How to Derive Polynomials from Given Conditions?

  • Thread starter Thread starter Mathman23
  • Start date Start date
  • Tags Tags
    Polynomials
Click For Summary
SUMMARY

This discussion focuses on deriving two third-degree polynomials, p(x) and q(x), based on specific conditions: p(0) = -1, p'(0) = 1, q(1) = 3, q'(1) = -2, q(0) = p(0), and q'(0) = p'(0). The polynomials are expressed in the form p(x) = a0 + a1x + a2x² + a3x³ and q(x) = b0 + b1x + b2x² + b3x³. The discussion reveals that while q(x) can be fully determined using the provided conditions, p(x) remains partially indeterminate without additional information regarding coefficients a2 and a3.

PREREQUISITES
  • Understanding polynomial functions and their degrees
  • Knowledge of derivatives and their application in polynomial equations
  • Familiarity with solving linear equations
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study polynomial interpolation techniques to understand how to derive polynomials from given conditions
  • Learn about the Fundamental Theorem of Algebra and its implications for polynomial roots
  • Explore methods for solving systems of linear equations, particularly in the context of polynomial coefficients
  • Investigate the use of symbolic computation tools like SymPy for polynomial manipulation and solving
USEFUL FOR

Students in mathematics or engineering fields, educators teaching polynomial functions, and anyone involved in algebraic problem-solving or mathematical modeling.

Mathman23
Messages
248
Reaction score
0
Hi

This is the character equation for a polynomial of degree where [itex]n \geq 0[/itex]

[tex]p(x) = a_0 x^{n} + a_{1} x^{n-1} + a_2 x ^{n-2} + \cdots + a_{n-1}x + a_{n}[/tex]

I'm presented with the following assignment:

Two polynomials [itex]\mathrm{p, q}[/itex] where n = 3. These polynomials can derived using the following conditions:

[tex]\begin{array}{ccc} p(0) = -1 & q(1) = 3 & p(0) = q(0) \\ p'(0) =1 & q'(1) = -2 & p'(0) = q'(0) \end{array}[/tex]

If was told by professor that these polynomials can we written in the following form.

[tex]p(x) = (a_{0} + a_{1})x^{n-1} + (a_{1})x^{n-2} + \cdots + (a_{n-1})x + a_{n}[/tex]

How do I apply formula to my assignment??

Thanks in advance

Sincerely
Fred
 
Physics news on Phys.org
Restating, you have two 3rd degree polynomials, p and q, such that:
p(0)= -1, p'(0)= 1, q(1)= 3, q'(1)= -2, q(0)= p(0)= -1, q'(0)= p'(0)= 1.

Okay, write p(x)= a0+ a1x+ a2x2+ a3x3 and
q(x)= b0+ b1x+ b2x2+ b3x3

p(0)= a0= -1 and p'(0)= a1= 1. Also, since q(0)= a(0), b0= -1 and since q'(0)= p'(0)= 1, b1= 1.

So far we have p(x)= -1+ x+ a2x2+ a3x3 and since that is all the information given about p(x), I don't see anyway of determining a2 or a3.

We have q(x)= -1+ x+ b2x2+ b3x3
so q(1)= -1+ 1+ b2+ b3= 3
and q'(x)= 1+ 2b22+ 3b2 so
q'(1)= 1+ 2b2+ 3b2= -2.

You can solve those two linear equations for b2 and b3, completely determining q(x) but, unless you are given more information, there is no way to completely determine p(x).

That's assuming that assignment was to determine the two polynomials! You never did say what the assignment was.
 
Hi

I'm told that the resulting two polynomials are:

I'm told that the resulting two polynomials of degree 3 are:

p(x) = (2 + s - 2t) x^3 + (3 + 2s - 3t) x^2 + s*x + t

q(x) = (-6 + s + 2t) x^3 + (9 - 2s - 3t) x^2 + s*x +t

where s,t belong to R.

But my textbook doesn't have any information on how one can arrive at the above result.

Maybe You Hall can provide me with a hint on howto obtain the above result ?

Sincerley and Best Regards,

Fred

HallsofIvy said:
Restating, you have two 3rd degree polynomials, p and q, such that:
p(0)= -1, p'(0)= 1, q(1)= 3, q'(1)= -2, q(0)= p(0)= -1, q'(0)= p'(0)= 1.

Okay, write p(x)= a0+ a1x+ a2x2+ a3x3 and
q(x)= b0+ b1x+ b2x2+ b3x3

p(0)= a0= -1 and p'(0)= a1= 1. Also, since q(0)= a(0), b0= -1 and since q'(0)= p'(0)= 1, b1= 1.

So far we have p(x)= -1+ x+ a2x2+ a3x3 and since that is all the information given about p(x), I don't see anyway of determining a2 or a3.

We have q(x)= -1+ x+ b2x2+ b3x3
so q(1)= -1+ 1+ b2+ b3= 3
and q'(x)= 1+ 2b22+ 3b2 so
q'(1)= 1+ 2b2+ 3b2= -2.

You can solve those two linear equations for b2 and b3, completely determining q(x) but, unless you are given more information, there is no way to completely determine p(x).

That's assuming that assignment was to determine the two polynomials! You never did say what the assignment was.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
48
Views
7K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K