So you want a dag?(Original post by Smaug123)
Kind of, but more flexible  I'm visualising something like the roots of a tree, stretching down, but sometimes meeting up and joining. Not just a straight binary tree, no meetingandjoining.

ukdragon37
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http://www.youtube.com/watch?v=zH64dlgyydM 
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Felix Felicis
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 31072013 23:47
Solution 280
Subbing in gives , which yields the desired series.
I can't quite justify the interval of convergence, though I can show that it converges if by the ratio test but not sure how to whenLast edited by Felix Felicis; 01082013 at 05:17. 
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 01082013 11:44
(Original post by Felix Felicis)
Solution 280
Spoiler:ShowClearly it is sufficient to prove convergence in the case x=1. Try arranging the series into a different form and then comparing it to a known series. 
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 02082013 00:14

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 02082013 13:01
Last edited by joostan; 02082013 at 13:05. Reason: Missing brackets. 
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 02082013 16:13

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 02082013 18:13
(Original post by joostan)
x
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 and .
I thought I had proven the generalisation though clearly I had made some mistake. I even tried n=5 separately and still got . Back to the drawing board:
Problem 298*
Evaluate for where the underbrace represents the number of times e appears in the exponentiation.Last edited by Jkn; 02082013 at 21:59. 
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 02082013 21:02
(Original post by FireGarden)
Problem 297***
Calculate the volume of the ellipsoid defined by
Where are positive constants.
Solution 297*
We know that the area enclosed by an ellipse is . The proof is part of the Alevel specification (and is of little interest).
Rearranging the equation of the ellipsoid bounding our volume, we get .
As the volume is a dimesional extension to area as area is to length, it can be easy inferred that where 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 , as can be seen by setting in the original equation and, using the above result, we get that . This gives:
Note that this equation has the required abc 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 ndimensional space. Let the ndimensional volume be denoted by .
We define our hypothetical region as the set of points satisfying . We will also assume that all solutions can be expressed in the form 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 which gives us the following relation.
Noting that the function we are integrating is even, letting , using Beta functions and finally the Gamma representation of the Beta function:
Noting that the only sensible way to proceed in to let (again, there are various ways we can justify this claim, such as verifying that it leads to the correct values for and , which it does):
Woo! This has to be one of the most beautiful results I have discovered independently!
Is this a wellknown result? Is it correct? What you you call ?
Solution 299 (weak corollary)
Setting gives the required result of .Last edited by Jkn; 02082013 at 23:11. Reason: Details. 
Smaug123
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 02082013 21:08
(Original post by Jkn)
Instead of using 'triple integration', I will derive the result in a way that could have been done by an Alevel student.
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 , 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 , for example, but that's too complicated, and it must be the simpler one.Last edited by Smaug123; 02082013 at 21:11. Reason: ETA 
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 02082013 21:18
(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?Spoiler:Show2.814748587862548678050445650343 78650553053524527015318777075354 827928\
30371377245089302614394292704623 80418879979374304513430442428837 653058\
41624608864703657720269273518028 54473294562709938722054449126147 174140\
72850162897807163139976305739619 91772662688059938897174929482584 337521\
46024533297279126519151045044185 28042192530641575921118846740422 514025\
52880502659077596731252318257658 69581747592075202571315222503434 314109\
57488248810598252366055487920295 82783810400699542540650976883269 006880\
95570952708140728142215404454333 62550599588037179595129037214288 176607\
42051167660894602290092141250264 57206649676739597221299574147120 814876\
17950725698627131132985739489683 45423998395693930390611697281265 995161\
24695244147204643630704755698803 98974608587570173500753575778619 442900\
36035501528596656715108841001854 37843977582718213581082496803166 157351\
82349156207907687737409162641325 28719697950723862588315707223213 822585\
29588073383812512325803443310313 53088986130946151406258352788894 008076\
424627477783602829528*10^1656469
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) 
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 02082013 21:45
(Original post by Smaug123)
How likely are we to accept the following?
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 , 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 , for example, but that's too complicated, and it must be the simpler one.
Oh btw, I may have just generalised the method to ndimensions. 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 ndimensional '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 isSpoiler:Show2.814748587862548678050445650343 78650553053524527015318777075354 827928\
30371377245089302614394292704623 80418879979374304513430442428837 653058\
41624608864703657720269273518028 54473294562709938722054449126147 174140\
72850162897807163139976305739619 91772662688059938897174929482584 337521\
46024533297279126519151045044185 28042192530641575921118846740422 514025\
52880502659077596731252318257658 69581747592075202571315222503434 314109\
57488248810598252366055487920295 82783810400699542540650976883269 006880\
95570952708140728142215404454333 62550599588037179595129037214288 176607\
42051167660894602290092141250264 57206649676739597221299574147120 814876\
17950725698627131132985739489683 45423998395693930390611697281265 995161\
24695244147204643630704755698803 98974608587570173500753575778619 442900\
36035501528596656715108841001854 37843977582718213581082496803166 157351\
82349156207907687737409162641325 28719697950723862588315707223213 822585\
29588073383812512325803443310313 53088986130946151406258352788894 008076\
424627477783602829528*10^1656469
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) 
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 02082013 22:00
(Original post by Smaug123)
How likely are we to accept the following?
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 , 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 , for example, but that's too complicated, and it must be the simpler one.
This is probably what you meant or had in mind, but I wrote it here for explicitness. 
Smaug123
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 02082013 22:04
(Original post by henpen)
This is probably what you meant or had in mind, but I wrote it here for explicitness. 
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 02082013 22:16
(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 
Smaug123
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 02082013 22:31
(Original post by henpen)
It would be enough for a physicist. Physicists' proofs are quite nice in a way entirely different to mathematical ones. 
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 02082013 23:08
(Original post by henpen)
There surely is an easier way: set of the variables equal to , then the extremal solution for the remaining th variable will be equal to that dimension's semimajor/minor/inbetween axis, . Thus the unit hypersphere was stretched in that dimension by . The stretch multiplies the volume by a factor that is independent of the other dimensions' stretching, and the volume of the unit hypersphere is (Problem 299* : prove this), so the volume of the stretched hypersphere is
This is probably what you meant or had in mind, but I wrote it here for explicitness.
So satisfying to rediscover something! Do you know when my result was established and by who?
Edit: How did you get the result for a the generalized ellipse btw? I can't find stuff about it anywhere online (did you read it somewhere and, if so, is the original derivation the same as mine?)
Problem 300*/***
Working within the base10 number system, find all 9digit numbers containing all digits except zero (a zeroless Pandigital number) such that the first n digits form a number that is divisible by n for all n.
[For example, 123456789 is of the correct form. We also note that 1 is divisible by 1, 12 is divisible by 2, 123 is divisible by 3 though, as 1234 is not divisible by 4, this number fails to meet the requirements.]
Can you generalise interestingly to number systems with other bases and/or in any other way? (bonusonly)Last edited by Jkn; 03082013 at 01:21. 
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 03082013 17:25
Solution 300
Let be the number such that is the first digit, is the second and so forth.
We have, from basic divisibility criteria the following set of equations:
From, Equations (1), (3), (5), (7), b, d, f, h are even and the rest are odd. e is clearly equal to 5, so the rest of the numbers must be either 1, 3, 7, or 9.
From (3), and the fact that c is odd, we have .
As d and f are even, we have .
as 4 is too small and 16 is too big to be the sum of distinct even numbers under 10.
This, coupled with the fact that d is 2 or 6 gives us, .
From (7) and the fact that we get .
But, as g is odd, we have . In fact, we have, .
This leaves b=2, when h = 6 and b = 6 when h = 2.
So, as far as the configuration of even integers is concerned we have,
Now, note that from (3), and .
Collating the pairs with the appropriate sets of even integers, we get the following:
.
Now, finally, look at (6). This and (2) will help us fix a and c and conclude.
Noting that , we have,
Now, remember that, when, (3), and .
So, notice that,
So, we have 4 sets of congruences. Notice that a and c are odd, so, we only have the integers to choose from.
But, note that for each set of congruences, one of these has already been used up by . So, one can only consider the other 3, narrowing down the bunch of solutions.
In fact, after some computation, it turns out that only the last set of equations is solvable with .
So, that leaves as the only thing left.
And gives us with just such an integer we want:
.
A quick check reveals this to be acceptable. 
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 04082013 15:37
(Original post by Jkn)
Edit: How did you get the result for a the generalized ellipse btw? I can't find stuff about it anywhere online (did you read it somewhere and, if so, is the original derivation the same as mine?
)
http://en.wikipedia.org/wiki/Gamma_function
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