- #1

ZeroPivot

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lim x->o sin(nx)/x = n for n E real numbers

i havent seen it in any books but it works.

i havent seen it in any books but it works.

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- Thread starter ZeroPivot
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- #1

ZeroPivot

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lim x->o sin(nx)/x = n for n E real numbers

i havent seen it in any books but it works.

i havent seen it in any books but it works.

- #2

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$$\lim_{y \rightarrow 0} \left( \frac{\sin(y)}{y/n}\right) = \lim_{y \rightarrow 0}\left( n \frac{\sin(y)}{y}\right) = n \left(\lim_{y \rightarrow 0} \frac{\sin(y)}{y}\right)$$

The limit inside the parentheses on the right hand side is standard, equal to ##1##. (Proof?)

- #3

Mark44

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It can be derived from this limit:lim x->o sin(nx)/x = n for n E real numbers

i havent seen it in any books but it works.

$$\lim_{x \to 0}\frac {sin(x)}{x} = 1$$

$$\lim_{x \to 0}\frac{sin(nx)}{x} = \lim_{x \to 0}\frac{n \cdot sin(nx)}{nx} $$

$$= n \lim_{u \to 0}\frac{sin(u)}{u} = n\cdot 1$$

- #4

ZeroPivot

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so im a genuise right for inventing it?

- #5

arildno

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Usually, potential candidates for the Fields Medal must wait to see if it is awarded to them.so im a genuise right for inventing it?

- #6

Mark44

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I've seen this problem in several calculus books, so I think you're premature in saying that you invented it.so im a genuise right for inventing it?

- #7

ZeroPivot

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I've seen this problem in several calculus books, so I think you're premature in saying that you invented it.

problem yes, but general standard solution no. they just say lim x->o sint/t = 1

my way is far more efficient.

- #8

Mark44

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- #9

ZeroPivot

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many people have done many things but since i have not seen anywhere

lim x->o sin(nx)/x = n

you can say that I invented it. its like you know dancing many people could have foxtrotted but its the person that calls it foxtrot and creates a perimeter of foxtrot he/she is the inventor.

The GREAT thing about my standard limit is that its FAR superior to the wellknown standard limit that is lim x->0 sinx/x = 1 this is far inferior to my limit.

- #10

arildno

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It is not "your" limit. It is a completely trivial corollary.

- #11

ZeroPivot

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It is not "your" limit. It is a completely trivial corollary.

show me where lim x->0 sin(nx)/x = n is written?

i feel alot of hostility here i dont know why, i thought this forum was about comradery and discussing ideas about science without persecution.

- #12

Office_Shredder

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Furthermore, not that it's terribly relevant, but your result is stated in this here yahoo answer:

http://answers.yahoo.com/question/index?qid=20090919181027AADfJkR

lim x->0 sin(nx)/nx = 1

A genuine mathematical discovery is when you can answer a problem that people have been unable to answer before; answering a problem that people simply haven't bothered to write down before isn't a discovery, it's just applying some math to a problem.

- #13

Mark44

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- #14

ZeroPivot

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All im saying is that the standard limit: lim x->0 sin(nx)/x = n is FAAAAAAR superior than lim x->0 sin(x)/x = 1

its not trivial at all. the yahoo link provided above is not my Standard Limit and its not as simple and elegant.

I think people here should be more open minded about discoveries you know alot of scientist work so far they become nearsighted to great solutions and things do slip the system.

- #15

Mark44

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You are confused if you think what has transpired here falls under the category of "hate." No one has said anything of a personal nature about you. We have made valid points that contradict your claim of inventing something.Ok we have a lot of haters here today.

As already stated, "your" limit is well known, and is an easy corollary of the sin(x)/x limit.All im saying is that the standard limit: lim x->0 sin(nx)/x = n is FAAAAAAR superior than lim x->0 sin(x)/x = 1

its not trivial at all. the yahoo link provided above is not my Standard Limit and its not as simple and elegant.

I think people here should be more open minded about discoveries you know alot of scientist work so far they become nearsighted to great solutions and things do slip the system.

Since you don't have any more to offer, I am closing this thread.

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