# Polynomial Interpolation example

I'm having trouble understanding this result (actually, this whole chapter of the Linear Algebra course really): (a) Where did these fractions come from? The previous sections do not indicate how we got a rational function.

(b) Is the answer given, $x^2 + 2x$ correct? When I multiplied these rational functions by the constants in question I got:

$P(x) = x^2 + 2x - \frac{7}{4}$

EDIT: For part (b) I figured out where I went wrong, just an algebraic error. $x^2 + 2$ is correctly given, I just misread the answer.

I feel like I have no idea what is going on here even though I just finished a class in linear algebra. Note by Jeff Folster
1 month ago

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- 4 weeks, 1 day ago

I believed you're referring to this problem.

The previous problem demonstrates how you can get the coefficients $3,11,27$.

If you substitute the three polynomials $Q_1(x), Q_2 (x) , Q_3 (x)$ as stated into $P(x)$, you get the desired $x^2 +2$.

- 4 weeks, 1 day ago

I know where the coefficients come from, but why are we dividing the polynomials by $(1 - 3)(1 - 5)$, etc.? This process isn't explained at all.

I figured out where my math went wrong, I did get $x^2 + 2$ after all.

- 4 weeks, 1 day ago

Let's say we want to interpolate a set of data points $(a_{1},b_{1}) , (a_{2}, b_{2}) , \dots , (a_{n},b_{n})$ with a polynomial. There are a few real-world applications where this is desirable, so this isn't just an academic exercise. But let's not get into all that now.

If we express the interpolating polynomial $p(x)$ as $p_{0} + p_{1} x + p_{2} x^2 + \dots+ p_{n-1} x^{n-1},$ then we'd have to solve the system of $n$ equations $p(a_{i}) = b_{i}$ for the coefficients of $p(x),$ which is messy and unpleasant.

A better way is to express $p(x)$ as a linear combination of polynomials $Q_{j}(x)$ that are carefully constructed so as to make the interpolation process as smooth as possible.

Imagine it were possible to define a polynomial $Q_{j}(x)$ that satisfies $Q_{j}(a_{i}) = 0$ if $i \neq j$ but $Q_{j}(a_j) = 1.$ In other words, $Q_{j}$ vanishes at all of the $a$'s except for $a_{j}$ where it's 1.

Assuming for the moment that it's possible to do this, then consider the polynomial $p(x) = b_{1} Q_{1}(x) + b_{2} Q_{2}(x) + \dots + b_{n}Q_{n}(x).$ If we plug in $a_{j},$ then we get $p(a_{j}) = b_{1} Q_{1}(a_{j}) + b_{2} Q_{2}(a_{j}) + \dots + b_{n}Q_{n}(a_{j}) = b_{1} \times 0 + \dots + b_{j} \times 1 + \dots + b_{n} \times 0 = b_{j}.$ So the linear combination of these hypothetical $Q$'s with weights given by the $b_{j}$'s, the $y$-coordinates of the data points, is precisely the interpolating polynomial we're after.

We just have to show that the $Q$'s exist. We can do this by constructing them. The polynomial $(x-a_{1}) (x-a_{2}) \dots (x-a_{j-1})(x-a_{j+1}) \dots (x-a_{n})$ is designed to be 0 at every $a$ except for $a_{j}.$ This is almost what we need, but the polynomial isn't 1 at $a_{j}.$

The simple fix is to take the polynomial above and then divide by $(a_{j}-a_{1}) (a_{j} -a_{2}) \dots (a_{j}-a_{j-1})(a_{j}-a_{j+1}) \dots (a_{j}-a_{n}).$ This gives us a formula for $Q_{j}(x)$:

$Q_{j}(x) = \frac{(x-a_{1}) (x-a_{2}) \dots (x-a_{j-1})(x-a_{j+1}) \dots (x-a_{n})}{(a_{j}-a_{1}) (a_{j} -a_{2}) \dots (a_{j}-a_{j-1})(a_{j}-a_{j+1}) \dots (a_{j}-a_{n})}.$

Hopefully, that explains why we divide by $(1- 3)(1-5)$ in the problem you reference above.

In the future, if you have concerns about a problem's wording/clarity/etc., you can report the problem. See how here.

Staff - 4 weeks, 1 day ago