heureka

What is the whole order of Fibonacci numbers?

\(\text{Golden Ratio } = \varphi \\ \varphi =\frac{ 1+\sqrt{5} } {2}\\ \varphi = 1.61803398875\dots\)

Fibonacci numbers:

\(f_n = \dfrac{ \varphi^n - (1-\varphi)^n } {\sqrt{5}}\)

Sequence below zero:

\(f_{-n} = (-1)^{n+1}\cdot f_n\)

\(\begin{array}{rrrrrrrrrrrrrrrrrrrr} \hline n &=& \dots &-6 &-5 &-4 &-3 &-2 &-1 &0 &1 &2 &3 &4 &5 &6 &\dots \\ \hline f_n &=& \dots &-8 &5 &-3 &2 &-1 &1 &0 &1 &1 &2 &3 &5 &8 &\dots \\ \hline \end{array} \)

Hi I was looking for steps to show z=sin(i ln(2)) + cos(i ln(2)) = 5/4 +3i/4

\(\boxed{~ \begin{array}{rcll} \sin{z} &=& \frac{1}{2i} \cdot \left( e^{ix} - e^{-ix} \right) \\ \cos{z} &=& \frac{1}{2} \cdot \left( e^{ix} + e^{-ix} \right) \\ \end{array} ~}\)

\(\begin{array}{rcll} \sin{(i\cdot ln(2))} &=& \frac{1}{2i} \cdot \left( e^{i\cdot i\cdot ln(2)} - e^{-i\cdot i\cdot ln(2)} \right) \\ &=& \frac{1}{2i} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right) \qquad i\cdot i = i^2 = -1\\ \cos{(i\cdot ln(2))} &=& \frac{1}{2} \cdot \left( e^{i\cdot i\cdot ln(2)} + e^{-i\cdot i\cdot ln(2)} \right) \\ &=& \frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right) \qquad i\cdot i = i^2 = -1\\ \\ \sin{(i\cdot ln(2))}+ \cos{(i\cdot ln(2))}&=& \frac{1}{2i} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right) +\frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right)\\ &=& \frac{1}{2i} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right) \cdot \frac{i}{i} +\frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right)\\ &=& \frac{i}{-2} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right) +\frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right)\\ &=& \frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right) - \frac{i}{2} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right)\\\\ e^{ ln(2)} = 2 && e^{- ln(2)} = \frac{1}{e^{ ln(2)}} = \frac12\\\\ \sin{(i\cdot ln(2))}+ \cos{(i\cdot ln(2))} &=&\frac{1}{2} \cdot \left( \frac12 + 2 \right) - \frac{i}{2} \cdot \left( \frac12 - 2 \right)\\ &=&\frac{5}{4} - \frac{i}{2} \cdot \left( -\frac32 \right) \\ &=&\frac{5}{4} + i \cdot \left( \frac34 \right) \end{array}\)

#4:

solution:

\(x=\pm\sqrt{972} \\ = \pm 18\sqrt{3}\\ =\pm 31.1769145362\)

#5:

solution:

\(x = \pm 36\)

9l9-8xl=2x+3 it saids extraneous solution? can someone teach me step by step to solve this?

\(\begin{array}{rcll} 9 |9-8x| &=& 2x+3 \qquad & | \qquad (\text{square both sides}) \\ 9^2 \cdot (9-8x)^2 &=& (2x+3)^2 \\ 81 \cdot (9^2-2\cdot 9 \cdot 8\cdot x +8^2x^2 ) &=& 2^2x^2 +2\cdot 2 \cdot 3\cdot x + 3^2 \\ 81 \cdot (81-144 x +64x^2 ) &=& 4x^2 + 12x + 9 \\ 81^2 - 81\cdot 144 x + 81\cdot 64x^2 &=& 4x^2 + 12x + 9 \\ 6561 - 11664x + 5184x^2 &=& 4x^2 + 12x + 9 \qquad & | \qquad -4x^2\\ 6561 - 11664x + 5180x^2 &=& 12x + 9 \qquad & | \qquad -12x\\ 6561 - 11676x + 5180x^2 &=& 9 \qquad & | \qquad -9\\ 6552 - 11676x + 5180x^2 &=& 0 \\ 5180x^2 - 11676x + 6552 &=& 0 \qquad & | \qquad :4\\ 1295x^2 - 2919x + 1638 &=& 0\\ \end{array}\)

\(\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}\)

\(\begin{array}{rcll} 1295x^2 - 2919x + 1638 &=& 0 \quad | \quad a=1295 \quad b = -2919 \quad c = 1638 \\ x &=& \dfrac{-(-2919) \pm \sqrt{(-2919)^2-4\cdot 1295 \cdot 1638} }{2\cdot 1295} \\ x &=& \dfrac{2919 \pm \sqrt{ 8520561-8484840 } }{2590} \\ x &=& \dfrac{2919 \pm \sqrt{ 35721 } }{2590} \\ x &=& \dfrac{2919 \pm 189 }{2590} \\\\ x_1 &=& \dfrac{2919 + 189 }{2590} \\ x_1 &=& \dfrac{3108 }{2590} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{1.2} \\\\ x_2 &=& \dfrac{2919 - 189 }{2590} \\ x_2 &=& \dfrac{2730}{2590} \\ x_2 &=& \dfrac{273}{259} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{1.\overline{054}}\\ \end{array}\)

\(\begin{array}{rcll} 2^{5299999} &=& 10^{5299999 \cdot \log_{10}{(2)}} \\ &=& 10^{5299999 \cdot 0.30102999566} \\ &=& 10^{1595458.67598910467} \\ &=& 10^{0.67598910467}\cdot 10^{1595458} \\ &=& 4.74230087900\cdot 10^{1595458} \\ \end{array}\)

\(\begin{array}{rcll} log_{10}(x+3)-log_{10}x &=& 2 \\ log_{10}( \frac{x+3}{x} ) &=& 2 \\ \frac{x+3}{x} &=& 10^2 \\ \frac{x+3}{x} &=&100 \\ x+3 &=& 100x \qquad | \qquad -x \\ 3 &=& 99x \\ 99x &=& 3 \qquad | \qquad :99 \\ x &=& \frac{3}{99} \\ x &=& \frac{1}{33} \\ x &=& 0.03030303030 \end{array}\)

\(\begin{array}{rcll} log_6(5x^2)-log_6(5) &=& 4 \\ log_6( \frac{5x^2} {5} ) &=& 4 \\ log_6( x^2 ) &=& 4 \\ x^2 &=& 6^4 \\ x &=& 6^2 \\ x &=& 36 \\ \end{array}\)

\(\begin{array}{rcll} log_3(x^2)-log_3(4) &=& 5 \\ log_3( \frac{x^2} {4} ) &=& 5 \\ \frac{x^2} {4} &=& 3^5 \\ \frac{x^2} {4} &=& 234 \\ x^2 &=& 4\cdot 234 \\ x^2 &=& 972 \\ x &=& 31.1769145362 \end{array}\)

\(\begin{array}{rcll} log_2(5)-log_2(4x) &=& 3 \\ log_2( \frac{5}{4x} ) &=& 3 \\ \frac{5}{4x} &=& 2^3 \\ \frac{5}{4x} &=& 8 \qquad | \qquad \cdot 4x \\ 5 &=& 8 \cdot 4x \\ 5 &=& 32x \\ 32x &=& 5 \qquad | \qquad :32 \\ x &=& \frac{5}{32} \\ x &=& 0.15625 \end{array}\)

\(\begin{array}{rcll} log_{12}(2-3m) &=& log_{12}(-5m+2) \\ 2-3m &=& -5m+2 \\ -3m &=& -5m \\ 5m-3m &=& 0 \\ 2m &=& 0 \qquad | \qquad : 2\\ m &=& 0 \\ \end{array}\)