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greysica Group Title

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 Group Title
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    • one year ago
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  2. lgbasallote Group Title
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    No.

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

    • one year ago
  4. GOODMAN Group Title
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    Harsh. @lgbasallote

    • one year ago
  5. lgbasallote Group Title
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    haha lol

    • one year ago
  6. GOODMAN Group Title
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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 Group Title
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    i have no idea

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

    • one year ago
  9. TuringTest Group Title
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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 Group Title
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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 Group Title
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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 Group Title
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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 Group Title
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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 Group Title
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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 Group Title
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    is that for A ?

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

    • one year ago
  18. greysica Group Title
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    ya :)

    • one year ago
  19. TuringTest Group Title
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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 Group Title
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    x=0 ?

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

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

    • one year ago
  23. TuringTest Group Title
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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 Group Title
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    devivative of 0 = 1 ?

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

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

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

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

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

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

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

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

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

    • one year ago
  35. TuringTest Group Title
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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 Group Title
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    f(n)(0)=1 ---------1 ?? 0!

    • one year ago
  37. TuringTest Group Title
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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 Group Title
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    1/0 * 1 ?

    • one year ago
  39. TuringTest Group Title
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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 Group Title
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    \[0!=1\]by definition

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

    • one year ago
  42. TuringTest Group Title
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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 Group Title
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    1/1 * 1= 1

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

    • one year ago
  45. TuringTest Group Title
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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 Group Title
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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 Group Title
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    (f^1)(0) ----------x^1 1!

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

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

    • one year ago
  50. TuringTest Group Title
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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 Group Title
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    *I think you have...

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

    • one year ago
  53. TuringTest Group Title
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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 Group Title
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    so 1^1 --------1 ?? 1!

    • one year ago
  55. TuringTest Group Title
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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 Group Title
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    1! = 2 right ? 1! means 1+1*1 = 2 ? like that ?

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

    • one year ago
  58. TuringTest Group Title
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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 Group Title
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    oh ..

    • one year ago
  60. TuringTest Group Title
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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 Group Title
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    i dunno :(

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

    • one year ago
  63. TuringTest Group Title
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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 Group Title
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    f (x) + f '(x) ?

    • one year ago
  65. TuringTest Group Title
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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 Group Title
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    f^0 (x) ?

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

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

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

    • one year ago
  71. TuringTest Group Title
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    yes

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

    • one year ago
  73. TuringTest Group Title
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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 Group Title
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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 Group Title
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    I'm so desperate and shame about this .. thank u for trying help me

    • one year ago
  76. TuringTest Group Title
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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 Group Title
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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 Group Title
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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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is replying to Can someone tell me what button the professor is hitting...

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