Solve e^x+2x-5=0: Step-by-Step Guide

  • Thread starter Luhter
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In summary, solving the equation e^x+2x-5=0 requires the use of the Newton-Raphson iteration method, which involves following the tangent to the graph of the equation to find approximate solutions. Another approach is to use the Lambert-W function, which gives all infinite and complex solutions to the equation. However, it is more important to understand the logical and hierarchical foundations of mathematics before delving into more complex concepts. Good pedagogy involves gradually introducing new concepts and connecting seemingly disparate ideas to develop a deeper understanding of mathematics.
  • #1
Luhter
8
0
How do you solve this?
[itex] e^x+2x-5=0 [/itex]

And i mean the algorithm
 
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  • #2
This is a transcendental equation, and there is no finite number of steps that will give you the exact answer.

Newton-Raphson iteration should work nicely to get an approximate answer, though.
 
  • #3
If you are unfamiliar with Newton-Raphson, the trick involved is to follow the tangent from one point on the graph to where it crosses the x-axis, in order tofind your new approximate solution.

AT the true solution (i.e, f(x)=0), the tangent at that point will, of course, ALSO cross the x-axis there.

Pick your first approximate value, [itex]x_{0}=1[/itex]

Then, [tex]f(x_{0})=e+2-5=e-3[/tex]
where [tex]f(x)=e^{x}+2x-5, f'(x)=e^{x}+2[/tex]

In order to generate better approximation values for "x", use the iteration scheme:
[tex]x_{n+1}=x_{n}-\frac{f(x_{n}}{f'(x_{n}},n\geq{0}[/tex]

Thus, from the above, we get:
[tex]x_{1}=1+\frac{3-e}{e+2}=\frac{5}{e+2}[/tex]
and so on..
 
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  • #4
cool .
Thanks alot
 
  • #5
Although you were interested in the algorithm, you could also solve it in terms of the special function [itex]W(z)[/itex], defined by [itex]z = W(z)\exp(W(z))[/itex]. This is called the Lambert-W function (or sometimes the polylog function). If you can get your equation to look like this form, you can solve for x in terms of the Lambert-W function

Let u = 2x - 5. Then,

[tex]u = -\exp\left(\frac{1}{2}(u+5)\right) \Rightarrow u = -e^{u/2)}e^{5/2}[/tex]

Dividing by -2 on both sides and moving the e^{u/2)} to the LHS gives

[tex](-u/2)e^{-u/2} = e^{5/2}/2,[/tex]

which is of the form we^w = z, and so -u/2 = W(e^(5/2)/2), which gives

[tex]x = \frac{5}{2} - W(e^{5/2}/2)[/tex].

Notes: W(z) is a multivaled function. For real arguments there are two braches, corresponding to two solutions.
 
  • #6
Apart from the minor advantage of already-tabulated approximate values of the Lambert function, I don't see any pedagogical advantage in that approach.
 
  • #7
arildno said:
Apart from the minor advantage of already-tabulated approximate values of the Lambert function, I don't see any pedagogical advantage in that approach.

Why not Arildno? It's a beautiful function and solves the equation more completely than a numerical approach since the solution in terms of the W-function gives all (infinite and complex) solutions to the problem and provides the student with a glimpse of the more global structure of equations from the perspective of Complex Analysis. In my opinion, by restricting math to real analysis, we inhibit the student from seeing the whole picture of the complex-analytic functions that real analysis is embedded in, and this complex picture greatly assists in understanding mathematics.
 
  • #8
There are many beautiful things&equations in the world, it is unpedagogical to show them before the student is mature enough.

Complex numbers?
Integral equations?

The best way to get there is to start by understanding how things work in the reals, and to understand how numerical approaches like the Newton-Raphson method is based on a highly intuitive and visual idea, that STILL fulfills all requirements of logical rigour.
 
  • #9
With all due respect Arildno (cus' I think you're a better mathematician than I'll ever be), I feel we are doing math education a disservice by waiting to teach Complex Analysis after the student has struggled to understand real analysis, then shock him/her with complex-analytic functions. It is my belief that we should re-vamp math education and integrate complex numbers, complex variables, complex analysis early in their education, high-school even. We should change Calculus textbooks to more incorporate this complex-analytic approach but do so in a manner that is still comprehensible to the student. By presenting this more global picture in math education, I believe much more progress in mathematics would be possible.
 
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  • #10
I disagree.

Your position seems to say that there is an inevitable "shock" to learn about complex numbers at university level.

To me, it was a pleasant.,. surprise.

The point is that all of us start out with a number of hazy ideas, that are not really placed in logical hierarchies.

That should be the basis for pedagogics!

Namely, how to gradually make maths students become aware of the crucial importance of orderly logical thought.

And that is best done by starting out with mathematically fruitful concepts that are STILL easily visualized, like the number line.

As students delve deeper into this, and other related concepts, they will meet non-trivial problems they will see DEMAND careful, logical thinking in order to solve, and they will become convinced of the necessity to develop an axiomatic approach to mathematics.

And once THAT foundation has been laid, it is actually fairly trivial to introduce OTHER, equally logical axiomatic systems that can be more fruitful than that they started out with.
 
  • #11
Ok Arildno. You know more than me about it. I just like complex analysis even though I'm not very good at it. It has made concepts in math more clear for me and I think it would clear up things for other students as well.
 
  • #12
jackmell said:
I just like complex analysis even though I'm not very good at it. It has made concepts in math more clear for me and I think it would clear up things for other students as well.
I love complex analysis as well, not the least its beauty, and how it <i>clarifies</i> and organizes seemingly disparate issues.

But, you ought to reflect on the following:
Why do some topics lend themselves so easily to be (mis-)understood as "separate issues"?

My answer to that is that, on an intuitive level, yet with logical rigour, those issues ARE separate, because you cannot see the hidden connections between them.

Those connections will, however, make themselves apparent, once you have understood each topic deeply enough on their own, i.e, that you have matured to move beyond a "thesis-antithesis" stage unto a "synthesis" stage. (I'm a Hegelian in disguise, but don't tell on me..)

In a way, good pedagogics is a re-enactment of the whole history of mathematics, without mentioning all the pit falls&blind alleys that people actually erred into.

A truly good teacher, though, should know about those blind alleys, so that he knows how to steer his pupils out of them as quickly as possible..
 
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1. What is the first step in solving e^x+2x-5=0?

The first step in solving this equation is to bring the constant term to the right side by adding 5 to both sides, resulting in e^x+2x=5.

2. How do you isolate the exponential term e^x in this equation?

To isolate the exponential term, we can take the natural logarithm (ln) of both sides, resulting in ln(e^x+2x)=ln(5). This simplifies to x+ln(e^2)=ln(5).

3. What is the next step after isolating e^x in the equation?

After isolating e^x, we can simplify the natural logarithm by using the property ln(e^y)=y, which means ln(e^2)=2. This leaves us with x+2=ln(5).

4. How do you solve for x in the final step?

To solve for x, we can subtract 2 from both sides, resulting in x=ln(5)-2. This gives us the exact value of x that satisfies the equation e^x+2x-5=0.

5. Can this equation be solved using a calculator?

Yes, this equation can be solved using a calculator. However, it is important to know the steps and concepts involved in solving it algebraically, as it is a fundamental part of understanding exponential functions and logarithms.

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