Log(x): monotonic increasing?
Arun Nagarkatti
increasing function. I was wondering if this is a valid enough proof,
or is there a bit more rigour involved?
(log(x) is to the base e, incidentally)
f(x) = log(x) defined for x > 0.
f'(x) = 1/x
f''(x) = -1/x^2.
The second derivative is negative, and this holds for all x > 0, thus
the function is non-decreasing, or monotonic increasing function.
Would this be a reasonable proof?
Thanks.
Neeraj.
spam...@nil.nil
> I've often needed to assume in some proofs that log(x) is a monotonic
> increasing function. I was wondering if this is a valid enough proof,
> or is there a bit more rigour involved?
> (log(x) is to the base e, incidentally)
> f(x) = log(x) defined for x > 0.
> f'(x) = 1/x
STOP HERE
> f''(x) = -1/x^2.
> The second derivative is negative, and this holds for all x > 0, thus
> the function is non-decreasing, or monotonic increasing function.
> Would this be a reasonable proof?
No. Consider log(x)-100x (this function is not monotonic increasing). Its
second derivative is -1/x^2 also. The second derivative tells you how the
function is BENDING (it is bending down - but a function can go down and
still bend down). The FIRST derivative tells you which way the function is
actually going. Just look at the first derivative. 1/x. Log (for real
analysis) is only defined for positive x. 1/x is positive. The function is
GOING up (it is increasing).
The first derivative tells you about increasing or decreasing (which way
the function is "going" - up or down).
The second derivative tells you about concavity/convexity (which way the
function is "bending" - up or down).
(a function can go up and bend downwards, go down and bend further
downwards, go down and bend upwards, go upwards and bend further upwards -
the first derivative tells you the direction and the second tells you
which way it is "bending")
Arun Nagarkatti
increasing function?
Robin Chapman
Arun Nagarkatti <arun.na...@ntlworld.com> wrote:
> So how do you suggest I go about proving this fact, that Log(x) is
monotonic
> increasing function?
If you really must use differentiation, then at least refrain
from repeating it,
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
Sent via Deja.com http://www.deja.com/
Before you buy.
Johannes H Andersen
Arun Nagarkatti wrote:
>
> So how do you suggest I go about proving this fact, that Log(x) is monotonic
> increasing function?
>
Use the definition of log(x) = integral(1/t)dt from 1 to x .
Hence: log(x2) - log(x1) = ( integral(1/t)dt from x1 to x2 ) > 0 for x2 > x1 >
0.
since the integrand is > 0 .
Johannes
David C. Ullrich
<arun.na...@ntlworld.com> wrote:
>So how do you suggest I go about proving this fact, that Log(x) is monotonic
>increasing function?
Read what he wrote! You need only show that the _first_
derivative is positive. Which is clear since the derivative is 1/x.
(We're only talking about x > 0 since that's the domain of log.)
Ronald Bruck
--Ron Bruck
In article <39FB8385...@ntlworld.com>, Arun Nagarkatti
<arun.na...@ntlworld.com> wrote:
:So how do you suggest I go about proving this fact, that Log(x) is
:
--
Due to University fiscal constraints, .sigs may not be exceed one
line.
Nate Eldredge
> So how do you suggest I go about proving this fact, that Log(x) is monotonic
> increasing function?
Prove that f'(x) is positive over the region of interest. (This
should not be hard :)
--
Nate Eldredge
neld...@hmc.edu
Doug Norris
>Would this be a reasonable proof?
What's wrong with just looking at f'(x)=1/x, and noticing that f'(x)>0 for
all x where log(x) is defined?
Doug
Doug Norris
>So how do you suggest I go about proving this fact, that Log(x) is monotonic
>increasing function?
Try actually reading what he wrote to you, particularly the part where he
explains how to prove Log(x) is monotone increasing.
Doug
Wim Benthem
<rjch...@my-deja.com> wrote:
>In article <39FB8385...@ntlworld.com>,
> Arun Nagarkatti <arun.na...@ntlworld.com> wrote:
>> So how do you suggest I go about proving this fact, that Log(x) is
>monotonic
>> increasing function?
>
>If you really must use differentiation, then at least refrain
>from repeating it,
Without differentiation you have to prove that y>x implies tha
log(y)>log(x).
log(y) = log(x + (y-x)) = log ( x (1 + (y-x)/x) ) =
log(x) + log (1+(y-x)/x)
if y>x then y-x is positive and (1+(y-x)/x) > 0 therefore
log(y)>log(x)
Wim
Zdislav V. Kovarik
Doug Norris <norr...@rintintin.colorado.edu> wrote:
To have more fun with it:
(1)
You can also use one of the possible definitions of the natural
logarithm:
log(x) = lim[n -> infinity] (x^(1/n) - 1) * n
knowing that x^(1/n) is increasing in x. Sure, all you get in the
limit is
"if 0 < x < y then log(x) <= log(y)"
then you have to establish that log is not constant.
(2)
Using the Mean Value Theorem for the logaritm and some elementary
algebra, you can prove
(*) that the expression
t = 1/log(w) - 1/(w-1)
which is defined for all w>0 with w different from 1, satisfies
0 < t < 1.
From this you can prove (setting w = b/a, a>0 and b >0 and
different)
log(b) - log(a) = (b - a) / (a + (b-a)*t)
so that you will have some insight about how fast log increases.
(Remark: It would be interesting to prove (*) without MVT. Anyone?)
Have fun, ZVK(Slavek).