Understanding q_n(x) Expansions

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In summary, the conversation discusses a factorization problem and the attempt to understand it. The solution involves expanding the factorization and finding the coefficient of x^n, which is determined to be 1. The conversation also mentions trying with different values of n and provides an example of the expanded factorization.
  • #1
Jbreezy
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Homework Statement



Hi so I'm not understanding my reading of this text.

Homework Equations



Below is what I don't understand.

The Attempt at a Solution



## q_n(x) = (x-a_1)(x-a_2)(x-a_3)...(x-a_n) ##
Now if we expanded this factorization by multiplying it out it should be clear the coefficient of ##x^n## should be one because it could only be derived by choosing x from each of the n bracket terms when composing the product.

So what I don't understand what this is telling me. Expand the factorization.
## q_n(x) = (x-a_1)(x-a_2)(x-a_3)...(x-a_n) ##

For the first three terms I got

##(-x^3a_3 + x^2a_2a_3 + x^2a_1a_3 - a_1a_2a_3x ) ...(x-a_n) ##
Not sure how to include the nth term. How to write it.

Would it be

##(a_3...a_nx^n - a_2a_3..a_nx^n -a_1a_3..a_nx^n + a_1a_2a_3..a_nx^n)##

seems like crap. I don't get what this is telling me!
 
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  • #2
It is telling you that the term with [itex]x^n[/itex] after doing the multiplication comes with a factor of 1 because it is the result of multiplying the n [itex]x's[/itex] and all of them have a factor of 1! And [itex]1^n=1[/itex].

Try with [itex]n=2[/itex] and then [itex]n=3[/itex].
 
  • #3
Jbreezy said:

Homework Statement



Hi so I'm not understanding my reading of this text.


Homework Equations



Below is what I don't understand.

The Attempt at a Solution



## q_n(x) = (x-a_1)(x-a_2)(x-a_3)...(x-a_n) ##
Now if we expanded this factorization by multiplying it out it should be clear the coefficient of ##x^n## should be one because it could only be derived by choosing x from each of the n bracket terms when composing the product.

So what I don't understand what this is telling me. Expand the factorization.
## q_n(x) = (x-a_1)(x-a_2)(x-a_3)...(x-a_n) ##

For the first three terms I got

##(-x^3a_3 + x^2a_2a_3 + x^2a_1a_3 - a_1a_2a_3x ) ...(x-a_n) ##



Not sure how to include the nth term. How to write it.

Would it be

##(a_3...a_nx^n - a_2a_3..a_nx^n -a_1a_3..a_nx^n + a_1a_2a_3..a_nx^n)##

seems like crap. I don't get what this is telling me!

I get ##(x-a_1)(x-a_2)(x-a_3)\cdots(x-a_n)\\=(x^2-(a_1+a_2)x+a_1a_2)(x-a_3)\cdots(x-a_n)\\=(x^3-(a_1+a_2+a_3)x^2+(a_1a_2+a_2a_3+a_1a_3)x-a_1a_2a_3)\cdots(x-a_n).##
 
  • #4
Mandelbroth said:
I get ##(x-a_1)(x-a_2)(x-a_3)\cdots(x-a_n)\\=(x^2-(a_1+a_2)x+a_1a_2)(x-a_3)\cdots(x-a_n)\\=(x^3-(a_1+a_2+a_3)x^2+(a_1a_2+a_2a_3+a_1a_3)x-a_1a_2a_3)\cdots(x-a_n).##

How did you x^3 with no coefficent in front of it. I think you rearranged inbetween the steps.
How did you get this.

##(x^3-(a_1+a_2+a_3)x^2+(a_1a_2+a_2a_3+a_1a_3)x-a_1a_2a_3)\cdots(x-a_n).##

More steps. I see how you got the middle two parts but how do you have the last term before the nth as just coefficients?
 
  • #5
DOnt; answer this question thanks cya
 

FAQ: Understanding q_n(x) Expansions

1. What is the purpose of qn(x) expansions?

Qn(x) expansions are used in mathematics to approximate functions or expressions that cannot be easily calculated or represented in a simpler form. They allow us to break down complicated functions into a series of simpler functions, making them easier to analyze and understand.

2. How are qn(x) expansions derived?

Qn(x) expansions are derived using a process called Gram-Schmidt orthogonalization. This involves finding a set of orthogonal functions that can be used as a basis for the expansion, and then determining the coefficients that best fit the original function.

3. What is the difference between qn(x) expansions and Taylor series?

The main difference between qn(x) expansions and Taylor series is the type of functions they are used to approximate. Taylor series are typically used for smooth, continuously differentiable functions, while qn(x) expansions can be used for more general functions that may not have those properties.

4. How do you determine the accuracy of a qn(x) expansion?

The accuracy of a qn(x) expansion can be determined by comparing it to the original function or expression it is approximating. The accuracy may also depend on the number of terms used in the expansion - the more terms, the more accurate the approximation will be.

5. What are some applications of qn(x) expansions?

Qn(x) expansions have many applications in mathematics, physics, and engineering. They can be used to approximate solutions to differential equations, calculate integrals, and analyze physical systems. They are also used in numerical methods to solve problems that cannot be solved analytically.

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