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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
avatar+118703 
+1

Do you mean ..?

 

(e5π/6+e4π/3)9(e5π/6e4π/3)9=[(e5π/6+e4π/3)(e5π/6e4π/3)]9=[(e25π/6e24π/3]9=[(e5π/3e8π/3]9=[e5π/3(1e3π/3]9=[e59π/3][(1eπ)]9=e15π[1+9(eπ)+(92)(eπ)2+(93)(eπ)3+(94)(eπ)4+(95)(eπ)5+(96)(eπ)6+(97)(eπ)7+(98)(eπ)8+(99)(eπ)9]=e15π[19(eπ)+(92)(eπ)2(93)(eπ)3+(94)(eπ)4(95)(eπ)5+(96)(eπ)6(97)(eπ)7+(98)(eπ)8(99)(eπ)9]=e15π[19eπ+(92)(e2π)(93)e3π+(94)(e4π)(95)5eπ+(96)e6π(97)e7π+9e8πe9π]

 

 

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
avatar+28 
0

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

twiistedwhyami  Nov 21, 2022
 #3
avatar+118703 
+2

(e5π6+e4π3)9=9n=0(9n)(e5πn6)(e8π(9n)6)=(90)(e5π06)(e8π(9)6)+(91)(e5π16)(e8π(8)6)+(92)(e5π26)(e8π(7)6)+(93)(e5π36)(e8π(6)6)+(94)(e5π46)(e8π(5)6)+(95)(e5π56)(e8π(4)6)+(96)(e5π66)(e8π(3)6)+(97)(e5π76)(e8π(2)6)+(98)(e5π86)(e8π(1)6)+(99)(e5π96)(e8π(0)6)=(e72π6)+9(e5π6)(e64π6)+(92)(e10π6)(e56π6)+(93)(e15π6)(e48π6)+(94)(e20π6)(e40π6)+(95)(e25π6)(e32π6)+(96)(e30π6)(e24π6)+(97)(e35π6)(e16π6)+9(e40π6)(e8π6)+(e45π6)=(e72π6)+9(e69π6)+(92)(e66π6)+(93)(e63π6)+(94)(e60π6)+(95)(e57π6)+(96)(e54π6)+(97)(e51π6)+9(e48π6)+(e45π6)=(e24π2)+9(e23π2)+(92)(e22π2)+(93)(e21π2)+(94)(e20π2)+(95)(e19π2)+(96)(e18π2)+(97)(e17π2)+9(e16π2)+(e15π2)

 

 

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
avatar+118703 
+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

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