Alan'space

카테고리

Here are some techniques for solving math problems.

Adding new term while keep the equation true

a=b−c+cba=bc×ca\begin{aligned} a &= b - c + c \\ \frac{b}{a} &= \frac{b}{c} \times \frac{c}{a} \\ \end{aligned}

Manipulating fractions

1−1n=n−1nnn−1=n−1n−1+1n−1=1+1n−11n(n+1)=1n−1n+1\begin{aligned} 1 - \frac{1}{n} &= \frac{n-1}{n} \\ \frac{n}{n-1} &= \frac{n-1}{n-1} + \frac{1}{n-1} = 1 + \frac{1}{n-1} \\ \frac{1}{n(n+1)} &= \frac{1}{n} - \frac{1}{n+1} \end{aligned}
Example: Limit of (1−1/n)n(1-1/n)^n as n→∞n \to \infty
(1−1n)n=(n−1n)n=(1nn−1)n=(11+1n−1)n=(11+1n−1)n−1⋅11+1n−1\begin{aligned} (1-\frac{1}{n})^n &= (\frac{n-1}{n})^n \\ &= (\frac{1}{\frac{n}{n-1}})^n \\ &= (\frac{1}{1+\frac{1}{n-1}})^n \\ &= (\frac{1}{1+\frac{1}{n-1}})^{n-1} \cdot \frac{1}{1+\frac{1}{n-1}}\\ \end{aligned}

Recall that:

e≔lim⁡n→∞(1+1n)ne \coloneqq \lim_{n \to \infty} (1+\frac{1}{n})^n

As n→∞n \to \infty, the left term goes to 1e\frac{1}{e} and the right term goes to 11. So:

lim⁡n→∞(1−1n)n=1e\lim_{n \to \infty} (1-\frac{1}{n})^n = \frac{1}{e}

Absolute values

For a,b∈Ra, b\in \mathbb{R}:

a∣=∣−a∣ab≤∣a∣∣b∣a2=∣a∣−∣a∣≤a≤∣a∣∀r>0,∣a∣≤r  ⟺  −r≤a≤r\begin{alignat*} |a| = |-a| \\ ab \leq |a||b| \\ \sqrt{a^2} = |a| \\ -|a| \leq a \leq |a| \\ \forall r > 0, |a| \leq r \iff -r \leq a \leq r \\ \end{alignat*}

From the triangle inequality, for a,b,c∈Ra, b, c \in \mathbb{R}:

∣a+b∣≤∣a∣+∣b∣∣a−b∣≤∣a∣+∣b∣∣a−b∣≤∣a−c∣+∣c−b∣\begin{align*} |a+b| &\leq |a| + |b| \\ |a-b| &\leq |a| + |b| \\ |a-b| &\leq |a-c| + |c-b| \\ \end{align*}