mersenneforum.org > Math Question about a power series
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2004-09-13, 16:28 #1 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 64110 Posts Question about a power series I was playing around with some power series stuff, and I wondered what I would get if I used cos(n) as the coefficient, and I was thinking it might not converge or it might be a pathological function or something .. I plugged it into excel and got some rough estimations, and it seemed to converge between -1 and 1, graphing a smooth curve I ended up plugging the series into Mathematica and after some tweaking, I got Sum(n:0 to inf) cos(n)x^n = (cos(1)x - 1) / (-x^2 + 2cos(1)x - 1) now .. how would you go about finding that out with pencil and paper?
 2004-09-13, 19:01 #2 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 3·373 Posts Assuming that x is a real variable: cos(n) = Re(e^(in)) , from Euler's formula. (Re indicates the real part of a complex number.) Therefore, your sum is Sum(n:0 to inf) Re((e^i)^n)*x^n. Assuming x^n is real, you can take the Re outside the sum and write this as the real part of a geometric series: Re(Sum(n:0 to inf) ((e^i)*x)^n = Re (1/(1-(e^i)*x) Now you just need to multiply by the complex conjugate and use the fact that e^i = cos(1) + i*sin(1) Hope this helps!

 Similar Threads Thread Thread Starter Forum Replies Last Post wildrabbitt Math 3 2016-08-16 08:38 R. Gerbicz Math 1 2015-02-10 10:56 grandpascorpion Math 23 2005-01-24 20:11 Gary Edstrom Puzzles 7 2003-07-03 08:32 Rosenfeld Puzzles 2 2003-07-01 17:41

All times are UTC. The time now is 01:27.

Sat Jun 25 01:27:34 UTC 2022 up 71 days, 23:28, 0 users, load averages: 0.86, 0.97, 1.10

Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔