MatLab, et al Users?

Hi, Eman,

you are challenging cr4? you are good at the calculation. but you still post this question, it must be very very difficult problem.

what is R? real number or radius? as well as S? area? or other thing?

What dont you try Mathematica, mathcad or this matlab by yourself?

a is a Real number; ie, a is an element of the set of Real numbers. It is not a complex number. In this equation it is a constant.

s is Real. Units of length. This equation was generated by a path integral.

I don't have Mathematica, MathCad or MatLab. Tried several online solvers, but they gagged on this one. I will be installing Maxima, the Open Source version of Macsyma on my Linux box, probably this weekend.

Can you post the path integral? may it has differnet integral result

another formula may easily draw.

why dont you use Ln symbol? allof your colleagues like to use the Log_e?

looks like this is the solution to a problem already! there is a quadratic root equation lurking within!

does 1/2a = 1/(2a) or 1/2a = a/2

so "a" is real but can "x" be complex then? and "s"?

Hello omw7,

1/2a = 1/(2a). Sorry about the ambiguity.

More strictly,

...(1/(2a))·log_e[...

or

...(log_e[...])/(2a)

a is Real.

s and x are Real, though it is possible they may take on complex values. Physically meaningful values are Real.

Cross multiplying by 2a and substituting 2ax=y

2as = y√(y²+1)+log_e[y+√(y²+1)]

is a simplification just too make it look easier (I have omitted the ± as the √ implies this)

s, x and y are all real (not complex numbers such as a + ib where i=√-1)

let also t=√(y²+1) to obtain a parametric equation

then 2as = yt + log_e(y+t)

this then needs plotting. i can't think of a way of seperating the log_e from the normal y value, indeed I don't think that it can be done except in special circumstances (I should add that I would love to know how to do it otherwise). Perhaps in using the expansion of log_e and then saying as y tends to 0 or some such.

not much help I'm afraid.

Your thinking is right, Im same of yours at first.

why dont you all use loge rather than Ln symbol?

It seems a special equation socalled exceed equation(?)(how to speak in englissh?). at first , the (m+(m2+1)^1/2 >0, where m=2a. and

I can't see whether the following leads anywhere, but for what it's worth:

If you substitute p/2=arcsinh(y) you get:
2.a.s = sinh(p/2).cosh(p/2) + p/2
a.s = sinh(p) + p/4

Simple mans solution use Excel. Notice that root(y²+1) is always bigger than y, so the second ± term must always be positive. I've plotted the two possible graphs of omw7's 2as.

Hi Europium

First I would rearrange the formula to end with x = f(x)

Then Iterate to find the values of x.

Do you have a compiler?

I have an old DOS program EUREKA by Borland here somewhere that would do it.

If values for s and a are stipulated, x can be found by trial and error. Otherwise, I don't know how to find a general expression for x.

Reminds me of the old ladder puzzle. Two walls are parallel and x feet apart. Two ladders of length 20' and 30' respectively rest at the base of each wall and lean on the opposite wall forming a sort of unsymmetrical 'X' shape. The intersection point is 10' above ground. Solve for x. Not so easy, but if you turn the problem around and say the walls are 10' apart...solve for height, it's a snap.

x=-1/2*(-exp(-(lambertw((16*a+1)^(1/2)/a^2*exp((16*a+1+2*s*a)/a^2))*a^2-16*a-1-2*s*a)/a^2)+(16*a+1)^(1/2))/a

Not sure what the Lambert's W function represents

but this is what Matlab gives me.

Chazl

Most of times, root expression is not too important.

In fact, you can simplization the equation as metioned above and then draw it trends, so that you can gt more univeral result.

try this by hand draw,

y axies marked by Ln and x axis marked by scale unit of me^n, where m=2ax; then you can seee the trend of m(x)

Hello Chazl,

Would you do a screen capture of MatLab's output and post it here as an image? If MatLab shows the intermediate steps to the solution, please post them as well.

Thanks!
-e

Hey Europium,

It did not show the intermediate steps but here is the screen capture.

Not sure how this is going to display.

>> syms s d x a
>> d=x*sqrt(4*a*2^2 + 1) + (1/2*a)*log(2*a*x + sqrt(4*a*2^2 + 1))-s

d =

x*(16*a+1)^(1/2)+1/2*a*log(2*a*x+(16*a+1)^(1/2))-s


>> solve (d,x)

ans =

-1/2*(-exp(-(lambertw((16*a+1)^(1/2)/a^2*exp((16*a+1+2*s*a)/a^2))*a^2-16*a-1-2*s*a)/a^2)+(16*a+1)^(1/2))/a


>>

Chazl

Hello Chazl,

Would you mind running this through MatLab again? Note that I've made slight changes to the equation as entered to make it conform to the original problem. The changes are marked in bold...

>> syms s d x a

as entered:

>> d=x*sqrt(4*a*2^2 + 1) + (1/2*a)*log(2*a*x + sqrt(4*a*2^2 + 1))-s

edited:

>> d=x*sqrt(4*a^2*x^2 + 1) + (1/(2*a))*log(2*a*x + sqrt(4*a^2*x^2 + 1))-s

alternatively [simpler]:

>> d=x*sqrt((2*a*x)^2 + 1) + (1/(2*a))*log(2*a*x + sqrt((2*a*x)^2 + 1))-s

d = ?

>> solve (d,x)

ans = ?

Thanks. I sure appreciate your help on this!

-e

Europium,

I see my error.. thanks for the correction.

I ran the corrected equation and cannot get a solution using Matlab.

This is my readout:

d =

x*(4*a^2*x^2+1)^(1/2)+1/2/a*log(2*a*x+(4*a^2*x^2+1)^(1/2))-s


>> solve (d,x)
??? Error using ==> solve
Unable to find closed form solution.

Error in ==> sym.solve at 49
[varargout{1:max(1,nargout)}] = solve(S{:});

Chazl

Chazl,

Thanks for all your help. This baby's gonna have to be solved numerically <sigh>.

Hello Chazl,

I suspected x(s) did not have a closed-form solution (the presence of the Lambert W function in the solution confirms this).

There is a very nice writeup about the Lambert W on Wiki, by the way. It is useful for solving equations of the form x = W(s)e^W(s). s(x) can be reduced to a similar exponential form suitable to this approach.

-e

PS: The Italians call it the Woo function. There is no 'W' in the Italian alphabet.

x=+-20

For all values of a and s? If only the world were so simple!

Hello Europium

Can't solve analytically for x if x (and powers of) are added to log_e(x plus powers of).

But if you post values for s and a I can find a specific solution for x using Mathcad.

Cheers

Codey

where a is a parameter. you can select it in the range of real number. but when a is negtive, it seems no solution.

Hi, Eman,

How did you all spend your new year day?

I try to show you the graphic as follow.

here is another for a =1 to 9