Comments on: Trigonometry Identities

By: Denaya Lesa

Denaya Lesa — Sun, 28 Dec 2008 14:00:25 +0000

Oh my GOD, I am also wrong in writing Rohedi’s formula for dy/dx+Py=Qy^n. This is the true solution:

y(x) = 1/[( 1/y0^(n-1 )-Q/P )e^{P(n-1)(x-x0)} + Q/P]^(1/(n-1))

Thx Mr.Pramono.

By: Pramono

Pramono — Sun, 28 Dec 2008 13:16:40 +0000

Apologize, there is small correction. Rohedi’s formula for BDE’s solution dy/dx+Py=Qy^n is of form

y(x)=[(P/2Q)^1/(n-1)][1+tanh(-P(n-1)(x-x0)/2 +atanh( (2Q/P)y0^(n-1) -1) ) ]^1/(n-1)

By: Pramono

Pramono — Thu, 25 Dec 2008 08:25:01 +0000

Of course the rohedi’s name that purposed by Nadya Fermega. He is the Head of ROHEDI LAboratory who has created the second form ol solving the Bernoulli DE :

y(x)=[(P/2Q)^1/(n-1)][1+tanh(P(n-1)(x-x0)+atanh( (2Q/P)y0^(n-1) -1) ) ]^1/(n-1)

Oh my God, indeed the above formulation is very nice. For detail explanation please visit to this address : http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=34 on the topic of Stable Modulation Technique (SMT).

By: Robert

Robert — Thu, 25 Dec 2008 00:22:57 +0000

Sorry, which one of Rohedi that your purpose who create the BDE’s Formula. Coz after searching at google there many rohedi’s name, such as nikita rohedi, cucun rohedi, yoedi rohedi, DR.Yossi Rohedi, etc, and finally of course the rohedi’s name on ROHEDI LAboratory. I think all rohedi’s name originate from Indonesia.

By: Nadya Fermega

Nadya Fermega — Wed, 24 Dec 2008 09:10:27 +0000

O yeah, I also ever read the nice explanation about the above Rohedi’s Formula when visiting to this address :
http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=34&st=0&sk=t&sd=a&start=20.
Are you Maduranish from Indonesian coutry???

By: Denaya Lesa

Denaya Lesa — Wed, 24 Dec 2008 09:01:53 +0000

Ho ho ho ho ho, Mr.Rohedi, I believe that the new breakthrough for your problem in creating the Euler Formula is only by using Rohedi’s Formula for solution the Bernoulli differential equation of constant coefficients : dy/dx + Py = Qy^n, for n is not equal 1, that is of form:

y(x) = 1/[( 1/y0^(n-1 )-Q/P )e^{P(n-1)(x-x0)} + Q/P]^(n-1)

with both initial values of x0 and y0. Let’s inspire to get it.

By: rohedi

rohedi — Wed, 24 Dec 2008 08:44:56 +0000

Hi Juliani, Nadya, and Denaya, the common way to get the Euler formula e^(ix)=cosx+isinx was performed by using Argand Diagram and Maclaurin Series. Let assume the two tools until now do not exist. My question, how to get the Euler Formula?

By: Juliani

Juliani — Tue, 23 Dec 2008 16:42:22 +0000

Nadya and Denaya, please stay on the topic. Back to Rohedi’s question, but let assume that the Euler formula also not exist up today, how to prove the trig identity?

By: Denaya Lesa

Denaya Lesa — Tue, 23 Dec 2008 16:06:40 +0000

Thanks Nadya for your hint, Next I apply square operation for the Euler Formula

[e^(ix)][e^(ix)] = [cosx+isinx][cosx+isinx]

e^(i2x) =[ (cosx)^2 - (sinx)^2] + i[2cos(x)sin(x)] , (1)

we know from the definition your Euler Formula

e^(i2x) =cos(2x) + i sin(2x) , (2)

ho ho ho…then from the equality real part of eq.(1) and eq.(2) we find

cos(2x)=(cosx)^2 – (sinx)^2

wihile from the equality imaginary part of eq.(1) and eq.(2) we find

sin(2x)=2sin(x)cos(x)

Okay Rohedi, this is also simple derivation, isn’t it.

By: Nadya Fermega

Nadya Fermega — Tue, 23 Dec 2008 15:26:17 +0000

Rohedi, let taking the absolute square of Euler formula e^(ix)=cosx+isinx,
where i=(-1)^(1/2). Then we will find…

[e^(ix)][e^(-ix)] = [cosx+isinx][cosx-isinx]

1 = (cosx)^2 + (sinx)^2

A simple proof, isn’t it.