Basic Math Aug 27, 2023 Quadratic Equation
a x 2 + b x + c = 0 x 2 + b a x + c a = 0 x 2 + b a x = − c a Forcibly create a perfect square on the left x 2 + b a x + ( b 2 a ) 2 = − c a + ( b 2 a ) 2 ( x + b 2 a ) 2 = − c a + ( b 2 a ) 2 ( x + b 2 a ) 2 = − c a + b 2 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b 2 a ± b 2 − 4 a c 4 a 2 \begin{align*}
ax^2 + bx + c = 0 \\
x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \\
x^2 + \frac{b}{a}x = -\frac{c}{a} \\
\text{Forcibly create a perfect square on the left} \\
x^2 + \frac{b}{a}x + \big(\frac{b}{2a}\big)^2 = -\frac{c}{a} + \big(\frac{b}{2a}\big)^2 \\
(x + \frac{b}{2a})^2 = -\frac{c}{a} + \big(\frac{b}{2a}\big)^2 \\
(x + \frac{b}{2a})^2 = -\frac{c}{a} + \frac{b^2}{4a^2} \\
(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2} \\
x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} \\
x = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} \\
\end{align*} a x 2 + b x + c = 0 x 2 + a b x + a c = 0 x 2 + a b x = − a c Forcibly create a perfect square on the left x 2 + a b x + ( 2 a b ) 2 = − a c + ( 2 a b ) 2 ( x + 2 a b ) 2 = − a c + ( 2 a b ) 2 ( x + 2 a b ) 2 = − a c + 4 a 2 b 2 ( x + 2 a b ) 2 = 4 a 2 b 2 − 4 a c x + 2 a b = ± 4 a 2 b 2 − 4 a c x = − 2 a b ± 4 a 2 b 2 − 4 a c Series
Geometric series
S n ≔ ∑ i = 0 n a r i = a + a r + a r 2 + a r 3 + ⋯ + a r n = a ( 1 − r n + 1 ) 1 − r \begin{align*}
S_n &\coloneqq \sum_{i=0}^{n} ar^i \\
&= a + ar + ar^2 + ar^3 + \dots + ar^n \\
&= \frac{a(1 - r^{n+1})}{1 - r}
\end{align*} S n : = i = 0 ∑ n a r i = a + a r + a r 2 + a r 3 + ⋯ + a r n = 1 − r a ( 1 − r n + 1 ) Derivation :
S n = a + a r + a r 2 + a r 3 + ⋯ + a r n r S n = a r + a r 2 + a r 3 + ⋯ + a r n + a r n + 1 \begin{align*}
S_n &= a + &&{\color{magenta}ar + ar^2 + ar^3 + \dots + ar^n} \\
rS_n &= &&{\color{magenta}ar + ar^2 + ar^3 + \dots + ar^n} + ar^{n+1} \\
\end{align*} S n r S n = a + = a r + a r 2 + a r 3 + ⋯ + a r n a r + a r 2 + a r 3 + ⋯ + a r n + a r n + 1 By subtracting the two equations, we get:
S n − r S n = a − a r n + 1 ( 1 − r ) S n = a ( 1 − r n + 1 ) S n = a ( 1 − r n + 1 ) 1 − r \begin{align*}
S_n - rS_n &= a - ar^{n+1} \\
(1-r)S_n &= a(1 - r^{n+1}) \\
S_n = \frac{a(1 - r^{n+1})}{1 - r}
\end{align*} S n − r S n ( 1 − r ) S n S n = 1 − r a ( 1 − r n + 1 ) = a − a r n + 1 = a ( 1 − r n + 1 ) Notice that this only works when r ≠ 1 r \neq 1 r = 1 r = 1 r = 1 r = 1 S n = a ( n + 1 ) S_n = a(n+1) S n = a ( n + 1 )
A good example to memorize:
∑ i = 1 ∞ r i = r + r 2 + r 3 + … = r 1 − r \begin{align*}
\sum_{i=1}^{\infty} r^i &= r + r^2 + r^3 + \dots \\
&= \frac{r}{1-r}
\end{align*} i = 1 ∑ ∞ r i = r + r 2 + r 3 + … = 1 − r r NOTE: You can simply set a=r in the previous formula.
A variant:
S ∞ ′ ≔ ∑ i = 0 ∞ i r i = r + 2 r 2 + 3 r 3 + ⋯ + n r n + … \begin{align*}
S'_\infty &\coloneqq \sum_{i=0}^{\infty} i r^{i} \\
&= r + 2r^2 + 3r^3 + \dots + nr^n + \dots \\
\end{align*} S ∞ ′ : = i = 0 ∑ ∞ i r i = r + 2 r 2 + 3 r 3 + ⋯ + n r n + … Derivation :
S ∞ ′ = r + 2 r 2 + 3 r 3 + 4 r 4 + ⋯ + n r n + … r S ∞ ′ = 2 r 2 + 2 r 3 + 3 r 4 + ⋯ + ( n − 1 ) r n + n r n + 1 + … \begin{align*}
S'_\infty &= r + &&{\color{magenta}2r^2 + 3r^3 + 4r^4 + \dots + nr^n} &&&{\color{magenta}+ \dots} \\
rS'_\infty &= &&{\color{magenta}\hphantom{2}r^2 + 2r^3 + 3r^4 + \dots + (n-1)r^n} &&&{\color{magenta}+ nr^{n+1} + \dots }\\
\end{align*} S ∞ ′ r S ∞ ′ = r + = 2 r 2 + 3 r 3 + 4 r 4 + ⋯ + n r n 2 r 2 + 2 r 3 + 3 r 4 + ⋯ + ( n − 1 ) r n + … + n r n + 1 + … By subtracting the two equations, we get:
S ∞ ′ − r S ∞ ′ = r + r 2 + r 3 + ⋯ + r n + … = r 1 − r \begin{align*}
S'_\infty - rS'_\infty &= r + {\color{magenta}r^2 + r^3 + \dots + r^n + \dots} \\
&= \frac{r}{1-r}
\end{align*} S ∞ ′ − r S ∞ ′ = r + r 2 + r 3 + ⋯ + r n + … = 1 − r r Thus,
( 1 − r ) S ∞ ′ = r 1 − r S ∞ ′ = r ( 1 − r ) 2 \begin{align*}
(1-r)S'_\infty &= \frac{r}{1-r} \\
S'_\infty &= \frac{r}{(1-r)^2}
\end{align*} ( 1 − r ) S ∞ ′ S ∞ ′ = 1 − r r = ( 1 − r ) 2 r