solve for x

Perdurat

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[math]\sin\left(\frac{\pi x}{2}\right)=\frac{4x}{\pi}[/math]
i know this resolves in a constant (0.6995<x<0.6996), is there a way to express this constant as in x = ...

thanks for your help
 
this resolves in a constant (0.6995<x<0.6996), is there a way to express this constant as in x = ...
Hi Perdurat. I think that value of x can only be approximated. The equation equates an algebraic function with a transcendental function. (Your question is similar to asking, "What is the exact value of pi".)

There are two additional solutions, one of which is an Integer. :)
[imath]\;[/imath]
 
[math]\sin\left(\frac{\pi x}{2}\right)=\frac{4x}{\pi}[/math]
i know this resolves in a constant (0.6995<x<0.6996), is there a way to express this constant as in x = ...

thanks for your help
Well there is, but are you sure that you want to see it?

[imath]\displaystyle x = \sqrt{\frac{56}{π^2} + \frac{1}{3 π^8 \left(\frac{2}{-69672960 π^{16} - 2128896 π^{18} + \sqrt{4854321355161600 π^{32} + 296652971704320 π^{34} + 11611416821760 π^{36}}}\right)^{1/3}} - 4032 π^4 \left(\frac{2}{-69672960 π^{16} - 2128896 π^{18} + \sqrt{4854321355161600 π^{32} + 296652971704320 π^{34} + 11611416821760 π^{36}}}\right)^{1/3}} \approx 0.6995[/imath]


[imath]\displaystyle x = -\sqrt{\frac{56}{π^2} + \frac{1}{3 π^8 \left(\frac{2}{-69672960 π^{16} - 2128896 π^{18} + \sqrt{4854321355161600 π^{32} + 296652971704320 π^{34} + 11611416821760 π^{36}}}\right)^{1/3}} - 4032 π^4 \left(\frac{2}{-69672960 π^{16} - 2128896 π^{18} + \sqrt{4854321355161600 π^{32} + 296652971704320 π^{34} + 11611416821760 π^{36}}}\right)^{1/3}} \approx -0.6995[/imath]

I hope that Mr.Otis is now convinced that this is my [imath]1000[/imath] complicated approximation 🙂

Note:
I have put [imath]=[/imath] there in purpose.
 
I'd be surprised if there were an exact analytical expression for the solution.
Just curious: where does this problem come from?
I was playing around with a graphing calculator (desmos)

starting off with:

[math]y=x^{2}[/math]
within the interval -1<x<+1, I tried to approximate the curve with a trigonometric function:

[math]y=-\cos\left(\frac{\pi}{2}x\right)+1[/math]
for the value x=-1, 0 and +1 both functions yield the same result (1, 0 and 1 resp.)

next I asked myselve for which value in between, the difference beween both functions would yield the biggest difference (derrivative):

[math](-\cos\left(\frac{\pi}{2}x\right)-x^{2}+1)dx/dt[/math]
this yields:

[math]\frac{\pi\sin\left(\frac{\pi x}{2}\right)}{2}-2x[/math]
since for a max and min, the derrivative yields 0 :

[math]\frac{\pi\sin\left(\frac{\pi x}{2}\right)}{2}-2x = 0[/math]
or:

[math]\sin\left(\frac{\pi x}{2}\right)=\frac{4x}{\pi}[/math]
expressing the distance of two parallel lines to the Y axis (+-X)

so now I was trying to find an expression for this value
 
Well there is, but are you sure that you want to see it?

[imath]\displaystyle x = \sqrt{\frac{56}{π^2} + \frac{1}{3 π^8 \left(\frac{2}{-69672960 π^{16} - 2128896 π^{18} + \sqrt{4854321355161600 π^{32} + 296652971704320 π^{34} + 11611416821760 π^{36}}}\right)^{1/3}} - 4032 π^4 \left(\frac{2}{-69672960 π^{16} - 2128896 π^{18} + \sqrt{4854321355161600 π^{32} + 296652971704320 π^{34} + 11611416821760 π^{36}}}\right)^{1/3}} \approx 0.6995[/imath]


[imath]\displaystyle x = -\sqrt{\frac{56}{π^2} + \frac{1}{3 π^8 \left(\frac{2}{-69672960 π^{16} - 2128896 π^{18} + \sqrt{4854321355161600 π^{32} + 296652971704320 π^{34} + 11611416821760 π^{36}}}\right)^{1/3}} - 4032 π^4 \left(\frac{2}{-69672960 π^{16} - 2128896 π^{18} + \sqrt{4854321355161600 π^{32} + 296652971704320 π^{34} + 11611416821760 π^{36}}}\right)^{1/3}} \approx -0.6995[/imath]

I hope that Mr.Otis is now convinced that this is my [imath]1000[/imath] complicated approximation 🙂

Note:
I have put [imath]=[/imath] there in purpose.
you're out of my league, by a few lightyears or so
 
Something tells me that @mario99 is joking. Wonder if that huge expression evaluates to close to 0.
I don't doubt the expression evaluates to the correct answer, but many expressions result in the same numerical value. I would like to see the approximation method that arrives at that specific expression.
 
If you say so, but I urge the OP not to trust the result until the process is verified.
option 1: the monster equation yields the result (+-0.6995...) (i do not understand the method)
option 2: mario typed in the equation in desmos or the like (from which I started) (a) and then added the monster as a joke
(a) this would be indirect proof that the earth is round, untill the monster is confirmed ligid, i shall remain confident that the earth is flat 8-)
 
option 1: the monster equation yields the result (+-0.6995...) (i do not understand the method)
option 2: mario typed in the equation in desmos or the like (from which I started) (a) and then added the monster as a joke
(a) this would be indirect proof that the earth is round, untill the monster is confirmed ligid, i shall remain confident that the earth is flat 8-)
Desmos is crazy. It gives that the two functions meet each other at [imath]x = 0.7[/imath]. I thought that he is so precise!

🥶
 
Nice one to get the point by using the derivative of each function.
anyway i calculated the monster : 2,75885829529689

this:
-not necessarily disproves that the earth is flat
-definitely proves your real name is not Mario Ramanujan
since a joke can only be considered to be a good joke if the laughter consumes more time than the telling,
the question remains: did I spent more time in xls calculating it, or did you spent more time in latex creating it
(for it=monster)







 
anyway i calculated the monster : 2,75885829529689

this:
-not necessarily disproves that the earth is flat
-definitely proves your real name is not Mario Ramanujan
since a joke can only be considered to be a good joke if the laughter consumes more time than the telling,
the question remains: did I spent more time in xls calculating it, or did you spent more time in latex creating it
(for it=monster)
Then, your calculator is not a good calculator. How did you calculate it?
 
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