Is $\sum_{i=1}^n [-1-x_i e^{-\alpha}]=0$ solvable analytically or numerically for $\alpha$?

Question

My own interpretation is that if one tries to solve the above by taking the logarithm, then one is not able to get rid of the logarithm, but rather it has to be approximated numerically.

Is this correct?

Exact solution ...$\alpha=- \ln \left(\frac{-n}{\sum x_i} \right)$ ... the moment you plug values in, you will have an approximate solution (unless you are really lucky). — Donald Splutterwit, Apr 01 '17 at 12:26
@DonaldSplutterwit But there's no logarithm of negative number? — mavavilj, Apr 02 '17 at 01:08

score 1 · Accepted Answer · answered Apr 01 '17 at 12:28

1

Its actually totally straight-forward $$\begin{align} 0 &= \sum_{i=1}^n(-1-x_ie^{-\alpha}) \\ &=-n-e^{-\alpha}\sum_{i=1}^nx_i \end{align}$$ This can be explicitly solved for alpha: $$ \alpha = \log\left(\frac{-1}{n}\sum_{i=1}^nx_i\right)$$

answered Apr 01 '17 at 12:28

Simon

5,061

But there's no logarithm of negative number? – mavavilj Apr 02 '17 at 01:05
2

@mavavilj If you take a look at the second expression you will notice that $n$ is positive and $e^{-\alpha}$. If you rewrite this expression to $\sum_{i=1}^nx_i = -\frac{n}{e^{-\alpha}}$ you will notice that the sum is negative. This means that argument of the logarithm function will always be positive. – Slaven Glumac Apr 02 '17 at 06:06

Is $\sum_{i=1}^n [-1-x_i e^{-\alpha}]=0$ solvable analytically or numerically for $\alpha$?

1 Answers1