# Is it possible to find function that contains every given point?

Let say we have a arbitrary number of given points and there is at least one function, for which every point lies on its graph. Is it possible to find that function using only X and Y coordinates of every given point?

Example:

We are given points $A(0,1)$; $B(0.27;0)$ and $C (3.73;0)$. All of these points lie on the graph of the function $f(x) = x^2 - 4x + 1$. Is it possible to find this function using only X and Y coordinates of the point A,B and C.

Note: The number of point is can be really big and also the exponent of the function.

• I assume by "function" you mean "formula"? A function that matches all points you specify and is zero everywhere else satisfies the question and counts as a function. May 29, 2013 at 10:38
• Chech my edit. Maybe it'll be clearer to you. May 29, 2013 at 10:48
• @RobertMastragostino, it doesn't always. The "given points" must have different $x$ coordinates to be part of a function. If they have, use Lagrange interpolation and you are set with a continuous function going through all. May 29, 2013 at 11:48

There are many such functions so in general it's not possible to recover the one you started with.

For example if you had a set of points $\{(n,n): n\in\mathbb N\}$ both the functions $f(x) = x$ and $g(x) = x+sin(\pi x)$ would pass through all the points.

If you have a finite number of points you can find a polynomial that passes through them all.

One way to find it would be the following algorithm.

Let the coordinates of the points be $(x_1,y_1), (x_2,y_2),\dots, (x_N,y_N)$

First we find a function $f_1$ that passes through $(x_1,y_1)$ that's easy. $F_1(x) = y_1$.

Now suppose we've constructed $F_n$ that passes through $(x_1,y_1), \dots,(x_n, y_n)$ (so far we've only done this for $n=1$).

Let $$g_n(x) = \prod_{i=1}^n (x-x_i)$$ then $g_n(x_i) = 0$ for every $i\leq n$. Therefore for every value of $\lambda$ the function $F_n(x_i) + \lambda g_n(x_i) = y_i$ for every $i\leq n$. So set $$\lambda_n = \frac{y_{n+1} - F_n(x_{n+1})}{g_n(x_{n+1})}$$ and $$F_{n+1}(x) = F_n(x) + \lambda_n g_n(x).$$

Then $F_{n+1}$ passes through $(x_i,y_i)$ for every $i\leq n+1$. So apply this step repeatedly until you get $F_N$, which passes through every point.

Let's try it with your equation I'm going to choose slightly different points so that we can use whole numbers. Set $$\begin{array}{rlrl} x_1 &=0 & y_1 &=1 \\ x_2 &=2 & y_2 &=-3 \\ x_3 &=5 & y_3 &=6 \end{array}$$

It's easy to check that $x^2-4x+1$ passes through all three points. So let's follow the recipe.

First we set $F_1(x) = y_1 = 1$ and $g_1(x) = (x-x_0) = x$.

Then we can calculate $$\begin{array}{rl}\lambda_1 &= \frac{y_2 - F_1(x_2)}{g_1(x_2)} \\&= -2\end{array}$$ Hence $$\begin{array}{rl}F_2(x) &= F_1(x) + \lambda_1 g_1(x) \\ &= 1 -2x \end{array}$$

We can check that $F_2(0)=1$ and $F_2(2)=-3$.

Next we have $g_2(x) = x(x-2) = x^2-2x$ and we can work out $$\begin{array}{rl}\lambda_2 &= \frac{y_3 - F_2(x_3)}{g_2(x_3)} \\ &= \frac{6 - (-3)}{40} \\&= 1\end{array}$$

So $\lambda_2 = -\frac 3{15}$ and we have

Then we have $$\begin{array}{rl}F_3(x) &= F_2(x) + \lambda_2 g_2(x) \\ &= (1-2x) + (x^2 - 2x) \\ &= x^2-4x+1 \end{array}$$

Then as we've run out of points we set $f(x) = F_3(x)$ and we are done.

• I need to find at least one function that satisfy the condition. Also by infinite number of point, i mean you may be given 10 000, 20 or 10 points. May 29, 2013 at 10:47
• @Stefan4024: So by "infinite", you mean "finite"? May 29, 2013 at 10:50
• Somehow. I'm don't speak good English, so i didn't know how to say that the number of points can be so big number. May 29, 2013 at 10:53
• I think the correct term would be an arbitrary number of points.
– Tim
May 29, 2013 at 10:53
• @Stefan4024 I've just updated this answer to give you a method for an arbitrary list of points.
– Tim
May 29, 2013 at 11:15

Suppose you have $n$ given points. Assume a polynomial function of order $n-1$ with unknown constant coefficients. You'll now have to solve simultaneously $n$ number of equations for coefficients that make the polynomial curve pass through the points.

For example in your case, there are $3$ given points $A(0,1)$, $B(0.27,0)$ and $C(3.73,0)$.

Now assume a polynomial of order $2$ with unknown constant coefficients.

$$y=A+Bx+Cx^2$$

Plug in the $3$ points to get $3$ equations in $A$, $B$ and $C$ and simultaneously solve them.

And indeed $A=1$, $B=-4$ and $C=1$.

This can be employed for any arbitrary set of points, though the process gets increasingly tedious with large number of points.