MATHEMATICA[solve numerical integration and find min value]

  • Mathematica
  • Thread starter shafieza_garl
  • Start date
  • #1

Main Question or Discussion Point

i want to find the minimum sum of this equation and find the t like using solver in Excel,but since the integration cannot be integrate directly,i didnt know how to use the numerical integration to apply for this problem.
i want to find the value :let say when n=2, t_0=0 and t_n=1, so i need to find t_1 that will make the equation become minimum value.I hope that anyone can help me.

Sum[Integrate[4.9*^6*E^(-0.03*t + 0.03*(t + Subscript[t, j]))*
((1 - 0.015*(6 + t))/E^(0.015*Subscript[t, j]) +
(-0.91 + 0.015*Subscript[t, j])/E^(0.015*t))^2,
{t, Subscript[t, -1 + j], Subscript[t, j]}], {j, 1, n}]
 

Answers and Replies

  • #2
30
0
write the equation you want to solve in normal way (like what written in your book) so i can help
 
  • #3
i had attached the equation.how can we use numerical integration in mathematica to solve the problem?or any other method can be used?
 

Attachments

  • #4
30
0
Let us try to solve it step by step
first I'll open the sum and get the following equation
http://www.3mints.info/upload/uploads/bbb65e0e17.jpg
and x is t1

press enter and get this answer
http://www.3mints.info/upload/uploads/f53bc6e9d8.jpg

now to find minimum you can take derivative and make it equal zero but i faced problem when trying solve function to zero i got the following massege

olve::tdep: The equations appear to involve the variables to be \
solved for in an essentially non-algebraic way

so i have to find another way . I plot the function to see the shape
http://www.3mints.info/upload/uploads/65a285154e.jpg

as you notice the minimum value of x is around 0.5 corresponding to arround 1 as a value for cos t.
I evaluated the function x at numbers between 0.4 to 0.6 with 0.00002 step and found that cos t = 0.894953
see figure
http://www.3mints.info/upload/uploads/116575a4af.jpg

and by iteration you can find value of x

x = 0.51219

please this is not for sure the right answer but it could give an idea
 

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