solve for x
- Thread starter Perdurat
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Otis
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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]
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 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
you're out of my league, by a few lightyears or soWell 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.
Thanks. I hope that this will serve your purpose.
BigBeachBanana
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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.
The only way to find out is to test it!
That is beyond the scope of this thread!
BigBeachBanana
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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
BigBeachBanana
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Legit?
People who tried to prove the Earth was flat often went in circles.
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!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
BigBeachBanana
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Ahh thanks, so the trick was to zoom in!
see attachmentDesmos 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.see attachment
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)
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Then, your calculator is not a good calculator. How did you calculate it?