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? one year ago one year ago

1. swissgirl

Every x has the formula of -5+jh

2. swissgirl

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

3. 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

4. Mikael

@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

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

6. swissgirl

I am stuck finding my intial points the x's

7. mahmit2012

it is always unique !

8. Mikael

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

9. mahmit2012

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

10. mahmit2012

The different methods give a unique solution.

11. Mikael

Lagrange $\neq$ Chebyshev

12. Mikael

Same degree - is critically import

13. Mikael

I see now @swissgirl solved I think:

14. swissgirl

ohhh ya??????

15. Mikael

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

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

17. Mikael

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

18. mahmit2012

Mikael Chebishov just gives you the fix points.

19. swissgirl

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

20. Mikael

So pls tell me - here you mean Lagrange ?

21. swissgirl

I guess cuz I need to use Neville's method

22. mahmit2012

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

23. Mikael

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

no the functionis bounded and continuous ==> bounded variation

25. Mikael

They must converge to it

26. swissgirl

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

27. swissgirl

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

28. mahmit2012

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29. mahmit2012

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30. mahmit2012

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31. mahmit2012

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32. mahmit2012

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

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34. mahmit2012

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35. swissgirl

Thanks @mahmit2012