- #1

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How do you solve this?

[itex] e^x+2x-5=0 [/itex]

And i mean the algorithm

[itex] e^x+2x-5=0 [/itex]

And i mean the algorithm

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- Thread starter Luhter
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- #1

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How do you solve this?

[itex] e^x+2x-5=0 [/itex]

And i mean the algorithm

[itex] e^x+2x-5=0 [/itex]

And i mean the algorithm

- #2

arildno

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Newton-Raphson iteration should work nicely to get an approximate answer, though.

- #3

arildno

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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..

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

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cool .

Thanks alot

Thanks alot

- #5

Mute

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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.

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arildno

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- #7

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approximatevalues 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

arildno

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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.

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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 text books 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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arildno

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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

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.

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- #12

arildno

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I love complex analysis as well, not the least its beauty, and how it <i>clarifies</i> and organizes seemingly disparate issues.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.

But, you ought to reflect on the following:

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,

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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