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# The Proof is Trivial! Watch

Here is an interesting problem I encountered not too long ago: it is tough in my opinion, but the solution is very short.

Problem 454**/***

Consider where finite and for all . The question is whether there exists s.t.:

and
2. (Original post by Lord of the Flies)

Here is an interesting problem I encountered not too long ago: it is tough in my opinion, but the solution is very short.

Problem 454**/***

Consider where finite and for all . The question is whether there exists s.t.:

and
Well, if you don't need to be injective,

and . Obviously, , so , and . Thus there is some for which this property holds.

I guess I've made a mistake or misinterpreted the question.
3. (Original post by henpen)
...
Indeed you have misunderstood the question: it asks whether for any there exists such an , not whether you can find both and satisfying the conditions (as an aside: is the successor cardinal of and nothing more. It seems you believe ; this is an undecidable assertion).
4. Since this thread loves integration:

Problem 455 **/***

Let . Show that

, for

Problem 456 **/***

For find the value of

To make it easier, you may assume both the Tonelli and Fubini theorem hold for both.
5. (Original post by Lord of the Flies)

Here is an interesting problem I encountered not too long ago: it is tough in my opinion, but the solution is very short.

Problem 454**/***

Consider where finite and for all . The question is whether there exists s.t.:

and

(Original post by henpen)
Well, if you don't need to be injective,

and . Obviously, , so , and . Thus there is some for which this property holds.

I guess I've made a mistake or misinterpreted the question.
Solution 454

As stated in the original problem, the solution is very short.

A very short solution does not allow time to consider different cases.

Threfore the answer is either yes or no for all such functions.

Since we have found a function for which the answer is yes, the answer must be yes for all such functions.
6. Problem 457**

Find all automorphisms of the real numbers

i.e. find all functions

such that

and

Now do the same of the complex numbers***
7. (Original post by james22)
Problem 457**

Find all automorphisms of the real numbers

i.e. find all functions

such that

and

Now do the same of the complex numbers***
Done now - thanks, LotF!

Spoiler:
Show

There are two real automorphisms: the constant zero one, and the identity. There is a third complex one: the conjugate.

If any , then for all y, so the constant function f(x) = 0 works; otherwise, the only zero-value must be . From now on, assume we're not the constant zero function, and hence if f(x) = 0 then x=0.
by the two identities, for all , so . Similarly for all integer .

Now we want . Have . Solving this for gives . Hence for all integer n.

We can now hit all the algebraics: , so for every algebraic . Actually, hmm - this doesn't take the multiple-roots aspect into account. Better just stick with the rationals at the moment, for which it does hold. We do have that is irrational, because means so .

We prove (thanks, LotF!) that when . Indeed, .

Since when , we have that is increasing. Indeed, consider . Have so , so . Hence if we let be an arbitrary real, and a sequence of rational approximations increasing to the limit , then we have increasing. Moreover, if we let be a sequence of rational approximations tending down to , then is decreasing. We have , and hence .

The complexes bit: because has solutions , so also does. This gives us a single extra possible automorphism from moving into the complexes. This is consistent because of the fact that the sum/product of conjugates is the conjugate of the product/sum.

EDIT: As james22 points out, this is only true if we assume that a restriction to the reals of a complex automorphism is an automorphism.
8. (Original post by Smaug123)
...
Hint: for . Two possible routes now: prove that a solution to Cauchy's eq. satisfying either 1. increasing or 2. there exists a real interval on which is bounded below [the former case is easier to deal with in my opinion], must be
9. Spoiler:
Show
(Original post by jjpneed1)
if you assume the power-mean inequality I will murder you
(Original post by Tarquin Digby)

By the power-mean inequality...

(Original post by Flauta)
Problem 439**

Find

Nothing particularly ground breaking, just thought the graph looked pretty cool
Solution 439**

I haven't seen the graph yet, but Wolfram agrees with my final result

Spoiler:
Show

Spoiler:
Show
I was careful to check for any discontinuities when making the Weierstrass substitution!

(Original post by Lord of the Flies)
...
10. (Original post by Smaug123)
Done now - thanks, LotF!

Spoiler:
Show

There are two real automorphisms: the constant zero one, and the identity. There is a third complex one: the conjugate.

If any , then for all y, so the constant function f(x) = 0 works; otherwise, the only zero-value must be . From now on, assume we're not the constant zero function, and hence if f(x) = 0 then x=0.
by the two identities, for all , so . Similarly for all integer .

Now we want . Have . Solving this for gives . Hence for all integer n.

We can now hit all the algebraics: , so for every algebraic . Actually, hmm - this doesn't take the multiple-roots aspect into account. Better just stick with the rationals at the moment, for which it does hold. We do have that is irrational, because means so .

We prove (thanks, LotF!) that when . Indeed, .

Since when , we have that is increasing. Indeed, consider . Have so , so . Hence if we let be an arbitrary real, and a sequence of rational approximations increasing to the limit , then we have increasing. Moreover, if we let be a sequence of rational approximations tending down to , then is decreasing. We have , and hence .

The complexes bit: because has solutions , so also does. This gives us a single extra possible automorphism from moving into the complexes. This is consistent because of the fact that the sum/product of conjugates is the conjugate of the product/sum.

Nice solution. Although you are wrong about the automorphisms of the complex numbers. It was a bit of a trick question because you have found all the contructablle automorphisms, but the set of automorphisms of is the same size as the power set of the reals.

EDIT: If you assume that the automorphisms must be continuous, you have them all.

EDIT 2: Your mistake with the complex numbers bit is you assume that the restriction of an automorphism of to is an automorphism of , this is not true. In your proof for you assumed that y^2>0 for all y, but this may not be true in .
11. (Original post by james22)
Nice solution. Although you are wrong about the automorphisms of the complex numbers. It was a bit of a trick question because you have found all the contructablle automorphisms, but the set of automorphisms of is the same size as the power set of the reals.

EDIT: If you assume that the automorphisms must be continuous, you have them all.

EDIT 2: Your mistake with the complex numbers bit is you assume that the restriction of an automorphism of to is an automorphism of , this is not true. In your proof for you assumed that y^2>0 for all y, but this may not be true in .
Ah, that did twinge in my mind at the time, but I dismissed it because I was thinking about increasing-ness I loved the problem, though!
12. (Original post by Smaug123)
Ah, that did twinge in my mind at the time, but I dismissed it because I was thinking about increasing-ness I loved the problem, though!
Try and find one that is not continuous (hint: you need the axiom of hoice, in particular zorns lemma).
13. Problem 458***

For show that

14. (Original post by Lord of the Flies)
Well done I'll type it out for you (credit to Izzy in the OP):

Hence the limit is

Replacing with obviously makes no difference.
how does one evaluate the last line to get ln2?
15. (Original post by Khallil)
Spoiler:
Show

Solution 439**

I haven't seen the graph yet, but Wolfram agrees with my final result

Spoiler:
Show

Spoiler:
Show
I was careful to check for any discontinuities when making the Weierstrass substitution!

nice solution though i think a more basic one would be to change the sin^2 in the denominator into 1-cos^2 and let u=cosu, then a little partial fractions and were done.
16. (Original post by newblood)
nice solution though i think a more basic one would be to change the sin^2 in the denominator into 1-cos^2 and let u=cosu, then a little partial fractions and were done.
I noticed that way but I really wanted to bash some of the algebra out in the integrand with the Weierstrass sub
17. Anyone like stats? Here are a couple of questions i particularly liked that id done recently [not very hard, especially for someone whos done any uni stats or S3/4 even]:

Problem 459**
A fair coin is tossed n times. Let un be the probability that the sequence of tosses never has two consecutive heads. Show that u_n =0.5u_(n-1)+0.25u_(n-2) .Find also u_n

Problem 460*
The radius of a circle has the exponential distribution with parameter lambda. What is the probability density function of the area of the circle.
18. (Original post by newblood)
How does one evaluate the last line to get ln 2?
19. (Original post by Lord of the Flies)
Why are you justified in swapping the order of limits here?
20. (Original post by james22)
Why are you justified in swapping the order of limits here?
if youre referring to the original integral, i think he also changed the signs as well to change the limits, just for ease of manipulation id imagine

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