# How to solve this equation for $y$?

I have the following equation that I would like to solve for $$y$$: $$\alpha=\sum_{i=1}^{n}\exp(-y\beta_i)$$ $$\alpha$$, $$y$$ and $$\beta_i$$ are all positive reals (the $$\beta_i$$'s are independent random numbers), and $$n$$ is a large positive integer. I'm unable to find a way to isolate $$y$$. Is this equation solvable?

Probability distribution of $$\beta_i$$:

$$\beta_i$$ is defined as $$\beta_i=\sqrt{2\omega_i}$$, where $$\omega_i$$ assumes a uniform probability distribution in some interval $$[\omega_a,\omega_b]$$, where $$0\leq\omega_a<\omega_b$$.

• No, in general this equation is not solvable for y. Espacially when $n$ is a large number. – Cornman Jun 17 '19 at 13:07
• As @Cornman said, you won't be able to get rid of that sum of exponentials – David Jun 17 '19 at 13:18
• If $\beta_i$ were all rational numbers, then this will become equivalent to solving a polynomial of a potentially enormous degree (probably much larger than $n$). Naturally, there is no formula for solving polynomials of degree $5$ or more, so you might be out of luck. – Theo Bendit Jun 17 '19 at 13:20
• hmm.. ok, thank you – Tofi Jun 17 '19 at 13:25
• It is worth noting that there is a unique real solution $y$ for any values of $\alpha$ and the $\beta_i$. Given $\alpha$ and the $\beta_i$ you can approximate $y$ with the help of a computer. An explicit expression for $y$ in terms of $\alpha$ and the $\beta_i$ is too much to ask, as the other comments explain. – Servaes Jun 17 '19 at 13:29

Updated post

After publishing my original post the author of the OP has specified the PDF for the $$\beta_i$$, also hereby he has assumed that all $$\beta_i$$ have the same distribution (which was my main simplifying assumption).

Here I take up this definition and update my solution.

EDIT (18.06.19): the following through derivation yields the corrected result which is even simpler than the previous one.

The distribution function for $$b$$ is from the definition $$w = \frac{1}{2}b^2$$ and $$f(w)$$ flat between $$w_a$$ and $$w_b$$ given from $$dw = b\; db$$ after normalization

$$f(b) = \frac{2 b}{b_b^2-b_a^2} = \frac{b}{w_b-w_a}$$

Now we can do the integral $$(2)$$ which gives

$$\alpha = n \int_{b_a}^{b_b} f(b) e^{- y b} \, db \\ = \frac{2 n \left(e^{- y b_a } (y b_a+1)-e^{-y b_b} (y b_b +1)\right)}{y^2 \left(b_b^2-b_a^2\right)}\tag{4a}$$

We have obtained a transcendental equation for $$y(\alpha)$$. This is best solved in the inverted form $$(4a)$$ for $$\alpha (y)$$.

In the limit of a very narrow distribution ($$w_a \to w_b (=w)$$) we get from $$(4a)$$

$$\alpha = n e^{-b_b y}\tag{4b}$$

which can be inverted to give the explicit solution

$$y = \frac{1}{b_b}\log(\frac{\alpha}{n})\tag{4c}$$

My original post

Making the simplifying assumptions that the $$\beta_i$$ have the same PDF $$f(\beta)$$ we can take the expectation value of the equation over all $$\beta_i$$ and get

$$E(\alpha) = \alpha = \sum_{i=1}^n E(\exp(- y \beta_i))= n E(\exp(- y \beta))\tag{1}$$

More explicitly we have for a continuous variable $$\beta$$

$$E(\exp(- y \beta))=\int_{0}^\infty f(\beta) \exp(- y \beta))^\,d\beta\tag{2}$$

Assuming for simplicity that $$\beta$$ is exponentially distributed, i.e.

$$f(\beta) = a \exp(- a \beta )\tag{3}$$

the integral can be done, and we obtain

$$\alpha = n \frac{a}{a + y}\tag{4}$$

which can be solved explicitly for $$y$$ to give

$$y = a (\frac{n}{\alpha}-1)\tag{5}$$

• Thank you, it looks like this equation is just too much! – Tofi Jun 17 '19 at 23:42
• @ Tofi Why "too much"? The specific problem is solved by equation $(4a)$ leaving just the standard problem of solving numerically a transcendental equation. – Dr. Wolfgang Hintze Jun 18 '19 at 8:06
• Yeah, but what I wanted was a nice analytical expression. By the way, the value I found for the PDF of $\beta$ is $\dfrac{\beta}{\omega_b-\omega_a}$, so is this the same as yours? – Tofi Jun 18 '19 at 11:36