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Suppose the function $u(x)$ interpolates $f(x)$ at $x_0,x_1,\cdots ,x_{n-1}$ and the function $v(x)$ interpolates $f(x)$ at $x_1,x_2,\cdots ,x_n$. Then find a function $F(x)$ which interpolates $f(x)$ at all the points $x_0,x_1,\cdots ,x_{n-1},x_n$.

Here , $f(x)=u(x_0)+(x-x_0)u(x_0,x_1)+(x-x_0)(x-x_1)u(x_0,x_1,x_2)+\cdots+(x-x_0)\cdots(x-x_{n-2})u(x_0,x_1,\cdots x_{n-1})$.

Again , $f(x)=v(x_1)+(x-x_1)v(x_1,x_2)+(x-x_1)(x-x_2)v(x_1,x_2,x_3)+\cdots+(x-x_1)\cdots(x-x_{n-1})v(x_1,x_2,\cdots ,x_{n})$.

But from here how I can find such a function ?

Update :

From given data , $u(x_i)=f(x_i)$ for $0\le i\le n-1$. Also , $v(x_i)=f(x_i)$ for $1\le i\le n$. Then , $(x-x_n)u(x_i)=(x-x_n)f(x_i)$ and $(x-x_0)v(x_i)=(x-x_0)f(x_i)$ for all $0\le i\le n.$Then subtracting , $$(x-x_n)u(x_i)+(x-x_0)v(x_i)=(x_0-x_n)f(x_i).$$So , $$\frac{x-x_n}{x_0-x_n}u(x)+\frac{x-x_0}{x_0-x_n}v(x)$$interpolates all the points $x_0,x_1,\cdots ,x_n$. Is it correct now ?

Empty
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  • You do not know anything about $f(x)=\cdots$. What you do know is that $f(x_i)=u(x_i)$ and $f(x_j)=v(x_j)$ for fitting $i,j$. You are asked to find a combination of $u$ and $v$ that works for all $x_i$. It might look something like this: (not a correct solution) $$(x-x_0)v(x)+(x-x_n)u(x)$$ – Arthur Oct 30 '15 at 07:37

1 Answers1

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$u(x)+{(v(x)-u(x))}{(x-x_0)(x-x_1)...(x-x_{n-1})\over(x_n-x_0)(x_n-x_1)...(x_n-x_{n-1})}$

is the function you want.

The above is a long one, amd I have come up with a short one as well:

$v(x){x-x_0\over x_n-x_0}+u(x){x_n-x\over x_n-x_0}$

cr001
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  • How...........? – Empty Oct 30 '15 at 08:01
  • For the long one, sub in any $x=x_0,x_1,...,x_{n-1}$ will give you $u(x)$ since right part is $0$ and sub in $x=x_n$ will give you $u(x)+v(x)-u(x)=v(x)$. For the short one, sub in anything other than $x_0, x_n$, since $u(x)=v(x)$ it will give you $v(x){x_n-x+x-x_0\over x_n-x_0}=v(x)$, sub in $x_n$ it will give $v(x)$ since right part is $0$ and sub in $x_0$ will give you $u(x)$ since the left part is $0$. – cr001 Oct 30 '15 at 08:09
  • Assuming that $u,v$ are polynomials of degree $n-1$, this solution has a rather high degree. Esp. since $u(x)-v(x)$ already has roots at $x_1,…,x_{n-1}$. – Lutz Lehmann Oct 30 '15 at 08:46
  • The short solution has degree $n$ which is definitely required for interpolating $n$ roots. The long one has high degree and yes it is not that good but it still work. – cr001 Oct 30 '15 at 08:48
  • @@cr001) See my update please and comment whether am I correct or wrong ? – Empty Nov 04 '15 at 14:12
  • You are almost correct except the second term need to be the negative of itself. Try to sub in $x=x_n$ and see what you get, in your current formula you will get $-v(x)$ instead of $v(x)$. – cr001 Nov 04 '15 at 17:08