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What kind of values would Sin(n) take where whatever (n) is ?

- #3

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All values from -1 to 1, including these values, and then you would raise all those values to ^(1/n), were n is all values from 1 to infinity including 1, still unsure why though

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http://www.wolframalpha.com/input/?i=lim+n-%3Einf+sin%28n%29^%281%2Fn%29

it claims that lim n->inf sin(n)^(1/n)=1

- #5

lanedance

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now do you need to prove it or just convince yourself qualitatively?

first what is the limit of

[tex] \lim_{n \to \infty}(-1)^{1/n}[/tex]

in fact for any number a, not zero, what is

[tex] \Lim_{n \to \infty}(a)^{1/n}[/tex]

another thing that is instructive to do might be to consider the magnitude in the limit

The problems with sin will be around the zero, or n = k.pi, for integers k. For large n the exponent will "squeeze" these points of divergence to an infinitesimal region

- #6

dynamicsolo

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It is correct to say that you will need the "Squeeze Theorem" in some way, because L'Hopital's Rule will be useless since [itex]\lim_{n \rightarrow \infty} \sin(n) [/itex] does not exist.

now do you need to prove it or just convince yourself qualitatively?

first what is the limit of

[tex] \lim_{n \to \infty}(-1)^{1/n}[/tex]

in fact for any number a, not zero, what is

[tex] \Lim_{n \to \infty}(a)^{1/n}[/tex]

another thing that is instructive to do might be to consider the magnitude in the limit

The problems with sin will be around the zero, or n = k.pi, for integers k. For large n the exponent will "squeeze" these points of divergence to an infinitesimal region

However, you will need to avoid using the problematic function [itex]a^{1/n}[/itex] , with a < 0 : it has a countably infinite number of undefined points, those being where n is even. (I notice, in fact, that Grapher won't even attempt y = (-1)

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lim n->inf a^(1/n) = 1

so

lim n->inf sin(n)^(1/n) = 1

but I wasn't so sure about that because lim n->inf sin(n) is undefined but would contain some value between -1<= sin(inf) <=1, at least I think

and ya interestingly enough for some reason (-1)^(1/inf) is -1 and not positive one, not really sure why, (-1)^(1/inf) should be equal to (-1)^0 and should be equal to one 1 but it's not and actually negative one, not really sure why it's -1 and not 1, 0^(1/inf)=0^0 is undefined as well.

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Dick

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- #9

Dick

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Why would you think lim n->inf (-1)^(1/n)=(-1)?? First of all you have to DEFINE what (-1)^(1/n) means since there are n different nth roots. By making a sufficiently bizarre choice I think you can make the limit come out to any complex number z such that |z|=1. The usual principal root choice would make it +1.

lim n->inf a^(1/n) = 1

so

lim n->inf sin(n)^(1/n) = 1

but I wasn't so sure about that because lim n->inf sin(n) is undefined but would contain some value between -1<= sin(inf) <=1, at least I think

and ya interestingly enough for some reason (-1)^(1/inf) is -1 and not positive one, not really sure why, (-1)^(1/inf) should be equal to (-1)^0 and should be equal to one 1 but it's not and actually negative one, not really sure why it's -1 and not 1, 0^(1/inf)=0^0 is undefined as well.

- #10

dynamicsolo

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In the intervals where this function

I will again point to the deficiencies of computer algebra/calculus systems in the fact that Wolfram Alpha actually claims a limit at infinity for this function. You really shouldn't take the word of a "chunk of code" too seriously...

- #11

lanedance

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I think if you make some assumptions you can start to justify a limit,

so following on from Dick and trying to deal with points on multivalued functions, if you put in the following caveats:

- consider finding the limit of the series: sin(n)^(1/n) for n = 1,2,3,...

- as exponentiating to a fraction is in general a multivalued operation, choose the principal root

the fact any integer n is never equal to k.pi removes the issues with zeroes of sin(n)

choosing the principal root means removing the issues with a multi-valued function

in the previous post I was not laying out a blue-print for a comprehensive proof, but asking the OP what they wanted to achieve, and hopefully giving some insights that should help find them to a way to approach the problem

so for the limit

[tex] \lim_{n \to \infty }(-1)^\frac{1}{n}[/tex]

why not consider [itex] -1 = e^{i \pi}[/itex], whilst acknowledging this is "choosing" a branch when the operation is multi-valued , which I'll return to later.

[tex]\lim_{n\to \infty }(-1)^\frac{1}{n}

=\lim_{n \to \infty }(e^{i \pi})^\frac{1}{n}

=\lim_{n \to \infty }e^{i\frac{\pi}{n}}

[/tex]

in this form the limit becomes clear

now in terms of the multivalued function, for the case [itex] -1^{\frac{1}{n}}[/itex] it is true we could choose [itex] -1 = e^{i (2k-1) \pi}[/itex] for k = 1,2,3... which gives [tex]\lim_{n\to \infty }(-1)^\frac{1}{n}

=\lim_{n \to \infty }(e^{i (2k-1) \pi})^\frac{1}{n}

=\lim_{n \to \infty }e^{i\frac{(2k-1) \pi}{n}} = 1, \ for \ \ finite \ \ k

[/tex]

It only really makes sense choosing k such that 0<(2k-1)<n, and if the series starts from 1, the natural choice is probably k=1. And for any choice of finite k the limit still tends to 1.

In fact if you think of it as a function [itex] f(x) = (-1)^{\frac{1}{x}}[/itex] then changing this k value would in fact introduce discontinuities into what is a continuous function with f(1) = -1. That forces us to set a finite k value, in effect choosing a single-valued branch of the function. All the paths have f(1) = -1, and appear to tend to 1 for large n.

If you consider plotting the function in the complex plane along the x axis, in a pseudo-3D sense, then I think they are are different "stretched spirals" in the complex plane with starting at f(1) = -1 converging on 1 for large x, with |f(x)|=1. The number of half revolutions of the x-axis is (2k-1).

All that said its been a while since I've played with multi-valued functions, so feel free to look for any mis-assumptions/inconsistencies.

Now if you consider that n is an integer, then we know sin(n) is non-zero and for any n we for some real a, with 0<a<1 we can can write:

[tex] sin(n) = a [/tex]

or

[tex] sin(n) = a(-1) [/tex]

so following on from Dick and trying to deal with points on multivalued functions, if you put in the following caveats:

- consider finding the limit of the series: sin(n)^(1/n) for n = 1,2,3,...

- as exponentiating to a fraction is in general a multivalued operation, choose the principal root

the fact any integer n is never equal to k.pi removes the issues with zeroes of sin(n)

choosing the principal root means removing the issues with a multi-valued function

in the previous post I was not laying out a blue-print for a comprehensive proof, but asking the OP what they wanted to achieve, and hopefully giving some insights that should help find them to a way to approach the problem

so for the limit

[tex] \lim_{n \to \infty }(-1)^\frac{1}{n}[/tex]

why not consider [itex] -1 = e^{i \pi}[/itex], whilst acknowledging this is "choosing" a branch when the operation is multi-valued , which I'll return to later.

[tex]\lim_{n\to \infty }(-1)^\frac{1}{n}

=\lim_{n \to \infty }(e^{i \pi})^\frac{1}{n}

=\lim_{n \to \infty }e^{i\frac{\pi}{n}}

[/tex]

in this form the limit becomes clear

now in terms of the multivalued function, for the case [itex] -1^{\frac{1}{n}}[/itex] it is true we could choose [itex] -1 = e^{i (2k-1) \pi}[/itex] for k = 1,2,3... which gives [tex]\lim_{n\to \infty }(-1)^\frac{1}{n}

=\lim_{n \to \infty }(e^{i (2k-1) \pi})^\frac{1}{n}

=\lim_{n \to \infty }e^{i\frac{(2k-1) \pi}{n}} = 1, \ for \ \ finite \ \ k

[/tex]

It only really makes sense choosing k such that 0<(2k-1)<n, and if the series starts from 1, the natural choice is probably k=1. And for any choice of finite k the limit still tends to 1.

In fact if you think of it as a function [itex] f(x) = (-1)^{\frac{1}{x}}[/itex] then changing this k value would in fact introduce discontinuities into what is a continuous function with f(1) = -1. That forces us to set a finite k value, in effect choosing a single-valued branch of the function. All the paths have f(1) = -1, and appear to tend to 1 for large n.

If you consider plotting the function in the complex plane along the x axis, in a pseudo-3D sense, then I think they are are different "stretched spirals" in the complex plane with starting at f(1) = -1 converging on 1 for large x, with |f(x)|=1. The number of half revolutions of the x-axis is (2k-1).

All that said its been a while since I've played with multi-valued functions, so feel free to look for any mis-assumptions/inconsistencies.

Now if you consider that n is an integer, then we know sin(n) is non-zero and for any n we for some real a, with 0<a<1 we can can write:

[tex] sin(n) = a [/tex]

or

[tex] sin(n) = a(-1) [/tex]

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- #12

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As n approaches infinity 1/n approaches 0 and anything to the 0 is equal to 1

- #13

dynamicsolo

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Well, not just anything: if the base "grows as fast" as the exponent is shrinking, you can get a constant thatAs n approaches infinity 1/n approaches 0 and anything to the 0 is equal to 1

- #14

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Ah, thankyou :) Sorry about that GreenPrint

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