# Impossible equation to solve?

by Jiachao
Tags: equation, impossible, solve
 P: 12 Hi, I know its possible to get an answer by graphing the curves and finding the point of intersection but I was wondering if there was a way to do them algebraically. sin(2x) = $$\frac{x}{2}$$ or $$e^{\frac{x}{8}}$$ = x Also, do these types of equations have a name?
 Sci Advisor P: 2,193 Your first equation is not solvable (in terms of elementary functions or any I know of). The equation simply is not written in a way in which it is possible to express it as a solution x=something. The second equation at first glance appears the same, but is actually solvable (although not in terms of elementary functions). The solution is: x=-8 W(-1/8) Where W is the Lambert W function. ( http://en.wikipedia.org/wiki/Lambert%27s_W_function ) I don't know of a name for these [unsolvable] equations, I'd call em non-algebraic (not sure if that's official though).
 Sci Advisor HW Helper PF Gold P: 12,016 They are called transcendental equations.
 P: 12 Impossible equation to solve? Thanks for the replies.
 P: 150 haha try this one. Let me know if you're interested in the problem, I actually got it from physicsforums.com. $$\left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6} = e + \pi$$ n=?
 P: 365 And what are the 0 and 6 ??
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,488 I assume that he means that formula evaluated between 6 and 0: its value at 6 minus its value at 0 is equal to $e+ \pi$.
 P: 150 of course. Its an integral remember? $$\int_0^{6}\sqrt{1 - (nx)^2}dx = e + \pi$$
 P: 12 I'm still in high school so is that possible to solve using my level of mathematics? After I subbed the numbers in, I did a quick sketch on graphmatica and it didn't show any intersections between $$y = \sqrt{1-36n^{2}} +\frac{arcsin (6n)}{n} -1$$ and y = 2e + 2pi so have I done something wrong already?
 P: 607 I agree that there is no solution to $$\left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6} = e + \pi$$ However, there is a solution to $$\int_0^{6}\sqrt{1 - (nx)^2}\,dx = e + \pi$$ camilus is mistaken in asserting that these are equal.
 P: 1 The equations in the first post can't be solved explicitly for x, but you could use a method of iteration to get as accurate a solution for x as you require.
P: 150
 Quote by g_edgar I agree that there is no solution to $$\left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6} = e + \pi$$ However, there is a solution to $$\int_0^{6}\sqrt{1 - (nx)^2}\dx = e + \pi$$ camilus is mistaken in asserting that these are equal.
wtf? explain..?

$$\int_0^{6}\sqrt{1 - (nx)^2}\dx = \left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6} = e + \pi$$

now you gotta find what value of n will make will make the integral equal to e + pi. I have found n to several digits, although Im under the impression that n is nonalgebraic.
HW Helper
P: 3,531
 Quote by camilus wtf? explain..?
$$\int_0^{6}\sqrt{1 - (nx)^2}\dx \neq \left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6}$$

Now, I can't actually explain because I don't know how to evaluate the integral, but I did go about it backwards and derived the result. They're not equal from what I can see.
 P: 607 the integral should be $$\int \sqrt{1-(nx)^2}\,dx =\frac{1}{2}\;x\;\sqrt{1-(nx)^2} + \frac{1}{2n}\;\arcsin({nx)$$ notice the extra $$x$$?
P: 2,751
 Quote by Mentallic $$\int_0^{6}\sqrt{1 - (nx)^2}\dx \neq \left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6}$$ Now, I can't actually explain because I don't know how to evaluate the integral, but I did go about it backwards and derived the result. They're not equal from what I can see.
It is relatively easy to show that:

$$\int \sqrt{1 - (nx)^2}\, dx = {1 \over 2} x \sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n}$$

Two methods come to mind.

Method 1.

Substitute $nx = \sin(u)[/tex] and it transforms into [itex]1/n \int \cos^2(u) du$. Then just use the trig identity cos^2(u) = 1/2 + 1/2 * cos(2u) and it's straight forward to get the above stated result.

Method 2. (A little more complicated but avoids need for trig identities and "trig of inverse trig" simplifications).

Note that $$\sqrt{1-(nx)^2} = 1/ \sqrt{1-(nx)^2} \, - \, (nx)^2 / \sqrt{1-(nx)^2}$$. The first term is pretty much a standard inverse sin integral and the second term is amenable to integration by parts. This is actually one of those integration by part problems where you end up with the original integral on both the left and right hand side of the equals sign and it simplifies down pretty easily (again to give the above stated result).

Edit. Yes thanks g_edgar. I didn't notice that missing x before.
HW Helper
P: 3,531
No uart, you are wrong here. g_edgar posted the correct integral and I was able to confirm it for myself by again differentiating the result.

 Quote by uart It is relatively easy to show that:...
Funny how it can be relatively easy to show the wrong answer
 P: 365 Actually, the task is to solve for n $$\sqrt{1-36n^{2}} +\frac{arcsin (6n)}{n} -\frac{1}{2}-\frac{arcsin(0)}{2n}= e+\pi$$ $$6n=sin(x)$$ $$n=\frac{sin(x)}{6}$$ $$\sqrt{1-sin^2(x)} + \frac{6x}{sin(x)} - \frac{1}{2} = e+ \pi$$ $$cos(x)+\frac{6x}{sin(x)}=e+\pi + \frac{1}{2}$$