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What do you get when you add exponential forms? (e^5pi/6 + e^4pi/3)^9 (e^5pi/6 - e^4pi/3)^9

0
186
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What do you get when you add exponential forms?

(e^5pi/6 + e^4pi/3)^9
(e^5pi/6 - e^4pi/3)^9

Nov 21, 2022

#1
+118617
+1

Do you mean ..?

$$(e^{5\pi/6} + e^{4\pi/3})^9 (e^{5\pi/6} - e^{4\pi/3})^9\\ =[(e^{5\pi/6} + e^{4\pi/3}) (e^{5\pi/6} - e^{4\pi/3})]^9\\ =[ (e^{2*5\pi/6} - e^{2*4\pi/3}]^9\\ =[ (e^{5\pi/3} - e^{8\pi/3}]^9\\ =[ e^{5\pi/3}(1 - e^{3\pi/3}]^9\\ =[ e^{5*9\pi/3}][(1 - e^{\pi})]^9\\ =e^{15\pi}[1+9 (- e^{\pi})+\binom{9}{2}(- e^{\pi})^2+\binom{9}{3}(- e^{\pi})^3+\binom{9}{4}(- e^{\pi})^4+\binom{9}{5}(- e^{\pi})^5+\binom{9}{6}(- e^{\pi})^6+\binom{9}{7}(- e^{\pi})^7+\binom{9}{8}(- e^{\pi})^8+\binom{9}{9}(- e^{\pi})^9]\\ =e^{15\pi}[1-9 (e^{\pi})+\binom{9}{2}(e^{\pi})^2-\binom{9}{3}(e^{\pi})^3+\binom{9}{4}(e^{\pi})^4-\binom{9}{5}( e^{\pi})^5+\binom{9}{6}(e^{\pi})^6-\binom{9}{7}(e^{\pi})^7+\binom{9}{8}(e^{\pi})^8-\binom{9}{9}(e^{\pi})^9]\\ =e^{15\pi}[1-9 e^{\pi}+\binom{9}{2}(e^{2\pi})-\binom{9}{3}e^{3\pi}+\binom{9}{4}(e^{4\pi})-\binom{9}{5}5e^{\pi}+\binom{9}{6}e^{6\pi}-\binom{9}{7}e^{7\pi}+9e^{8\pi}-e^{9\pi}]\\$$

LaTex

(e^{5\pi/6} + e^{4\pi/3})^9  (e^{5\pi/6} - e^{4\pi/3})^9\\
=[(e^{5\pi/6} + e^{4\pi/3})  (e^{5\pi/6} - e^{4\pi/3})]^9\\
=[ (e^{2*5\pi/6} - e^{2*4\pi/3}]^9\\
=[ (e^{5\pi/3} - e^{8\pi/3}]^9\\
=[ e^{5\pi/3}(1 - e^{3\pi/3}]^9\\
=[ e^{5*9\pi/3}][(1 - e^{\pi})]^9\\
=e^{15\pi}[1+9 (- e^{\pi})+\binom{9}{2}(- e^{\pi})^2+\binom{9}{3}(- e^{\pi})^3+\binom{9}{4}(- e^{\pi})^4+\binom{9}{5}(- e^{\pi})^5+\binom{9}{6}(- e^{\pi})^6+\binom{9}{7}(- e^{\pi})^7+\binom{9}{8}(- e^{\pi})^8+\binom{9}{9}(- e^{\pi})^9]\\
=e^{15\pi}[1-9 (e^{\pi})+\binom{9}{2}(e^{\pi})^2-\binom{9}{3}(e^{\pi})^3+\binom{9}{4}(e^{\pi})^4-\binom{9}{5}( e^{\pi})^5+\binom{9}{6}(e^{\pi})^6-\binom{9}{7}(e^{\pi})^7+\binom{9}{8}(e^{\pi})^8-\binom{9}{9}(e^{\pi})^9]\\
=e^{15\pi}[1-9 e^{\pi}+\binom{9}{2}(e^{2\pi})-\binom{9}{3}e^{3\pi}+\binom{9}{4}(e^{4\pi})-\binom{9}{5}5e^{\pi}+\binom{9}{6}e^{6\pi}-\binom{9}{7}e^{7\pi}+9e^{8\pi}-e^{9\pi}]\\

Nov 21, 2022
#2
+28
0

No, I meant for them to be two seperate equations.

twiistedwhyami  Nov 21, 2022
#3
+118617
+2

$$\displaystyle (e^\frac{5\pi}{6} + e^\frac{4\pi}{3})^9\\ =\displaystyle \sum_{n=0}^9 \binom{9}{n}(e^\frac{5\pi\; n}{6}) (e^\frac{8\pi (9-n)}{6})\quad \\ = \binom{9}{0}(e^\frac{5\pi\; 0}{6}) (e^\frac{8\pi (9)}{6}) + \binom{9}{1}(e^\frac{5\pi\; 1}{6}) (e^\frac{8\pi (8)}{6}) + \binom{9}{2}(e^\frac{5\pi\; 2}{6}) (e^\frac{8\pi (7)}{6}) + \binom{9}{3}(e^\frac{5\pi\; 3}{6}) (e^\frac{8\pi (6)}{6}) + \binom{9}{4}(e^\frac{5\pi\; 4}{6}) (e^\frac{8\pi (5)}{6}) + \binom{9}{5}(e^\frac{5\pi\; 5}{6}) (e^\frac{8\pi (4)}{6}) + \binom{9}{6}(e^\frac{5\pi\; 6}{6}) (e^\frac{8\pi (3)}{6}) + \binom{9}{7}(e^\frac{5\pi\; 7}{6}) (e^\frac{8\pi (2)}{6}) + \binom{9}{8}(e^\frac{5\pi\; 8}{6}) (e^\frac{8\pi (1)}{6}) + \binom{9}{9}(e^\frac{5\pi\; 9}{6}) (e^\frac{8\pi (0)}{6})\\ = (e^\frac{72\pi }{6}) + 9(e^\frac{5\pi}{6}) (e^\frac{64\pi }{6}) + \binom{9}{2}(e^\frac{10\pi}{6}) (e^\frac{56\pi }{6}) + \binom{9}{3}(e^\frac{15\pi}{6}) (e^\frac{48\pi }{6}) + \binom{9}{4}(e^\frac{20\pi}{6}) (e^\frac{40\pi }{6}) + \binom{9}{5}(e^\frac{25\pi }{6}) (e^\frac{32\pi }{6}) + \binom{9}{6}(e^\frac{30\pi}{6}) (e^\frac{24\pi }{6}) + \binom{9}{7}(e^\frac{35\pi}{6}) (e^\frac{16\pi }{6}) +9 (e^\frac{40\pi}{6}) (e^\frac{8\pi }{6}) + (e^\frac{45\pi}{6}) \\ = (e^\frac{72\pi }{6}) + 9 (e^\frac{69\pi }{6}) + \binom{9}{2}(e^\frac{66\pi}{6}) + \binom{9}{3}(e^\frac{63\pi}{6}) + \binom{9}{4}(e^\frac{60\pi}{6}) + \binom{9}{5}(e^\frac{57\pi }{6}) + \binom{9}{6}(e^\frac{54\pi}{6}) + \binom{9}{7}(e^\frac{51\pi}{6}) +9 (e^\frac{48\pi}{6}) + (e^\frac{45\pi}{6}) \\ = (e^\frac{24\pi }{2}) + 9 (e^\frac{23\pi }{2}) + \binom{9}{2}(e^\frac{22\pi}{2}) + \binom{9}{3}(e^\frac{21\pi}{2}) + \binom{9}{4}(e^\frac{20\pi}{2}) + \binom{9}{5}(e^\frac{19\pi }{2}) + \binom{9}{6}(e^\frac{18\pi}{2}) + \binom{9}{7}(e^\frac{17\pi}{2}) +9 (e^\frac{16\pi}{2}) + (e^\frac{15\pi}{2}) \\$$

LaTex

\displaystyle (e^\frac{5\pi}{6} + e^\frac{4\pi}{3})^9\\
=\displaystyle \sum_{n=0}^9    \binom{9}{n}(e^\frac{5\pi\; n}{6})  (e^\frac{8\pi (9-n)}{6})\quad \\
= \binom{9}{0}(e^\frac{5\pi\; 0}{6})  (e^\frac{8\pi (9)}{6})
+ \binom{9}{1}(e^\frac{5\pi\; 1}{6})  (e^\frac{8\pi (8)}{6})
+ \binom{9}{2}(e^\frac{5\pi\; 2}{6})  (e^\frac{8\pi (7)}{6})
+ \binom{9}{3}(e^\frac{5\pi\; 3}{6})  (e^\frac{8\pi (6)}{6})
+ \binom{9}{4}(e^\frac{5\pi\; 4}{6})  (e^\frac{8\pi (5)}{6})
+ \binom{9}{5}(e^\frac{5\pi\; 5}{6})  (e^\frac{8\pi (4)}{6})
+ \binom{9}{6}(e^\frac{5\pi\; 6}{6})  (e^\frac{8\pi (3)}{6})
+ \binom{9}{7}(e^\frac{5\pi\; 7}{6})  (e^\frac{8\pi (2)}{6})
+ \binom{9}{8}(e^\frac{5\pi\; 8}{6})  (e^\frac{8\pi (1)}{6})
+ \binom{9}{9}(e^\frac{5\pi\; 9}{6})  (e^\frac{8\pi (0)}{6})\\

= (e^\frac{72\pi }{6})
+ 9(e^\frac{5\pi}{6})  (e^\frac{64\pi }{6})
+ \binom{9}{2}(e^\frac{10\pi}{6})  (e^\frac{56\pi }{6})
+ \binom{9}{3}(e^\frac{15\pi}{6})  (e^\frac{48\pi }{6})
+ \binom{9}{4}(e^\frac{20\pi}{6})  (e^\frac{40\pi }{6})
+ \binom{9}{5}(e^\frac{25\pi }{6})  (e^\frac{32\pi }{6})
+ \binom{9}{6}(e^\frac{30\pi}{6})  (e^\frac{24\pi }{6})
+ \binom{9}{7}(e^\frac{35\pi}{6})  (e^\frac{16\pi }{6})
+9 (e^\frac{40\pi}{6})  (e^\frac{8\pi }{6})
+ (e^\frac{45\pi}{6}) \\

= (e^\frac{72\pi }{6})
+ 9 (e^\frac{69\pi }{6})
+ \binom{9}{2}(e^\frac{66\pi}{6})
+ \binom{9}{3}(e^\frac{63\pi}{6})
+ \binom{9}{4}(e^\frac{60\pi}{6})
+ \binom{9}{5}(e^\frac{57\pi }{6})
+ \binom{9}{6}(e^\frac{54\pi}{6})
+ \binom{9}{7}(e^\frac{51\pi}{6})
+9 (e^\frac{48\pi}{6})
+ (e^\frac{45\pi}{6}) \\

Nov 23, 2022
edited by Melody  Nov 23, 2022
#4
+118617
+2

the second one is the same except every second term is negative. You can work out if I meant the odd terms or the even ones.

Nov 23, 2022