Polynomial question from spivak's calculus

[osnarf]

Question from spivak's calculus - 3rd edition - chapter 3, question 6(a).

Homework Statement

If x₁, ..., x_n are distinct numbers, find a polynomial function f_i of degree n - 1 which is 1 at x_i and 0 at x_j for j =/= i. Hint, the product of all (x - x_i) for j =/= i, is 0 at x_j if j =/= i. (This product is usually denoted by see below, the symbol (capital pi) playing the same role for products that sigma plays for sums.)

Homework Equations

n
\prod(x - x_j)
j = 1
j =/= i

The Attempt at a Solution

I really don't even know where to start, the question is really confusing. If someone could just explain the question better that would be really appreciated. I'm confused by just about every word:

-x₁,...,x_n <--- Are these just random numbers? or are they increments along the x axis? or ?
-why does the function have a subscript (...find a polynomial function f_i)?
-which x is x_i?
-which x is x_j?
-when he says (x - x_i), I would assume x is not a distinct number, but the variable x, and that x_i is meant to represent one of each of the distinct x's each time and that they are multiplied together, but later he asks for a specific function value at x_i. I don't understand that.
-why did he need to write j =/= i twice, about 5 words apart (not trying to be sarcastic I'm sure there's a reason i just don't know why).
-in response to the capital-Pi eqation: so x_j is x₁?

Thanks for helping, I'm sure I can figure out the problem once i understand what it is. If someone could just explain it to me I would really appreciate it.

 

[micromass]

The x_1,...,x_n are just random numbers.

Maybe you'll understand it better if I gave you an example. Consider the numbers 1 and 2.
The question asks that we construct two polynomials:
1) We want to construct a polynomial f such that f(1)=1 and f(2)=0. An example of such a polynomial is f(x)=-x+2
2) We also want a polynomial f such that f(1)=0 and f(2)=1. An example of such a polynomial is f(x)=x-1.


We can also consider the numbers 0,1,2. Then the question asks us to construct three polynomials. One of these polynomials will be an f such that f(0)=1, f(1)=0 and f(2)=0. An example of such a polynomial is f(x)=\frac{1}{2}x^2-\frac{3}{2}x+1.
The other two polynomials can also be constructed, but I'll leave that to you...

 

[osnarf]

Sorry, I'm still confused. I understand how to construct polynomials to fit specifications, it is the abstraction of the question that is really getting me. And i just peeked in the solutions manual and the answer isn't in standard polynomial form its in the form of a capital-pi divided by another capital pi. Anyhow, in regards to your examples, why is the question asking us to make multiple polynomials?

Thanks for your quick response.

 

[micromass]

You're just given x_1,x_2,...,x_n real numbers and you're asked to construct a polynomail f such that f(x_1)=1,f(x_2)=0,...,f(x_n)=0. That's all they're asking from you here.

The answer will be in capital-pi notation because that is the easiest way to represent the polynomail in.

What I suggest is that you take some examples and construct the requested polynomial. For example, take x_1=0,x_1=2,x_2=-1. Can you now build a polynomial such that f(0)=1, f(2)=0 and f(-1)=0??
Once you've found that, try to handle the general case. You'll see immediately why the capital-pi notation comes in.

 

[osnarf]

So the value of the function at one of the x's needs to be 1, and at all the others it will be 0?
Could you please explain to me how your getting that from the notations. I have a feeling the rest of the book is going to be just like this and if i don't get the notation down I'm going to be frustrated very much for the next few months.

 

[osnarf]

I understand part of it now. Because the degree is n - 1 there would have to be n - 1 zeros. I overlooked that. And I understand why it is in capital-pi notation now, after playing around with it. We are getting somewhere, :P.

Okay, now how about the hint.

It says the product of all (x - x_i), for j =/= i, is 0 at x_j if j =/= i.

x_j is the zero of the function, so in your example with x₁ = 1 and x₂ = 0, if we multiplied all the (x - x_i), where j =/= i, together, we would have only one so it would be (x - x₁) = (x - 1). It says this product is 0 at x_j if j =/= i, so if i were to plug x_j into this: (x_j - 1) = (2 - 1) = 1 =/= 0, so clearly I'm still confused.

 

[osnarf]

I've come to the conclusion there is a typo in the hint and it is supposed to read (x - x_j), if I'm wrong correct me please it would make a lot more sense.

---------------

thank you micro, figured it out. i was making that problem way harder than it was

 

[micromass]

Yes, I think you are correct and that there is a typo in the hint. Nice catch!

 

[diligence]

I've come to the conclusion there is a typo in the hint and it is supposed to read (x - x_j), if I'm wrong correct me please it would make a lot more sense.

---------------

thank you micro, figured it out. i was making that problem way harder than it was

zomfg meetooooo.

kinda pissed that I've been fretting about this for quite some time and come to find out there's a freakin typo!