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## Main Question or Discussion Point

Is there a way to simplify this? Is this a known function?

tan(arccos(((x/y)-1/y )/ ((x/y)+ 1/y)))*((x/2)-1/2)=sqrt(x)

tan(arccos(((x/y)-1/y )/ ((x/y)+ 1/y)))*((x/2)-1/2)=sqrt(x)

- Thread starter JeremyEbert
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- #1

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Is there a way to simplify this? Is this a known function?

tan(arccos(((x/y)-1/y )/ ((x/y)+ 1/y)))*((x/2)-1/2)=sqrt(x)

tan(arccos(((x/y)-1/y )/ ((x/y)+ 1/y)))*((x/2)-1/2)=sqrt(x)

- #2

tiny-tim

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Hi Jeremy!

Start with tan(arccos(z)) …

Start with tan(arccos(z)) …

what would that be?

- #3

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oh yea...z = (x-1)/(x+1)Hi Jeremy!

Start with tan(arccos(z)) …

what would that be?

is there a simple explination for this:

e^(-2/n) ~ (n-1)/(n+1)

where is in the inverse natural log constant 2.71828182845904523536028747135266249....

- #4

tiny-tim

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what are you talking about?oh yea...z = (x-1)/(x+1)

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what are you talking about?

just saying the messy part of my original equation is:

"((x/y)-1/y )/ ((x/y)+ 1/y)"

and it basically equals this:

(x-1)/(x+1) which is the z part of tan(arccos(z)) right?

- #6

tiny-tim

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oh Ijust saying the messy part of my original equation is:

"((x/y)-1/y )/ ((x/y)+ 1/y)"

and it basically equals this:

(x-1)/(x+1) which is the z part of tan(arccos(z)) right?

that was so difficult to read that I didn't recognise it!

ok, now go back to tan(arccos(z)) … for any z … what would that be?

(alternatively, (x-1)/(x+1) is a fairly familiar formula …

if A = (x-1)/(x+1), what does (A-1)/(A+1) equal?)

- #7

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I see... tan(arccos(z)) = sqrt(1-z^2)/zoh Isee…

that was so difficult to read that I didn't recognise it!

ok, now go back to tan(arccos(z)) … for any z … what would that be?

(alternatively, (x-1)/(x+1) is a fairly familiar formula …

if A = (x-1)/(x+1), what does (A-1)/(A+1) equal?)

and

if A = (x-1)/(x+1) then (A-1)/(A+1) = 1/-x or (A+1)/(A-1)=x

what about e^(-2/x) converging to A?

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- #8

tiny-tim

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Let's do one thing at a time …what about e^(-2/x) converging to A?

now solve tan(arccos((x-1)/(x+1)))

- #9

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well I know that tan(arccos((x-1)/(x+1))) = sqrt(x)/((x/2)+1/2)Let's do one thing at a time …

now solve tan(arccos((x-1)/(x+1)))

but I'm sure thats not what your looking for...hints? sorry... I'm new at this.

- #10

tiny-tim

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Is there a way to simplify this? Is this a known function?

tan(arccos(((x/y)-1/y )/ ((x/y)+ 1/y)))*((x/2)-1/2)=sqrt(x)

I assumed you wanted to prove the equation in your first post …well I know that tan(arccos((x-1)/(x+1))) = sqrt(x)/((x/2)+1/2)

but I'm sure thats not what your looking for...hints? sorry... I'm new at this.

have you worked out how to

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I have not worked out how to prove this. I have never done a proof before. This is a small peice to a large puzzle I''ve been working on. Thank you so much for helping me so far, I'm understanding things much better. Please continue showing me how to prove this.I assumed you wanted to prove the equation in your first post …

have you worked out how toprovetan(arccos((x-1)/(x+1))) = sqrt(x)/((x/2)+1/2) ?

- #12

tiny-tim

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Carry on from …

I see... tan(arccos(z)) = sqrt(1-z^2)/z

- #13

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Sorry for the delay. My furnace went out and it was -3 here.... Fun times.Carry on from …

I'm going to try and explain the equation that I am looking for. I think I have a understanding of whats going on with the tan(arccos(z)) = sqrt(1-z^2)/z. Its

obvious that its just the pythagorean theorem with the hypotenuse=1 and adjecent=z.

I have an attachment (http://3.bp.blogspot.com/-5UhMF-uGw...AFQ/oDdl_oSXPM0/s1600/prime-+squares+edit.png) that is a visual representation of the first part showing the sqrt(1-z^2)/z piece.

Basically my visualization of this equation is showing me that in the case of 5:

5=3+2

3^2 - 2^2 = 5

3^2 - 1^2 = 8

3^2 - 0 = 9

and here is a link to the whole system:

http://4.bp.blogspot.com/_u6-6d4_gs.../bdPIJMIFTLE/s1600/prime-+square+12a+zoom.png

in the case of 9:

9=5+4

5^2 - 4^2 = 9

5^2 - 3^2 = 16

5^2 - 2^2 = 21

5^2 - 1^2 = 24

notice the primes in the link I provided.

basically i want to show that primes have no other congruence to a square besides the (p-1)/(p+1) relationship.

make any sense???

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- #14

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interseting...I see... tan(arccos(z)) = sqrt(1-z^2)/z

and

if A = (x-1)/(x+1) then (A-1)/(A+1) = 1/-x or (A+1)/(A-1)=x

what about e^(-2/x) converging to A?

e^(-2/n) ~ (n-1)/(n+1)

and the 2/n part here:

http://en.wikipedia.org/wiki/RMP_2/n_table

whats the connection?

- #15

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more interesting prime links. notice the decoded table highlights the primesinterseting...

e^(-2/n) ~ (n-1)/(n+1)

and the 2/n part here:

http://en.wikipedia.org/wiki/RMP_2/n_table

whats the connection?

http://rmprectotable.blogspot.com/

- #16

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as n reaches infinity e^(-2/n) will equal (n-1)/(n+1)

here is a table of the base bart of my equation notice columns h and b and thier equivalence in column i:

http://4.bp.blogspot.com/-heoUDug-LwM/TVx-czNOsTI/AAAAAAAAAFY/HWIeStwwCKU/s1600/RMP2n.png

a visual of the equation:

http://3.bp.blogspot.com/-5UhMF-uGw...AFQ/oDdl_oSXPM0/s1600/prime-+squares+edit.png

and a visual with some primes hilighted:

http://4.bp.blogspot.com/_u6-6d4_gs.../bdPIJMIFTLE/s1600/prime-+square+12a+zoom.png

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