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greysica

  • 2 years ago

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

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  1. greysica
    • 2 years ago
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  2. lgbasallote
    • 2 years ago
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    No.

  3. greysica
    • 2 years ago
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    okay .. :(

  4. GOODMAN
    • 2 years ago
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    Harsh. @lgbasallote

  5. lgbasallote
    • 2 years ago
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    haha lol

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

  7. lgbasallote
    • 2 years ago
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    i have no idea

  8. greysica
    • 2 years ago
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    yeah yeah , at least try if u can .. :)

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

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

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

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

  13. greysica
    • 2 years ago
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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 ?

  14. TuringTest
    • 2 years ago
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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

  15. greysica
    • 2 years ago
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    is that for A ?

  16. sami-21
    • 2 years ago
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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.

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

  18. greysica
    • 2 years ago
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    ya :)

  19. TuringTest
    • 2 years ago
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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 \(n=0\)

  20. greysica
    • 2 years ago
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    x=0 ?

  21. TuringTest
    • 2 years ago
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    not not x=0

  22. greysica
    • 2 years ago
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    |dw:1344008695732:dw|

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

  24. greysica
    • 2 years ago
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    devivative of 0 = 1 ?

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

  26. TuringTest
    • 2 years ago
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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 \(f^{(0)}(0)\) ?

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

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

  29. greysica
    • 2 years ago
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    f(0) ?

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

  31. greysica
    • 2 years ago
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    so f(0)=a^x ?

  32. greysica
    • 2 years ago
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    f(0)=a^0

  33. greysica
    • 2 years ago
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    f(0)=1 ?

  34. greysica
    • 2 years ago
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    or f(0)= a ?

  35. TuringTest
    • 2 years ago
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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?

  36. greysica
    • 2 years ago
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    f(n)(0)=1 ---------1 ?? 0!

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

  38. greysica
    • 2 years ago
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    1/0 * 1 ?

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

  40. TuringTest
    • 2 years ago
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    \[0!=1\]by definition

  41. TuringTest
    • 2 years ago
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    so try again, what does it simplify too?

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

  43. greysica
    • 2 years ago
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    1/1 * 1= 1

  44. TuringTest
    • 2 years ago
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    correct, that is the first term :)

  45. TuringTest
    • 2 years ago
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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.

  46. TuringTest
    • 2 years ago
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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

  47. greysica
    • 2 years ago
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    (f^1)(0) ----------x^1 1!

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

  49. greysica
    • 2 years ago
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    0/1 * 1 ?

  50. TuringTest
    • 2 years ago
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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 \(f(x)=a^x\)

  51. TuringTest
    • 2 years ago
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    *I think you have...

  52. greysica
    • 2 years ago
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    n^th ? what does mean ?

  53. TuringTest
    • 2 years ago
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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.

  54. greysica
    • 2 years ago
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    so 1^1 --------1 ?? 1!

  55. TuringTest
    • 2 years ago
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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 \(f(x)=a^x\) ?

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

  57. greysica
    • 2 years ago
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    or 1 ! = 1*1 = 1 ?

  58. TuringTest
    • 2 years ago
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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

  59. greysica
    • 2 years ago
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    oh ..

  60. TuringTest
    • 2 years ago
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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

  61. greysica
    • 2 years ago
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    i dunno :(

  62. greysica
    • 2 years ago
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    ax^x-1 ?

  63. TuringTest
    • 2 years ago
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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

  64. greysica
    • 2 years ago
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    f (x) + f '(x) ?

  65. TuringTest
    • 2 years ago
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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

  66. greysica
    • 2 years ago
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    f^0 (x) ?

  67. greysica
    • 2 years ago
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    ughh .. okay .. thanks ,,,

  68. TuringTest
    • 2 years ago
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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

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

  70. greysica
    • 2 years ago
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    a^x ln(a)

  71. TuringTest
    • 2 years ago
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    yes

  72. greysica
    • 2 years ago
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    what is the meaning of In ?

  73. TuringTest
    • 2 years ago
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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)

  74. greysica
    • 2 years ago
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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 :(

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

  76. TuringTest
    • 2 years ago
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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

  77. greysica
    • 2 years ago
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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 :)

  78. TuringTest
    • 2 years ago
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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 :)

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