- #1

JeremyEbert

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

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- Thread starter JeremyEbert
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- #1

JeremyEbert

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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

JeremyEbert

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Hi Jeremy!

Start with tan(arccos(z)) …

what would that be?

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

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

what are you talking about?

- #5

JeremyEbert

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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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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?

oh I

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

JeremyEbert

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oh 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?)

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?

Last edited:

- #8

tiny-tim

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

Let's do one thing at a time …

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

- #9

JeremyEbert

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Let's do one thing at a time …

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

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.

- #10

tiny-tim

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

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.

I assumed you wanted to prove the equation in your first post …

have you worked out how to

- #11

JeremyEbert

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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) ?

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.

- #12

tiny-tim

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Carry on from …

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

- #13

JeremyEbert

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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

JeremyEbert

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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?

interseting...

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

JeremyEbert

- 204

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interseting...

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?

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

http://rmprectotable.blogspot.com/

- #16

JeremyEbert

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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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