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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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swissgirlBest ResponseYou've already chosen the best response.0
Every x has the formula of 5+jh
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
I am just not sure what i wld plug in for my j and h
 one year ago

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

MikaelBest 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
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
I dont really need help finding the polynomials. There is a method for that
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
I am stuck finding my intial points the x's
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
it is always unique !
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
Lagrange of specified degree IS unique . But if Not lagrange or not specific degree  MULTIPLIQUE !
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
My bamboo is going to be install, so I will tell you .
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
The different methods give a unique solution.
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
Lagrange \[ \neq \] Chebyshev
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
Same degree  is critically import
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
I see now @swissgirl solved I think:
 one year ago

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

swissgirlBest ResponseYou've already chosen the best response.0
Read the question the x's are derived from the formula 5+jh
 one year ago

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

mahmit2012Best ResponseYou've already chosen the best response.2
Mikael Chebishov just gives you the fix points.
 one year ago

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

MikaelBest ResponseYou've already chosen the best response.0
So pls tell me  here you mean Lagrange ?
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
I guess cuz I need to use Neville's method
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
Mikael it is not different. The assumption gives fix points.
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
I 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
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
no the functionis bounded and continuous ==> bounded variation
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
They must converge to it
 one year ago

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

swissgirlBest ResponseYou've already chosen the best response.0
I posted the question on MSE maybe someone will have an answer
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
dw:1347835065357:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
dw:1347835285217:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
dw:1347835529207:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
dw:1347835583070:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
I 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.
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
dw:1347835972434:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
dw:1347836054752:dw
 one year ago
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