Easy approximations of log2(3/2) and log2(4/3)

Take 1/x. The integral of 1/x is ln x. Take the integral from x=2 to
x=4. Take the points x=2, x=3 and x=4. The whole integral is ln 4 - ln
2= ln (4/2)= ln (2) which is nice because I need to divide that out
anyway.

Take tangents at 2.5 and 3.5. Draw this as a diagonal line between 2
and 3, and another between 3 and 4 (It doesn't really matter what slope
you use, but the derivatives at these points make pretty nice slopes).

Now you obtain trapezoids with volume 2/5 and 2/7. This is 14/35 and
10/35. The whole thing is approximated by 24/35 (ln 2). Since we need
to divide by ln 2 to get log(base2) we get (14/35)/(24/35) and (10/35)/
(24/35) which is 7/12 and 5/12, which of course is the 12t-et
approximation of log(base2) of 3/2 and 4/3 respectively.

Can you provide graphical aids along with the explanations?

----- Original Message -----
From: "Paul G Hjelmstad" <paul.hjelmstad@us.ing.com>
To: <tuning@yahoogroups.com>
Sent: 27 Mart 2006 Pazartesi 20:13
Subject: [tuning] Easy approximations of log2(3/2) and log2(4/3)

> Take 1/x. The integral of 1/x is ln x. Take the integral from x=2 to
> x=4. Take the points x=2, x=3 and x=4. The whole integral is ln 4 - ln
> 2= ln (4/2)= ln (2) which is nice because I need to divide that out
> anyway.
>
> Take tangents at 2.5 and 3.5. Draw this as a diagonal line between 2
> and 3, and another between 3 and 4 (It doesn't really matter what slope
> you use, but the derivatives at these points make pretty nice slopes).
>
> Now you obtain trapezoids with volume 2/5 and 2/7. This is 14/35 and
> 10/35. The whole thing is approximated by 24/35 (ln 2). Since we need
> to divide by ln 2 to get log(base2) we get (14/35)/(24/35) and (10/35)/
> (24/35) which is 7/12 and 5/12, which of course is the 12t-et
> approximation of log(base2) of 3/2 and 4/3 respectively.
>
>
>

--- In tuning@yahoogroups.com, "Ozan Yarman" <ozanyarman@...> wrote:
>
> Can you provide graphical aids along with the explanations?

I'll post a pretty picture to my Files Section. (on tuning-math,
Files, Paul Hj.)

Here's a crude attempt (Hit Reply to see it correctly)

|*
| *
| | | | *
| | | | |
0-------1-------2-------3-------4

Imagine the curve connecting the dots. (It's not really a straight
line). Find the tangents to the curve at 2 1/2 and 3 1/2, with
slope d/dx (1/x). Draw diagonal with slopes -x^-2 at these points,
which I is -4/25 and -4/49 respectively - which gives a nice
fit to the curve, but the exact slope isn't too important, because,
with trapezoids, the area is just going to be the same as if you
drew a horizontal line across at those tangent points.

That's all there is to it!

>
>
> ----- Original Message -----
> From: "Paul G Hjelmstad" <paul.hjelmstad@...>
> To: <tuning@yahoogroups.com>
> Sent: 27 Mart 2006 Pazartesi 20:13
> Subject: [tuning] Easy approximations of log2(3/2) and log2(4/3)
>
>
> > Take 1/x. The integral of 1/x is ln x. Take the integral from
x=2 to
> > x=4. Take the points x=2, x=3 and x=4. The whole integral is ln
4 - ln
> > 2= ln (4/2)= ln (2) which is nice because I need to divide that
out
> > anyway.
> >
> > Take tangents at 2.5 and 3.5. Draw this as a diagonal line
between 2
> > and 3, and another between 3 and 4 (It doesn't really matter what
slope
> > you use, but the derivatives at these points make pretty nice
slopes).
> >
> > Now you obtain trapezoids with volume 2/5 and 2/7. This is 14/35
and
> > 10/35. The whole thing is approximated by 24/35 (ln 2). Since we
need
> > to divide by ln 2 to get log(base2) we get (14/35)/(24/35) and
(10/35)/
> > (24/35) which is 7/12 and 5/12, which of course is the 12t-et
> > approximation of log(base2) of 3/2 and 4/3 respectively.
> >
> >
> >
>

--- In tuning@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@...> wrote:

You can do the same with endpoint-determined trapezoids

Connect y=1/2 to y=1/3, and y=1/3 to y=1/4, giving 10/17 and
7/17, which are the log2 approximations for 3/2 and 4/3 in 17t-ET...

> Take 1/x. The integral of 1/x is ln x. Take the integral from x=2
to
> x=4. Take the points x=2, x=3 and x=4. The whole integral is ln 4 -
ln
> 2= ln (4/2)= ln (2) which is nice because I need to divide that out
> anyway.
>
> Take tangents at 2.5 and 3.5. Draw this as a diagonal line between
2
> and 3, and another between 3 and 4 (It doesn't really matter what
slope
> you use, but the derivatives at these points make pretty nice
slopes).
>
> Now you obtain trapezoids with volume 2/5 and 2/7. This is 14/35
and
> 10/35. The whole thing is approximated by 24/35 (ln 2). Since we
need
> to divide by ln 2 to get log(base2) we get (14/35)/(24/35) and
(10/35)/
> (24/35) which is 7/12 and 5/12, which of course is the 12t-et
> approximation of log(base2) of 3/2 and 4/3 respectively.

>

Paul, can you direct me to the file you mentioned in case I missed it?

Also, an excel spreadsheet could help, since I'm a math doofus.

Cordially,
Oz.

----- Original Message -----
From: "Paul G Hjelmstad" <paul.hjelmstad@us.ing.com>
To: <tuning@yahoogroups.com>
Sent: 30 Mart 2006 Per�embe 1:39
Subject: [tuning] Re: Easy approximations of log2(3/2) and log2(4/3)

> --- In tuning@yahoogroups.com, "Ozan Yarman" <ozanyarman@...> wrote:
> >
> > Can you provide graphical aids along with the explanations?
>
> I'll post a pretty picture to my Files Section. (on tuning-math,
> Files, Paul Hj.)
>
> Here's a crude attempt (Hit Reply to see it correctly)
>
>
>
> |*
> | *
> | | | | *
> | | | | |
> 0-------1-------2-------3-------4
>
> Imagine the curve connecting the dots. (It's not really a straight
> line). Find the tangents to the curve at 2 1/2 and 3 1/2, with
> slope d/dx (1/x). Draw diagonal with slopes -x^-2 at these points,
> which I is -4/25 and -4/49 respectively - which gives a nice
> fit to the curve, but the exact slope isn't too important, because,
> with trapezoids, the area is just going to be the same as if you
> drew a horizontal line across at those tangent points.
>
> That's all there is to it!
>
>

--- In tuning@yahoogroups.com, "Ozan Yarman" <ozanyarman@...> wrote:
>
> Paul, can you direct me to the file you mentioned in case I missed
it?
>
> Also, an excel spreadsheet could help, since I'm a math doofus.
>
> Cordially,
> Oz.

Sorry, haven't done it yet. Actually, plan to do the graph in Excel...

>
> ----- Original Message -----
> From: "Paul G Hjelmstad" <paul.hjelmstad@...>
> To: <tuning@yahoogroups.com>
> Sent: 30 Mart 2006 Perþembe 1:39
> Subject: [tuning] Re: Easy approximations of log2(3/2) and log2(4/3)
>
>
> > --- In tuning@yahoogroups.com, "Ozan Yarman" <ozanyarman@> wrote:
> > >
> > > Can you provide graphical aids along with the explanations?
> >
> > I'll post a pretty picture to my Files Section. (on tuning-math,
> > Files, Paul Hj.)
> >
> > Here's a crude attempt (Hit Reply to see it correctly)
> >
> >
> >
> > |*
> > | *
> > | | | | *
> > | | | | |
> > 0-------1-------2-------3-------4
> >
> > Imagine the curve connecting the dots. (It's not really a straight
> > line). Find the tangents to the curve at 2 1/2 and 3 1/2, with
> > slope d/dx (1/x). Draw diagonal with slopes -x^-2 at these points,
> > which I is -4/25 and -4/49 respectively - which gives a nice
> > fit to the curve, but the exact slope isn't too important,
because,
> > with trapezoids, the area is just going to be the same as if you
> > drew a horizontal line across at those tangent points.
> >
> > That's all there is to it!
> >
> >
>