I am trying to solve and verify the below equation using Mathematica.

$\frac{\partial}{\partial b} \sum_{i=1}^n y_i \log\left[ \frac{\sum_{i=1}^n y_i}{1-\prod_{i=1}^n (1-b)^{Exp[x_i\beta]}} \right]$

I tried to solve this on paper and I was able to obtain the following solution:

$\sum_{i=1}^n \left( \frac{(1-b)^{Exp[x_i \beta]-1} Exp[x_i \beta] y_i \prod_{i=1}^n (1-b)^{Exp[x_i \beta]} \sum_{i=1}^n\frac{Exp[x_i\beta]}{1-b}}{1-\prod_{i=1}^n(1-b)^Exp[x_i\beta]} \right)$

I usually verify my equations using Mathematica by plugging in numbers. However, in this case, Mathematica is unable to solve the first equation. So, I appreciate if any of you can share any information on this.

Thank you in advance.

  • 1
    $\begingroup$ Can you show us your attempt to calculate the derivative? One thing is that your notation is wierd... you seem to be using the i variable in two places at once (outside the log, and inside the log). One of these should probably be a sum over j. $\endgroup$ – bill s Dec 13 '18 at 1:53
  • 1
    $\begingroup$ People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful $\endgroup$ – Michael E2 Dec 13 '18 at 2:27
  • $\begingroup$ It is unclear what is being asked. I do not see any equations in the post, only a derivative of a sum. Can you clarify? You tagged this as "differential-equations", but I see no differential equation in the question. Please remove unrelated tags. Are you just trying to compute a derivate (and not solve a differential equation)? $\endgroup$ – Szabolcs Jan 21 '19 at 15:59

Give n a numerical value, then Mathematica will evaluate the derivative and you can check it against your result. Try several values of n to convince yourself the your result is correct. I coded your initial equation (but not your derived result) and calculated the derivative for n=2:

f = Sum[y[i] Log[Sum[y[i], {i, 1, n}]/(
    1 - Product[(1 - b)^Exp[x[i] β], {i, 1, n}])], {i, 1, n}];
D[f /. n -> 2, b]

the result is

-(((1 - b)^(-1 + E^(β x[1]) + 
E^(β x[2])) (E^(β x[1]) + E^(β x[2])) y[1])/(
1 - (1 - b)^(E^(β x[1]) + E^(β x[2])))) - ((1 - b)^(-1 +
E^(β x[1]) + E^(β x[2])) (E^(β x[1])
+ E^(β x[2])) y[2])/(1 - (1 - b)^(E^(β x[1]) + E^(β x[2])))
| improve this answer | |
  • $\begingroup$ Thank you. I generalized it and verified it based on your suggestion. $\endgroup$ – Lucky Dec 13 '18 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.