Primitive of Pointwise Sum of Functions/Proof 1
Jump to navigation
Jump to search
Theorem
Let $f_1, f_2, \ldots, f_n$ be real functions which are integrable.
Then:
- $\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
Proof
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
$\map P 1$ is true, as this just says:
- $\ds \int \map {f_1} x \rd x = \int \map {f_1} x \rd x$
Basis for the Induction
$\map P 2$ is the case:
\(\ds \int \map {\paren {f_1 \pm f_2} } x \rd x\) | \(=\) | \(\ds \int \paren {\map {f_1} x \pm \map {f_2} x} \rd x\) | Definition of Pointwise Addition of Real-Valued Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \map {f_1} x \rd x \pm \map {f_2} x \rd x\) | Linear Combination of Primitives |
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x$
Then we need to show:
- $\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_{k + 1} } x \rd x$
Induction Step
This is our induction step:
\(\ds \) | \(\) | \(\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {\map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x + \map {f_{k + 1} } x} \rd x\) | Definition of Pointwise Addition of Real-Valued Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x + \int \map {f_{k + 1} } x \rd x\) | Basis for the Induction | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x + \int \map {f_{k + 1} } x \rd x\) | Induction Hypothesis |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N: \ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
$\blacksquare$