# Taylor series cos third order [closed]

I need some help to develop the Taylor series on the following: With the signals:

x1(t)=A1Cos(w1t)

x2(t)=A2Cos(w2t)

x3(t)=A3Cos(w3t)


and a system

y(t)=k1x(t)+k3x3(t)


and considerint this

x(t)=x1(t)+x2(t)+x3(t)


I have to develop the Taylor series and I started developing but I think I'm missing many parts. Please, can you help me? • In-built function names should start with Upper case. Here Cos. Function definition using := Please refer to documentation on Functions & Series. reference.wolfram.com/language/tutorial/DefiningFunctions.html and reference.wolfram.com/language/ref/Series.html – Lotus Feb 28 '17 at 10:48
• The formula is incorrect: unbalanced braces and new variable ($\omega$ without an index). – m0nhawk Feb 28 '17 at 10:49
• Please include working mathematica code. – Kuba Feb 28 '17 at 10:54
• I'm doing it by hand, sorry – R user Feb 28 '17 at 10:57
• I'm voting to close this question as off-topic because, according to the OP's comment, it is not about Mathematica. – Szabolcs Feb 28 '17 at 13:14

Not sure about you are asking. Consider reformulate your question and certainly put some Mathematica code. You can you use as a template the input below:

x1[t_,w1_] := A1 Cos[w1 t]

x2[t_,w2_] := A2 Cos[w2 t]

x3[t_,w3_] := A3 Cos[w3 t]

(*This is the standard way to define functions in Mathematica. Note that
the arguments are included within brackets and not parentheses, note
also the underscore and the semicolon. Contrary to other languages you
can ommit the symbol of multiplication; use a space instead.*)


Then, for instance,

y[t_, w1_, w2_, w3_] =
k1 x1[t, w1] + k2 x2[t, w2] + k3 x3[t, w3]
(* A1 k1 Cos[t w1] + A2 k2 Cos[t w2] + A3 k3 Cos[t w3] *)

Series[y[t, w1, w2, w3], {t, 0, 5}]
(* SeriesData[t, 0, {
A1 k1 + A2 k2 + A3 k3, 0,
Rational[1, 2] (-A1 k1 w1^2 - A2 k2 w2^2 - A3 k3 w3^2), 0,
Rational[1, 24] (A1 k1 w1^4 + A2 k2 w2^4 + A3 k3 w3^4)}, 0, 6, 1] *)

• Thank @dimitris. I'll use your answer as a template – R user Feb 28 '17 at 11:48
• You are welcome:-)! – Dimitris Feb 28 '17 at 11:52
• Sorry but now I am more puzzled:-)! You completely change your question. And now my answer appears like "out of space". – Dimitris Feb 28 '17 at 13:53

I leave the first answer as it is. May be it is still useful for you. But I think you will not avoid the "closing" of the question.

http://mathematica.stackexchange.com

is an an amazing site but you cannot learn the very basics of Mathematica syntax from it. It's not its purpose.

Just googling with keywords like: basic mathematica, mathematica intro, beginner's mathematica etc

In any case I post a new reply more relevant to the new query (I hope:-)!). Still it is not clear what you are looking about.

Let

eq1 = (B Cos[x])^3 == 1/4 (3 B^3 Cos[x] + B^3 Cos[3 x]);


Then you can use Reduce in order to solve it for x.

Reduce[eq1, x]
(* (C ∈ Integers && x == π + 2 π C) ||
B == 0 || (-π + x)/(2 π) ∉ Integers *)


Then substitute x to

eq2 = Cos[x] + Cos[y] == 2 Cos[x/2 + y/2] Cos[x/2 - y/2];


and solve it in the same way for y.

Lastly, do the substitutions in the third equation

k(A Cos[x]+B Cos[y]+C Cos[z])^3


via expression/.{x->...,y->...}.

Hope that this helps.

• I forgot to mention wolfram's own intro wolfram.com/broadcast/screencasts/handsonstart – Dimitris Feb 28 '17 at 14:12
• Now after the new editing it is the second reply that seems irrelevant:-!) – Dimitris Feb 28 '17 at 14:14
• Thanks again @dimitris, very helpful. In fact, I'm looking for a cos(2A-B) ... type expression. BTW, I don't know what happen with my new edition of the question.. – R user Feb 28 '17 at 14:45