Simplifying Square Roots: Unraveling the Mystery of the Square Root Function

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Discussion Overview

The discussion revolves around the simplification of the square root function within the context of trigonometric identities and their relationships to algebraic expressions. Participants explore the function tan(arccos(z)) and its implications, as well as connections to exponential functions and prime numbers.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants inquire about simplifying the expression tan(arccos(((x/y)-1/y )/ ((x/y)+ 1/y)))*((x/2)-1/2)=sqrt(x).
  • There is a suggestion that tan(arccos(z)) can be expressed as sqrt(1-z^2)/z.
  • One participant proposes that z = (x-1)/(x+1) and questions the simplification of the expression ((x/y)-1/y )/ ((x/y)+ 1/y) to (x-1)/(x+1).
  • Another participant expresses uncertainty about proving the relationship tan(arccos((x-1)/(x+1))) = sqrt(x)/((x/2)+1/2).
  • There are references to the convergence of e^(-2/n) to (n-1)/(n+1) as n approaches infinity, with a request for clarification on this connection.
  • Visual representations are shared to illustrate the relationships between primes and squares, linking back to the original mathematical expressions discussed.

Areas of Agreement / Disagreement

Participants express various viewpoints and uncertainties regarding the simplification of the functions and the relationships between the mathematical expressions. No consensus is reached on the proofs or simplifications proposed.

Contextual Notes

Some participants mention the difficulty in recognizing certain algebraic forms and the potential for multiple interpretations of the relationships discussed. There is also a reliance on visual aids to convey complex ideas, which may not fully capture the underlying mathematical principles.

Who May Find This Useful

This discussion may be of interest to those exploring advanced mathematical concepts, particularly in trigonometry, algebra, and number theory, as well as individuals looking for insights into the relationships between different mathematical functions.

JeremyEbert
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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)
 
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Hi Jeremy! :wink:

Start with tan(arccos(z)) …

what would that be? :smile:​
 
tiny-tim said:
Hi Jeremy! :wink:

Start with tan(arccos(z)) …

what would that be? :smile:​

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

is there a simple explanation for this:

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

where is in the inverse natural log constant 2.71828182845904523536028747135266249...
 
JeremyEbert said:
oh yea...z = (x-1)/(x+1)

what are you talking about? :confused:
 
tiny-tim said:
what are you talking about? :confused:


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?
 
JeremyEbert said:
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! :biggrin:

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?)
 
tiny-tim said:
oh I see …

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

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

Let's do one thing at a time …

now solve tan(arccos((x-1)/(x+1)))
 
tiny-tim said:
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 that's not what your looking for...hints? sorry... I'm new at this.
 
  • #10
JeremyEbert said:
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)
JeremyEbert said:
well I know that tan(arccos((x-1)/(x+1))) = sqrt(x)/((x/2)+1/2)

but I'm sure that's 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) ? :confused:
 
  • #11
tiny-tim said:
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) ? :confused:

I have not worked out how to prove this. I have never done a proof before. This is a small piece 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
Carry on from …
JeremyEbert said:
I see... tan(arccos(z)) = sqrt(1-z^2)/z
 
  • #13
tiny-tim said:
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 what's 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?
 
Last edited:
  • #14
JeremyEbert said:
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?
 
  • #16
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 their 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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