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anonymous
 4 years ago
Construct a sequence of interpolating values \(Y_n, to,f(1 + \sqrt{10})\), where \(f(x) = (1 + X^2)^{1}\) for
\(5 \leq X \leq 5\), as follows: For each n = 1,2, ... ,10, let h = 10/n and \(Y_n = P_n(1 + \sqrt{10})\),
where Pn(x) is the interpolating polynomial for f(x) at the nodes \(x_0^n, x_1^n,…,x_n^n\) and
\(x_j^n= 5 + jh\), for each j = 0, 1,2, ... ,n. Does the sequence {Y_n} appear to converge to
\(f(1 + \sqrt{10})\)
How would i set up this sequence?
anonymous
 4 years ago
Construct a sequence of interpolating values \(Y_n, to,f(1 + \sqrt{10})\), where \(f(x) = (1 + X^2)^{1}\) for \(5 \leq X \leq 5\), as follows: For each n = 1,2, ... ,10, let h = 10/n and \(Y_n = P_n(1 + \sqrt{10})\), where Pn(x) is the interpolating polynomial for f(x) at the nodes \(x_0^n, x_1^n,…,x_n^n\) and \(x_j^n= 5 + jh\), for each j = 0, 1,2, ... ,n. Does the sequence {Y_n} appear to converge to \(f(1 + \sqrt{10})\) How would i set up this sequence?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Every x has the formula of 5+jh

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am just not sure what i wld plug in for my j and h

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Like i just need to figure out what my x's wld be but with these h's and j's I am getting confused

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@swissgirl Not that I ever dealt with interpolating polynomials much, BUT 1) They are definitely NOT unique  even I know of at least 23 completely different such interpolating polynomials  Lagrange, Bezier curves http://en.wikipedia.org/wiki/B%C3%A9zier_curve, and Chebyshev polynomials 2) They are very oscillating beasts  don't behave well when forced too much

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I dont really need help finding the polynomials. There is a method for that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am stuck finding my intial points the x's

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is always unique !

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lagrange of specified degree IS unique . But if Not lagrange or not specific degree  MULTIPLIQUE !

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My bamboo is going to be install, so I will tell you .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The different methods give a unique solution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lagrange \[ \neq \] Chebyshev

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Same degree  is critically import

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I see now @swissgirl solved I think:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You find ur interp.ing values by simple 1st or 2nd degree Taylor approxim. THEN you costruct your Lagrange polyn. or whatever

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Read the question the x's are derived from the formula 5+jh

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well I tried. Anyway , for me it very clear that the words "THE interpolating polynomial of degree 10" are ill defined.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mikael Chebishov just gives you the fix points.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ya maybe I am slow idk this question is confusing. Thanks @Mikael for trying :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So pls tell me  here you mean Lagrange ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess cuz I need to use Neville's method

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mikael it is not different. The assumption gives fix points.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I vaguely remember tha on compact interval they do converge in most norms to the function  unless of course the function has unbounded variation. And this may be here because of vertic asymptote

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no the functionis bounded and continuous ==> bounded variation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0They must converge to it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The sequence i dont think converges but you wld only be able to see that if u knew ur starting points

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I posted the question on MSE maybe someone will have an answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1347835065357:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1347835285217:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1347835529207:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1347835583070:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess e can not solve it directly. So I guess it is not going to be zero because f(x) at interval [5,5] has no Tylor polynomial. and it just converge for interval with radios one around a fix point.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1347835972434:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1347836054752:dw
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