Geometric y=1/x thinking.. (A(a) = int[1, a] [1/x] dx; why is A(a) + A(b) = A(ab)?)

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apple2357

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I am trying to get my head around thinking about this observation geometrically:

Take f(x) = 1/x and find the area using integration between say x=1 and x= a, lets call that A(a)
Why is it the case that A(a) + A(b) = A(ab) ?

I can see this comes directly from the properties of logarithms but can't see what is taking place geometrically?
What is about f(x) = 1/x that enables this property to hold? So far my thinking has gone along the lines of stretches and transformations but can't quite put my finger on it.

Is anyone particularly good at thinking geometrically and can offer me some help here?
 
apple2357 said:
I am trying to get my head around thinking about this observation geometrically:
Take f(x) = 1/x , why is it the case that A(a) + A(b) = A(ab) ?
Define \(\displaystyle A(x) = \int_1^x {\frac{1}{t}dt}\) now suppose that \(\displaystyle a>0~\&~b>0\) consider \(\displaystyle A(ab)\).
Setup preliminaries. Let \(\displaystyle u=\tfrac{t}{a},~du=\tfrac{1}{a}dt~\&~\begin{array}{*{20}{c}}t&\|& 1&{ab}\\\hline u&\|& 1&b\end{array}\)
\(\displaystyle \begin{align*}A(ab) &= \int_1^{ab} {\frac{1}{t}dt} \\&=\int_1^{a} {\frac{1}{t}dt} +\int_a^{ab} {\frac{1}{t}dt}\\&=\int_1^{a} {\frac{1}{u}du} +\int_a^{b} {\frac{1}{u}du}\\& =A(a)+A(b) \end{align*}\)
Click to expand...
Click to expand...
 
If I'm understanding you correctly, you've defined a function A(a) like so:

\(\displaystyle \int\limits_{1}^{a} \frac{1}{x} \: dx\)

Assuming that's correct, then we want to investigate the following:

\(\displaystyle A(a) + A(b) \stackrel{?}{=} A(ab)\)

To do so, let's assume that \(a > 1\) and \(b > 1\), and use the Fundamental Theorem of Calculus:

\(\displaystyle A(a) = \int\limits_{1}^{a} \frac{1}{x} \: dx = \ln(a) - \ln(1) = \ln(a)\)

\(\displaystyle A(b) = \int\limits_{1}^{b} \frac{1}{x} \: dx = \ln(b) - \ln(1) = \ln(b)\)

\(\displaystyle A(a) + A(b) = \ln(a) + \ln(b)\)

Then we can use one of the rules of logarithms to see that:

\(\displaystyle A(a) + A(b) = \ln(a) + \ln(b) = \ln(ab) = A(ab)\)
 
Thats nice and certainly justifies it algebraically, what does this mean geometrically though?
PKA you have used u= t/a ? Does that imply some kind of stretch taking place?
 
apple2357 said:
I am trying to get my head around thinking about this observation geometrically:

Take f(x) = 1/x and find the area using integration between say x=1 and x= a, lets call that A(a)
Why is it the case that A(a) + A(b) = A(ab) ?

I can see this comes directly from the properties of logarithms but can't see what is taking place geometrically?
What is about f(x) = 1/x that enables this property to hold? So far my thinking has gone along the lines of stretches and transformations but can't quite put my finger on it.

Is anyone particularly good at thinking geometrically and can offer me some help here?
Click to expand...
You're looking for a geometrical explanation, rather than calculus, right?

If you look carefully at what pka did (and fix a little typo), you can see the geometrical interpretation.

The property of y = 1/x you ask for is the fact that a horizontal stretch by a factor of k is equivalent to a vertical stretch by a factor of k. (That is, replacing x with x/k to get y = 1/(x/k) = k/x is the same as multiplying by k.)

So the area from a to ab is the area from 1 to b, multiplied by a (because it has been stretched horizontally by that factor) and then divided by a (because it has been compressed vertically by the same factor) -- that is, it has the same area. So the area from 1 to ab is the area from 1 to a plus the area from a to ab, and the latter equals the area from 1 to b.
 
Dr.Peterson said:
You're looking for a geometrical explanation, rather than calculus, right?

If you look carefully at what pka did (and fix a little typo), you can see the geometrical interpretation.

The property of y = 1/x you ask for is the fact that a horizontal stretch by a factor of k is equivalent to a vertical stretch by a factor of k. (That is, replacing x with x/k to get y = 1/(x/k) = k/x is the same as multiplying by k.)

So the area from a to ab is the area from 1 to b, multiplied by a (because it has been stretched horizontally by that factor) and then divided by a (because it has been compressed vertically by the same factor) -- that is, it has the same area. So the area from 1 to ab is the area from 1 to a plus the area from a to ab, and the latter equals the area from 1 to b.
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Brilliant. That's exactly what i have been strugglng to articulate. I knew it was something to do with the fact that a horizontal stretch of k was equivalent to a vertical stretch of k ( which is unique to y=1/x) that was key here.
 
I'm glad you followed what I said, because (due to time constraints) I wrote it tersely enough that I wasn't sure I followed it myself! If anyone needs a longer, illustrated version that makes it clearer, I can work on that!
 
Can't say i have fully resolved it but i know it i was on the right tracks with my thinking on stretching in two directions. i feel i need to do a picture with coordinates to really understand it.
So yes, i haven't quite understood it well enough if i had to explain it just yet!
 
apple2357 said:
Thats nice and certainly justifies it algebraically, what does this mean geometrically though?
PKA you have used u= t/a ? Does that imply some kind of stretch taking place?
Click to expand...
This is the exact proof that Leonard Gillman gives in his 1973 edition of Calculus pages 272f. In that whole text is concept of the integral is based upon the geometric idea of betweenest.
 
apple2357 said:
Can't say i have fully resolved it but i know it i was on the right tracks with my thinking on stretching in two directions. i feel i need to do a picture with coordinates to really understand it.
So yes, i haven't quite understood it well enough if i had to explain it just yet!
Click to expand...
Here are some pictures that may help, with a specific example. First, here is what we want to show, that ln 6 = ln 2 + ln 3; i.e. that the area from 1 to 6, which is the area from 1 to 2 plus the area from 2 to 6, is equal to the area from 1 to 2 plus the area from 1 to 3 (shown in the second picture):
FMH115301a.png

So let's see that the area from 1 to 3 equals the area from 2 to 6.
Here we start with the area from 1 to 3 and stretch it horizontally by a factor of 2, doubling its area:
FMH115301b.png
Now we compress this vertically by a factor of 2, returning the area to what we started with:
FMH115301c.png

It isn't perfect, but does it give the idea?
 
Yes it gives the idea. Though it is hard to see how the shaded area ( from 2 to 6) in your second pic above is twice the area from 1 to 3 graphically but by stretching horizontally, it must be?
 
pka said:
This is the exact proof that Leonard Gillman gives in his 1973 edition of Calculus pages 272f. In that whole text is concept of the integral is based upon the geometric idea of betweenest.
Click to expand...

Thats really interesting. Is there a screenshot i can see of this text anywhere? The text costs $90 on amazon!
 
apple2357 said:
Thats really interesting. Is there a screenshot i can see of this text anywhere? The text costs $90 on amazon!
Click to expand...
Are you not near a university library? Or look for a used copy in fair to good condition.
 
You are right. Seen a copy in the states from ebay for $7 but some postage!
But first i will try and find one in a library!
 
Just realised i dont get understand PKA post above,

How line 2 moves to line 3, if du= 1/a dt, where has the 'a' gone to ?
 
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