Expert: Paul Klarreich - 10/19/2011

Question
QUESTION: Hello Sir,
         Do you know a cubic polynomial that approximates cosine or sine "reasonably well" over entire period i.e 0 to 2pi.Or is there a method of getting one?
WHERE "reasonably well" could be having a max error less than or equal to 25% of the value of the function at any point FOR EXAMPLE.
The known:There are good approximations in the -pi/2 to pi/2 range.Thanks.

ANSWER: Questioner:jean
Country:Gambia
Category:Advanced Math
Private:No
Subject:cubic approximation
Question:Hello Sir,
         Do you know a cubic polynomial that approximates cosine or sine "reasonably well" over entire period i.e 0 to 2pi.Or is there a method of getting one?
WHERE "reasonably well" could be having a max error less than or equal to 25% of the value of the function at any point FOR EXAMPLE.
The known:There are good approximations in the -pi/2 to pi/2 range.Thanks.
..............................................
Let's consider  y = sin x  over [0,2 pi].  This 'sort of' looks like a cubic graph if we put the zeroes at x = 0, pi, 2pi.  So it would be of the form:

f(x) = A(x - 0)(x - pi)(x - 2pi)

Now you would want to make it go through the max and min points:

x = pi/2, y = 1
and
x = 3pi/2,  y = -1.

So f(pi/2) = A(p/2 - 0)(pi/2 - pi)(pi/2 - 2pi) = 1

A(p/2)(- pi/2)(-3pi/2) = 1

A(1)(- 1)(-3) (pi/2)^3 = 1

3 A (pi/2)^3 = 1

A = 1/[3  (pi/2)^3 ] = 1/11.6

So  y = x(x - pi)(x - 2pi)/ 11.6  might be a nice approximation.


---------- FOLLOW-UP ----------

QUESTION: Thanks .How would one prove a system of quadratic equations is linear independent?Thanks.

ANSWER: ANSWER: Questioner:jean
Country:Gambia
Category:Advanced Math
Private:No
QUESTION: Thanks .How would one prove a system of quadratic equations is linear independent?Thanks.
......................................
I don't think the question is meaningful.  Please define linearLY independent for quadratic equations.

---------- FOLLOW-UP ----------

QUESTION: Hello Again,
         Definition:for Q0(x),Q1(x),.....,Qn(x) where Qi(x) are quadratic polynomials,such that, without loss of generality (in terms of the notation i.e Qi(x) is just the quadratic in the ith position on the right hand side  of the   expression below)
Qj(x)!=Q0(x)*a0+Q1(x)*a1+........+Qn(x)*an;
!= is not equal to;
Thanks.

Answer
Questioner:jean
Country:Gambia
Category:Advanced Math
Private:No
QUESTION: Thanks .How would one prove a system of quadratic equations is linear independent?Thanks.
---------- FOLLOW-UP ----------

QUESTION: Hello Again,
         Definition:for Q0(x),Q1(x),.....,Qn(x) where Qi(x) are quadratic polynomials,such that, without loss of generality (in terms of the notation i.e Qi(x) is just the quadratic in the ith position on the right hand side  of the   expression below)
Qj(x)!=Q0(x)*a0+Q1(x)*a1+........+Qn(x)*an;
!= is not equal to;
Thanks.
.......................................
So, when you said "a system of quadratic equations", you meant "a system of quadratic POLYNOMIALS".

(Carelessness in using vocabulary is a good way to fail in mathematics.)

If  <ai, bi, ci> is the vector associated with the Q.P.:

ai x^2 + bi x + ci,

then a system of at least three of these:

a1   b1   c1
a2   b2   c2
a3   b3   c3

is independent if the determinant of that matrix is non-zero.

[look up your standard linear algebra stuff]