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Partially differentiate an improper integral with a summation?

GROUPS:

I am new to Mathematica with experience in MATLAB, and I am frustrated with the long-winded introductory material focusing on syntax and functions irrelevant to my task at hand; I am asking for your help because I have not been able to find the documentation answering these questions. Given equations 1-4, I am trying to verify equations 7-11, 13-15: equations 1-3 defining P, m, and sigma equation 4 defining D enter image description here

[Edit: Regarding equations 13-15, partial derivatives of D, these forums are apparently not compatible with Windows 10 Microsoft Edge and image URLs get deleted erroneously when I Publish edits to the post, and won't let me include more than three images.]

This should be easy to do -- simply define the functions and execute a differentiation command -- but I am struggling both to learn the syntax, GUI, and functions to use.

Here is the code I have written so far trying to verify equation 11:

In[63]:= du = (Sum[v*d^(1/w), i])^w
Out[63]= (d^(1/w) i v)^w

In[64]:= m = (du - T)/(s*T)
Out[64]= (-T + (d^(1/w) i v)^w)/(s T)

In[65]:= D[1/Sqrt[2*Pi]*Integrate[Exp[-x^2/2], {x, -Infinity, m}], w]
Out[65]= (E^(-((-T + (d^(1/w) i v)^w)^2/(
  2 s^2 T^2))) (d^(1/w) i v)^w (-(Log[d]/w) + 
   Log[d^(1/w) i v]))/(Sqrt[2 \[Pi]] s T)

It appears the sum is incorrect: What is this factor of 'i'? I am wondering whether I must use the product of two matrices since the summation is a sum in the physics sense, using subscripts, not in the technical mathematical sense as I see in the Wolfram documentation (in which the summation variable is explicit as a base or exponent).

I intend to continue reading and searching the documentation to try to accomplish this task myself, but I hope you will answer sooner so that I can proceed with my research sooner. Thank you for your time.

POSTED BY: Daniel Bridges
Answer
1 month ago

Browsing documentation and trying various commands like natural input, the user can learn Mathematica through trial-and-error.

POSTED BY: Daniel Bridges
Answer
1 month ago

I think this is correct, although for the partial derivative for d and v one must keep in mind the other terms in the sum are 0.

proposed solution

POSTED BY: Daniel Bridges
Answer
1 month ago

Group Abstract Group Abstract