Linear Function
A linear function is a function f which satisfies
f(x+y)=f(x)+f(y)
and
f(alpha x)=alpha f(x)
for all x and y in the domain, and all scalars alpha.
A continuous linear function must have the form f(x) = ax. Discontinuous linear functions look dreadful.
To be more specific, I am going to discuss real valued functions of one real variable, i.e. f: RR, where R is, as usual, the set of all real numbers. Such a function is called linear provided the following condition holds:
(*) For every two real x1 and x2, f(x1 + x2) = f(x1) + f(x2)
Assuming that the function f is also continuous I plan to show that f(x) = ax for some real a. Please note that if indeed f(x) = ax then a = f(1) which provides a starting point for the proof. But first let me note that (*) contains an unknown which, as we are going to establish, is equal to f(x) = ax. In other words, (*) serves as an example of a functional equation - an equation whose unknown is a function
Proof
The proof proceeds in several steps.
1)x is 0. f(0) = f(0 + 0) = f(0) + f(0) = 2f(0).
Therefore f(0) = 2f(0) and finally f(0) = 0.
2)x is negative.
Let x be negative, e.g., let x + y = 0, where y is positive; so that -x = y. Then
0 = f(0) = f(x + y) = f(x) + f(y).
Therefore f(-x) = f(y) = -f(x).
3)x is an integer.
We have f(2) = f(1 + 1) = f(1) + f(1) = 2f(1). By induction, assume f(k - 1) = (k - 1)f(1). Then
f(k) = f(1 + (k-1)) = f(1) + (k-1)f(1) = kf(1).
Let's denote a = f(1). We have shown that for all integers n, f(n) = an.
4)x is rational
First of all, for any integer n0, we have 1 = n/n. Then, as before, a = f(1) = f(n/n) = nf(1/n). Hence, f(1/n) = a/n = a(1/n). For p = m/n we similarly have
f(p) = f(m/n) = mf(1/n) = m*a/n = a(m/n) = ap.
5)x is irrational
Any irrational number r can be approximated by a sequence of rational numbers pi. The closer pi is to r, the closer api is to ar. However, since api = f(pi) and assuming f continuous we must necessarily get f(r) = ar.
Continuity of the function is quite essential as it's possible to show [Ref. 1, 2] that the graph of any discontinuous solution to (*) is dense in the plane R2. For the sake of reference, the graph of a function f: RR is defined as a set of pairs (x, y), i.e. elements of R2 such that y = f(x). Formally, graph(f) = {(x, y)R2: y = f(x)}.
U can understand by this example:
My question is about the standard form of a linear equation. I already
know it's ax+by=c, but I need to know what the variables mean and if
there are any conditions to them. I tried looking in a textbook called
_College Algebra_, but they gave this other fomula, ax+b=o. I already
know a little about how it can be converted into slope-intercept form
but I'm still a little fuzzy on its relation to ax+by=c.
Also, what is it about something function f f(x)+ mx+b ?
The numbers represented by a, b, and c don't have meanings like m, the
slope, and b, the y-intercept, in the slope-intercept form y = mx + b.
Notice that if you multiply all three numbers a, b, and c by
the same amount, you get a different equation that has the SAME graph
- that is, an equivalent equation. For instance,
2x - 3y = 7
and
4x - 6y = 14
are both equivalent to the slope-intercept form
y = (2/3)x - 7/3
In general, the equation ax + by = c is equivalent to the slope-
intercept form
y = (-a/b)x + c/b
You see that the slope is -a/b and the y-intercept is c/b. It's the
ratios of a and c to b that have meaning.
Why, then, do we bother with the "standard form"? Isn't the slope-
intercept form better? It's simpler, and the numbers have meaning.
Well, the advantage that the standard form has over slope-intercept
form is that EVERY linear equation can be written in standard form,
but not every linear equation can be written in slope-intercept form.
For instance, this is a
linear equation:
x = 5
This is the equation of a straight line - in particular, a vertical
line. Its slope is "infinite"; that is, it is not a number. You can't
write it in the form y = mx + b if m is not a number. Just try it!
Thus, the standard form is important when you are trying to describe a
line that may be vertical. If you know the slope, and it is a number
(not infinite), then the slope-intercept form is fine. But suppose you
are describing a rotating line: at some time it will be vertical, and
then the slope-intercept form won't work. You can either make that a
special case, or you can use the standard form so the same description
will work all the time.
I don't know exactly what you saw about functions. Perhaps you meant
to type
f(x) = mx + b
This is a linear function. You can think of a function as a machine
that takes in a number, does some sort of work on it, and puts it out
as another number at the other end of the machine. In this case, you
put in some number x; the function multiplies it by m, then adds b to
the result, and puts out this new number, which is called f(x).
If you give that new number (the output of the function) the name y,
then you have your familiar linear equation in slope-intercept form:
y = mx + b.
A big difference between the function and the equation is that a
function is only allowed to put out ONE number for each number that
goes in. If you think about a vertical line such as x = 5, you can't
make this into a function. You can only put in the number 5, and ANY
number can come out. (The point (5,y) is on the line for ANY value of
y.) A function isn't allowed to do this. Therefore, a linear FUNCTION
is never graphed by a vertical line. The slope-intercept form can
describe any linear function.
A linear EQUATION, on the other hand, may be graphed as a vertical
line; the slope isn't always a number, so the slope-intercept form
doesn't always work. That's why you need the standard form for a
linear equation
I can tell u more with graphs but its not possible here!!