swissgirl
  • swissgirl
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?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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swissgirl
  • swissgirl
Every x has the formula of -5+jh
swissgirl
  • swissgirl
I am just not sure what i wld plug in for my j and h
swissgirl
  • swissgirl
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

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More answers

anonymous
  • anonymous
@swissgirl Not that I ever dealt with interpolating polynomials much, BUT 1) They are definitely NOT unique - even I know of at least 2-3 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
swissgirl
  • swissgirl
I dont really need help finding the polynomials. There is a method for that
swissgirl
  • swissgirl
I am stuck finding my intial points the x's
anonymous
  • anonymous
it is always unique !
anonymous
  • anonymous
Lagrange of specified degree IS unique . But if Not lagrange or not specific degree - MULTIPLIQUE !
anonymous
  • anonymous
My bamboo is going to be install, so I will tell you .
anonymous
  • anonymous
The different methods give a unique solution.
anonymous
  • anonymous
Lagrange \[ \neq \] Chebyshev
anonymous
  • anonymous
Same degree - is critically import
anonymous
  • anonymous
I see now @swissgirl solved I think:
swissgirl
  • swissgirl
ohhh ya??????
anonymous
  • anonymous
You find ur interp.-ing values by simple 1-st or 2-nd degree Taylor approxim. THEN you costruct your Lagrange polyn. or whatever
swissgirl
  • swissgirl
Read the question the x's are derived from the formula -5+jh
anonymous
  • anonymous
Well I tried. Anyway , for me it very clear that the words "THE interpolating polynomial of degree 10" are ill defined.
anonymous
  • anonymous
Mikael Chebishov just gives you the fix points.
swissgirl
  • swissgirl
Ya maybe I am slow idk this question is confusing. Thanks @Mikael for trying :)
anonymous
  • anonymous
So pls tell me - here you mean Lagrange ?
swissgirl
  • swissgirl
I guess cuz I need to use Neville's method
anonymous
  • anonymous
Mikael it is not different. The assumption gives fix points.
anonymous
  • anonymous
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
anonymous
  • anonymous
no the functionis bounded and continuous ==> bounded variation
anonymous
  • anonymous
They must converge to it
swissgirl
  • swissgirl
The sequence i dont think converges but you wld only be able to see that if u knew ur starting points
swissgirl
  • swissgirl
I posted the question on MSE maybe someone will have an answer
anonymous
  • anonymous
|dw:1347835065357:dw|
anonymous
  • anonymous
|dw:1347835285217:dw|
anonymous
  • anonymous
|dw:1347835529207:dw|
anonymous
  • anonymous
|dw:1347835583070:dw|
anonymous
  • anonymous
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.
anonymous
  • anonymous
|dw:1347835972434:dw|
anonymous
  • anonymous
|dw:1347836054752:dw|
swissgirl
  • swissgirl
Thanks @mahmit2012

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