- 11 May '04 00:12The series:

1 - 1/3 + 1/5 - 1/7 +... = pi/4

is widely known. Besides just integrating 1/(1 + t^2) from t=1 to 0 and then integrating the equivalent series, prove this. I saw an old STEP problem that guides one through a clever proof, so I'd like to see more. I'll post mine in a few days if no-one beats me to it, but I suppose there are better. - 12 May '04 08:42

What properties can we assume pi has? Do you want a geometric argument?*Originally posted by royalchicken***The series:**

1 - 1/3 + 1/5 - 1/7 +... = pi/4

is widely known. Besides just integrating 1/(1 + t^2) from t=1 to 0 and then integrating the equivalent series, prove this. I saw an old STEP problem that guides one through a clever proof, so I'd like to see more. I'll post mine in a few days if no-one beats me to it, but I suppose there are better. - 12 May '04 17:09

Any type of argument will do.*Originally posted by Acolyte***What properties can we assume pi has? Do you want a geometric argument?**

I don't understand the first question in this context; even some of the more complicated arguments involving Fourier series don't really use properties of pi other than the value of circular functions at rational multiples of pi, all of which are derivable from the definition. If you mean 'should the argument include a proof that sin pi = 0' or something like that, then no. - 12 May '04 17:33

The definition of sine:*Originally posted by royalchicken***The series:**

1 - 1/3 + 1/5 - 1/7 +... = pi/4

is widely known. Besides just integrating 1/(1 + t^2) from t=1 to 0 and then integrating the equivalent series, prove this. I saw an old STEP problem that guides one through a clever proof, so I'd like to see more. I'll post mine in a few days if no-one beats me to it, but I suppose there are better.

Sin(x) = x - x^3/3! + x^5/5! - x^7/7! +...

Sin(1) = 1 - 1/3! + 1/5! - 1/7! +...

Looks like it, but i'm still working on how to get closer... - 15 May '04 18:22Arctan you mean i assume?

It's expansion around 0 is this:

arctan(z) = 1- z/3 + z^3/5 -...

So arctan(1) =1 - 1/3 + 1/5 -...

Also, Arctan is a one-on-one function (on [0, pi) ) and the inverse of tan. We know that tan(pi/4) = 1, so arctan(1) = pi/4.

I'm not sure why arctan should be expanded around 0 though... - 15 May '04 19:13

Arguments involving Fourier series sound a bit circular, if you'll pardon the pun. What I meant was, are we defining pi as the smallest postive real x such that 1 - x^2/2! + ..... tends to -1 (or something similar involving power series), or are we defining it as the perimeter of the unit circle?*Originally posted by royalchicken***Any type of argument will do.**

I don't understand the first question in this context; even some of the more complicated arguments involving Fourier series don't really use properties of pi other than the value of circular functions at rational multiples of pi, all of which are derivable from the definition. If you mean 'should the argument include a proof that sin pi = 0' or something like that, then no. - 15 May '04 21:40

Uhm, the perimeter of the unit circle is what we commonly call 2pi. Why not make pi half the perimeter of the unit circle, though if we're being completely formal I see no reason why we can't make pi about 6.3 and find series for ''pi''/8 .*Originally posted by Acolyte***Arguments involving Fourier series sound a bit circular, if you'll pardon the pun. What I meant was, are we defining pi as the smallest postive real x such that 1 - x^2/2! + ..... tends to -1 (or something similar involving power series), or are we defining it as the perimeter of the unit circle?** - 15 May '04 23:41

Replace perimeter by area.*Originally posted by royalchicken***Uhm, the perimeter of the unit circle is what we commonly call 2pi. Why not make pi half the perimeter of the unit circle, though if we're being completely formal I see no reason why we can't make pi about 6.3 and find series for ''pi''/8 .**