Question

Express the peice-wise function: \[
f(x) =
\begin{cases}
g(x)& x>a\\
h(x) &x 12 years ago

Answer 1

The heaviside function is \[
H(t) =
\begin{cases}
1&t>0\\
0&t<0
\end{cases}
\]

Answer 2

If we put in \(t-a\) we get: \[
H(t-a) =
\begin{cases}
1&t>a\\
0&t<a
\end{cases}
\]

Answer 3

Multiplying by \(g(t)\) we get: \[


g(t)H(t-a) =
\begin{cases}
g(t)&t>a\\
0&t<a
\end{cases}

\]

Answer 4

Another thing we could do is: \[
f(t)+[g(t)-f(t)]H(t-a)
\]

Answer 5

So one simple thing we could do is: \[
h(t)H(a-t) + g(t)H(t-a)
\]

Answer 6

Another thing we could do is: \[
f(t) =h(t)+[g(t)-h(t)]H(t-a)
\]

Answer 7

I'm producing a method of expressing peice-wise functions without using \(\large\{\) notation but instead by using \(H(t)\).

Answer 8

Suppose we have \[
f(t) = \begin{cases}
k(t)& t>a \\
m(t)& b<t<a\\
n(t)& t<b

\end{cases}
\]If we let \[
h(t) = \begin{cases}
k(t)& t>a \\
m(t)& t<a

\end{cases}
\]Then we can say that: \[
f(t) = \begin{cases}
h(t)& t>b \\
n(t)& t<b

\end{cases}
\]

Answer 9

Thus, by induction, we could construct any piece-wise function, no matter how complicated, with Heaviside functions.