Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

swissgirl

  • 3 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?

  • This Question is Closed
  1. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Every x has the formula of -5+jh

  2. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am just not sure what i wld plug in for my j and h

  3. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  4. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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

  5. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I dont really need help finding the polynomials. There is a method for that

  6. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am stuck finding my intial points the x's

  7. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    it is always unique !

  8. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Lagrange of specified degree IS unique . But if Not lagrange or not specific degree - MULTIPLIQUE !

  9. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    My bamboo is going to be install, so I will tell you .

  10. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The different methods give a unique solution.

  11. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Lagrange \[ \neq \] Chebyshev

  12. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Same degree - is critically import

  13. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I see now @swissgirl solved I think:

  14. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohhh ya??????

  15. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You find ur interp.-ing values by simple 1-st or 2-nd degree Taylor approxim. THEN you costruct your Lagrange polyn. or whatever

  16. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Read the question the x's are derived from the formula -5+jh

  17. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well I tried. Anyway , for me it very clear that the words "THE interpolating polynomial of degree 10" are ill defined.

  18. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Mikael Chebishov just gives you the fix points.

  19. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ya maybe I am slow idk this question is confusing. Thanks @Mikael for trying :)

  20. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So pls tell me - here you mean Lagrange ?

  21. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I guess cuz I need to use Neville's method

  22. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Mikael it is not different. The assumption gives fix points.

  23. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  24. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no the functionis bounded and continuous ==> bounded variation

  25. Mikael
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    They must converge to it

  26. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The sequence i dont think converges but you wld only be able to see that if u knew ur starting points

  27. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I posted the question on MSE maybe someone will have an answer

  28. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1347835065357:dw|

  29. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1347835285217:dw|

  30. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1347835529207:dw|

  31. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1347835583070:dw|

  32. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  33. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1347835972434:dw|

  34. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1347836054752:dw|

  35. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks @mahmit2012

  36. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy