Sequence of Interpolating Values

[Hero1]

Construct a sequence of interpolating values \(y_n\) to \(f(1 +\sqrt{10})\), where \(f(x)= \frac{1}{1+x^2 }\) for \(−5≤x≤5\), as follows: For each \(n = 1,2,…,10\), let \(h =\frac{10}{n}\) and \(y_n= P_n (1+\sqrt{10})\), where \(P_n(x)\) is the interpolating polynomial for \(f(x)\) at nodes \(x_0^n,x_1^n,…,x_n^n=−5+jh\), for each \(j=0,1,2,…,n\). Does the sequence \({y_n }\) appear to converge to \(f(1+\sqrt{10} )\)?

 

[Sudharaka]

Construct a sequence of interpolating values \(y_n\) to \(f(1 +\sqrt{10})\), where \(f(x)= \frac{1}{1+x^2 }\) for \(−5≤x≤5\), as follows: For each \(n = 1,2,…,10\), let \(h =\frac{10}{n}\) and \(y_n= P_n (1+\sqrt{10})\), where \(P_n(x)\) is the interpolating polynomial for \(f(x)\) at nodes \(x_0^n,x_1^n,…,x_n^n=−5+jh\), for each \(j=0,1,2,…,n\). Does the sequence \({y_n }\) appear to converge to \(f(1+\sqrt{10} )\)?

Hi Hero, :)

I am not very clear about your question. Do you have to construct interpolating polynomials for each, \(\frac{10}{n}\) where \(n=1,2,\cdots,10\) separately?

Kind Regards,
Sudharaka.

 

[chisigma]

Construct a sequence of interpolating values \(y_n\) to \(f(1 +\sqrt{10})\), where \(f(x)= \frac{1}{1+x^2 }\) for \(−5≤x≤5\), as follows: For each \(n = 1,2,…,10\), let \(h =\frac{10}{n}\) and \(y_n= P_n (1+\sqrt{10})\), where \(P_n(x)\) is the interpolating polynomial for \(f(x)\) at nodes \(x_0^n,x_1^n,…,x_n^n=−5+jh\), for each \(j=0,1,2,…,n\). Does the sequence \({y_n }\) appear to converge to \(f(1+\sqrt{10} )\)?

That is a classical 'example' of 'divergence' of a interpolating polynomial with equidistant points that was 'discovered' by the German Mathematician C.D.T. Runge. See...

Runge's phenomenon - Wikipedia, the free encyclopedia

The 'error function' is given by...

$\displaystyle f(x)-p(x)= \frac{f^{(n+1)}(\xi)}{(n+1)!}\ \prod_{k=1}^{n+1} (x-x_{k})$ (1)

... where $\xi$ is in the definition interval of f(*). In general is the behavior of the error at the edges of interval that causes divergence...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

That is a classical 'example' of 'divergence' of a interpolating polynomial with equidistant points that was 'discovered' by the German Mathematician C.D.T. Runge. See...

Runge's phenomenon - Wikipedia, the free encyclopedia

The 'error function' is given by...

$\displaystyle f(x)-p(x)= \frac{f^{(n+1)}(\xi)}{(n+1)!}\ \prod_{k=1}^{n+1} (x-x_{k})$ (1)

... where $\xi$ is in the definition interval of f(*). In general is the behavior of the error at the edges of interval that causes divergence...

Kind regards

$\chi$ $\sigma$

A good method to avoid the ‘Runde’s phenomenon’is to avoid to use equidistant point and to interpolate in the so called ‘Chebysheff-Gauss-Lobatto’ points given by …

$\displaystyle x_{k}= - \cos \frac{k\ \pi}{n}\,\ k=0,1,…,n$ (1)


For the details see…

http://mathdl.maa.org/images/upload_library/4/vol6/Sarra/Chebyshev.html

... where a very interesting 'animation' at the end of section 5.1 shows the better performance of the CGL points approach respect to the 'spontaneous' equidistant points approach...

Kind regards

$\chi$ $\sigma$

 

[CellCoree]

Yes, the sequence \({y_n}\) appears to converge to \(f(1+\sqrt{10})\). As \(n\) increases, the value of \(h\) decreases, resulting in a larger number of nodes for the interpolating polynomial. This allows for a more accurate approximation of \(f(1+\sqrt{10})\), leading to a convergence of the sequence to the true value. Additionally, since \(f(x)\) is a continuous function, the interpolating polynomial will also be continuous and therefore the sequence \({y_n}\) will approach \(f(1+\sqrt{10})\) in a smooth manner.