# Square root simplification.

JeremyEbert
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)

## Answers and Replies

Homework Helper
Hi Jeremy! what would that be? JeremyEbert
Hi Jeremy! 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....

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

what are you talking about? JeremyEbert
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?

Homework Helper
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 see

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?)

JeremyEbert
oh I see

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?

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

Let's do one thing at a time …

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

JeremyEbert
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.

Homework Helper
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)
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 prove tan(arccos((x-1)/(x+1))) = sqrt(x)/((x/2)+1/2) ? JeremyEbert
I assumed you wanted to prove the equation in your first post …

have you worked out how to prove tan(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.

Homework Helper
Carry on from …
I see... tan(arccos(z)) = sqrt(1-z^2)/z

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

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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JeremyEbert
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?

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

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