Dammit, now that I think about it I was obviously wrong.(Original post by matt2k8)
That isn't true; e.g. take and to be for even n, and for odd . Now doubting my proof for the powers thing...
I wonder if it is true that if f(x) is a continuous function which is preserves signs, and f(x)<=x for all x, then sum of x_n converges implies sum of f(x_n) converges.

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 31122013 00:17

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 31122013 00:50
(Original post by james22)
Dammit, now that I think about it I was obviously wrong.
I wonder if it is true that if f(x) is a continuous function which is preserves signs, and f(x)<=x for all x, then sum of x_n converges implies sum of f(x_n) converges.
I think when restricting to just continuity, something almost piecewise linear would mean examples like the last one workLast edited by matt2k8; 31122013 at 02:35. 
Lord of the Flies
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 31122013 03:42
(Original post by james22)
I wonder if it is true that if is a continuous function which is preserves signs, and for all then converges implies converges.
Guidance (spoiler alert):
Spoiler:ShowThe claim in 428 is true, and hence the answer to both parts of 420 is no. The former is substantially tougher to prove.Last edited by Lord of the Flies; 31122013 at 04:08. Reason: spoiler alert 
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 31122013 12:33
(Original post by Lord of the Flies)
False in general. Take everywhere except where and consider
Guidance (spoiler alert):
Spoiler:ShowThe claim in 428 is true, and hence the answer to both parts of 420 is no. The former is substantially tougher to prove. 
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 31122013 17:38
Problem 434*
Find the area of a triangle with angles drawn on a unit sphere.
I've always liked this one. 
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 01012014 18:38
(Original post by Lord of the Flies)
False in general. Take everywhere except where and consider
Guidance (spoiler alert):
Spoiler:ShowThe claim in 428 is true, and hence the answer to both parts of 420 is no. The former is substantially tougher to prove.Spoiler:ShowAh that makes a lot more sense... I think I have a way to do it when the series is allowed to be complex valued and the complement of S is finite, but I don't know if my construction extends to real valued/ having infinite S complement (my idea hinges around things like for suitable ), or the time/motivation to check it :P 
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 01012014 19:57
Does anyone even update the OP anymore?

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 01012014 20:34
(Original post by IceKidd)
Does anyone even update the OP anymore? 
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 03012014 15:37
Last edited by Arithmeticae; 03012014 at 15:40. 
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 03012014 15:51
so theres infinitely many but thats the right thinking 
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 03012014 15:52
(Original post by IceKidd)
technically that is just one value (principal value i guess you could say)
so theres infinitely many but thats the right thinking 
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 03012014 15:59
(Original post by majmuh24)
Yeah, but the original question said given it is real
For example if you had the argument as 5pi/2 then it is e^(5pi/2)
what makes you think that isnt real? 
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 03012014 16:03
(Original post by IceKidd)
they would still also be real....
For example if you had the argument as 5pi/2 then it is e^(5pi/2)
what makes you think that isnt real? 
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 03012014 16:05
(Original post by majmuh24)
It is, but that just happened to be the first one I got because I cba to put lots of different values in, but I get what you mean because you can just go up in intervals of 2pi every time and you would still get the same value as all these values correspond to i. 
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 04012014 15:07
(Original post by Tarquin Digby)
It doesn't have infinitely many values, saying i^i = e^5pi/2 or whatever is technically incorrect. When taking complex exponents, you are supposed to use the principal argument in the interval (pi,pi]. 
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 04012014 15:12
(Original post by majmuh24)
So I was right with 0.2078 then my solution used the principal argument of i=e^(i*pi/2)? 
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 05012014 14:21
i^i does have infinitely many values because e^x is a multi valued function. For instance e^0.5 = sqrt(e) and sqrt(e).
i^i = e^(i Log i) = e^(i*( ln(i) + 2Pi*k*i ))
"Here ln(x) is the principle value of the logarithm whereas Log(x) is the SET of possible values of the logarithm. I.e. Log(z) = ln(z) +2Pi*k*i where k is an integer. "
So continuing with our derivation:
i^i = e^(i*( ln(i) + 2Pi*k*i )) = e^(i*(i*Pi/2 + 2*Pi*k*i)) = e^((Pi(2k + 1/2))
Of which e^(5Pi/2) is only one value
Now as an interesting follow on question, what are the possible values of 1^sqrt(2) ? (It is not JUST 1 btw) 
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 05012014 14:54
(Original post by Tarquin Digby)
It doesn't have infinitely many values, saying i^i = e^5pi/2 or whatever is technically incorrect. When taking complex exponents, you are supposed to use the principal argument in the interval (pi,pi].
(Original post by Elie Bergman)
i^i does have infinitely many values because e^x is a multi valued function. For instance e^0.5 = sqrt(e) and sqrt(e).
i^i = e^(i Log i) = e^(i*( ln(i) + 2Pi*k*i ))
"Here ln(x) is the principle value of the logarithm whereas Log(x) is the SET of possible values of the logarithm. I.e. Log(z) = ln(z) +2Pi*k*i where k is an integer. "
So continuing with our derivation:
i^i = e^(i*( ln(i) + 2Pi*k*i )) = e^(i*(i*Pi/2 + 2*Pi*k*i)) = e^((Pi(2k + 1/2))
Of which e^(5Pi/2) is only one value
Now as an interesting follow on question, what are the possible values of 1^sqrt(2) ? (It is not JUST 1 btw)
This is an absolutely convergent series everywhere and is singlevalued. The square root by definition implies the positive root which is why is always written in the cases when both are used.Last edited by DJMayes; 05012014 at 14:59. 
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 05012014 15:09
(Original post by Tarquin Digby)
You're not allowed to do that with say the inverse trig functions though. I'd be wrong if I said arcsin(1) = 5pi/2, wouldn't I? Why is it ok with complex logs? I thought "multivalued functions" were illegal.
I cannot just say because of the periodicity; I need to state the arbitrary integer multiple of unless you restrict the range. You are correct in saying that it is customary to just use the principal value, but there are many cases (e.g. finding complex roots using DeMoivre's) where this will miss solutions, and so it is not technically incorrect to state that is multivalued. If you were using explicit properties of a welldefined function this may be different but that is not the case here. This simply isn't a function, but the multivalues are certainly legal. 
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 05012014 15:44
(Original post by Tarquin Digby)
You're not allowed to do that with say the inverse trig functions though. I'd be wrong if I said arcsin(1) = 5pi/2, wouldn't I? Why is it ok with complex logs? I thought "multivalued functions" were illegal.
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