Sunday, October 11, 2015

Understanding simplification of summation



I am reading the following proof and I am having trouble understanding this line of reasoning.

$$ \sum_{i=1}^n i^k \leq \sum_{i=1}^nn^k = n* n^k $$



However, I do not understand how the second summation simplifies to $n* n^k$. Where is the $i$ in the second summation to iterate? And without it, how can we even sum anything?


Answer



$$\sum_{i=1}^nn^k = \underbrace{n^k+ \cdots + n^k}_{n\text{ times}} = n^k(\underbrace{1+ \cdots + 1}_{n\text{ times}}) = n^k\cdot n$$






Identities to remember: $$\begin{array}{lr}\sum^n_{i=1} 1 = n & \text{(add $1$ $n$ times)} \\ \sum^n_{i=1} ka_i = k\sum^n_{i=1} a_i & \text{(factoring out a constant)}\end{array}$$ Putting those together you get $$\sum^n_{i=1} k = k\sum_{i=1}^n 1 = kn$$ In this case $n^k$ is just a constant (with respect to $i$).


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...