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anonymous

  • 5 years ago

find the bound on the magnitude of the error if we approximate sqrt 2 using the taylor approximation of degree three for sqrt 1+x about x=0

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  1. anonymous
    • 5 years ago
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    I hate approximations >.<

  2. anonymous
    • 5 years ago
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    Let me see.

  3. anonymous
    • 5 years ago
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    k ty

  4. anonymous
    • 5 years ago
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    i just dont know where to start or what to do

  5. anonymous
    • 5 years ago
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    hey anwara can u plz look at my problem too when u r done here

  6. anonymous
    • 5 years ago
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    I think I am almost there :)

  7. anonymous
    • 5 years ago
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    ty u so much i really appreciate it

  8. anonymous
    • 5 years ago
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    Do you know how to find the Taylor expansion of a function?

  9. anonymous
    • 5 years ago
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    can you please showme

  10. anonymous
    • 5 years ago
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    Hmm, Do you have the formula of finding the Taylor expansion?

  11. anonymous
    • 5 years ago
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    We're just looking for the third degree expansion, that's the first four terms.

  12. anonymous
    • 5 years ago
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    okay my only question is what do i plug into x

  13. anonymous
    • 5 years ago
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    That's good. So, we have the first four terms of the expansion: \[\sqrt{1+x}=1+{x \over 2}-{x^2 \over 8}+{x^3 \over 16}+....\] We are looking for the approximation of sqrt(2), that's sqrt(1+1). So, we will plug x=1, and then we get: \[\sqrt{1+1}=\sqrt{2}=1+{1 \over 2}-{1 \over 8}+{1 \over 16}={23 \over 16}\]

  14. anonymous
    • 5 years ago
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    hey anwara can u plz look at my problem too when u r done here 14 minutes ago

  15. anonymous
    • 5 years ago
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    okay so now what do i do with this value

  16. anonymous
    • 5 years ago
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    Yeah sure rsaad2.

  17. anonymous
    • 5 years ago
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    its on work done

  18. anonymous
    • 5 years ago
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    Now this is the approximated value of sqrt(2), we're looking now for the error in this value.

  19. anonymous
    • 5 years ago
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    Oh I don't really know what the bound on the magnitude of the error exactly is.

  20. anonymous
    • 5 years ago
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    for the taylor series expansion did u use the binomial series to figure it out

  21. anonymous
    • 5 years ago
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    now can u plz look at my problme plz

  22. anonymous
    • 5 years ago
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    Well, I didn't. But, you could, it's actually easier to be found by the binomial series.

  23. anonymous
    • 5 years ago
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    now can u plz look at my problem

  24. anonymous
    • 5 years ago
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    so basically i would plug in sqrt of 1

  25. anonymous
    • 5 years ago
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    Yep.

  26. anonymous
    • 5 years ago
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    okay so u dont know how do the rest of the problem

  27. anonymous
    • 5 years ago
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    Not really. I don't know what the bound on the magnitude of the error is. If you could provide me with a formula or any information regarding that, I might be able to help.

  28. anonymous
    • 5 years ago
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    if you click on this link you will see it http://math.jasonbhill.com/courses/fall-2010-math-2300-005/lectures/taylor-polynomial-error-bounds

  29. anonymous
    • 5 years ago
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    its under this topic Theorem 10.1 Lagrange Error Bound 

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