I am still working on Spivak calculus and at problem 6 of the third chapter, I came across an exercise that has been confusing me for a couple hours. I understand the solution (well, I understand how I got there but I have no idea what my solution represents). If someone could enlighten me it would be so great.
Question:
If $x_1,....,x_n$ 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\neq i $.
Solution:
If I understand well, there is only one $x_i$ such that $fi(x) = 1$ (correct me if I'm wrong)
Moreover, for every $x_j$ such that $j \neq i$, the function must be equal to 0.
Therefore, the function should look like $k(x - x_1)(x-x_2)...(x-x_j)$ admitting $i \neq 1$ and $i \neq 2$
So, $f_i(x) = k \prod_{j = 1, j \neq i} ^n (x - x_j)$
And, if I want $f_i(x)$ to be $1$ at $x_i$, then I must fund a $k$ that allows it:
$f(x_i) = 1 = k\prod_{j = 1, j \neq i} ^n (x_i - x_j) \Rightarrow k =( \prod_{j = 1, j \neq i} ^n (x_i - x_j))^-1$
Finally, we find that
$f_i(x) = \prod_{j = 1, j \neq i} ^n \frac{(x - x_j)}{(x_i - x_j)}$
My questions
How does such a function pass many times by $0$ but only once does it equal $1$. I mean I think the other 1 would be for f(complex number) but still. If someone could enlighten me on how this is possible since when I make the graph the function is continue and passes by 1 many times but if I try to find the roots by lowering the function by 1 it says that the only real roots is 1..
Basically, I'd really like to know what kind of function I just constructed and why does it behave in such a weird way, and if i'm right to assume there is only 1 $x_i$ for which $f_i(x) = 1$
Thank you so much
What Spivak asserts is that, among the numbers $x_1,x_2,\ldots,x_n$, there is one and only one number ($x_i$) at which the function $f_i$ takes the value $1$ and that all other numbers (the $x_j$'s with $j\neq i$) are zeros of $f_i$. He never says that there is no other real (or complex) number $x$ such that $f(x)=1$.