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blackstreet23

  • one year ago

Approximate tan inverse (0.1) to three decimal place accuracy using the Maclaurin series tan inverse x

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  1. blackstreet23
    • one year ago
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    @PlasmaFuzer

  2. blackstreet23
    • one year ago
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    @ganeshie8

  3. blackstreet23
    • one year ago
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    @pooja195

  4. blackstreet23
    • one year ago
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    @Shalante

  5. blackstreet23
    • one year ago
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  6. blackstreet23
    • one year ago
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    I did this so far The book says that the error to 3 decimal places is \[(0.1)- \frac{ 0.1^3 }{ 3 }\]

  7. blackstreet23
    • one year ago
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    however when i put (0.1^3)/3 in my calculator the answer is 3.3333 * 10^-4 that is less that 0.5 * 10^-3

  8. blackstreet23
    • one year ago
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    so why could that be the answer? wouldn't it just be 0.1 ?

  9. blackstreet23
    • one year ago
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    @Hero

  10. IrishBoy123
    • one year ago
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    the question asks you to do it using the maclaurin for cos x is that what you did?

  11. blackstreet23
    • one year ago
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    yes

  12. IrishBoy123
    • one year ago
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    |dw:1442179486992:dw| terrible graphics i know but it seems you plugged your numbers into the expansion for\(\tan^{-1}x\).

  13. freckles
    • one year ago
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    I wonder why it says to approximate arctan(0.1) using the maclaurin series for cos(x) and here it says use maclaurin series for arctan(x) I think we would want to use the series for arctan(x) since we are apporximating arctan( something)

  14. blackstreet23
    • one year ago
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    yeah i did.

  15. freckles
    • one year ago
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    but it really does say cos(x)?

  16. blackstreet23
    • one year ago
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    tan inverse (0.1)

  17. freckles
    • one year ago
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    You know I'm trying to tell you that there is conflicting information here right?

  18. blackstreet23
    • one year ago
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    ohhh I see

  19. blackstreet23
    • one year ago
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    my bad the one i copied was wrong

  20. blackstreet23
    • one year ago
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    but the question here is right

  21. freckles
    • one year ago
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    ok this question makes more sense to me :) Approximate tan inverse (0.1) to three decimal place accuracy using the Maclaurin series tan inverse x \[0.1-\frac{0.1^3}{3} \text{ is a pretty good approximation for } \arctan(0.1) \\ 0.1-\frac{0.1^3}{3} \approx 0.099\overline{6} \\ \text{ while } \arctan(0.1) \text{ to 6 decimal spots is } 0.09966\]

  22. blackstreet23
    • one year ago
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    Why is the the answer they have on the back of the book is (0.1)−(0.1^3/3)

  23. blackstreet23
    • one year ago
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    but the way I learned is by taking each monomial of the series and substituting the value of x

  24. blackstreet23
    • one year ago
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    and (0.1^3)/3 is less that 5*10^-4

  25. freckles
    • one year ago
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    yep but then you have to add the terms in the series

  26. freckles
    • one year ago
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    where does 5*10^(-4) come from ?

  27. blackstreet23
    • one year ago
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    from a rule that says that the decimal places they are asking you for is equal to 0.5*10^-n where n is the number of decimal places

  28. blackstreet23
    • one year ago
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    so i just simplified to 5*10^-4

  29. blackstreet23
    • one year ago
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    and i need to find a term that is smaller than 5*10^-4 in the series

  30. blackstreet23
    • one year ago
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    @PlasmaFuzer

  31. freckles
    • one year ago
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    oh I don't remember that I would have honestly found arctan(0.1) first which is approximately 0.0996 from the calculator then added the terms in the series until I got with in digits of that number \[\arctan(x)=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1} \\ \arctan(x)=(-1)^0 \frac{x^{2(0)+1}}{2(0)+1}+(-1)^{1}\frac{x^{2(1)+1}}{2(1)+1}+(-1)^{2}\frac{x^{2(2)+1}}{2(2)+1} \\ + \cdots +(-1)^k \frac{x^{2k+1}}{2k+1}+ \cdots \\ \text{ so let's see if one term is enough } \\ \arctan(x) \approx x \\ \arctan(0.1) \approx 0.1 \text{ this is a really bad approximation } \\ \text{ how about two terms } \\ \arctan(x) \approx x- \frac{x^3}{3}\] \[\arctan(0.1) \approx 0.1-\frac{0.1^3}{3} =.099\overline{6} \text{ this is an awesome approximation }\]

  32. freckles
    • one year ago
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    the approximation only gets better if we add more terms onto that

  33. IrishBoy123
    • one year ago
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    |dw:1442182723650:dw|

  34. anonymous
    • one year ago
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    @blackstreet23 Hello... Freckles and irishboy have it right here... That formula you have listed (0.5*10^-n)I dont recall (though I think I can guess where it comes from), but if your issue is you want more accuracy simply add one more term to the approximating series no?

  35. blackstreet23
    • one year ago
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    well i guess i will ask this one to my professor. Thanks guys for your help!

  36. anonymous
    • one year ago
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    Sorry I couldn't help Im not sure what you are hung up on as far as the problem is concerned

  37. blackstreet23
    • one year ago
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    ohh no problem! i know you did your best :)

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