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Given some sum, $\displaystyle\sum \ln x_i = k $ We have $\ln \prod x_i = k$

I've always found this relation to be really interesting. I saw it used once in a linear algebra proof but I haven't seen it since. Are there any other interesting uses of this trick?

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Yes, finding product expansion for sums, especially infinite sums, can sometimes shed new light on them. For example, consider the zeta function, defined for complex numbers with $\Re(z)>1$ by

$\zeta(s)=\sum_{n=0}^\infty \frac{1}{n^s}.$

It is not hard, though not trivial, to show this "factors" as one would expect, giving the infinite product

$\zeta(s)=\prod_p^\infty \left(1-\frac{1}{p^s}\right)$

where the product is taken over all primes $p$.

This shows that the zeta function has no zeros for $\Re(z)>1$, something which is not obvious from examining the sum representation.

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    In your first formula I think you either want $\zeta(s)$ or $n^z$.2012-06-08
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I often use the series of the logarithm to reduce computational effort when computing logarithms. It is usually easier to compute many small logarithms then one large one, so I often exploit this by cutting up the number I want to compute.

This is a simple way to compute $\log x$. We have

$\log x=\log \left(\frac {x}{2}\right)+\log 2$

repeating this process $n$ times with a pre-computed $\log 2$, we have

$\log x=\log \left(\frac{x}{2^{n}}\right)+n\log 2$

We can then quickly compute $\log \left(\frac{x}{2^{n}}\right)$ with a Taylor's Series, its Chebyshev economized counterpart, or a different, more efficient method. I usually choose $n$ as a number that makes $\left(\frac{x}{2^{n}}\right) \approx 1$ to make it easier to estimate in a series.


For example, I may compute $\log 10$ efficiently using the above basic method. Say we know $\log 2$ to $10$ places, i.e. $0.6931471805$. Letting $n=3$ (making $\left(\frac{10}{2^{n}}\right) = 1.25$),

$\log 10=\log \left( \frac{10}{8} \right)+3 * 0.6931471805=\log \left( \frac{5}{4} \right)+2.0794415415$

Already, note that $2.0794415415$ is a decent estimate of $\log 10$. Using 5 terms of the basic Taylor's series expnsion of $\log$, we have

$\log \left( \frac{5}{4} \right)\approx \left( \frac{5}{4} -1\right)-\frac{1}{2}\left( \frac{5}{4} -1\right)^2+\frac{1}{3}\left( \frac{5}{4} -1\right)^3-\frac{1}{4}\left( \frac{5}{4} -1\right)^4+\frac{1}{5}\left( \frac{5}{4} -1\right)^5 \approx 0.22317708\bar{3}$

accurate to 4 decimal places. Together, we have

$\log 10\approx 0.223177083+2.0794415415=2.3026186245$

accurate again to 4 decimal places.