Solving Tricky Equations: Is There a Way to Solve Them Algebraically?

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In summary, the conversation discussed the possibility of solving equations algebraically, specifically equations involving sin(2x) and e^{\frac{x}{8}}. It was mentioned that the first equation is not solvable in terms of elementary functions, while the second equation can be solved using the Lambert W function. These types of equations are called transcendental equations. The conversation then shifted to a specific integral equation and the methods of solving it. It was concluded that the equation is still transcendental and can be solved numerically.
  • #1
Jiachao
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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) = [tex]\frac{x}{2}[/tex]


or [tex]e^{\frac{x}{8}}[/tex] = x

Also, do these types of equations have a name?
 
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  • #2
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's_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).
 
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  • #3
They are called transcendental equations.
 
  • #4
Thanks for the replies.
 
  • #5
haha try this one. Let me know if you're interested in the problem, I actually got it from physicsforums.com.

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

n=?
 
  • #6
And what are the 0 and 6 ??
 
  • #7
I assume that he means that formula evaluated between 6 and 0: its value at 6 minus its value at 0 is equal to [itex]e+ \pi[/itex].
 
  • #8
of course. Its an integral remember?

[tex] \int_0^{6}\sqrt{1 - (nx)^2}dx = e + \pi[/tex]
 
  • #9
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

[tex] y = \sqrt{1-36n^{2}} +\frac{arcsin (6n)}{n} -1 [/tex]

and y = 2e + 2pi

so have I done something wrong already?
 
  • #10
I agree that there is no solution to

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


However, there is a solution to

[tex] \int_0^{6}\sqrt{1 - (nx)^2}\,dx = e + \pi[/tex]

camilus is mistaken in asserting that these are equal.
 
  • #11
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.
 
  • #12
g_edgar said:
I agree that there is no solution to

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


However, there is a solution to

[tex] \int_0^{6}\sqrt{1 - (nx)^2}\dx = e + \pi[/tex]

camilus is mistaken in asserting that these are equal.

wtf? explain..?

[tex] \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[/tex]

now you got to find what value of n will make will make the integral equal to e + pi. I have found n to several digits, although I am under the impression that n is nonalgebraic.
 
  • #13
camilus said:
wtf? explain..?

[tex]\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}[/tex]

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.
 
  • #14
the integral should be

[tex]
\int \sqrt{1-(nx)^2}\,dx =\frac{1}{2}\;x\;\sqrt{1-(nx)^2} +
\frac{1}{2n}\;\arcsin({nx)
[/tex]

notice the extra [tex]x[/tex]?
 
  • #15
Mentallic said:
[tex]\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}[/tex]

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:

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

Two methods come to mind.

Method 1.

Substitute [itex]nx = \sin(u)[/tex] and it transforms into [itex]1/n \int \cos^2(u) du[/itex]. 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 [tex]\sqrt{1-(nx)^2} = 1/ \sqrt{1-(nx)^2} \, - \, (nx)^2 / \sqrt{1-(nx)^2} [/tex]. 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.
 
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  • #16
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.

uart said:
It is relatively easy to show that:...
Funny how it can be relatively easy to show the wrong answer :tongue:
 
  • #17
Mentallic said:
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.


Funny how it can be relatively easy to show the wrong answer :tongue:

Yes it's a relatively easy integral. I posted two ways to do it (and have done it many times in the past) but no I didn't actually do the calculations this time as the expression looked correct (as in it looked that same as what I've got in the past when I've actually done the integral rather than just describing the method to use). Sorry I didn't notice that the "x" was missing in the expression you posted.
 
  • #18
Actually, the task is to solve for n

[tex]\sqrt{1-36n^{2}} +\frac{arcsin (6n)}{n} -\frac{1}{2}-\frac{arcsin(0)}{2n}= e+\pi
[/tex]

[tex]6n=sin(x)[/tex]

[tex]n=\frac{sin(x)}{6}[/tex]

[tex]\sqrt{1-sin^2(x)} + \frac{6x}{sin(x)} - \frac{1}{2} = e+ \pi[/tex]

[tex]cos(x)+\frac{6x}{sin(x)}=e+\pi + \frac{1}{2}[/tex]
 
  • #19
Дьявол said:
Actually, the task is to solve for n
Yes we knew that. :) And with the corrected equation that would be,

[tex]3 \sqrt{1-36n^2} + {\frac {\sin^{-1}(6n)} {2n} = \pi + e [/tex]

Which is still a transendental equation. (that is, no closed form solution in terms of standard functions). Numerically n=0.0617 to 3 significant figures.
 
  • #20
uart said:
Yes we knew that. :)


And with the corrected equation that would be,

[tex]3 \sqrt{1-36n^2} + {\frac {\sin^{-1}(6n)} {2n} = \pi + e [/tex]

Which is still a transendental equation. (that is, no closed form solution in terms of standard functions). Numerically n=0.0617 to 3 significant figures.

Look at my way of solving, it's much easier. Yes, I know that it is still transcendental equation.
 
  • #21
uart said:
Yes it's a relatively easy integral. I posted two ways to do it (and have done it many times in the past) but no I didn't actually do the calculations this time as the expression looked correct (as in it looked that same as what I've got in the past when I've actually done the integral rather than just describing the method to use). Sorry I didn't notice that the "x" was missing in the expression you posted.

Ahh ok, I see now. Your methods to solve them seemed unflawed, however since you insisted on the wrong answer, I wasn't sure what to think anymore :smile:
 

1. What is an impossible equation to solve?

An impossible equation to solve is an equation that has no real solutions. This means that there is no value that can be substituted for the variable(s) that will make the equation true.

2. How do you know if an equation is impossible to solve?

An equation is impossible to solve if it leads to a contradiction or an absurdity. This can happen when there are no solutions that satisfy all the given conditions or when the solutions lead to nonsensical or impossible values.

3. Can an impossible equation ever have a solution?

No, an impossible equation will never have a solution. This is because the equation itself is based on contradictory or impossible conditions, making it impossible to find a valid solution.

4. Are there any strategies for solving impossible equations?

No, there are no strategies for solving impossible equations. Since these equations have no real solutions, there is no mathematical method or strategy that can be used to find a solution.

5. What is the significance of studying impossible equations?

Studying impossible equations can help us understand the limitations and boundaries of mathematics. It also highlights the importance of carefully defining and setting up equations in order to avoid contradictions and impossible scenarios.

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