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94 lines
3.6 KiB
TeX
94 lines
3.6 KiB
TeX
\newsection
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\section{Introduction}
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\subsection{Sufficiency \& Necessity}
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\begin{definition}[]{Sufficiency}
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A condition $P$ is called \textit{sufficient} for $Q$ if knowing $P$ is true is enough evidence to conclude that $Q$ is true.
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This is equivalent to saying $Q \Rightarrow P$.
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\end{definition}
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\begin{definition}[]{Necessity}
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A condition $P$ is called \textit{necessary} for $Q$ if $Q$ cannot occur unless $P$ is true, but doesn't imply that $Q$ is true, only that it is false if $P$ is false.
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This is equivalent to saying $P \Rightarrow Q$
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\end{definition}
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\subsection{Asymptotic Growth}
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$f$ grows asymptotically slower than $g$ if $\displaystyle\lim_{m \rightarrow \infty} \frac{f(m)}{g(m)} = 0$.
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We can remark that $f$ is upper-bounded by $g$, thus $f \leq \tco{g}$ and we can say $g$ is lower bounded by $f$, thus $g \geq \tcl{f}$.
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If two functions grow equally fast asymptotically, $\tct{f} = g$
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\subsection{Runtime evaluation}
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Identify the basic operations (usually given by the task), then count how often they are called and express that as a function in $n$.
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It is easier to note that in sum notation, then simplify that sum notation into a formula not containing any summation symbols.
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% ────────────────────────────────────────────────────────────────────
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\subsection{Tips for Converting Summation Notation into Summation-Free Notation}
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\subsubsection{Identify the Pattern:}
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\begin{itemize}
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\item Examine the summand.
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\item Look for patterns related to the index variable (usually $i$, $j$, etc.). Is it a linear function, a power of $i$, a combination?
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\end{itemize}
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\subsubsection{Arithmetic Series Formula}
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If the summand is a simple arithmetic progression (e.g., $a + bi$ where $a$ and $b$ are constants), use the formula:
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\[
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\sum_{i=m}^{n} (a + bi) = (n - m + 1)\left(a + b\frac{m + n}{2}\right)
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\]
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\subsubsection{Power Rule for Sums}
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\begin{itemize}
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\item For sums involving powers of $i$, you can use the following pattern:
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\[
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\sum_{i=1}^{n} i^k = \frac{n^{k+1}}{k+1}
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\]
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\item Remember that this rule only applies when the index starts at 1.
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\end{itemize}
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\subsubsection{Telescoping Series}
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Look for terms in consecutive elements of the summand that cancel out, leaving a simpler expression after expanding. This is particularly helpful for fractions and ratios.
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\subsubsection{Geometric Series Formula}
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For sums involving constant ratios (e.g., $a \cdot r^i$ where $r$ is the common ratio), use:
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\[
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\sum_{i=0}^{n} a \cdot r^i = a \frac{1 - r^{n+1}}{1-r}
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\]
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\subsubsection{Gaussian Formula}
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If $S$ is an arithmetic series with $n$ terms, then $S = \frac{n}{2} * (a + 1)$
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\subsubsection{Examples}
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The only other way (other than learning these tips) in which you are going to get better at this is by parctising.
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Work through examples, starting with simpler ones and moving towards more complex expressions.
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\fhlc{Aquamarine}{Example:}
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Let's convert the summation: $\sum_{i=1}^{5} i$
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\begin{enumerate}
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\item \textbf{Pattern:} The summand is simply $i$, which represents a linear arithmetic progression.
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\item \textbf{Arithmetic Series Formula:} Applying the formula with $a = 1$, $b = 1$, $m = 1$, and $n = 5$:
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\[
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\sum_{i=1}^{5} i = (5 - 1 + 1)\left(1 + 1 \cdot \frac{1 + 5}{2}\right) = 5 \cdot 3 = 15
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\]
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\end{enumerate}
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Therefore, the summation evaluates to $15$.
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\subsection{Specific examples}
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\begin{align*}
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\frac{n}{\log(n)} \geq \Omega(\sqrt{n}) \Leftrightarrow \sqrt{n} \leq \tco{\frac{n}{\log(n)}}
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\end{align*}
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