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# The Proof is Trivial! Watch

1. Just came up with a really nice result (can I blag it as a theorem? )

Problem 152**

Given that and for real numbers such that , find a condition on S for which P is irrational.

Spoiler:
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Note: To find a complete condition on P would require solving one of "the great unsolved problems" in mathematics. For this problem, you must find all conditions on P that are you can and then prove this assertion from first principles
2. For anyone taking Edexcel Statistics 4...

Problem 153
*/**

Prove that a t-distribution with degree of freedom is asymptotically equivalent a the Normal distribution for sample size such that , stating the parameters of the Normal distribution to which it is equivalent.

Hint:
Spoiler:
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Would be awesome if someone derived this from definitions, but if not:

It can be shown that , where denotes the probability density function of a t-distribution with degree of freedom

Note: You may not use standard values/results for the gamma function. First principles por favor

Spoiler:
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Also... Is this so-called hint just an over complication? Perhaps....
3. Problem 154 / *

Let be positive, non-consecutive integers. Let

for .

Prove that, for all , the interval contains a perfect square.
4. I did some really nice problems today and thought I would share them in an attempt to get people interested in this thread again:

Problem 155*

Find (without calculus) a fifth degree polynomial such that is divisible by and is divisible by .

Spoiler:
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Not too bad but is very reminiscent of a creation STEP question, but without being permitted (let alone led into) the simple differentiation approach. So is going to be good STEP preparation!

Problem 156*/**/***

Evaluate , where a is a real number, in the case such that .

Problem 157*/**/***

Evaluate

Problem 158*/**/***

Evaluate

and finally, because there's no point in me simply typing my solution up (especially since no-one seems too interested anyway)...

Problem 159**

Evaluate , where n is a positive integer, in terms of a combination of a finite number of elementary functions.
Problem 154 / *

Let be positive, non-consecutive integers. Let

for .

Prove that, for all , the interval contains a perfect square.
Solution 154

As they are non-consecutive integers, the difference between to consecutive terms in at least 2.

For the size of the interval,

Similarly, .

From this we can deduce that .

Therefore, as the interval is always greater than or equal to and the smallest term in the interval is always greater than or equal to , it follows trivially that the interval is always larger than or equal to the difference between any two square numbers.

Therefore, the interval must contain a perfect square.

Edit: Where's this problem from btw? It was rather neat
6. Solution 156

We denote . Clearly, . Furthermore, for , . Now, . Hence, we obtain , implying for .
Let . So, . Therefore, for , we have .

Solution 157

We let , and then . Thus obtaining . More generally, .

Solution 158

For , we define . is continuous, with respect to , for . Thus . Thence, we have .

Solution 159

Notice .
We need to treat the cases only.
If , differentiating ( times), we get . The case is done analogically.
Hence, .

By the way, Borwein have proved far more general result.
Fix complex numbers .
Define , ; , and .
It follows that .
Hence in our case, when , we have .

May I ask, why are we not allowed to use calculus in problem 155 in order to solve it in a minute?
Spoiler:
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It is enough to note that is divisible by and .
May I ask, why are we not allowed to use calculus in problem 155 in order to solve it in a minute?
Spoiler:
Show
It is enough to note that is divisible by and .
Because the way which uses calculus was part of/similar to a previous STEP question that has obviously been solved many times on this forum.
8. Accccch why did I have to miss the integrals? Well done on 159 Mlad, that's a nice solution.
Spoiler:
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Solution 156

We denote . Clearly, . Furthermore, for , . Now, . Hence, we obtain , implying for .
Let . So, . Therefore, for , we have .

Solution 157

We let , and then . Thus obtaining . More generally, .

Solution 158

For , we define . is continuous, with respect to , for . Thus . Thence, we have .

Solution 159

Notice .
We need to treat the cases only.
If , differentiating ( times), we get . The case is done analogically.
Hence, .

By the way, Borwein have proved far more general result.
Fix complex numbers .
Define , ; , and .
It follows that .
Hence in our case, when , we have .

May I ask, why are we not allowed to use calculus in problem 155 in order to solve it in a minute?
Spoiler:
Show
It is enough to note that is divisible by and .
He returns

You used the same method I did for Problem 158 but the rest I did differently. I used differentiation under the integral sign for 156 which pops the solution out almost instantly (yours looks a little complicated!) Really like the elegance in 157 (I simply used the same approach for that one - was fairly simple but surely took a little longer than yours!)

And I don't see what's going on in 159. Where does the first line come from? And how does it link to the second and third? The differentiation thing doesn't seem to make sense (though I'm sure it does), could you clarify it for me My solution looks so different to yours, it's in you form of two symmetrical series expression involving all the square numbers (and no factorials!) (I checked it though and it does work!)

And yeah, as said above, it's far too easy and boring to do it that way

Please post a load of really nice integrals! (None of that crap involving 2013s and stuff where you're never going to come across a similar thing )

This one is really nice:

Problem 160*/**/***

Evaluate
10. (Original post by Lord of the Flies)
Accccch why did I have to miss the integrals? Well done on 159 Mlad, that's a nice solution.
Post some! Post loads! I hereby declare the 29th May TSR Integration Day
11. Very nice Mlad! I'm too slow again
solution 156

the second integral can be seen to be zero by expanding the logarithm (|a|>1 guarantees convergence) and each term integrated is still just a function of e^ix which is 2pi periodic.
<insert rigour here>
Not sure if this works but I thought I'd give it a go
12. Solution 160

Alternate solution to previous integrals:

Solution 158

Solution 157

13. (Original post by ben-smith)
Very nice Mlad! I'm too slow again
solution 156

the second integral can be seen to be zero by expanding the logarithm (|a|>1 guarantees convergence) and each term integrated is still just a function of e^ix which is 2pi periodic.
<insert rigour here>
Not sure if this works but I thought I'd give it a go
Looks decent to me!

I might as well add my solution into the soup ...

Solution 156 (3)

Let

Using the substitution :
,
for |a|>1.

...also, , for |a|<1 (as the arctangent would take a different value).
Spoiler:
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Or I suppose you could just say

So as f(1)=0, for |a|>1.
14. (Original post by Lord of the Flies)
Solution 160

Solution 157
Exactly what I did

160 is from the Putnam exam (undergrad comp for the USA) and only 20 of the top 200 scorers got more than 2/10 marks on that question! No idea why this method isn't widely taught/used!

POST SOME INTEGRALS!!!!!!!!!!
15. (Original post by Jkn)
As f(0)=0, for |a|>1.
This doesn't work, the integral clearly does not converge. You need to show that f(1) = 0 and integrate from 1 to a.
16. (Original post by Lord of the Flies)
This doesn't work, the integral clearly does not converge. You need to show that f(1) = 0 and integrate from 1 to a.
Typo fixed!
17. Problem 161**/***

Let k be an integer greater than 1. Suppose and for

Evaluate
18. Problem 162 *

If I drop a ball off the Eiffel tower facing east. Where does it land? (Neglect air resistance)
19. (Original post by bananarama2)
Problem 162 *

If I drop a ball off the Eiffel tower facing east. Where does it land? (Neglect air resistance)
On the ground below. (?!?!?)
20. (Original post by Jkn)
Solution 162

On the ground below. (?!?!?)
Directly below?

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