Ach, this thread is overloaded with calculus!
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.:

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(Original post by Lord of the Flies)
Ach, this thread is overloaded with calculus!
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 . Obviously, , so , and . Thus there is some for which this property holds.
I guess I've made a mistake or misinterpreted the question.Last edited by henpen; 25032014 at 23:04. 
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(Original post by henpen)
... 
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 01042014 10:37
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. 
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 05042014 22:19
(Original post by Lord of the Flies)
Ach, this thread is overloaded with calculus!
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.:
(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.
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. 
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 05042014 22:26
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*** 
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 05042014 23:20
(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***
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 zerovalue 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 multipleroots 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.
Last edited by Smaug123; 06042014 at 10:47. Reason: Done 
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 05042014 23:36
(Original post by Smaug123)
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 06042014 01:11
Spoiler:Show(Original post by jjpneed1)
if you assume the powermean inequality I will murder you
(Original post by Flauta)
Problem 439**
Find
Nothing particularly ground breaking, just thought the graph looked pretty cool
I haven't seen the graph yet, but Wolfram agrees with my final result
Spoiler:Show
(Original post by Lord of the Flies)
...Last edited by Khallil; 06042014 at 08:21. 
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 06042014 02:56
(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 zerovalue 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 multipleroots 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: 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 .Last edited by james22; 06042014 at 03:05. 
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 06042014 10:48
(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 . 
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(Original post by Smaug123)
Ah, that did twinge in my mind at the time, but I dismissed it because I was thinking about increasingness I loved the problem, though! 
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Last edited by Indeterminate; 08042014 at 23:56. 
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 15042014 21:23
(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. 
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 15042014 21:33
(Original post by Khallil)
Spoiler:Show
Solution 439**
I haven't seen the graph yet, but Wolfram agrees with my final result
Spoiler:Show
Loving your new avatar 
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 15042014 21:36
(Original post by newblood)
nice solution though i think a more basic one would be to change the sin^2 in the denominator into 1cos^2 and let u=cosu, then a little partial fractions and were done. 
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 15042014 22:04
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_(n1)+0.25u_(n2) .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.Last edited by newblood; 15042014 at 22:32. 
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 15042014 22:19
(Original post by newblood)
How does one evaluate the last line to get ln 2? 
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 15042014 22:33
(Original post by james22)
Why are you justified in swapping the order of limits here?
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