Can some one help me to solve the following equation for x :
It actually popped out of a physics problem that I was solving and is still not solved.
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Can some one help me to solve the following equation for x :
It actually popped out of a physics problem that I was solving and is still not solved.
Sorry Faizan. In your solution, 2y = ln[sqrt(m) +t] / [sqrt(m) + t] leads to 2y = ln(1) = 0
So y = 0 leads to 2t - 1 = 0 or t = 1/2
That leads to x = [1-4m]/4pA.
But it's worse than that because I think you have made a sign error in line 13:
e^2y = ln { [sqrt(m) + t] / [sqrt(m) - t)] } - the green "-"
You will not be able to isolate t. Too bad about the minus sign. Otherwise it would have worked out.
Wasi: tanh[f(y)] = g(y) cannot be solved for y unless f(y) is a constant. Maybe there is some kind of convoluted Lambert function solution.
That's my take, unless I have missed something obvious.
@Ian Fowler Any solution will be ok even if it uses either lambert W function or some non - elementary function. Until the solution is computable it is okay for me.
@Ian Fowler
Thanks for pointing out the typo error The updated solution is here.
I hate to be the fly in the ointment again, but here's throwing caution to the wind.
In line 15 you have some of the form : e^2y = B.
In log form this become: 2y = ln(B),
not 2y = B which what you have in line 17
You can even take the route of e^2y = e^ln(B) , which is unnecessary, but it still leads to 2y = ln(B). So line 17 is missing the ln function.
The bottom line is you have 2y = ln( a function of y) which does not have an elementary solution. Maybe there is some kind of Lambert function solution but that's the best you have.
Wasi: You might try using the infinite Maclauren series for arc_tanh(x) and use a computer to grind out some desired level of accuracy and then go from there or the Lambert function possibility. I think those 2 are your only shot.
Faizan: I don't mean to flog the point but you missed the 2nd part of my response.
In line 15 you have: e^2y = B where B = [sqrt(m) +t] / [sqrt(m) - t]
Then on line 17 you have : 2y = B
This is incorrect as line 15 implies that: 2y = ln(B)
So everything that follows doesn't work.
Wasi: I have discovered another problem:
line 15 is O.K.
I assume p is a density, A is a cross-sectional? area and x is a length?
At any rate, assuming m,p,A are positive line 15 does not have a real solution for positive x as [sqrt(m) +sqrt(m+pAx] / [sqrt(m) - sqrt(m+pAx)] will be negative since
sqrt(m+pAx) > sqrt(m).
This leads directly to ln(negative) which is not real.
Or you can look above:
e^2y can never be negative so e^2y = B(negative) has no real solution.
So I need to know more about I,m,p,A,x actually represent. It looks to me like a physics problem. I need information on the domain (possible x values). Because if x is positive then there is no real solution.
Actually the original equation looked something like this:
where,
A1,A2 : Area
M : Mass
x : height
C : Some constant
t : time
ρ : Density
We need to find function x(t)
What's the heck is h? This significantly changes changes the whole complexity of the question.
it is height
I mistakenly wrote both h and x for the same thing😅
Ignore everything above.
Move the inverse hyperbolic tangent term to the RHS
Square both sides
Multiple the 4 in
Subtract 4M
Divide by 4*roh*A1