Interpolating polynomials

7670 /www/nrich/html/content/id/7670/ 1 0000-00-00T00:00:00 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Often you'll be given a polynomial and are asked to find what its value is at some values of $x$. In this problem, we'll ask you to go the other way around: we give you a set of co-ordinates with distinct $x$ values, and you try to find a polynomial that goes through all of them. First you will prove that such a polynomial exists, then you'll prove some results about the uniqueness of that polynomial. <h3 id="existence">Existence</h3> First, draw some points on a graph to convince yourself of the following: <ul> <li>for any two points, you can draw a straight line between them.</li> <li>for three points, you can't always fit them on a straight line, but you might be able to fit them to a quadratic.</li> <li>you can draw four points that a quadratic couldn't possibly fit, but maybe a cubic would do.</li> </ul> Now for the algebra: if I give you two points $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1\not=x_2$, you should be able to find the equation of a straight line going through both. The standard way of doing this is to state that the equation of a straight line is $y=mx+c$, calculate $m$ via the formula \[m=\frac{y_2-y_1}{x_2-x_1}\] then calculate $c$ via \[c=y_1-mx_1\] and you're done. (Why is it important that $x_1\not=x_2$?) I don't like this method, because it doesn't give me any clues as to how I might fit a curve to three points, or four points, or anything else. Here's an alternative way of thinking about the problem: <ol style="list-style-type: decimal"> <li> Find an equation that will fit the first point. The most obvious such would be $y=y_1$. </li> <li> Now, find out by how much you miss the second point, say $d=y_2-y_1$. </li> <li> How can you modify your existing equation so that it hits the second point, without ruining the fact it goes through the first point? Well, if you add any equation to yours which is zero at $x_1$, then the result will have the same value at $x_1$, but may be different elsewhere. </li> <li> Straight lines that are zero at $x_1$ are of the form $m(x-x_1)$. What value of $m$ do we need? Well, take $m=1$ and see what happens at $x_2$: our graph has moved by $D=x_2-x_1$. Clearly if we take $m=2$ it'll move by $2D$, or $m=\frac{1}{2}$ would move it by $\frac{1}{2}D$. We want to move it by $d$, so we take $m=\frac{d}{D}$. We then have $y=y_1+\frac{d}{D}(x-x_1)$, which we can expand and simplify if we so wish, and we're done. </li> </ol> Can you generalise this approach to find a quadratic that goes through (1,0), (4,6), and (2,-2)? Can you find a cubic that goes through these points and (0,-2)? Can you find a general procedure for fitting $n$ co-ordinates? What kind of graph will you need to use? <h3 id="uniqueness">Uniqueness</h3> Draw some graphs of some quadratics to convince yourself that if a quadratic is zero in three places, it must be zero everywhere (equivalently, no quadratic crosses the $x$-axis three times). Then, using the Factor Theorem or otherwise, prove it. (Here by quadratic we just mean $y=ax^2+bx+c$ where any of $a$, $b$ or $c$ may be zero). Hence prove that if two quadratic curves are equal in three places, they must be equal everywhere. So the quadratic you found that went through (1,0), (4,6), and (2,-2) is the only quadratic that does so. Generalise your results to polynomials of any degree. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This problem required making logical arguments that come together to form a proof. It can be hard to put all the steps in the right place but we received great solutions from Michael from the UK, Kaan from the German Highschool, Istanbul, Turkey and Alex from England. They all used very similar ideas and Michael produced a very well written and formatted word document explaining the ideas. The full solution is in PDF format at the following link: <a href="/content/id/7670/Interpolating%20Polynomials%20Soln1.pdf">Interpolating Polynomials Soln1.pdf</a> Thanks for the fantastic ideas you sent in!</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This problem is an extension of secondary ideas, intended primarily for the keen and those considering mathematics at university. There are many applications of the idea of interpolating polynomials, but this problem is more about presenting the ideas of existence and uniqueness proofs, as well as giving students an intuition for the different ways graphs can be manipulated. <h3 id="related-resources">Related resources</h3> <ul> <li>If students struggle manipulating graphs, there are some simpler problems in that area: <a href="/773">Parabolic Patterns</a>, <a href="/791">Parabolas Again</a>, <a href="/802">Cubics</a>, <a href="/6481">Tangled Trig Graphs</a>, <a href="/7120">Quadratic Transformations</a>.</li> </ul> <h3 id="discussion-ideas">Discussion ideas</h3> You can also discuss how a polynomial of degree $n$ can be defined in two different ways – either as the $n+1$ coefficients of powers of $x$, or the values of the polynomial at $n+1$ distinct inputs. In linear algebra terminology, such polynomials belong to an $n+1$-dimensional vector space. A more intuitive notion of dimension that may be more suitable at this level is a measure of "free"-ness: add a dimension for each (real-valued) free variable and subtract one for each constraint.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> To find four points that a quadratic couldn't possibly fit, remember that quadratics only have one turning point. To find the quadratic that fits three points, make sure you understand how you can add and subtract graphs, and what happens to the result. Don't try to fit all three points at once &ndash; fit two and then &ldquo;fix&rdquo; your line to fit the third. Uniqueness: The Factor Theorem states that if $p$ is a polynomial and $p(a) = 0$, then there is a polynomial $q$ such that $p(x) = (x-a)q(x)$. What does this mean about the degrees of $p$ and $q$? Finally, to prove two polynomials are equal, try proving their difference is zero. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3 id="existence">Existence</h3> <ul> <li>The quadratic method is essentially the same as the linear method; join two of the points by a line and then use $m(x-1)(x-4)$ for varying values of $m$ to hit the third. The quadratic that goes through (1,0), (4,6), and (2,-2) is \[y=2x^2-8x+6\] The cubic that goes through those points and (0,-2) is \[y=x^3-5x^2+6x-2\]</li> <li>You need a polynomial of degree $n-1$. The general procedure is by induction: assume you have $(x_1,y_1)\dots(x_n,y_n)$, and that you can produce a degree $n-2$ polynomial $p$ that fits the first $n-1$ points. Then let $q(x)=\Pi_{k=1}^{n-1}(x-x_k)$, and the polynomial \[p(x)+q(x)\frac{y_n-p(x_n)}{q(x_n)}\] is degree $n-1$ and fits all $n$ points.</li> </ul> <h3 id="uniqueness">Uniqueness</h3> Suppose $p(x)$ is a polynomial of degree $n$ which is zero on $n+1$ distinct points $x_1\dots x_{n+1}$. The Factor Theorem states that $p(x)=(x-x_1)\dots(x-x_{n+1})q(x)$ for some polynomial $q$. But this has degree $\ge n+1$ unless $q$ is zero, so $p$ is zero. Suppose $r$ and $s$ are polynomials of degree n, and $r(x_i)=s(x_i)$ for $i=1\dots n+1$, $x_i$ distinct. Then $r(x)-s(x)$ is a polynomial degree $n$ which is zero in $n+1$ places, so it is zero, so $r=s$. </mdoxml> 2 5 0 0 0 0 1 Interpolating polynomials Given a set of points (x,y) with distinct x values, find a polynomial that goes through all of them, then prove some results about the existence and uniqueness of these polynomials. 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