0
$\begingroup$

I have difficulty concluding the derivatives of the following:

$$y=\ln\bigg(\sum_{j=1}^ng_j\cdot e^{xE_j}\bigg)$$ and $$y=x\cdot \sum_{j=1}^n \bigg(g_j \cdot \frac{N}{\sum_{j=1}^ng_j\cdot e^{xE_j}}\cdot e^{xE_j}\cdot E_j\bigg)$$ Note that $x$ does not change during the summation. Are these possible to solve? If so, how?

EDIT: I have made a mistake when copying the second formula, I have corrected it.

$\endgroup$
  • $\begingroup$ What derivatives are you taking? $\frac{\partial y}{\partial E_j}$, or $\frac{\partial y}{\partial g_j}$, presumably not $\frac{dy}{dx}$ since you're suggesting $x$ is constant. Is this something from statistical mechanics by any chance? $\endgroup$ – snulty May 7 at 23:47
  • $\begingroup$ He is saying that $x$ is the variable to differentiate wrt, but the series index is not $x$. Here what I'd do is say $e^y = \sum$ and then you can differentiate both sides wrt $x$ (implicit differentiation), giving $\frac{dy}{dx}e^y = \frac{d}{dx}\Sigma$. To differentiate the series, just differentiate each term and sum them together. $\endgroup$ – George Dewhirst May 8 at 0:35
  • $\begingroup$ For the second one I'd guess the answer is just $y=N$ with derivative zero, but seeing as there is probably $N = N_j$ i.e. $N$ varying, what you should do is take $(\sum_{j=1}^n g_j e^{xE_j})^{-1}$ out as a multiple, and use the product rule of differentiation. $\endgroup$ – George Dewhirst May 8 at 0:38
  • $\begingroup$ @GeorgeDewhirst For the 1st formula, if $\frac{dy}{dx}e^y = \frac{d}{dx}\Sigma$ then I get $\frac{dy}{dx} = \frac{1}{x} \cdot \sum_{j=1}^n(g_j \cdot E_j \cdot e^{xE_j})$. But a different source is telling me otherwise saying that the derivative of $\ln(g(x))$ is $\frac{g'(x)}{g(x)}$, thus $\frac{dy}{dx} = \frac{\sum_{j=1}^n (g_j \cdot E_j e^{xE_j})}{\sum_{j=1}^n (g_j \cdot e^{xE_j})}$. For the second formula, I made a mistake; the summation itself should be multiplied by $x$ as well (I corrected it in the OP). The $N$ is a constant and does not vary during the summation. $\endgroup$ – JohnnyGui May 8 at 17:15
1
$\begingroup$

If $y =\ln\bigg(\sum_{j=1}^ng_j\cdot e^{xE_j}\bigg) $ then, since $(\ln(f(x))' =\dfrac{f'(x)}{f(x)} $,

$\begin{array}\\ y' &=\dfrac{(\sum_{j=1}^ng_j e^{xE_j})'}{\sum_{j=1}^ng_je^{xE_j}}\\ &=\dfrac{\sum_{j=1}^ng_jE_j e^{xE_j}}{\sum_{j=1}^ng_je^{xE_j}}\\ \end{array} $

For your second $y$, you need to make the index in the inner sum different than the index in the outer sum.

$\begin{array}\\ y &=x \sum_{j=1}^n \bigg(g_j \dfrac{N}{\sum_{k=1}^ng_ke^{xE_k}} e^{xE_j}\bigg)\\ &=N\sum_{j=1}^n \bigg(g_j \dfrac{xe^{xE_j}}{\sum_{k=1}^ng_ke^{xE_k}} \bigg)\\ &=N\sum_{j=1}^n g_jy_j(x)\\ \end{array} $

where $y_j(x) = \dfrac{xe^{xE_j}}{\sum_{k=1}^ng_ke^{xE_k}} $.

Then

$\begin{array}\\ y_j'(x) &= \left(\dfrac{xe^{xE_j}}{\sum_{k=1}^ng_ke^{xE_k}}\right)'\\ &= \dfrac{(\sum_{k=1}^ng_ke^{xE_k})(xe^{xE_j})'-(\sum_{k=1}^ng_ke^{xE_k})'(e^{xE_j})}{(\sum_{k=1}^ng_ke^{xE_k})^2}\\ &= \dfrac{(\sum_{k=1}^ng_ke^{xE_k})((xE_j+1)e^{xE_j})-(\sum_{k=1}^ng_kE_ke^{xE_k})(e^{xE_j})}{(\sum_{k=1}^ng_ke^{xE_k})^2}\\ &= e^{xE_j}\dfrac{(xE_j+1)\sum_{k=1}^ng_ke^{xE_k}-\sum_{k=1}^ng_kE_ke^{xE_k}}{(\sum_{k=1}^ng_ke^{xE_k})^2}\\ \end{array} $

Now put this in $y'(x) =\sum_{j=1}^n g_jNy_j'(x) $, do any possible simplifications, correct any errors I may have made, and you are done.

$\endgroup$
  • $\begingroup$ Thanks a lot. I'm so sorry, I have a mistake in the OP, the second formula within the summation should be multiplied by $E_j$ such that $y=x\cdot \sum_{j=1}^n (g_j \cdot \frac{N}{\sum_{j=1}^ng_j\cdot e^{xE_j}}\cdot e^{xE_j}\cdot E_j)$. Not sure if this would change the whole derivation. $\endgroup$ – JohnnyGui May 8 at 19:40
  • $\begingroup$ Nope. Just changes the final summation. The $y_j$ remain the same. $\endgroup$ – marty cohen May 9 at 2:14
  • $\begingroup$ Such that $y'(x)=\sum_{j=1}^n E_j g_jNy_j'(x)$? $\endgroup$ – JohnnyGui May 9 at 16:00
  • $\begingroup$ Yup. If $E-J$ is there, use it. $\endgroup$ – marty cohen May 9 at 19:54
  • 1
    $\begingroup$ I think your $x$ in the denominator of $y'_j(x)$ should not be there. $\endgroup$ – JohnnyGui May 9 at 20:57

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.