MHB Sequence of Interpolating Values

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The discussion focuses on constructing a sequence of interpolating values \(y_n\) for the function \(f(x) = \frac{1}{1+x^2}\) at the point \(1+\sqrt{10}\) using equidistant nodes. Participants highlight that using equidistant points can lead to Runge's phenomenon, where the interpolating polynomial diverges, especially at the edges of the interval. The error function associated with this divergence is discussed, emphasizing the behavior of the error at the interval's boundaries. A suggested solution to mitigate this issue is to use Chebyshev-Gauss-Lobatto points instead of equidistant points for better convergence. Overall, the conversation underscores the challenges and solutions related to polynomial interpolation in numerical analysis.
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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 \(f(1+\sqrt{10} )\)?
 
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Hero said:
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.
 
Hero said:
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 said:
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

$\chi$ $\sigma$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...