Summation is Linear/Scaling of Summations
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Theorem
Let $\tuple {x_1, \ldots, x_n}$ and $\tuple {y_1, \ldots, y_n}$ be finite sequences of numbers of equal length.
Let $\lambda$ be a number.
Then:
- $\ds \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$
Proof
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
- $\ds \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \lambda \sum_{i \mathop = 1}^1 x_i\) | \(=\) | \(\ds \lambda x_1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^1 \lambda x_i\) |
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\ds \lambda \sum_{i \mathop = 1}^k x_i = \sum_{i \mathop = 1}^k \lambda x_i$
from which it is to be shown that:
- $\ds \lambda \sum_{i \mathop = 1}^{k + 1} x_i = \sum_{i \mathop = 1}^{k + 1} \lambda x_i$
Induction Step
This is the induction step:
\(\ds \lambda \sum_{i \mathop = 1}^{k + 1} x_i\) | \(=\) | \(\ds \lambda \paren {\sum_{i \mathop = 1}^k x_i + x_{k + 1} }\) | Definition of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \sum_{i \mathop = 1}^k x_i + \lambda x_{k + 1}\) | Distributive Property | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^k \lambda x_i + \lambda x_{k + 1}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^{k + 1} \lambda x_i\) | Definition of Summation |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \N_{> 0}: \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$
$\blacksquare$