 |
 |
Natural Logarithms Calculus
Name: Scott
Status: other
Age: 50s
Location: N/A
Country: N/A
Date: N/A
Question:
I have forgotten how to differentiate and integrate natural
logarithms (if I ever knew) and my current statistics class requires it.
In particular, I am supposed to know how to integrate these functions:
a * e^(-b*x) = -a/b*e^(-bx)
a * x * e^(-b*x) = (-a/b^2 - -a*x/b) * e^(-b*x)
a * x^2 * e^(-b*x) = a * ( -2/b^3 - 2*x/b - x^2/b) * e^(-b*x)
a * x * e^(-b*x^2) = -a/(2*b) * e^(-b * x^2)
The fact that I have the answers in no way implies I understand how the
integration proceeds.
Replies:
What you have is not the integral per se but the antiderivative.
The difference is key. Doing the integral is well-defined
algorithmically, something you can easily program a computer to do,
but finding the antiderivative is black magic, an art, impossible to
program.
The integral of a function can be found by drawing a graph of the
function and determining the area enclosed by the curve, the x-axis,
and vertical lines at the upper and lower limits of integration. You
can see how you might program a computer to do this: divide the area
into tiny vertical strips. Approximate each strip as a rectangle of
fixed (tiny) width dx and height f(x) = the value of the function at
the point x. Sum up the areas dA(x) = f(x) dx of all these little
strips,
A(x1,x2) = Sum_{x from x1 to x2} dA(x) = Sum_{x1,x2} f(x) dx (1)
In the limit that the width of each strip becomes infinitismal,
dx -> 0, then this sum is defined to be the integral, and indeed the
symbol for an integral is an elongated, stylized ``S'' for ``sum.''
Well and good. But you recognize immediately that this has nothing
to do with what you are after, with your equations above. So let us
consider a more complicated problem: is there, in fact, ANOTHER
function F(x), such that
A(x1,x2) = F(x2) - F(x1) ? (2)
That is, is there a function which when evaluated between the
limits specified tells us the result of doing the integral? Common
sense suggests that in general the answer should be yes. After all,
presumably there is a unique answer to the question of what the area
under the function is between 0 and x, and that is really all you need
to define a function: a one-to-one mapping between x coordinate and
the value of the function F(x).
This function would be called the antiderivative, because it
``undoes'' the action of taking the derivative. That is, it is clear
(and is the fundamental theorem of integral calculus) that
d F
--- = f(x) (3)
d x
The derivative of F is f itself. Why so? Consider increasing the
upper limit of the integral every so slightly, by some tiny fraction
dx'. Clearly the change in area is given by
d F
d A = --- (x2) dx' (4)
d x
and from our algorithm above it must also be given by the area of
the extra tiny strip we have tacked onto the end of the area defined
by the curve and the original limits,
d A = f(x2) dx' (5)
Comparing (4) and (5) establishes (3).
Note, however, that as soon as you proceed to two- and
three-dimensional functions, like those in physics that describe
motion in three dimensions, then all bets are off, and indeed in this
more general case it is typical that there is most definitely NOT an
antiderivative for the generic multidimensional function f(x,y,z).
Indeed, the entirety of thermodynamics can be said, in a sense, to
proceed from the mere assertion that the equations of state
(mathematical relations describing observed correlations between
pressure, volume, and temperature) must have an antiderivative, called
the free energy. But I digress.
How do you find an antiderivative for the generic function f(x)? No
one knows! There is no recipe for constructing it, only a check once
you have a candidate. Given a function f(x), we do not know how to
construct the antiderivative F(x), only how to verify, using (3),
whether F is or is not the antiderivative of f.
We do not navigate blindly, however. Experience teaches that most
functions are combinations of simpler functions, and a few rules will
let you construct the antiderivative of the complex function out of
some combination of the antiderivatives of the simpler functions,
which can be worked out by trial and error.
For example, inspection alone serves to prove that if
f(x) = x^n
that is x raised to the power n, then a natural candidate for the
antiderivative is
F(x) = (n+1)^(-1) x^(n+1).
Is this the only possible candidate? This gets into questions of
uniqueness which I am not qualified to address, so let's just assume
the answer is yes. (Weasel, weasel, weasel! Too true, alas. But we
press on nonetheless. . .)
Now a polynomial is a complex function, but composed entirely of
powers, e.g.:
f(x) = a x^2 + b x + c (6)
One rule of integral calculus is that you may integrate terms of a
sum independently, and sum the results, that is the antiderivative of
a sum of functions is the sum of the antiderivatives of each function,
viz.:
F(x) = a 3^(-1) x^3 + b 2^(-1) x^(2) + c x (7)
You can take the derivative of (7) and of each term therein and
verify not only that (7) is the antiderivative of (6), but that EACH
TERM of (7) is the antiderivative of the corresponding term in (6).
There are many rules like this. Constructing an antiderivative is a
question of (1) application of the rules, (2) manipulation of the
original function, and (3) recognition of the known antiderivatives of
simple functions. For example, let's look at your three functions:
d F
--- = f(x) = a e^(-b x) (8)
d x
First we manipulate (2) by making the substitution y = - b x,
d F d y
--- --- = a e^y (9)
d y d x
I've used the chain rule from differential calculus on the
left-hand side. We need to insert dy/dx = -b, like so,
d F
--- = (-a/b) e^y (10)
d y
Now we must recognize that the antiderivative of f(x) = e^x is
F(x) = e^x,
F(y) = (-a/b) e^y (11)
and undo our manipulation,
F(x) = (-a/b) e^(- b x) (12)
All done. This is your first equation. Now let us look at the
second,
d F
--- = f(x) = x e^(-b x) (13)
d x
We see here a function we can handle, multiplied by another. It's time
to use one of the rules, the product rule in particular, which says,
d d g d f
--- f(x) g(x) = f --- + --- g (14)
d x d x d x
If we let df/dx on the RHS be the function of which we already know how
to take the antiderivative, e^(-b x), then we have:
d d g
--- [ (-a/b) e^(-b x) g(x) ] = -(a/b) e^(- b x) --- + e^(- b x) g(x) (15)
d x d x
This is progress??? Well, yes it is. Let us now set g(x) equal to
the damned newcomer, g(x) = x. That will make the last term on the RHS
equal to our problem child the RHS of (13). We also need dg/dx to stick
into the first term on the RHS of (15), but dg/dx = 1, so we have
d
--- [ (-a/b) x e^(-b x) ] = -(a/b) e^(- b x) + x e^(- b x) (16)
d x
Rearranging:
d F d
x e^(- b x) = --- = --- [ (-a/b) x e^(-b x) ] + (a/b) e^(- b x) (17)
d x d x
Can you see it coming? Now look at the RHS of (17). The
antiderivative of the first term is obvious -- it's what's inside the
square brackets. And the antiderivative of the second term we already
know, because it's almost exactly our old friend from (8) above. That is,
using (12):
d F d d
--- = --- [ (-a/b) x e^(-b x) ] + (1/b) --- [ (-a/b) e^(- b x) ] (18)
d x d x d x
or, using the rule that the sum of antiderivatives is the
antiderivative of a sum,
d F d
--- = --- [ (-a/b) x e^(-b x) - (a/b^2) e^(- b x) ] (19)
d x d x
Well, we don't need to be a genius to realize that F must be just
exactly what is inside the square brackets. And if you factor out the
exponentials you get your second equation.
The third of your equations can be solved similarly, but you need
to ``integrate by parts'' (use the product rule) twice.
The fourth of your equations is more troublesome, because the
antiderivative of e^(-x^2) is called the error function Erf(x), and
cannot be written down in terms of simpler functions (like x raised to
powers, or logs, sines and cosines, that kind of thing). So in general
integrals of functions containing e^(-x^2) can't be easily done. This
one in particular is easy, however, by good luck:
d F
--- = a x e^(-b x^2) (20)
d x
Let y = - b x^2. Then
d F d y
--- --- = a x e^y (21)
d y d x
I need dy/dx = - 2 b x, which gives:
d F
--- = - 0.5 (a/b) e^y (22)
d y
and I use the known antiderivative of e^y,
F(y) = - 0.5 (a/b) e^y (23)
and undo the manipulation,
F(x) = - 0.5 (a/b) e^(- b x^2) (24)
That's your fourth equation. This was lucky, however. If the x
were not in front, it would not have canceled on both sides in the
transition from (21) to (22), and we would have found the problem
insoluble. If it had been an x^2 instead of an x, we would also have
been stuck.
You will find a standard college calculus textbook, like Thomas and
Finney, chock-a-block full of fun stuff like this. It's worth
studying, because while statistics is good fun and on occasion useful,
calculus underlies *all* of modern physics, and a great deal else
besides -- modern economics, derivative securities -- who knows?
perhaps sex, death, and the meaning of life itself. . .
Or maybe not. Good luck.
Dr. Grayce
Click here to return to the Mathematics Archives
| |
Update: June 2012
|
|
