# solve for x

#### Perdurat

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

I'd be surprised if there were an exact analytical expression for the solution.
Just curious: where does this problem come from?

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]

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

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:

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

$y=-\cos\left(\frac{\pi}{2}x\right)+1$
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):

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

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

$\frac{\pi\sin\left(\frac{\pi x}{2}\right)}{2}-2x = 0$
or:

$\sin\left(\frac{\pi x}{2}\right)=\frac{4x}{\pi}$
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

you're out of my league, by a few lightyears or so
Thanks. I hope that this will serve your 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.

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.

Wonder if that huge expression evaluates to close to 0.
The only way to find out is to test it!

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 is beyond the scope of this thread!

That is beyond the scope of this thread!
If you say so, but I urge the OP not to trust the result until the process is verified.

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

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
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!

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!
Zoom in more.

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!

see attachment

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?