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The Proof is Trivial!

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Original post by matt2k8
That isn't true; e.g. take xn=(−1)n/nx_n=(-1)^n/n and yny_n to be 1/n1/n for even n, and 00 for odd nn. Now doubting my proof for the powers thing...


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.
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'm now doubting it for even taking powers; conditional convergence is a tricky thing to preserve, given that it relies on not just the signs and the decay of the sequence, but the order the sequence is taken in. Also because there is no obvious target limit the only real tool is if the sequence is Cauchy, but that's tricky to check when you don't have absolute convergence.

I think when restricting to just continuity, something almost piecewise linear would mean examples like the last one work
(edited 10 years ago)
Original post by james22
I wonder if it is true that if ff is a continuous function which is preserves signs, andf(x)≤x f(x)\leq x for all x,x, then ∑xn\sum x_n converges implies ∑f(xn)\sum f(x_n) converges.


False in general. Take f(x)=xf(x)=x everywhere except (0,1)(0,1) where f(x)=x2f(x)=x^2 and consider xn=1n(−1)nx_{n}=\frac{1}{n}(-1)^n

Guidance (spoiler alert):

Spoiler

(edited 10 years ago)
Original post by Lord of the Flies
False in general. Take f(x)=xf(x)=x everywhere except (0,1)(0,1) where f(x)=x2f(x)=x^2 and consider xn=1n(−1)nx_{n}=\frac{1}{n}(-1)^n

Guidance (spoiler alert):

Spoiler



That explains why I couldn't prove it then. I'll start looking for counter examples now. I'm guessing we need something with bigger +ve terms and smaller -ve terms which converges, but when taking powers of 2013 the smaller terms becomes so small the larger terms dominate and diverge.
Reply 2644
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henpen
Problem 434*

Find the area of a triangle with angles α,β,γ\alpha, \beta, \gamma drawn on a unit sphere.

I've always liked this one.
Original post by Lord of the Flies
False in general. Take f(x)=xf(x)=x everywhere except (0,1)(0,1) where f(x)=x2f(x)=x^2 and consider xn=1n(−1)nx_{n}=\frac{1}{n}(-1)^n

Guidance (spoiler alert):

Spoiler



Spoiler

Reply 2646
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IceKidd
Does anyone even update the OP anymore?
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und
OP
16
Original post by IceKidd
Does anyone even update the OP anymore?

There are reasons I don't. The first is that, since the URL's take up a lot of space in the posts, the original post has run out of space and the second post is about to. The second reason is that it's extremely time-consuming and I no longer have the time to update it. Some time ago jack.hadamard made a script that generated a list of questions, but this missed quite a few questions and solutions.
Original post by Omghacklol
Problem 429 *
Evaluate ii i^i , given that it is real.


i=eiπ2ii=e(i2π2)ii=e−π2ii≈0.2078 if ii is real\begin{aligned} &i=e^{i \frac{ \pi }{2}}\\ \\ &i^i=e^{(i^2 \frac{ \pi }{2})}\\ \\ &i^i=e^{- \frac{ \pi }{2}}\\ \\ &i^i \approx 0.2078 \: \text{if} \: i^i \: \text{is real} \end{aligned}
(edited 10 years ago)
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IceKidd
Original post by majmuh24
i=eiπ2ii=e(i2π2)ii=e−π2ii≈0.2078 if ii is real\begin{aligned} &i=e^{i \frac{ \pi }{2}}\\ \\ &i^i=e^{(i^2 \frac{ \pi }{2})}\\ \\ &i^i=e^{- \frac{ \pi }{2}}\\ \\ &i^i \approx 0.2078 \: \text{if} \: i^i \: \text{is real} \end{aligned}


technically that is just one value (principal value i guess you could say)

Unparseable latex formula:

\begin{aligned} &i=e^{i (\frac{ \pi }{2}+2n\pi})\\ \\



so theres infinitely many but thats the right thinking
Original post by IceKidd
technically that is just one value (principal value i guess you could say)

Unparseable latex formula:

\begin{aligned} &i=e^{i (\frac{ \pi }{2}+2n\pi})\\ \\



so theres infinitely many but thats the right thinking


Yeah, but the original question said given it is real :tongue:
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IceKidd
Original post by majmuh24
Yeah, but the original question said given it is real :tongue:


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?
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?


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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IceKidd
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.


yeah thats right. but remember you would get different values of i^i, infinitely many obviously
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].


So I was right with 0.2078 then my solution used the principal argument of i=e^(i*pi/2)?
Original post by majmuh24
So I was right with 0.2078 then my solution used the principal argument of i=e^(i*pi/2)?


Yep, in general zw=ewlog⁡zz^w=e^{w\log{z}} where log⁡z\log{z} is the principal complex logarithm defined as log⁡z=ln⁡∣r∣+iθ\log{z}=\ln|r|+i\theta where −π<θ≤π-\pi<\theta\le\pi.
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)
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].


It is not technically incorrect at all. The correct statement is to say that it is multivalued and give all values unless you are told to restrict arguments to a specific interval, or to state that it is some value when restricted to this interval. There is nothing wrong with the multivalued answer.

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)


The exponential function is not multi-valued; it is the logarithm that is multi-valued. Your example involving the square root of e is nonsense, and using the power series definition you can quite easily see what you are saying is wrong:

ex=∑n=0∞xnn! e^x = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n!}

This is an absolutely convergent series everywhere and is single-valued. The square root by definition implies the positive root which is why ± \pm is always written in the cases when both are used.
(edited 10 years ago)
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 "multi-valued functions" were illegal.


You would be. What I am saying is that you must either specify your range so that it is single-valued, or state that is multivalued.

I cannot just say sin−1(1)=π2 sin^{-1}(1) = \dfrac{\pi}{2} because of the periodicity; I need to state the arbitrary integer multiple of 2π 2 \pi 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 De-Moivre's) where this will miss solutions, and so it is not technically incorrect to state that is multi-valued. If you were using explicit properties of a well-defined function this may be different but that is not the case here. This simply isn't a function, but the multivalues are certainly legal.
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 "multi-valued functions" were illegal.


Arcsin(x) is typically defined to be the unique number, y, in the interval [-pi/2,pi/2] such that sin(y)=x. With the complex logorithm it is less easy to pick a definition to be used in all cases, because different times need different definitions. Typically it is defined on all of C except the negative reals and 0, and defined to be 0 at 1 (this is enough to completely define it), but it could equally be defined on all of C except the positive imaginary numbers.

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