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how to get an analytical solution

Asked by Darcy Chou on 26 Jun 2018
Latest activity Commented on by Ameer Hamza
on 26 Jun 2018
syms tp t
NN=((exp(-1.45*10^(-4)*tp^1.274))^1.5)/2
z=4/3*pi*NN*(t-tp)^3
Z=int(z,tp,0,t)
output is
Z =
int((2*pi*exp(-(2674777890687885*tp^(637/500))/18446744073709551616)^(3/2)*(t - tp)^3)/3, tp, 0, t)
that is not an analytical solution, then how can I get the analytical solution?
If t=10 ,how to get the numerical solution?
Thanks for reading.

  1 Comment

You cannot expect analytic solutions when you use exponents that are floating point numbers, as calculus does not define integrals with floating point exponents.
You should not be asking for an analytic integral for any expression that involves any floating point constants at all. If you want to express uncertainty in value (floating point values need to be considered as uncertain) then replace the floating point numbers with rationals plus a symbolic uncertainty.

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3 Answers

Answer by Ameer Hamza
on 26 Jun 2018
 Accepted Answer

The existence of an analytical solution is not always necessary. I cannot prove mathematically but it might be the case that this function does not have an analytical solution at all. It might also be the limitation of MATLAB symbolic engine. In the first case, there is not much you can do other than resorting to the numerical solution. In the second case, you may try some other symbolic math engine. As far as the numerical result is concerned, you can get it as with integral().
syms tp t
NN=((exp(-1.45*10^(-4)*tp^1.274))^1.5)/2;
z=4/3*pi*NN*(t-tp)^3;
Z=int(z,tp,0,t);
Z_function = matlabFunction(z);
Z_numerical = integral(@(tp) Z_function(10, tp), 0, 10);

  2 Comments

Thank you, your comments are helpful. It is not necessary to get an analytical solution. And the problem is solved using numerical method. Have a nice day!
You are welcome.

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KSSV
Answer by KSSV
on 26 Jun 2018

syms tp t
NN=((exp(-1.45*10^(-4)*tp^1.274))^1.5)/2 ;
z=4/3*pi*NN*(t-tp)^3 ;
Z=int(z,0,t)
subs(Z,'t',10)

  4 Comments

Show 1 older comment
KSSV
on 26 Jun 2018
I am considering tp as a constant. I think it is a constant.
No, the syntax
Z=int(z,tp,0,t)
means integrate z(tp) over tp = 0 to tp = t . tp is not a constant in the integral.
t might be a constant, but tp is not.
Z=int(z,tp,0,t)
in this syntax, z is the integrand and tp is the variable of integration,
Z is a definite integral whose upper limit is the variable t.

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Answer by Walter Roberson
on 26 Jun 2018

You have
NN=((exp(-1.45*10^(-4)*tp^1.274))^1.5)/2
Remember that exp(A*x)^B is the same as exp(A*B*x), so you can bring the 1.5 inside, forming
NN=exp(-1.5*1.45*10^(-4)*tp^1.274)/2
Now introduce a symbolic variable A and rewrite this as
syms tp nonnegative
syms A positive
NN = exp(-A*tp^sym(1.274)) / 2
for A = 1.5*1.45*10^(-4) but leave A as symbolic at the moment.
Then look at
z=4/3*pi*NN*(t-tp)^3
and see that this can be rewritten as
z = B * exp(-A*tp^sym(1.274)) * (t-tp)^3
for B = 4/3*pi/2
Now you want int(z, tp, 0, t)
which is int(B * exp(-A*tp^sym(1.274)) * (t-tp)^3, tp, 0, t)
which is B * int(exp(-A*tp^sym(1.274)) * (t-tp)^3, tp, 0, t)
And it turns out that MATLAB can calculate
temp = int(exp(-A*tp^sym(1.274)) * (t-tp)^3, tp, 0, t)
fairly quickly.
So the overall result is 4/3*pi/2 * subs(temp, A, 1.5*1.45*10^(-4))

  3 Comments

Thank you for your kind answer!
As mentioned above, we can get that Z is a function of t,
and then I want
Y=1-exp(-Z)
y=-diff(Y)
How can I get the min of y(t) and the relative value of t when t > 0 using
fminbnd()
I'm sincerely looking forward to your reply.
Oooo... I just discovered that I had a bracket in the wrong place when I was doing the int(). MATLAB cannot calculate temp easily after all :(
Wolfram Alpha does produce a solution for indefinite integral of the form
int(exp(A*tp^1.274)*(t-tp)^3, tp)
in term of gamma function. Although gamma function is itself defined in term of an integral for most cases.
I cannot find the required definite integral because its calculation requires a pro license. But I guess maple or Mathematica might be able to give a solution for the definite integral. Although I am not sure whether such a solution will be useful for any practical purpose.
Also from @Darcy comment, it appears that the result will later be used in a numerical optimization problem. Since you are eventually using a numerical optimization algorithm, it might be better to also use numerical integration techniques instead of searching for a closed form solution.

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