1. Jan 9, 2007

### michellelover

1. The problem statement, all variables and given/known data
Can someone please solve: -.5x^2=ln x in detail. Thanks. It would be nice if I could get an answer about now? This is like due tomorrow. Thanks in advance.

2. Relevant equations

3. The attempt at a solution

2. Jan 9, 2007

### AKG

Show some work.

3. Jan 9, 2007

### Tom1992

as far as I can see, there is no exact analytic solution:

ln(x)=(x-1)-(x-1)^2/2+(x-1)^3/3-...

by taylor's expansion, so we have an infinite series equal to -0.5x^2. does your teacher want a numerical solution? if so, use the runge-kutta method. but if this is a pre-differential equations course, then you can try newton's method.

ok, my first day in this forum, and my dad caught me spending too much time posting here, sorry i'm forced to cut down on the visits and do more reading, as much as i like solving problems here...

Last edited: Jan 9, 2007
4. Jan 9, 2007

### chemhelper

If you can get away without showing work, you can plug it into a graphing calculator (which would be the easiest and most accurate way).

Rearrange to ln x + .5x^2 = 0

Set y=ln x + .5x^2

go to the solver and solve for x and you should get 0.753089

Last edited by a moderator: Jan 20, 2007
5. Jan 9, 2007

### michellelover

i need to show my work.

6. Jan 10, 2007

### AKG

Yes, you do need to show your work.

7. Jan 10, 2007

### Gib Z

Its getting late here so I may be incorrect, but we could use the Taylor Series Expansion for The Natural Logarithm, as Tom mentioned. It would also be an easier calculation if you used the other expression,

$$\ln (x+1)=\sum_{n=0}^{infinity} \frac {-1^n \cdot x^{n+1}}{n+1}$$.

We could truncate the expression at the degree to which you can solve. It is possible to solve the polynomials roots analytically for up to degree 5, but I think you will find it gets difficult after Quadratics (a substitution will not make it reducible to a quadratic).

To be even more correct, you would express the rest of the series in the form of a Lagrange/Cauchy Remainder term.

We could also use a a variety of fixed-point iteration methods, Such as halving the interval, newtons method, Halleys method (similar to newtons but with conditional cubic convergence), or the runge-kutta method.

For an extremely advanced method, use halving the interval until you have the root of f(x)=ln x + 0.5x^2 over a small closed interval. The interval must be small enough so that its Lipschitz constant, or maximum slope between any 2 points over that interval is less than 1. Then you can increase the rate of convergence of the iteration sequence with a convergence acceleration method, such as Euler Transforms, Kummer transforms, Richardson extrapolation, the e-algorithm, the Levin u-transform, Wilf-Zeilberger-Ekhad method, or the Aitken's delta-squared process (known as the Steffensen's method when applied to fixed point iterations such as this).

The last paragraph is quite difficult to carry out however, and seeing the time of your post, I am too late. Just incase you are bothered, these advanced methods have amazing rates of convergence, up to 17.83^-n with n iterations. That truly is extra ordinarily fast. In fact this just gave me an idea to calculate a sequence for pi, that will be very quick indeed. Not as quick as a professional mathematicians though, but o well.

I hope I helped.

Last edited: Jan 10, 2007
8. Jan 10, 2007

### HallsofIvy

If I remember correctly, there was another thread, on either this or another forum, that asked to show that the families of curves, defined by y= (-1/2)x2+ k and y= ln x+ k are orthogonal. Of course, that's true because for the first y'= -x and for the second y= 1/x: the product of the slopes at every point (x,y) is -1.

However, that person also seemed to be under the impression that he had to find the points of intersection! (-1/2)x2= ln x gives the x-value for the intersection of y= (-1/2)x2+ k and y= ln x+ k for the same k. In fact, the two families intersect at all (x,y) in the right half-plane (where ln x is defined).

In other words, I suspect that the original poster does NOT have to solve that equation at all.

9. Jan 11, 2007

### dextercioby

Sides, the Taylor expansion of ln(x+1) converges, iff x<=1. So it can't be used when possibly larger values of "x" occur.

To the OP. Try solving it by graph intersection method.

Daniel.

10. Jan 11, 2007

### Gib Z

O well yes, but in this case we know it fulfills the conditions. If the teacher wishes all work to be shown and justified, we could have used say, halving the interval until the root was known fulfill the condition that x<=1.