finding an infinity serie for a

[Perdurat]

given:
[math]A=2\left(\frac{1}{8k+1}-\frac{1}{8k+4}\right)[/math][math]B=\frac{1}{8k+1}-\frac{1}{8k+5}[/math][math]C=\frac{1}{8k+1}-\frac{1}{8k+6}[/math][math]\pi=\sum _{k=0}^{\infty }\:\frac{1}{16^{k}}\left(A+B+C\right) (Plouffe)[/math]
problem:
[math]\int _0^{\frac{\pi }{2}}sin\left(x\right)-\int _{a\left(\frac{\pi }{2}\right)^2}^{\frac{\pi }{2}}-a\left(x-\frac{\pi }{2}\right)^2+1=0[/math]
can a be written as an infinity serie?

 

[mario99]

What is the original problem that created all of these: infinite series and the integral? Also what is Plouf fe?

 

[mario99]

can a be written as an infinity serie?

If you want to find an infinite series for:

[imath]\displaystyle a\left(x-\frac{\pi}{2}\right)[/imath]

Yes, I have an idea to do that, but you have to write the integral correctly as it is missing [imath]dx[/imath]. And you don't need the infinite series of [imath]\pi[/imath] to do that.

 

[Perdurat]

If you want to find an infinite series for:

[imath]\displaystyle a\left(x-\frac{\pi}{2}\right)[/imath]

Yes, I have an idea to do that, but you have to write the integral correctly as it is missing [imath]dx[/imath]. And you don't need the infinite series of [imath]\pi[/imath] to do that.

What is the original problem that created all of these: infinite series and the integral? Also what is Plouf fe?

WHO is Plouffe : Simone Plouffe (mathematician) : the series provided for (my personal favorite) was found using a computer program (and reverse engineering its value), what is interesting about it is the fact that it is a so called spigot algorythm (you can use it to calculate the n-th digit of pi in base16 format) so it is an infinite serie but one that doesn't overshoot and undershoots, like Tailor's (sin,cos) hence easy to scale (if you have a multicore processor, and/or GPU)
to the contrairy of your statement i am not a genius, certainly not in the field of mathematics, the PDF I attached in the previous post (Ramanuan)
doesn't mean I understand it. you understand it. The only thing I have is is a hunch, that the calculation method you proposed is valid, because what is looked for resides within the rectangle (-1,+1),(+1, -1).Like I said, i'll keep it for a rainy day, to read.

perception:
Because knowing of ohm's law has a positive impact on the proliferate of a multimeter, I spent something close to a year to research the concept of multimeter = human, therefore ohm's law =? (quality: the ability to observe)... expanding quality : observe,move,regenerate,...
challenging the the expansion (all animals are equal some are more equal than others,Einstein=right, therefore not Einstein=wrong,...)
let quality be a classification, for which the members express themselve with a common nominator: amplitude
let quantity be a classification, for which the members express themselve with 2 common nominators: phase, timeframe
to allow for compressing of above, lets introduce a molecule: serotonine
within a human it is found in two variants:
the grey variant : synthesised in the brain to provide for intelligence (where to find food?)
the brown variant : synthesised in the intestines as a respond to food
spooky action from a distance?
(not all humans are evenly well equiped to only use the grey variant to provide for intelligence)
phase, timeframe,objectivity:
-3/4->0 : the human imune system is shared with the womb
0->4:excessive stress (rapid growth, short sleep/awake cycle, developping essential social cue's,...)->no longterm memory required
25 : biological adulthood
33 : ramanujan ceased to exist, that number was the reason I took interest in the man
club 27 was to noisy to investigate, or 3x3x3 -> 3x11 = upping the ante

compairing the curve of x^2 with the curve of sin(x):

[math]\left(-\cos x\left(\frac{\pi}{2}x\right)+1\ \right)vs\left(x^{2}\right)\to0.6995[/math]
[math]\left(\sin\left(x\right)\right)vs\left(-s\left(x-\frac{\pi}{2}\right)^{2}+1\right)\to a,b[/math]
a) find the scalefactor s, such that p1(0,0) and p2(pi,0) are points on the quadratic equation curve (desmos->+-0.405)
b) find the scalefactor s, such that the surface area defined by the X-axis and the goniometric function
matches that defined by the X-axis and the quadratic function (desmos->+-0.43)

(I am affraid, having exercised integral calculus, for the last time some 30 years ago ,only one braincel remains to nag me that I forgot)

 

[mario99]

Thank you for the lengthy explanation. But I am not a fan of reading, so I will go straight to the point. Did you consider to write the integral as this:

[imath]\displaystyle \int_{0}^{x} \sin t \ dt = \int_{a\left(\frac{\pi}{2}\right)^2}^{x} a\left(t - \frac{\pi}{2}\right)^2 \ dt[/imath]

For now, we will ignore the [imath]1[/imath] just to give you a taste of my idea.

By the way, it seems that you have misunderstood my reply in the previous post. I really meant that you are a Genius and I was not a Genius because my answer was based on using the technology while your remarkable answer was based on a research you have done by your own. Not anyone can easily figure out from where I came up with the numbers in the approximation, but you did. This thing, which seems little, is enough to call you a genius.

 

[Perdurat]

I am not sure if i understand the concept, give me a day or two, to chew on it (watch some youtube movies to refresh on (un)bound integrals, derivatives, antiderivatives)

 

[mario99]

I am not sure if i understand the concept, give me a day or two, to chew on it (watch some youtube movies to refresh on (un)bound integrals, derivatives, antiderivatives)

What do you think will happen if we take the derivative of both sides?

[imath]\displaystyle \frac{d}{dx} \left(\int_{0}^{x} \sin t \ dt\right) = \frac{d}{dx} \left(\int_{a\left(\frac{\pi}{2}\right)^2}^{x} a\left(t - \frac{\pi}{2}\right)^2 \ dt \right)[/imath]

 

[Perdurat]

What do you think will happen if we take the derivative of both sides?

[imath]\displaystyle \frac{d}{dx} \left(\int_{0}^{x} \sin t \ dt\right) = \frac{d}{dx} \left(\int_{a\left(\frac{\pi}{2}\right)^2}^{x} a\left(t - \frac{\pi}{2}\right)^2 \ dt \right)[/imath]

tilt, I'm gonna hit the sack, reboot tomorrow

 

[Perdurat]

What do you think will happen if we take the derivative of both sides?

[imath]\displaystyle \frac{d}{dx} \left(\int_{0}^{x} \sin t \ dt\right) = \frac{d}{dx} \left(\int_{a\left(\frac{\pi}{2}\right)^2}^{x} a\left(t - \frac{\pi}{2}\right)^2 \ dt \right)[/imath]

the way I understand it (correct me if I am wrong):
-connecting first the real world problem to a mathematical statement (...=..., ...<..., ...>..., , ...<=..., ...>=...)
-because dx/dy (2D) is something that belongs to a paper, real matematicians use dt^2 (1D),
-it begins to dawn on me why you would have things like the integegral of an integral
-because constants wouldn't survive the voyage they are kept in the real world
-putting the x as the upperbound on both sides leaves the equality statement intact
[math]\int\limits^{x}_{0} \sin\left(t\right) \, \mathrm{d}t=\int\limits^{x}_{a\left(\frac{\pi}{2}\right)^{2}}a\left(t-\frac{\pi}{2}\right)^{2}dt[/math][math]-\cos\left(x\right)-1+C=\frac{a \left(x - \frac{\pi}{2}\right)^{3}}{3}+C[/math][math]-\cos\left(x\right)-1=\frac{a \left(x - \frac{\pi}{2}\right)^{3}}{3}[/math][math]ax\left(8x^{3}-4{\pi}x^{2}+12{\pi}x-{\pi}^{3}\right)=8\cos\left(x\right)+8[/math]
rabbit<goat<monster?

 

[mario99]

the way I understand it (correct me if I am wrong):
-connecting first the real world problem to a mathematical statement (...=..., ...<..., ...>..., , ...<=..., ...>=...)
-because dx/dy (2D) is something that belongs to a paper, real matematicians use dt^2 (1D),
-it begins to dawn on me why you would have things like the integegral of an integral
-because constants wouldn't survive the voyage they are kept in the real world
-putting the x as the upperbound on both sides leaves the equality statement intact
[math]\int\limits^{x}_{0} \sin\left(t\right) \, \mathrm{d}t=\int\limits^{x}_{a\left(\frac{\pi}{2}\right)^{2}}a\left(t-\frac{\pi}{2}\right)^{2}dt[/math][math]-\cos\left(x\right)-1+C=\frac{a \left(x - \frac{\pi}{2}\right)^{3}}{3}+C[/math][math]-\cos\left(x\right)-1=\frac{a \left(x - \frac{\pi}{2}\right)^{3}}{3}[/math][math]ax\left(8x^{3}-4{\pi}x^{2}+12{\pi}x-{\pi}^{3}\right)=8\cos\left(x\right)+8[/math]
rabbit<goat<monster?

Oops, I thought [imath]a[/imath] is a function of [imath]\displaystyle t - \frac{\pi}{2}[/imath] like [imath]\displaystyle a\left(t - \frac{\pi}{2}\right)[/imath]

Is [imath]a[/imath] a constant?

 

[Perdurat]

Is a a constant?

well, once we found them, they becomes...
[math]A=sin(x)\>\>vs\>\>B=(-a(x-\frac{\pi}{2})^{2}+1)\\ for\>\> a_1(\approx0.405)\longrightarrow\>A \cap B=(0,0),(\frac{\pi}{4},1),(0,\frac{\pi}{2})\\ for\>\> a_2(\approx0.43)\longrightarrow\>A \cap B=(x_{1},y),(\frac{\pi}{4},1),(x_{2},y)\\ for\>\> a_1\longrightarrow\>C=\int\limits^{\frac{\pi}{2}}_{0} sin(t) \, \mathrm{d}t- \int\limits^{x_2}_{x_1}(-a_1(x-\frac{\pi}{2})^{2}+1) \, \mathrm{d}x\\ for\>\> a_2\longrightarrow\>\int\limits^{\frac{\pi}{2}}_{0} sin(t) \, \mathrm{d}t= \int\limits^{x_2}_{x_1}(-a_2(x-\frac{\pi}{2})^{2}+1) \, \mathrm{d}x-C\\ resolve\>for\>\> x_1=\>?[/math]

∫0xsint dt=∫a(2π)2xa(t−2π)2 dt

For now, we will ignore the 111 just to give you a taste of my idea

cliffhanger?

 

[mario99]

∫0xsint dt=∫a(2π)2xa(t−2π)2 dt

For now, we will ignore the 111 just to give you a taste of my idea

cliffhanger?

Now I will show you my idea without ignoring [imath]1[/imath]

[imath]\displaystyle \frac{d}{dx} \left(\int_{0}^{x} \sin t \ dt\right) = \frac{d}{dx} \left(\int_{a\left(\frac{\pi}{2}\right)^2}^{x} a\left(t - \frac{\pi}{2}\right)^2 + 1\ dt \right)[/imath]

This gives:

[imath]\displaystyle \sin x = a\left(x - \frac{\pi}{2}\right)^2 + 1[/imath]


[imath]\displaystyle \sum_{k = 0}^{\infty} \frac{(-1)^kx^{1+2k}}{(1 + 2k)!} = a\left(x - \frac{\pi}{2}\right)^2 + 1[/imath]


[imath]\displaystyle a = \frac{\sum_{k = 0}^{\infty} \frac{(-1)^kx^{1+2k}}{(1 + 2k)!} - 1}{\left(x - \frac{\pi}{2}\right)^2}[/imath]