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The Proof is Trivial! Watch

1. (Original post by james22)
I don't quite understand the statement "4k-dimensional Hyperkahler manifold", since all Hyperkahler manifolds have dimension 4k. It's like me saying "consider an associative group".
I adressed this issue in the following article, if you are interested.
2. (Original post by Lord of the Flies)
I adressed this issue in the following article, if you are interested.
Ach! I hadn't realized that you could apply the tools of differential geometry, particularly those properties of HyperKahler spaces, to Categories of Linear Functionals. That's good stuff, my dear friend.

I have a feeling that you can also consider this from the point of view of Fibre Bundle representational maps in Homological Algebra. Hmm, interesting.
3. All this discussion reminds me of the Bourbaki group.
4. (Original post by MW24595)
Ach! I hadn't realized that you could apply the tools of differential geometry, particularly those properties of HyperKahler spaces, to Categories of Linear Functionals. That's good stuff, my dear friend.

I have a feeling that you can also consider this from the point of view of Fibre Bundle representational maps in Homological Algebra. Hmm, interesting.
In fact, a recent breakthrough in combinatorial geometry (Borislav, 2013) makes it possible to draw an isomorphism between theorem spaces of categories of linear functionals and Hyperkahler spaces. It's a remarkably intricate and complex argument to show it though.
5. (Original post by Lord of the Flies)
I adressed this issue in the following article, if you are interested.
Pierre Woodman
6. Problem 427***

Evaluate
7. (Original post by james22)
Problem 427***

Evaluate
Spoiler:
Show

Is it possible to get anything better than a power series answer? I get , which I am confident of up to multiples of 2. This power series does look vaguely familiar my STEP days though...
8. Does anyone mind if I post a question that spoils the answer to 420?
9. (Original post by matt2k8)
Spoiler:
Show

Is it possible to get anything better than a power series answer? I get , which I am confident of up to multiples of 2. This power series does look vaguely familiar my STEP days though...
I don't know, when I did this I didn't get any better than a power series (I forget exactly what my series was, but it looked similar), I was wondering if anyone could find a closed form.
10. (Original post by james22)
I don't know, when I did this I didn't get any better than a power series (I forget exactly what my series was, but it looked similar), I was wondering if anyone could find a closed form.
Numerical approximations don't seem to give any hint of a candidate exact answer, but I believe that when I saw it before, that power series arose from the series for (1-x)^(-1/2) or similar.
11. Problem 428**(*):

If does there always exist a sequence of reals such that congerges if and only if ?
12. Problem 429 *
Evaluate , given that it is real.
13. Problem 430**

Prove that is irrational.

Problem 431*

Prove that where
14. (Original post by Jooooshy)
Problem 430**

Prove that is irrational.

Problem 431*

Prove that where
i know how to do these but not bothered to type up.

Guideline 430

Assume e is rational and can be expressed as p/q where gcd(p,q)=1. Then write out the series expansion e and equate it to p/q. Then mukltiply everything by q! to get rid of the fraction on LHS.

The LHS is now an integre ((q-1)!p) and the RHS until the qth term is an integer. Now for the remained of this on the rhs which is infinite we have 1/(q+1) +1/(q+1)(q+2) and so on. This is less than the series with first term 1/(q+1) and the same common ratio as the first term. The sum of which is 1/2, so the RHS is a positive numer plus a positive number less than half.

Cleaarly the RHS is not an integer as it is the sum of two positive numbgers, one of which is an integer while the other is not. However, the LHS is an integer hence we have reached a contradiction so e must be irrational.

Guideline 431

Area of triangle is 1/2*bh. For any triangle with sides a,b,c this can be shown to be 1/2*bcsinA.

So Area=0.5bcsinA. But from the formula of cosines we know that a^2=b^2+c^2-2bcCosA, so 2bcCosA=b^2+c^2-a^2. To get sin A in a similar form we simply multiply the earlier equation by 4 to get 4Area=2bcsinA.

Aquaring and adding and using the relation sin^2A+cos^2A=1 the rest is just a case of algebraic manipulation (which can be simplified using s=(a+b+c)/2 - the semiperimeter).
15. Problem 432*

Prove that a real, symmetric matrix has real eigenvalues

Problem 433**

Prove that the only two prime Catalan numbers are and

Also prove that
16. (Original post by Nebula)
Problem 428**(*):

If does there always exist a sequence of reals such that congerges if and only if ?
Idea for 428:
Spoiler:
Show

If the second part of (420) is true, then by the same argument it'll be true that if converges, so does - so the answer is no, it fails for everything except sets of the form .
17. (Original post by matt2k8)
Idea for 428:
Spoiler:
Show

If the second part of (420) is true, then by the same argument it'll be true that if converges, so does - so the answer is no, it fails for everything except .
Not true, consider the set {3,4,5,...} and the sequence 1/n^(1/2). This converges iff s is in the set.

I think it may work for all sets with 1 in though, since increasing s should only make the series converge faster.
18. (Original post by james22)
Not true, consider the set {3,4,5,...} and the sequence 1/n^(1/2). This converges iff s is in the set.

I think it may work for all sets with 1 in though, since increasing s should only make the series converge faster.
Ah yeah, my argument actually shows for any sequence, the set where it works is of the form {k, k+1, k+2 , k+3 , ...}
19. (Original post by matt2k8)
Ah yeah, my argument actually shows for any sequence, the set where it works is of the form {k, k+1, k+2 , k+3 , ...}
What did you do? I've been trying to prove a more general case that if the sum of x_n converges and y_n is a sequences with each term having modulus less than the modulus of the coresponding x_n, and the sign of x_n is the sign of y_n, then the sum of y_n converges. It seems obvious but I can't get it.
20. (Original post by james22)
What did you do? I've been trying to prove a more general case that if the sum of x_n converges and y_n is a sequences with each term having modulus less than the modulus of the coresponding x_n, and the sign of x_n is the sign of y_n, then the sum of y_n converges. It seems obvious but I can't get it.
That isn't true; e.g. take and to be for even n, and zero for odd . Now doubting my proof for the powers thing... thinking it may even in fact be false. You'd need some fairly exotic example though (for example, no alternating series can possibly work)

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