Searching for a solution for a long time

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In summary: Does somebody on PF know how to prove this?Re-express the problem in the form:x = ln (3/x)If you can find a solution, it'd be great if you could explain to me how you came about it in simple (if possible) explanation.In summary, the problem is to find the value of x when x is expressed in terms of the function ln(3/x). There are two possible methods to solve the equation: the method of successive substitutions and the half-interval method. Both methods are iterative, and it is difficult to tell how convergent or divergent they are going to be. However, if you can find a solution, it
  • #1
spriggster
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Hello all, I'm new to the forums (signed up to ask this question- this isn't a homework question). My friend gave me this equation a while back and all of my math teachers that I've asked have been stumped. I'm currently a sophomore in high school with no calculus experience (currently Algebra 2 honors), so this problem threw me off. If you can find a solution, it'd be great if you could explain to me how you came about it in simple (if possible) explanation.

The problem:

x(e^x) = 3
Solve for x

Thanks a lot guys!
 
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  • #2
What do you mean by "solve"? Do you just want to know the value of x? If so you can find the answer graphically by plotting the function e^x and the function 3/x on the same graph. The solution will be where they intersect, and it has a value of about 1.05. If you want to write x in terms of "elementary functions" (sines, cosines, polynomials, etc), that is not possible. However, there is a defined function that will give x. It is called the "Lambert W" function or the "Product Log" function. Here is the Wikipedia page on it.
 
  • #3
phyzguy said:
If you want to write x in terms of "elementary functions" (sines, cosines, polynomials, etc), that is not possible.

Does somebody on PF know how to prove this?
 
  • #4
Re-express the problem in the form:

x = ln (3/x)

You can solve this equation by the method of successive substitutions. This is an iterative method, where you have a starting guess, x = x0, and then you calculate successive values of x (on the left side of the equation) using previous values of x (on the right side of the equation). The successive substitution equation is:

[tex]x_{n+1}=ln \left(\frac{3}{x_n}\right)[/tex]

You keep repeating the calculation until x stops changing. Try doing this with an initial guess for x of x0 = 1. You can do this on a spreadsheet or on a hand calculator.

Another iterative method that can be used is the half-interval method (aka bisection method). Look this up on Google.

Chet
 
  • #6
micromass said:

Only a little motivational question: I happen to have an AMS book Algebraic Groups and Differential Galois Theory By Crepo and Hajto on my shelf, but I haven't read it almost at all. Do you think that reading it will help in this direction?

Chestermiller said:
Re-express the problem in the form:

x = ln (3/x)

You can solve this equation by the method of successive substitutions. This is an iterative method, where you have a starting guess, x = x0, and then you calculate successive values of x (on the left side of the equation) using previous values of x (on the right side of the equation). The successive substitution equation is:

[tex]x_{n+1}=ln \left(\frac{3}{x_n}\right)[/tex]

Writing fixed point iterations like this is dice game, because it is difficult to tell how convergent or divergent they are going to be. For example, your reply certainly left open the question why not write the problem in form

[tex]
x = ae^{-x}
[/tex]

and start iterating

[tex]
x_{n+1} = ae^{-x_n}
[/tex]
 
  • #7
jostpuur said:
Only a little motivational question: I happen to have an AMS book Algebraic Groups and Differential Galois Theory By Crepo and Hajto on my shelf, but I haven't read it almost at all. Do you think that reading it will help in this direction?



Writing fixed point iterations like this is dice game, because it is difficult to tell how convergent or divergent they are going to be. For example, your reply certainly left open the question why not write the problem in form

[tex]
x = ae^{-x}
[/tex]

and start iterating

[tex]
x_{n+1} = ae^{-x_n}
[/tex]
True. It helps to have some experience. But there is a of determining whether the successive substitution interation is going to be convergent or not.

Chet
 

FAQ: Searching for a solution for a long time

1. How long is considered a long time when searching for a solution?

The length of time considered to be "long" when searching for a solution can vary depending on the situation and the complexity of the problem. It could range from a few days to several months or even years.

2. What are some common reasons for a prolonged search for a solution?

Some common reasons for a prolonged search for a solution include the complexity or novelty of the problem, lack of available resources or information, and the need for multiple trial and error attempts.

3. How can I stay motivated while searching for a solution for a long time?

Staying motivated during a long search for a solution can be challenging, but it can be helpful to break the problem into smaller, more manageable tasks and celebrate small successes along the way. It can also be beneficial to take breaks and engage in activities that help reduce stress and maintain a positive mindset.

4. Is it common to encounter roadblocks or setbacks when searching for a solution?

Yes, it is common to encounter roadblocks or setbacks when searching for a solution. It is important to expect and anticipate these challenges and have a plan in place for how to overcome them. It can also be helpful to seek advice or input from others who may have experience with similar problems.

5. What are some effective strategies for finding a solution when I have been searching for a long time?

Some effective strategies for finding a solution after a prolonged search include reassessing the problem and approaching it from a different angle, seeking advice or collaboration from others, and taking breaks to allow for fresh perspectives and new ideas. It can also be helpful to keep an open mind and be willing to try new approaches or solutions.

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