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greysica

Hey I just met u and this is crazy .. Here my problems , Help me maybe ? :)) prove approximation of Mac Laurin Polinom from this function

  • one year ago
  • one year ago

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  1. greysica
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    • one year ago
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  2. lgbasallote
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    No.

    • one year ago
  3. greysica
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    okay .. :(

    • one year ago
  4. GOODMAN
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    Harsh. @lgbasallote

    • one year ago
  5. lgbasallote
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    haha lol

    • one year ago
  6. GOODMAN
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    So help her already. I dont know what that is, lol. Can you atleast try? @lgbasallote

    • one year ago
  7. lgbasallote
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    i have no idea

    • one year ago
  8. greysica
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    yeah yeah , at least try if u can .. :)

    • one year ago
  9. TuringTest
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    I'm not sure what you mean, you want to derive the three series expansions above?

    • one year ago
  10. TuringTest
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    @greysica do I understand that you just want to see how the above statements are derived?

    • one year ago
  11. greysica
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    I have to find the way how it works as a function based on mac laurin polinom

    • one year ago
  12. TuringTest
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    I can derive the mcLauren series from the function, but that doesn't sound like what you want.

    • one year ago
  13. greysica
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    hmm , I'm not sure actually about question for my homework , but would u like show me how it works based on ur opinion ?

    • one year ago
  14. TuringTest
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    the Taylor expansion of a function about x=0 is\[f(x)=\sum_{n=0}^\infty{f^{(n)}(0)\over n!}x^n\]where \(f^{(n)}(0)\) is the \(n^{th}\) derivative of the function at x=0

    • one year ago
  15. greysica
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    is that for A ?

    • one year ago
  16. sami-21
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    @greysica what Turing Test has mentioned above is general formula for Taylor series about a=0 also Known as Maclaurin series.

    • one year ago
  17. sami-21
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    @greysica you need Maclaurin Polynomials of the given functions??

    • one year ago
  18. greysica
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    ya :)

    • one year ago
  19. TuringTest
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    ok @greysica you there? I'll walk you though it must you must participate the formula for the series expansion of a function about \(x=0\) is\[f(x)=\sum_{n=0}^\infty{f^{(n)}(0)\over n!}x^n\]where \(f^{(n)}(0)\) is the \(n^{th}\) derivative of \(f(x)\) at the value \(x=0\) so let's start with the first function, \(f(x)=a^x\) what is the first term of the sequence according to the formula? i.e. when \(n=0\)

    • one year ago
  20. greysica
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    x=0 ?

    • one year ago
  21. TuringTest
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    not not x=0

    • one year ago
  22. greysica
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    |dw:1344008695732:dw|

    • one year ago
  23. TuringTest
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    what is \(f^{(0)}(x)\) when \(f(x)=a^x\) that is, what is the 0th derivative?

    • one year ago
  24. greysica
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    devivative of 0 = 1 ?

    • one year ago
  25. greysica
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    I dont understand about derivative actually .. is that ax^x-1 ?

    • one year ago
  26. TuringTest
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    no, the \(0^{th}\) derivative means we don't take the derivative at all; it's just the function so \(f^{(0)}(x)=f(x)=a^x\) now what then is \(f^{(0)}(0)\) ?

    • one year ago
  27. TuringTest
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    you say you don't know how to take the derivative of a function?

    • one year ago
  28. greysica
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    ya , I dont know how , I really confuse about this subject

    • one year ago
  29. greysica
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    f(0) ?

    • one year ago
  30. TuringTest
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    yes, and what is f(0) in this case?

    • one year ago
  31. greysica
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    so f(0)=a^x ?

    • one year ago
  32. greysica
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    f(0)=a^0

    • one year ago
  33. greysica
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    f(0)=1 ?

    • one year ago
  34. greysica
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    or f(0)= a ?

    • one year ago
  35. TuringTest
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    f(x)=a^x f(0)=1 correct what about the rest of the parts of the formula?\[T_n(a^x)={f^{(n)}(0)\over n!}x^n\] we agree that \(f^{(n)}(0)=1\) right? now where else in this formula do we need to plug in zero to get the first term?

    • one year ago
  36. greysica
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    f(n)(0)=1 ---------1 ?? 0!

    • one year ago
  37. TuringTest
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    right should have written f^(0)(0) but still, that's right so what does all that simplify to?

    • one year ago
  38. greysica
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    1/0 * 1 ?

    • one year ago
  39. TuringTest
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    \[f^{(0)}(0)=f(0)=a^0=1\] and \[x^0=1\]true, but\[0!\neq0\]

    • one year ago
  40. TuringTest
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    \[0!=1\]by definition

    • one year ago
  41. TuringTest
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    so try again, what does it simplify too?

    • one year ago
  42. TuringTest
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    \[T_0(a^x)={f^{(0)}(0)\over 0!}x^0={a^0\over0!}x^0=?\]

    • one year ago
  43. greysica
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    1/1 * 1= 1

    • one year ago
  44. TuringTest
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    correct, that is the first term :)

    • one year ago
  45. TuringTest
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    so let's write that down somewhere because we will need it later:\[T_0(a^x)=1\]now we try to find \(T_1(a^x)\) so now we use the formula\[T_n(a^x)={f^{(n)}(0)\over n!}x^n\]with \(n=1\) what do you get? (this should take you a moment to figure out, i'll brb.

    • one year ago
  46. TuringTest
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    in case you have forgotten your derivatives use this list http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Derivatives_Reduced.pdf and tell me what you get for the next term like I said, I'm going to the store, brb

    • one year ago
  47. greysica
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    (f^1)(0) ----------x^1 1!

    • one year ago
  48. TuringTest
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    and what is \(f^{(1)}(0)\) for \(f(x)=a^x\) ?

    • one year ago
  49. greysica
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    0/1 * 1 ?

    • one year ago
  50. TuringTest
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    I don't think you have forgotten what I said that \(f^{(n)}(x)\) means it means the \(n^{th}\) derivative of the function \(f(x)\) so what is \(f^{(1)}(x)\) considering that \(f(x)=a^x\)

    • one year ago
  51. TuringTest
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    *I think you have...

    • one year ago
  52. greysica
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    n^th ? what does mean ?

    • one year ago
  53. TuringTest
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    it means the n-th number derivative n=1 is the first derivative n=2 is the second derivative n=3 is the third derivative etc.

    • one year ago
  54. greysica
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    so 1^1 --------1 ?? 1!

    • one year ago
  55. TuringTest
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    first you need to find the correct derivative of the function since n=1 that means we need the first derivative of the function what is the first derivative of \(f(x)=a^x\) ?

    • one year ago
  56. greysica
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    1! = 2 right ? 1! means 1+1*1 = 2 ? like that ?

    • one year ago
  57. greysica
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    or 1 ! = 1*1 = 1 ?

    • one year ago
  58. TuringTest
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    no, the question we need to answer is: if\[f(x)=a^x\]what is\[f^{(1)}(x)=f'(x)=?\]there should be no number in the answer to the question I am asking you

    • one year ago
  59. greysica
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    oh ..

    • one year ago
  60. TuringTest
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    ignore every other part of the formula for now, we are just trying to figure out the numerator part at the moment

    • one year ago
  61. greysica
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    i dunno :(

    • one year ago
  62. greysica
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    ax^x-1 ?

    • one year ago
  63. TuringTest
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    look on the list I gave you http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Derivatives_Reduced.pdf find your function though if you have problems taking the derivative of functions this is going to be nearly impossible for you

    • one year ago
  64. greysica
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    f (x) + f '(x) ?

    • one year ago
  65. TuringTest
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    no sorry I can't really help you if you can't take the derivative you may need to retake calculus 1 if this is a problem

    • one year ago
  66. greysica
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    f^0 (x) ?

    • one year ago
  67. greysica
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    ughh .. okay .. thanks ,,,

    • one year ago
  68. TuringTest
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    the derivative formula you want is under the part of the list I gave you that says "common derivatives" top row, third column

    • one year ago
  69. TuringTest
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    what is \[\frac d{dx}(a^x)\]?

    • one year ago
  70. greysica
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    a^x ln(a)

    • one year ago
  71. TuringTest
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    yes

    • one year ago
  72. greysica
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    what is the meaning of In ?

    • one year ago
  73. TuringTest
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    natural logarithm (base e)\[\huge\log_e(a)=\ln a\]so what you just found above from the chart is the first derivative of the function is\[f^{(1)}(x)\]so what is\[f^{(1)}(0)\]? (plug in zero into what you found for the derivative)

    • one year ago
  74. greysica
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    @TuringTest I think I will ask about this to my senior .. I really really lost about this subject .. I dontknow why I'm so stupid :(

    • one year ago
  75. greysica
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    I'm so desperate and shame about this .. thank u for trying help me

    • one year ago
  76. TuringTest
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    Taylor series are tough for many people, so don't feel bad. I was watching the MIT multivariable calc lecture, and when the teacher said they needed Taylor series to solve a problem the whole class was like "NO!" hahaa so you're not alone, but you won't get there unless you make sure you understand derivatives very well, so work on that in the meantime first. Don't feel ashamed, just try to identify your problem areas and don't give up! good luck :D

    • one year ago
  77. greysica
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    yeah !! Thank u very much :) I'll try my best ! I have calculus 2 test in the first week of september , wish can do it properly !! I hope I can get at least B :)

    • one year ago
  78. TuringTest
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    ...and you are NOT stupid, so don't say that :P just review your derivatives you're welcome, and good luck :)

    • one year ago
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