# Find$f$ s.t. $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$.

Find a function where $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$.

I got $\dfrac16(x-3)(x-2)(x-1)\pi$ as a start to get rid of $\pi$.

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Do you know about Lagrange interpolation? – Git Gud Mar 29 '14 at 18:40
How is $$\frac16(1-3)(1-2)(1-1)\pi = 2$$? – Sabyasachi Mar 29 '14 at 18:42
@GitGud I think it is quite powerful to discover Langrange interpolation. And the question here is on the brink - so don't look it up, work it out. What do you have to do to deal with $3$ the way you dealt with $\pi$? – Mark Bennet Mar 29 '14 at 18:43
@Sabyasachi The key thing is that the expression in the question is zero for $x=1,2,3$ and $\pi$ when $x=4$ - that is a very good start. How to find something which is zero when $x=1,3,\pi$ and $4$ when $x=2$ is another question. Then how to put the pieces together, then realise you have discovered Lagrange interpolation - and understand it. – Mark Bennet Mar 29 '14 at 18:47
@MarkBennet (+1)I understand that. I was just pointing OP that he can cleverly deal with the rest like he deals with the $\pi$ – Sabyasachi Mar 29 '14 at 18:51

Layout: (As hinted by Mark Bennet in the comments).

Find a polynomial $P_1$ such that $P_1(1)=2$ and $P_1(2)=P_1(3)=P_1(4)=0$.

Find a polynomial $P_2$ such that $P_2(2)=4$ and $P_2(1)=P_2(3)=P_2(4)=0$.

Find a polynomial $P_3$ such that $P_3(3)=6$ and $P_3(1)=P_3(2)=P_3(4)=0$.

Find a polynomial $P_4$ such that $P_4(4)=\pi$ and $P_4(1)=P_4(2)=P_4(3)=0$.

The polynomial $P_1+P_2+P_3+P_4$ as the desired property.

You already found $P_4$, the rest should follow easily.

Alternatively, take the ansatz $P(x)=\alpha x^3+\beta x^2+\gamma x+\delta$.

You get $\begin{cases} \alpha +\beta +\gamma +\delta &=2\\ 8\alpha +4\beta +2\gamma+\delta &=4\\ 27\alpha +9\beta +3\gamma +\delta &=6\\ 64\alpha +16\beta +4\gamma +\delta &=\pi\end{cases}$ and you can solve a system, if you're patient enough.

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Remainder Factor Theorem works much better in this case. – Calvin Lin Mar 29 '14 at 19:31

[I'm assuming by function you mean cubic polynomial, otherwise this is simple.]

Observe that the first three equations are satisfied by the function $2x$.

Hence, the polynomial $f(x) - 2x$ has roots $x=1, 2, 3$, or that $f(x) = 2x + A(x-1)(x-2)(x-3)$.

Finally, set $x=4$, we get $\pi = 8 + A\times 3\times 2 \times 1$, or that $A = \frac{\pi-8}{6}$. Hence,

$$f(x) = 2x + \frac{\pi - 8} {6} ( x-1)(x-2)(x-3).$$

Lagrange Interpolation would be the general method to approach problems like this. This solution exploits a property from the selected values.

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Nicest solution, with simplest result. – Ruslan Mar 29 '14 at 19:44
Nice.${{{{}}}}$ – Git Gud Mar 29 '14 at 20:29

Your way of dealing with $\pi$ is very smart, but apparently you only took it because $\pi$ is irrational, and you are expecting to get the "nicer" values of $2,4,6$ in different ways. Here's something you can do, use seperate terms that are zero at $1,2,\pi$ and $6$ at $3$. Try it for yourself. Add all the terms. You already have the requisite insight.

$$\frac{(x-3)(x-2)(x-1)\pi}{6}+\frac{(x-\pi)(x-1)(x-2)6}{2(3-\pi)}+\frac{(x-\pi)(x-1)(x-3)4}{(2-\pi)\times-1}\\+\frac{(x-2)(x-3)(x-\pi)2}{(-1)(-2)(1-\pi)}$$

Simplyfying,

$$\frac{(x-3)(x-2)(x-1)\pi}{6}+\frac{3(x-\pi)(x-1)(x-2)}{(3-\pi)}+\frac{4(x-\pi)(x-1)(x-3)}{(\pi-2)}+\\\frac{(x-2)(x-3)(x-\pi)2}{2(1-\pi)}$$

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@seaturtles fair enough. – Sabyasachi Mar 29 '14 at 18:51

The Lagrange Interpolation method that Git Gud mentions is a way to find a polynomial that goes through a specified (finite) set of points. Suppose we want a polynomial function $p(x)$ such that $p(x_1)=y_1, p(x_2)=y_2, \ldots, p(x_n)=y_n$, where each $x_i$ is distinct from the others.

Then the following polynomial fulfills the conditions:

$$\sum_{i=1}^n{y_i\frac{(x_1-x)(x_2-x)\cdots (x_{i-1}-x)(x_{i+1}-x)\cdots (x_n-x)}{(x_1-x_i)(x_2-x_i)\cdots (x_{i-1}-x_i)(x_{i+1}-x_i)\cdots (x_n-x_i)}}$$

The idea is that we want to create a polynomial $p_i$ that is equal to 1 when evaluated at $x_i$, but zero when evaluated at the other $x_j$ values. Thus, our desired polynomial would then just be $\sum_{i=1}^n{y_ip_i(x)}$. The polynomial $\frac{(x_1-x)(x_2-x)\cdots (x_{i-1}-x)(x_{i+1}-x)\cdots (x_n-x)}{(x_1-x_i)(x_2-x_i)\cdots (x_{i-1}-x_i)(x_{i+1}-x_i)\cdots (x_n-x_i)}$ is precisely such a $p_i$.

Try using this method to create the polynomial: you've already figured out one of the terms in the sum, so just finish it off!

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