Leibniz series

[bugzpodder]

1-1/3+1/5-1/7+1/9-...=pi/4, according to Leibniz series for arctan expansion, also known as the Gregory's formula.

here is my modification (for the sum above) of euler's proof for zeta(2). see if this makes sense.
sin(x)-1=0
gives solutions for x=pi/2+2pik, or x=pi/2, 5pi/2, 9pi/2,... and x=-3pi/2, -7pi/2,-11pi/2

sin(x)-1=-1+x-x^3/3!+x^5/5!-... by the taylor series

if we write sin(x)-1=0 as -(1-x/r1)(1-x/r2)...=0

we get sin(x)-1=-(1-2x/pi)(1+2x/(3pi))(1-2x/(5pi))(1+2x/(7pi))...=-1+x-x^3/3!+x^5/5!-...

so the coefficient in front of x should be the same:

or -2/pi+2/(3pi)-2/(5pi)+2/(7pi)-...=-1
1-1/3+1/5-1/7=pi/2 ??? similar, but twice as large as pi/4
whats wrong with my logic?

[bugzpodder]

no one is familiar with this??

[sql_lall]

hey, i was just wondering abut a few of the steps:
1: "if we write sin(x)-1=0 as -(1-x/r1)(1-x/r2)...=0 "
what are r1, r2, r3 etc???

2: "-2/pi+2/(3pi)-2/(5pi)+2/(7pi)-...=-1"
how did you get this (though i guess if i knew #1, i would know this)

[bugzpodder]

if the roots to an equation is:

r1,r2,r3,...

we have

(x-r1)(r-r2)(x-r3)=0 correct?

divide the first bracket by -r1, second by -r2, third by -r3

(1-x/r1)(1-x/r2)(1-x/r3)=0

and you can verify that 1-x/r1=0 gives you x=r1, etc..., provided all the roots are non-zero

[bugzpodder]

the second part is just the coefficients infront of x in both equations

[krtxmrtz]

Originally posted by bugzpodder
sin(x)-1=-(1-2x/pi)(1+2x/(3pi))(1-2x/(5pi))(1+2x/(7pi))...=-1+x-x^3/3!+x^5/5!-...

I don't think this is an identity, in other words, it's not true for all x, only for a few values r_i which are the roots. Therefore I don't think the coefficients can't be equated to those of the Taylor seies.

[bugzpodder]

why not? you didn't give me a satisfactory reason.

in RHS eventually most of the terms will converge to one. so if you aren't convinced, you should try a value for x.

we'll just evaluate x=pi/4 for:
-(1-2x/pi)(1+2x/(3pi))(1-2x/(5pi))(1+2x/(7pi))

=-(1-1/2)(1+1/6)(1-1/10)(1+1/14)
=-(1/2)(7/6)(9/10)(15/14)
=-3*3/(2*2*2*2)
=-9/16=-0.5625

we have: sin(pi/4)-1=sqrt(2)/2-1=-0.2928932188

you are right, some does mess up, and it has to do with the pi/4 and pi/2 thing. if we divide -0.5625/2 is -0.28125 which is very close to -0.2929

if you still aren't convinced, lets try a smaller value (converges faster), say pi/6

=-(1-1/3)(1+1/9)(1-1/15)(1+1/21)
=-(2/3)(10/9)(14/15)(22/21)
=-(2*10*14*22)/(3*9*15*21)
=-(2*2*2*22)/(3*9*3*3)
=-0.7242798354

and sin(pi/6)-1=-0.5

and -0.7242798354/2=0.3621
which is fairly good to me.

well maybe not (converging faster)

i'll try pi/3

=-(1-2/3)(1+2/9)(1-2/15)(1+2/21)
=-(1/3)(11/9)(13/15)(23/21)
=-0.3867

sin(pi/3)-1=-0.13397459

and -0.3867/2=-0.1933568

now the best way to determine this is to verify this with a graph. i suggest you try it and tell me what you think. i am going to try it too now

[bugzpodder]

by the way, if the two polynomials have the same roots, and one of their coefficient is the same, then they must be the same equation, although things maybe a little different with infinite roots

[krtxmrtz]

Put 4 textboxes in a form and a command button and try this out (I hope there's no mistake in the code):

Const Pi = 3.141592654
Private Sub Command1_Click()
    Dim x As Double, z As Double
    x = Trim(Text2.Text)
    z = 2 * x / Pi
    N = Trim(Text1.Text)
    Result = 1
    For i = 1 To N
        Result = Result * (1 - z / (2 * i - 1))
    Next
    Result = 1 + Result
    Text3.Text = Str(Result)
    Text4.Text = Str(Sin(x))
End Sub