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View Full Version : Maths Equation needing solving
Hey
I'm a matric maths pupil and I need some help solving this equation for x. I certainly do not know enough laws to solve this and neither does my maths teacher.
0.5^x = x
(^ = to the power of)
I know (by graphically drawing this there is only one possible x value and it must lie between 0 and 1 but I cannot solve it. Please someone write the law and a complete proof for me it would be much appreciated.
this is a transcendental equation. which means that you cannot solve for the root using elementary functions, like radicals, logarithms, etc..
if you want to find the approximate solution, with numerical methods, we can surely help you with that, but there is no elementary equation to express the exact solution....
Please Do, just explain how you got to the answer. By simple trial and error (only could be done because I knew the answer to lie between 0 and 1, I got 0.64....... but could obviously not get it specifically. Please just show you would get this amount. And by all means any way possible.
Originally posted by SciBoy
Please Do, just explain how you got to the answer. By simple trial and error (only could be done because I knew the answer to lie between 0 and 1, I got 0.64....... but could obviously not get it specifically. Please just show you would get this amount. And by all means any way possible.
easiest method is newtons method. newton s method says if you want to find a root for f(x), use an iteration:
x<sub>n+1</sub>=x<sub>n</sub> - f(x<sub>n</sub>)/f'(x<sub>n</sub>)
in this case, i want to find the roots of (1/2)<sup>x</sup>-x. actually, i will use the log of that equation, for simplicity: f(x)=ln x + x*ln 2. then f'(x) = 1/x + ln 2. so then:
x<sub>n+1</sub>=x<sub>n</sub> - (x<sub>n</sub> + x<sub>n</sub>*ln 2)/(1/x<sub>n</sub> + ln 2).
finally, pick a value for x<sub>0</sub>, and grab your calculator. i used a value of 1, and my iteration converged in 2 or 3 steps.
my final answer is x = 0.641185744505. you can check that, it should solve your original equation.
Thanks for the answer, it was what I needed.
However, the idea is new to me (obviously) being in high school, I do not know how interations work. Could you please explain how the nature of your subscript works? What is n? You meantioned x to subscript 0 but I do not understand this. Subscripts as far as my knowledge goes is just a labbelling technique for certain variables? Sorry about being so daft, but this equation intrigues me
Originally posted by SciBoy
Thanks for the answer, it was what I needed.
However, the idea is new to me (obviously) being in high school, I do not know how interations work. Could you please explain how the nature of your subscript works? What is n? You meantioned x to subscript 0 but I do not understand this. Subscripts as far as my knowledge goes is just a labbelling technique for certain variables? Sorry about being so daft, but this equation intrigues me
You can get a fairly good idea of the solution by simply graphing both sides of the equation & seeing where they intersect: see attached.
I know (by graphically drawing this there is only one possible x value and it must lie between 0 and 1 but I cannot solve it
Thanks for a clearer graph than what I drew, but I really need a algabraic why to understand this, I need the precisist answer possible.
everneo 03-14-03, 03:48 PM The value for x is restricted by the represention of reals in computers if u go for numerical methods hence approximation. could not say really any exact value exists for x.
thanks. But what i'd really like is lethes answer explained more clearly
Originally posted by SciBoy
thanks. But what i'd really like is lethes answer explained more clearly
There are a lot of different methods you can use. He mentioned Newton's method. Search google on the topic.
i d love to have a thread about newton s method. but i m leaving for spring break in tow hrs! sorry i didn t get to finish this.
remind be when i get back and sober up, and i will explain all!
HallsofIvy 03-16-03, 01:02 PM If you have not studied calculus, at least how to calculate derivitives then an explanation of Newton's method won't help.
However, here is a very simple method, although not as fast as Newton's method- halving:
You want to solve the equation (1/2)^x= x which is the same as
(1/2)^x- x= 0. You already know that if you try x=0 you get
(1/2)^0- 0= 1 which is greater than 0. If you try x= 1, you get
(1/2)^1- 1= -1/2 which is less than 1: therefore there is a solution between 0 and 1. Since you don't know exactly where, try the simplest: halfway. Halfway between 0 and 1 is 1/2. Trying x= 1/2 you get (1/2)^(1/2)- 1/2= 0.207 (approx). Since that is positive, we know there is a solution between 1/2 and 1 (where the value was negative.) Halfway between 1/2 and 1 is
3/4: (1/2)^(3/4)- 3/4= -.155: negative! Now we know there must be a solution between 1/2 and 3/4 so we try half-way between: (0.5+ 0.75)/2= 0.625. (1/2)^(0.625)- 0.625= .023. That's positive so there must be a solution between 0.5 and 0.625. Keep cutting the interval in half until you get as close as you wish.
BloodSuckingGerbile 03-16-03, 02:38 PM The idea of Newtons' method is not that hard to grasp.
It's in a way similar to HallsofIvys' method of halving.
The basic Idea behind the method is that the tangent of a function is always between the point A (see attached picture) and the point X<sub>n</sub>, the x value of the point at which the tangent touches the curve of the function.
I assume you know how to derive functions and that a derivative of a function in X<sub>n</sub> equals to the slope of the tangent at that point.
You are probably familiar with the equation of a strait line:
f(X)-f(X<sub>0</sub>)=m*(X-X<sub>0</sub>)
where m is the slope and (X<sub>0</sub>,f(X<sub>0</sub>)) is a point on the line.
Knowing that, and that the derivative of a function at point X equals m (the slope),
f'(X)=m
the equation of the tangent line to the function f(x) at point (X<sub>0</sub>,f(X<sub>0</sub>))becomes:
f(X)-f(X<sub>0</sub>)=f'(X)*(X-X<sub>0</sub>)
Now, as you see from the picture attached, the point you want to find is the point A. This is the point where the curve of the function intersects the X axis. At that point, f(X), the Y value of that function equals 0.
f(X)=0
as you can see, the point X<sub>n+1</sub> is between the point A and X<sub>n</sub>, closer to A. Therefore, if you can find the value of X<sub>n+1</sub>, you are closer to the answer, which is the X value of point A.
How to find the value of X<sub>n+1</sub>? Easy. At that point, the Y value of the tangent equation equals 0. f(X)=0. Thus,
f(X)=0
X=X<sub>n+1</sub>
m=f'(X<sub>0</sub>)=f'(X<sub>n</sub>)
X<sub>0</sub>=X<sub>n</sub>
f(X<sub>0</sub>)=f(X<sub>n</sub>)
0-f(X<sub>n</sub>)=f'(X<sub>n</sub>)*(X<sub>n+1</sub>-X<sub>n</sub>)
Simple algebra gives you:
X<sub>n+1</sub>=X<sub>n</sub>-[f(X<sub>n</sub>)/f'(X<sub>n</sub>)]
That's it. Next, the X<sub>n+1</sub> you've just found becomes the next X<sub>n</sub>, and that way you find X<sub>n+2</sub> and X<sub>n+3</sub>. That way you get to the answer pretty quickly.
<B><U>Example:</U> Calculating the value of Pi</B>
Pi is the root of the equation
Sin(X)=0
thus, f(X)=Sin(X).
Next step - finding X<sub>n+1</sub>:
Assume X<sub>n</sub>=3. Then:
X<sub>n+1</sub>=3-[Sin(3)/Sin'(3)] = 3-[Sin(3)/Cos(3)]=3-tan(3)=3.142546543.
Now, X<sub>n+1</sub>=3.142546543. Next step - finding X<sub>n+2</sub>.
X<sub>n+2</sub>=3.142546543-tan(3.142546543)=3.141592653.
And so on.
In the case of your equation, what lethe did, he took
f(X) = X-0.5<sup>x</sup>
as the function and used Newtons' method to find the answer.
Anyway, I hope I helped. I have to go now. Bye
Thanks, a lot, in fact ironically, we started doing derivatives a few days ago and its easy to grasp so you explanations helped a lot.
Thanks for the help guys.
errandir 07-12-03, 06:32 PM Hey, Sciboy. Sorry for the late response, but, if you're still interested, I will try to identify the meaning of these subscripts.
Basically, n stands for the iteration. You have to start somewhere (with some value for x), but, at the starting point, you have not iterated anything, so the subscript (n) is 0, standing for zero iterations.
After you make the calculation, you see that the LHS of the equation has a subscript (n+1) that represents the iteration if you carry out the calculation again. After one calculation, this subscript will be 1. Notice that, in order to refine the value, you ITERATE the equation, meaning, feed the output back into the input. So, the first iteration takes x<sub>1</sub> on the RHS and spits out x<sub>n+1</sub> (x<sub>2</sub>) on the LHS.
If you want to further refine the value, you REITERATE the equation, using x<sub>2</sub> on the RHS. x<sub>3</sub> results. In principle, this can be done indefinitely.
Thanks!
Ok, that once daunting explanation of leethes now is understandable. Thanks for the input guys.
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