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# The Proof is Trivial! watch

1. (Original post by Felix Felicis)
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(Original post by Zakee)
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Thanks guys.

Now back to STEP.
2. (Original post by Lord of the Flies)
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Ach, doing my best to ignore this thread but I love series and integrals too much. Here are alternative solutions to 140 and 142:

Solution 140

Differentiation under the integral sign!

Everything falls into place now:

Solution 142

We can evaluate the sum directly:

The trick is now to use that near to get rid of a bunch of stuff:

As

Hence

Furthermore, observe that:

gives:

So, for instance:

Lol, my solution involves inverse fourier transforming everything in sight. Needless to say, I am not an elegant mathematician.
3. (Original post by Lord of the Flies)
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To top it off, you presented the answers really well. Très bien!
4. (Original post by ben-smith)
Needless to say, I am not an elegant mathematician.
Pf, nonsense Ben! Your solution to problem 69 was very neat for instance.

(Original post by shamika)
To top it off, you presented the answers really well. Très bien!
If only I could present things as neatly on paper... Mais cimer!
5. Problem 145**

Let be the set of all polynomials of degree at most 2.

Find such that for all
6. I'm back with an awesome problem... ENJOY!

Problem 146**

Evaluate
7. (Original post by Jkn)
I'm back with an awesome problem... ENJOY!

Problem 146**

Evaluate
So.. you mean the sum of the reciprocals of all the primes? 'cause if so..

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that sum diverges..
8. (Original post by FireGarden)
Problem 145**
I like the problem, so...

Let denote the set of all univariate polynomials, in , of degree atmost .

Define for .

The goal is to find such that for all , where .

Notice that by assumption. In addition, we have:

i) for a constant and polynomial .
ii) for polynomials and .

and .

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Let .

Hence, .

Using, for , the polynomials , we get the following system.

The symmetry in the values of implies that this is a Toeplitz matrix.

The special case, and , produces the following system

which has the solution .
9. Problem 147 / ***

Assume is a set containing numbers. Let and denote the sum and product, respectively, of all the elements in the set .

Let be the set .

i) Find an expression for .

ii) Prove that .

Note that .
10. (Original post by FireGarden)
So.. you mean the sum of the reciprocals of all the primes? 'cause if so..

Spoiler:
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that sum diverges..
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Yes. No prove it ...don't look it up though!
11. (Original post by Jkn)
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Yes. No prove it ...don't look it up though!
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I'm thinking proof by contradiction? I feel like the you write the LHS as a fraction and the right is a real number...erm I'm so bad at maths
12. (Original post by bananarama2)
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I'm thinking proof by contradiction? I feel like the you write the LHS as a fraction and the right is a real number...erm I'm so bad at maths
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You need to find a property of convergent series that can be applied in order to create the contradiction! Writing the left hand side as a fraction would require 2 sigmas and a capital pi so I don't know whether or not this would simplify the problem. This problem is extremely difficult though so don't feel too downtrodden (I found the proof online but didn't actually do it myself ) There may exist an easy proof using sophisticated methods but the proof I read is elementary and incredibly beautiful.
13. Solution 147

Part i). Consider . Hence, . I include .

Part ii). We use induction. The base case is obvious. Suppose that our result is true for all . We prove it for .
Split our sum into two parts:
.

Remark: I denoted in part ii).

Solution 146

The th prime is less than for . Hence the series diverges.
Solution 146

The th prime is greater than for . Hence the series diverges.
Reminds me of this:

Spoiler:
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You're going to have explain how you got that and prove it And also why it implies that a series diverges

Here's an easier one. Undergrads will find it child's play but it's a really nice result so I'll post it anyway:

Problem 148**

Evaluate
15. I shall use the same method as when I did in a previous solution.

Solution 148

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In fact we can carry on to higher powers, however getting to the 2nd line demands considerably more algebra.

It would be interesting if we could figure out a closed form for
16. I really must learn about Laplace transforms.
17. (Original post by bananarama2)
I really must learn about Laplace transforms.
Have a look at this. It's a bit slow at times but it is clear and he's quite a fun lecturer!
18. (Original post by Jkn)
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Reminds me of this:

Spoiler:
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You're going to have explain how you got that and prove it And also why it implies that a series diverges
I am too busy these days, and I considered the problem superficially.
I need an upper bound.
Here is one sufficient upper bound - for , we have . Now, it is obvious that ; the last series diverges.
19. (Original post by Lord of the Flies)
Have a look at this. It's a bit slow at times but it is clear and he's quite a fun lecturer!
Indeed, that was quite good! I skipped bits of working and examples though
20. (Original post by bananarama2)
Indeed, that was quite good! I skipped bits of working and examples though
Ha, yes that's what I meant by "slow"

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