Mathematics Thread

[TheNextFeliks]

I am in Geometry. It is not hard but the proofs are a pain in the neck.

 

[Ickathu]

On (6) it's easy to see that (tan(pi/5) + i)^10 = (tan(pi/5) - i)^10 but the two sides are complex conjugates. For them to be equal they must both be real.

Therefore tan(pi/5) +i is a 10th root of a real number. But its squared magnitude is tan^2(pi/5) + 1 = sec^2(pi/5), so therefore:

(tan(pi/5) + i)/sec(pi/5) = cos(2n pi/10) + i sin(2n pi/10)

and so cos(pi/5) = sin(2n pi/10) which is easy to solve, one finds n=2:

sin (2x2 pi/10) = sin( 2 pi/5) = cos( pi/2 - 2 pi/5) = cos(pi/5), as desired.

hmmm. I looked in the answer book and they said something along the lines of "One can look at this equation and reasonably assume that the answer is n = 3."
And that's the answer they gave. :/

I am in Geometry. It is not hard but the proofs are a pain in the neck.

I hated geometry. Darn proofs...

 

[CarlBrannen]

Ickathu, it's easy to do the calculation with a calculator.

(a) Compute the angle of tan(pi/5) + i.
(b) Compute the angle of tan(pi/5) - i.
(c) Subtract. Should give you a multiple of 2 pi/10.

The difference between the n=2 and n=3 answers amounts to a minus sign in the imaginary part. So if I got it wrong (seems likely), look for a sign error.

*edit*

Yep, I get n=3. The angle of tan (pi/5) +- i = +- (pi/2 - pi/5) = +- 3 pi/10, so the overall angle is 3 pi/10 - -3pi/10 = 6pi/10 and n=3.

 

[Ickathu]

Ickathu, it's easy to do the calculation with a calculator.

(a) Compute the angle of tan(pi/5) + i.
(b) Compute the angle of tan(pi/5) - i.
(c) Subtract. Should give you a multiple of 2 pi/10.

The difference between the n=2 and n=3 answers amounts to a minus sign in the imaginary part. So if I got it wrong (seems likely), look for a sign error.

*edit*

Yep, I get n=3. The angle of tan (pi/5) +- i = +- (pi/2 - pi/5) = +- 3 pi/10, so the overall angle is 3 pi/10 - -3pi/10 = 6pi/10 and n=3.

Oh yes, I forgot to mention that we are not allowed to use calculators on these tests oh well, even the teacher and math pro who just finished his college degree in engineering and mathematics wasn't able to solve this one but I see what you did, so that clarifies it some.

 

[IanTheCuber]

Gemetry Guide to Noobs

This is a geometry guide for the average noob.

Spoiler

Area of a triangle: 1/2bh
Area of a circle: (pi)r[SUP]2[/SUP]
Circumference of a circle: 2(pi)r or (pi)d

I'll continue it later

 

[5BLD]

This is a geometry guide for the average noob.

Spoiler

Area of a triangle: 1/2bh
Area of a circle: (pi)r[SUP]2[/SUP]
Circumference of a circle: 2(pi)r or (pi)d

I'll continue it later

Can you prove these? Namely the area of the circle one, I'm interested in your meffod.

 

[vcuber13]

heres the one i know:

Spoiler

cut circle along radius
unwrap it to make a triangle
dimensions are r by 2pi r
A=2pi*r^2 /2
=pi*r^2

 

[5BLD]

thats not a proof

 

[vcuber13]

how isnt it?

http://en.wikipedia.org/wiki/Area_of_a_circle#Triangle_method

 

[5BLD]

prove its a triangle

 

[vcuber13]

how about this then:

\( \int_0^r \! 2\pi x \, \mathrm{d} x = \pi r^2 \)

 

[5BLD]

Yes, that was the proof i was thinking of. You didn't prove it was a triangle for me for the other one though.

 

[vcuber13]

i know, the only thing i can think of is this:
y=2pi*r
y'=2pi
therefore it increases linearly and will make a triangle

 

[5BLD]

Calculator/Math Thread

You can do that even with polar functions? Ooh

 

[brandbest1]

I feel like bumping this thread cuz I'm a math noob.

How do you factor x^n + y^n with a factor of x + y?

 

[vcuber13]

it varies with n

 

[TheNextFeliks]

Geometry:
Sides of a triangle are 4x+6, 2x+1, and 6x-1. 6x-1 is the longest. For which values of x is the triangle obtuse. I keep applying quadratic formula and getting a negative in the discriminant.

EDIT: Nevermind, I multiplied wrong.

 

[5BLD]

Calculator/Math Thread

Right guys. Harmonic series from 1 to k closed form? Is there? Also on a related note how do you guys prove zeta2= pi^2/6? This is my way i thought of today though its not really mine cuz i think i saw it somewhere last year.

Spoiler

ye got ye expansion of sinx:
sinx=x-x^3/3!+x^5/5!-x^7/7!+x^9/9!-...

(sinx)/x=1-x^2/3!+x^4/5!-x^5/6!+...

cuz polynomial, you can write it using its roots. (remembering tis periodic in x, so roots at x=npi, where n element Z but not zero)
(sinx)/x=(1-x/pi)(1+x/pi)(1-x/(2pi))(1+x/(2pi))(...
sinx/x=(1-x^2/pi^2)(1-4x^2/pi^2)(...

i forgot how newton's identities let you multiply it out and collect x^2 terms. But someone told me you can. But yeah so the coefficient of x^2 in the series expansion will be therefore:

-1/pi^2-1/4pi^2-1/9pi^2-...
=-(1/pi^2+1/4pi^2+1/9pi^2+...)
=-1/pi^2 sigma k=1 to inf 1/k^2 = -1/pi^2 zeta2

looking at the original mclaurin series of (sinx)/x the coefficient of the x^2 tis -1/3!=-1/6

-1/pi^2 zeta2= -1/6
zeta2= pi^2/6

 

[ducttapecuber]

Well, since the this thread was bumped....

I currently take Algebra 1, which is (sadly) the highest level of math my school offers. Next year I am taking Geometry Honors IB (aka standard geometry course with extra enrichment, application, and more work) and I think I am also taking trig or algebra 2, not sure.
I want to go into a physics field, which requires a lot of math. I enjoy math and science a lot!

 

[MTGjumper]

Right guys. Harmonic series from 1 to k closed form? Is there? Also on a related note how do you guys prove zeta2= pi^2/6? This is my way i thought of today though its not really mine cuz i think i saw it somewhere last year.

Spoiler

ye got ye expansion of sinx:
sinx=x-x^3/3!+x^5/5!-x^7/7!+x^9/9!-...

(sinx)/x=1-x^2/3!+x^4/5!-x^5/6!+...

cuz polynomial, you can write it using its roots. (remembering tis periodic in x, so roots at x=npi, where n element Z but not zero)
(sinx)/x=(1-x/pi)(1+x/pi)(1-x/(2pi))(1+x/(2pi))(...
sinx/x=(1-x^2/pi^2)(1-4x^2/pi^2)(...

i forgot how newton's identities let you multiply it out and collect x^2 terms. But someone told me you can. But yeah so the coefficient of x^2 in the series expansion will be therefore:

-1/pi^2-1/4pi^2-1/9pi^2-...
=-(1/pi^2+1/4pi^2+1/9pi^2+...)
=-1/pi^2 sigma k=1 to inf 1/k^2 = -1/pi^2 zeta2

looking at the original mclaurin series of (sinx)/x the coefficient of the x^2 tis -1/3!=-1/6

-1/pi^2 zeta2= -1/6
zeta2= pi^2/6

That's precisely how Euler did it. But you haven't justified being able to manipulate some of your infinite series