica" :P I'd forgotten the term "directed acyclic", although actually it comes up fairly often.

Reply 2063

Felix Felicis

13

Solution 280

\displaystyle \begin{aligned} \frac{d}{dx} \text{arsinh} x = (1+x^2)^{- \frac{1}{2}} & = 1 + \left( - \tfrac{1}{2} \right) x^2 + \frac{ \left( - \frac{1}{2} \right) \left( - \frac{3}{2} \right) (x^2)^2}{2!} + \cdots \quad (\text{binomial expansion}) \\ & = 1 + \sum_{r=1}^{\infty} \left( \frac{(-1)^{r} x^{2r}}{r! \cdot 2^{r}} \prod_{i=0}^{r-1} (2i + 1) \right) \\ & = 1 + \sum_{r=1}^{\infty} \left( \frac{(-1)^{r} x^{2r}}{r! \cdot 2^{r}} \cdot \frac{(2r-1)!}{(r-1)! \cdot 2^{r-1}} \right) \\ & = 1+ \sum_{r=1}^{\infty} \left( \frac{(-1)^{r} x^{2r}}{4^{r}} \cdot \frac{2 \cdot (2r)!}{2r \cdot r! \cdot (r-1)!} \right)\\ & = \sum_{r=0}^{\infty} \frac{(-1)^{r} x^{2r}}{4^{r}} \cdot \frac{(2r)!}{(r!)^{2}} \\ \implies \text{arsinh} x & = \int \sum_{r=0}^{\infty} \frac{(-1)^{r} x^{2r}}{4^{r}} \cdot \frac{(2r)!}{(r!)^{2}} \ dx \\ & = \boxed{\sum_{r=0}^{\infty} \frac{(-1)^{r} (2r)! x^{2r+1}}{4^{r}(2r+1)(r!)^2} + \mathcal{C}} \end{aligned}

Subbing in

x = 0

gives

\mathcal{C} = 0

, which yields the desired series.

I can't quite justify the interval of convergence, though :confused:

I can show that it converges if

|x| < 1

by the ratio test but not sure how to when

|x| = 1

(edited 10 years ago)

Reply 2064

Original post by Felix Felicis

Solution 280

Nice stuff, bro! :biggrin:

I can't quite justify the interval of convergence, though :confused:

I can show that it converges if

|x| < 1

by the ratio test but not sure how to when

|x| = 1

Spoiler

Reply 2065

henpen

Problem 296**

Find

\int_0^{\pi} e^{\cos(x)}\cos(\sin(x)) dx

Quite a pretty one.

Spoiler

Reply 2066

joostan

13

Original post by henpen

Problem 296**

Find

\int_0^{\pi} e^{\cos(x)}\cos(\sin(x)) dx

Quite a pretty one.

Spoiler

Solution 296:

Unparseable latex formula:

I=\displaystyle\int_0^{\pi} e^{\cos(x)}\cos(\sin(x)) dx = \Re \left(\displaystyle\int_0^{\pi} e^{e^{ix}} \ dx \right)[br]\Rightarrow I= \Re \left(\displaystyle\int_0^{\pi} \displaystyle\sum_{r=0}^{\infty} \dfrac{(e^{ix})^r}{r!} \ dx \right)=\Re \left(\displaystyle\int_0^{\pi} 1+\dfrac{e^{ix}}{1!}+\dfrac{e^{2ix}}{2!}+. . . +\dfrac{e^{ikx}}{k!}+. . . \right) \ dx[br]\Rightarrow I =\Re \left[x+\dfrac{e^{ix}}{i \times 1!}+\dfrac{e^{2ix}}{2i \times 2!}+. . .+ \dfrac{e^{ikx}}{ki\times k!}+ . . .\right]^{\pi}_0[br]\Rightarrow I=\Re \left[x+\dfrac{1}{i}\displaystyle\sum_{r]^{\pi}_0[br]\Rightarrow I=\Re \left( \left( \pi-i \displaystyle\sum_{r=1}^{\infty} \dfrac{(e^{i\pi})^r}{r\times r!}\right)-\left(0-i\displaystyle\sum_{r=1}^{\infty} \dfrac{(e^{i0})^r}{r\times r!}\right) \right)[br]\Rightarrow I =\Re \left( \pi- i \left( \displaystyle\sum_{r=1}^{\infty} \dfrac{(-1)^r}{r\times r!} - \displaystyle\sum_{r=1}^{\infty} \dfrac{(1)^r}{r\times r!}\right) \right)[br]\Rightarrow I=\pi

(edited 10 years ago)

Reply 2067

FireGarden

14

Problem 297***

Calculate the volume of the ellipsoid defined by

V: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} \leq 1

Where

a,b,c

are positive constants.

Reply 2068

Original post by henpen

I wonder if

\int_{-\pi}^{\pi} e^{e^{e^{ix}}} dx

is doable.

Original post by joostan

x

Interesting stuff guys!

I have been able to do the first few cases of the generalisation and a pattern has emerged. I thought I had been able to prove the result below but, whilst Wolfram agrees with me for n=2,3,4 it doesn't for n=5.

Essentially

\displaystyle \Re \int_0^{\pi} e^{e^{e^{ix}}} \ dx = \pi e

and

\displaystyle \Re \int_0^{\pi} e^{e^{e^{e^{ix}}}} \ dx = \pi e^e

.

I thought I had proven the generalisation though clearly I had made some mistake. I even tried n=5 separately and still got

Unparseable latex formula:

\pi e^e^e

. Back to the drawing board:

Problem 298*

Evaluate

Unparseable latex formula:

\displaystyle \Re \int_0^{\pi} \underbrace{{e^{e^{\cdots}^{ix}}}}_n \ dx

for

n \ge 2

where the underbrace represents the number of times e appears in the exponentiation.

(edited 10 years ago)

Reply 2069

Original post by FireGarden

Problem 297***

Calculate the volume of the ellipsoid defined by

V: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} \leq 1

Where

a,b,c

are positive constants.

Instead of using 'triple integration', spherical coordinates and/or complicated parametrization, I will derive the result in a way that could have been done by an A-level student.

Solution 297*

We know that the area enclosed by an ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

is

\pi ab

. The proof is part of the A-level specification (and is of little interest).

Rearranging the equation of the ellipsoid bounding our volume, we get

\frac{x^2}{\left( a \sqrt{1-\frac{z^2}{c^2}} \right)^2}+\frac{y^2}{\left( ab\sqrt{1-\frac{z^2}{c^2}} \right)^2}=1

.

As the volume is a dimesional extension to area as area is to length, it can be easy inferred that

\displaystyle V=\int_a^b f(z) \ dz

where

f(z)

is the area of the ellipse contained by the surface generated at each z coordinate and a and b are the boundaries on the volume enclosed by the ellipsoid. These boundaries correspond to

z= \pm c

, as can be seen by setting

x=y=0

in the original equation and, using the above result, we get that

f(z) = \pi \left( a \sqrt{1-\frac{z^2}{c^2}} \right) \left( b \sqrt{1-\frac{z^2}{c^2}} \right)=\pi ab \left(1-\frac{z^2}{c^2} \right)

. This gives:

\displaystyle V = \pi ab \int_{-c}^c 1-\frac{z^2}{c^2} \ dz =\frac{4}{3} \pi abc \ \square

Note that this equation has the required a-b-c symmetry as well as the property that it reduces to the volume of a sphere when a=b=c.

Generalization:

We can now extend this method to generalise for 'volume' in n-dimensional space. Let the n-dimensional volume be denoted by

V_n

.

We define our hypothetical region as the set of points satisfying

\displaystyle \sum_{i=1}^n \frac{x_i^2}{r_i^2} \le 1

. We will also assume that all solutions can be expressed in the form

V_n = k \prod_{i=1}^n r_i

where k is a positive constant to be found (this assumption can be justified as valid in a number of ways).

Using our previous method, we note that the boundary can be expressed as

\displaystyle \sum_{i=1}^{n-1} \frac{x_i^2}{\left( r_i \sqrt{1-\frac{x_n^2}{r_n^2}} \right)^2} =1

which gives us the following relation.

Noting that the function we are integrating is even, letting

x \mapsto r_n \sqrt{x}

, using Beta functions and finally the Gamma representation of the Beta function:

\displaystyle \begin{aligned} V_n = \int_{-r_{n}}^{r_{n}} k \prod_{i=1}^{n-1} r_i \sqrt{1-\frac{x_n^2}{r_n^2}} \ dx_n = V_{n-1} \int_{-r_{n}}^{r_{n}} \left(1-\frac{x_n^2}{r_n^2} \right)^{\frac{n-1}{2}} \ dx_n = 2 V_{n-1} \int_0^1 (1-x)^{\frac{n-1}{2}} \frac{r_n dx}{2 \sqrt{x}} \end{aligned}

\displaystyle = r_n V_{n-1} \int_0^1 x^{- \frac{1}{2}} (1-x)^{\frac{n-1}{2}} \ dx = r_n V_{n-1} B \left(\frac{1}{2} , \frac{n+1}{2} \right) = V_{n-1} \frac{r_n \sqrt{\pi} \Gamma \left( \frac{n+1}{2} \right)}{\Gamma \left(1+\frac{n}{2} \right)}

Noting that the only sensible way to proceed in to let

V_0=1

(again, there are various ways we can justify this claim, such as verifying that it leads to the correct values for

V_2

and

V_3

, which it does):

\displaystyle \Rightarrow V_n = V_0 \prod_{i=1}^n r_i \frac{\sqrt{\pi} \Gamma \left( \frac{n+1}{2} \right)}{\Gamma \left(1+\frac{n}{2} \right)} = \pi^{n/2} \frac{r_1 r_2 \cdots r_n}{\Gamma \left(1+\frac{n}{2} \right)} \ \square

Woo! This has to be one of the most beautiful results I have discovered independently! :biggrin:

Is this a well-known result? Is it correct? What you you call

V_n

?

Solution 299 (weak corollary)

Setting

r_1=r_2= \cdots r_n=1

gives the required result of

\frac{\pi^{n/2}}{\Gamma \left(1+\frac{n}{2} \right)}

.

(edited 10 years ago)

Reply 2070

Original post by Jkn

Instead of using 'triple integration', I will derive the result in a way that could have been done by an A-level student.

How likely are we to accept the following? :tongue:

Volume has dimension [L]^3. The only dimensioned quantities we have are a,b,c, each of dimension [L].
Since the volume clearly depends on a, b and c, and is also symmetric on transposing a,b,c, the answer is obviously

f abc

, where f is a dimensionless function to be determined.
But since the only dimensionless quantities are constants and ratios of a,b,c (and we know that f should be invariant on cycling a,b,c, so ratios are out), we have f constant.
Now, a sphere has a=b=c, and volume 4/3 pi a^3. So f=4/3 pi.

ETA: Yes, I know that f could be

\dfrac{(a+b+c)^2}{a^2+b^2+c^2}

, for example, but that's too complicated, and it must be the simpler one.

(edited 10 years ago)

Reply 2071

Original post by Jkn

Could you verify/falsify this for me for a few values of n greater than or equal to 5 (on Mathematica) just to put my mind at ease? :colondollar:

Hmm, I'm not sure Mathematica is agreeing for n=5 - with 1000 digits of precision, it claims that the integral is

Spoiler

That is, 2*10^1656469.
I'm running it with two million digits of precision - hopefully that'll make sure we're on the right order of magnitude. (For reference, your predicted value is 1.19829*10^7)

Reply 2072

Original post by Smaug123

How likely are we to accept the following? :tongue:

Volume has dimension [L]^3. The only dimensioned quantities we have are a,b,c, each of dimension [L].
Since the volume clearly depends on a, b and c, and is also symmetric on transposing a,b,c, the answer is obviously

f abc

, where f is a dimensionless function to be determined.
But since the only dimensionless quantities are constants and ratios of a,b,c (and we know that f should be invariant on cycling a,b,c, so ratios are out), we have f constant.
Now, a sphere has a=b=c, and volume 4/3 pi a^3. So f=4/3 pi.

ETA: Yes, I know that f could be

\dfrac{(a+b+c)^2}{a^2+b^2+c^2}

, for example, but that's too complicated, and it must be the simpler one.

I considered that but the symmetry condition and a=b=c evaluation is not sufficient to directly deduce a formula for the volume. Also, I'm certain that it constitutes A-level knowledge as it would easily fit in to a C4 or FP3 textbook as an 'extension' exercise. For example, you plot the graph of area against 3rd co-ordinate, etc.. In fact, I am definitely right as this understanding is required for both M3 and M5. :tongue:

Oh btw, I may have just generalised the method to n-dimensions. I had to use beta functions but I have reduced it to a collection of series that, if solved, will yield the necessary recurrence relation between n-dimensional 'volume' and 'n+1'-dimensional 'volume' (hypervolume? no idea..)

Original post by Smaug123

Hmm, I'm not sure Mathematica is agreeing for n=5 - with 1000 digits of precision, it claims that the integral is

Spoiler

That is, 2*10^1656469.
I'm running it with two million digits of precision - hopefully that'll make sure we're on the right order of magnitude. (For reference, your predicted value is 1.19829*10^7)

Ah that's what Wolfram gave. Hmm, that is rather annoying! Then again it is worth noting how spookily close that value is to

Unparseable latex formula:

\pi e^e^e^e

(or

Unparseable latex formula:

e^e^e^e

for that matter). :tongue:

Reply 2073

henpen

Original post by Smaug123

How likely are we to accept the following? :tongue:

Volume has dimension [L]^3. The only dimensioned quantities we have are a,b,c, each of dimension [L].
Since the volume clearly depends on a, b and c, and is also symmetric on transposing a,b,c, the answer is obviously

f abc

, where f is a dimensionless function to be determined.
But since the only dimensionless quantities are constants and ratios of a,b,c (and we know that f should be invariant on cycling a,b,c, so ratios are out), we have f constant.
Now, a sphere has a=b=c, and volume 4/3 pi a^3. So f=4/3 pi.

ETA: Yes, I know that f could be

\dfrac{(a+b+c)^2}{a^2+b^2+c^2}

, for example, but that's too complicated, and it must be the simpler one.

There surely is an easier way: set

n-1

of the

n

variables equal to

0

, then the extremal solution for the remaining

k

th variable will be equal to that dimension's semi-major/minor/in-between axis,

r_k

. Thus the unit hypersphere was stretched in that dimension by

\frac{r_k}{1}

. The stretch multiplies the volume by a factor that is independent of the other dimensions' stretching, and the volume of the unit hypersphere is

\frac{\pi^{n/2}}{ \Gamma \left(1+\frac{n}{2} \right)}

(Problem 299* : prove this), so the volume of the stretched hypersphere is

\displaystyle \frac{\pi^{n/2}}{ \Gamma \left(1+\frac{n}{2} \right)} \prod_{k=1}^nr_k

This is probably what you meant or had in mind, but I wrote it here for explicitness.

Reply 2074

Original post by henpen

This is probably what you meant or had in mind, but I wrote it here for explicitness.

No, I wasn't being entirely serious - I was doing to the volume of an ellipsoid what we did in lectures to Kepler's Laws :tongue:

Reply 2075

henpen

Original post by Smaug123

No, I wasn't being entirely serious - I was doing to the volume of an ellipsoid what we did in lectures to Kepler's Laws :tongue:

It would be enough for a physicist. Physicists' proofs are quite nice in a way entirely different to mathematical ones.

Reply 2076

Original post by henpen

It would be enough for a physicist. Physicists' proofs are quite nice in a way entirely different to mathematical ones.

If you come from the "don't worry, be happy" school of mathematical rigour, as one of my other lecturers said…

Reply 2077