I want the complete expansion of log(a+b)=?
for ex
log(a*b)=log a + log b;
log(a/b)=lob a - log b;
Similar to this, is there any expansion for log(a+b)???
Thanks in advance...
A:
Why would you ever want to do this? The property that log (a*b) = log a + log b is only useful because it transforms a multiplication operation into an addition operation. log (a+b) already involves only an addition, so it makes no sense to have any further expansion.
Of course you can always use one of the several series for computing logarithms, but the fastest way would be to simply compute log (a+b) directly. For that matter, on most computers, even log (a*b) is going to be faster than log a + log b, since the latter involves an extra logarithm operation.
casablanca
2010-10-20 04:54:27
+3
A:
In general, one doesn't expand out log(a + b); you just deal with it as is. That said, there are occasionally circumstances where it makes sense to use the following identity:
log(a + b) = log(a * (1 + b/a)) = log a + log(1 + b/a)
(In fact, this identity is often used when implementing log in math libraries).
Stephen Canon
2010-10-20 06:32:48
And, we can also use the Taylor's Series for expanding ln(1+b/a)...
wiz kid
2010-10-20 06:34:16
@wiz kid: assuming `b` is smaller in magnitude than `a`, of course. (Otherwise, swap them).
Stephen Canon
2010-10-20 06:36:14
@Stephen: Thanks a lot buddy :)
wiz kid
2010-10-20 06:36:46
This makes sense only when log a is easy to calculate - which is rare. However if you want log10 or log2, the identity still holds, and the simple cases where log a is nice are easier to identify.
Michael Anderson
2010-10-20 06:38:27
@Michael Anderson: In the typical usage for a math library, one has a table with precomputed pairs of `a` and `log a` stored to high accuracy.
Stephen Canon
2010-10-20 06:41:22