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anonymous

  • one year ago

Assume that \sin(x) equals its Maclaurin series for all x. Use the Maclaurin series for sin(6 x^2) to evaluate the integral (integral [0, 0.78]) sin(6 x^2)dx . Your answer will be an infinite series. Use the first two terms to estimate its value.

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  1. anonymous
    • one year ago
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    \[\int\limits_{0}^{0.78}\sin(6x^2)dx\]

  2. anonymous
    • one year ago
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    @mathstudent55

  3. IrishBoy123
    • one year ago
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    writing the Maclaurin sin expansion in terms of t: \(\large sin(t) = \Sigma_{0}^{\infty} \frac{(-1)^n }{(2n+1)! }\ t^{2n+1}\) it seems they are happy for you to stuff \(t = 6 x^2\) in and go from there to use an actual Mac series for \(sin(6 x^2)\), you'd need to go \(f(0) = 0\) \(f'(x) = (sin(6 x^2))' = ...\) and \(f'(0) = ....\) so \(x \ f'(0) = ...\)

  4. anonymous
    • one year ago
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    Just plug it in directly?

  5. anonymous
    • one year ago
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    What do you think @sourwing

  6. IrishBoy123
    • one year ago
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    "Just plug it in directly?" yes, that is what the question mandates for #1, plug it in term by terms and come up with a nice summation with a \(\Sigma\) in it! for #2, generate a few terms and hope you get the first two quickly. [if you have a crack at the proper Mac expansion for that function, you will be busy for a while :p]

  7. anonymous
    • one year ago
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    So basically \[\sum_{n = 0}^{\infty} \frac{ -1^n }{ (2n+1)! } * (6x^2)^{2n+1}\], and solve for n = 0 and n=1?

  8. IrishBoy123
    • one year ago
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    sounds good

  9. anonymous
    • one year ago
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    Well what do I do to estimate the value of the series?

  10. anonymous
    • one year ago
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    I ask because I still have an x term in my result

  11. IrishBoy123
    • one year ago
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    if you have an 'x' term in your result, well that is a good thing. :p

  12. anonymous
    • one year ago
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    Okay lol. Sorry I'm clueless on the procedure here. Should I be evaluating the limit at both of the solutions? They are N_0 = -6x^2 and N_1 = -216x^2

  13. IrishBoy123
    • one year ago
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    now, forget about the \(\Sigma\) and just evaluate the first 2 terms you are not summing, you are just using the first 2 terms of the series make sense?!?! you need x terms because you have to integrate

  14. IrishBoy123
    • one year ago
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    nothing to do with limits, either. just knock out the first 2 terms of the [bogus] expansion

  15. anonymous
    • one year ago
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    So get the first two terms in the series, but how do I use that to estimate the value of the series?

  16. IrishBoy123
    • one year ago
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    |dw:1439852188969:dw|

  17. IrishBoy123
    • one year ago
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    |dw:1439852553035:dw|

  18. anonymous
    • one year ago
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    \[6x^2-x^5\]??

  19. IrishBoy123
    • one year ago
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    |dw:1439852806038:dw|

  20. anonymous
    • one year ago
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    -1/6 * 6x^5

  21. anonymous
    • one year ago
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    -6x^5/6

  22. anonymous
    • one year ago
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    -x^5

  23. IrishBoy123
    • one year ago
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    you are *not* summing a series, or *valuing* a series you are hacking an integral; by approximating the integrand to a hack of its Mac expansion

  24. anonymous
    • one year ago
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    brb I have to go feed goats

  25. anonymous
    • one year ago
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    @Kainui @Hero I'm sorry, but I still don't understand how to "hack the integral"? I'm still not quite getting what IrishBoy is trying to have me do here. Could you or one of the other moderators help here?

  26. anonymous
    • one year ago
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    Maclaurin series of \(sin(6x^2) \) is \(sin(6x^2) =\sum_{n=0}^{\infty}\dfrac{(-1)^n (6x^2)^{2n+1}}{(2n+1)!}\) \[\int_0^{0.78} sin(6x^2)dx=\int_0^{0.78}\sum_{n=0}^{\infty}\dfrac{(-1)^n(6x^2)^{2n+1}}{(2n+1)!}dx\\=\sum_{n=0}^{\infty}\int_0^{0.78}\dfrac{(-1)^n(6x^2)^{2n+1}}{(2n+1)!} dx\] This integral is easy to take, right? After take out the integral, you just calculate the first 3 term of it (approximate theorem) and you get the answer.

  27. anonymous
    • one year ago
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    oh, your problem asks for the first 2 terms, not 3 as I suggested. That is easier.!!

  28. anonymous
    • one year ago
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    I'm sorry but nothing about this seems easy to me. I hate to keep bothering you but is there any way you can force me to understand this? @Crazyandbeautiful

  29. anonymous
    • one year ago
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    OH, I assumed you know the formula of Maclaurin series of sin (6x^2) If you don't know it, you can derive it from Maclaurin series of sinx, ok?

  30. anonymous
    • one year ago
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    \(sin x =\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}\) ok?

  31. anonymous
    • one year ago
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    Okay

  32. anonymous
    • one year ago
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    so, just replace x by 6x^2 to get Maclaurin series of sin(6x^2)

  33. anonymous
    • one year ago
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    ok?

  34. anonymous
    • one year ago
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    Alright

  35. anonymous
    • one year ago
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    now, you want to take int of sin (6x^2) from 0 to 0.78 , right? Just take integral both sides.

  36. anonymous
    • one year ago
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    The left hand side is \(\int_0^{0.78} sin(6x^2)dx\)

  37. anonymous
    • one year ago
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    The right hand side is integral of the sum (...) right?

  38. anonymous
    • one year ago
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    Yes or NO? where don't you get?

  39. anonymous
    • one year ago
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    I'm just writing this in my notes as we go, sorry

  40. anonymous
    • one year ago
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    Yes

  41. anonymous
    • one year ago
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    And we know that integral of sum = sum of integral, right?

  42. anonymous
    • one year ago
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    Okay

  43. anonymous
    • one year ago
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    hence the right hand side is \[\sum_{n=0}^{\infty} \int_0^{0.78}(-1)^n\dfrac{(6x^2)^{2n+1}}{(2n+1)!}\] ok?

  44. anonymous
    • one year ago
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    Okay, I'm with you

  45. anonymous
    • one year ago
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    Now, calculate the integral. Can you?

  46. anonymous
    • one year ago
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    I can try.

  47. anonymous
    • one year ago
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    But what do I do with the fact that this has both n and x in it?

  48. anonymous
    • one year ago
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    Don't forget \((6x^2)^{2n+1}= 6^{2n+1}*x^{4n+2}\)

  49. anonymous
    • one year ago
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    hey, just x is the variable, n is a number!! nothing to do with n, take it as constant

  50. anonymous
    • one year ago
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    you take integral w.r.t.x (dx) right? how n works as variable?

  51. anonymous
    • one year ago
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    We will use n for the far left summation, not for integral.

  52. anonymous
    • one year ago
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    Okay

  53. anonymous
    • one year ago
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    It turns to \[\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!}\int_0^{0.78} x^{4n+2}dx\]

  54. anonymous
    • one year ago
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    ok?? take integral, what do you get?

  55. anonymous
    • one year ago
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    |dw:1439861021631:dw|

  56. anonymous
    • one year ago
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    \[\frac{ x ^{4n+3} }{ 4n+3 }\]???

  57. anonymous
    • one year ago
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    where are your limits?

  58. anonymous
    • one year ago
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    Sorry, evaluated from 0 to 0.78

  59. anonymous
    • one year ago
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    ok, =?

  60. anonymous
    • one year ago
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    Just sub .78 and 0 for x and subtract?

  61. anonymous
    • one year ago
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    0 for x =0, just plug x =0.8 in, ok?

  62. anonymous
    • one year ago
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    oh, 0.78

  63. anonymous
    • one year ago
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    So it's just \[\frac{ .78^{4n+3} }{ 4n+3 }\] ??

  64. anonymous
    • one year ago
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    So, now you have \[\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}\]

  65. anonymous
    • one year ago
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    ok ?

  66. anonymous
    • one year ago
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    Yes

  67. anonymous
    • one year ago
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    You need just 2 terms!! easy!! n =0, plug in, n =1, plug in add them together, done.

  68. anonymous
    • one year ago
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    That's it? Just add the resulting two terms together?

  69. anonymous
    • one year ago
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    Yes, dat sit. If you want, you can add you and me to get 4 terms. :)

  70. anonymous
    • one year ago
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    show me your result, please.

  71. anonymous
    • one year ago
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    Working on it now

  72. anonymous
    • one year ago
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    -0.745188?

  73. anonymous
    • one year ago
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    I got 0.0457318875

  74. anonymous
    • one year ago
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    ok, let's do together

  75. anonymous
    • one year ago
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    I'm absolutely sure that you're right, I just did mine mostly in my head...

  76. anonymous
    • one year ago
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    \[\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}\] \[(-1)^0\dfrac{6^{2*0+1}}{(2*0+1)!} *\dfrac{0.78^{4*0+3}}{4*0+3}+(-1)^1\dfrac{6^{2*1+1}}{(2*1+1)!} *\dfrac{0.78^{4*1+3}}{4*1+3}\]

  77. anonymous
    • one year ago
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    \[-1^{0}*\frac{ 6^{2(0)+1} }{ (2(0)+1)! }*\frac{ 0.78^{4(0)+3} }{ 4(0)+3 }\]

  78. anonymous
    • one year ago
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    Oh you're much faster at this than I am

  79. anonymous
    • one year ago
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    hahaha... I just copy and paste the previous post, then change the numbers. hehehe

  80. anonymous
    • one year ago
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    Well for n = 0, term by term, the first one would equal 1 * 6/1* .78^3/3 which I said totals 0.949104

  81. anonymous
    • one year ago
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    and your first term = \(\dfrac{6}{1!}\dfrac{0.78^3}{3}\) right? = \(2*0.78^3=0.949104\) Yes,

  82. anonymous
    • one year ago
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    The second term is ??

  83. anonymous
    • one year ago
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    For n = 1 I said -1* 6^3/6 * .78^7/7 which I thought should be -0.90337211

  84. anonymous
    • one year ago
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    \(-\dfrac{6^3}{3!}\dfrac{0.78^7}{7}=-0.9033721125\) Yes, again,

  85. anonymous
    • one year ago
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    add them together , you get??

  86. anonymous
    • one year ago
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    0.045732

  87. anonymous
    • one year ago
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    yup

  88. anonymous
    • one year ago
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    So I did just mess something up in my head the first time.

  89. anonymous
    • one year ago
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    easy, right?

  90. anonymous
    • one year ago
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    Thank you so much Crazyandbeautiful, you are literally the most beautiful person in the world to me right now

  91. anonymous
    • one year ago
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    haahaha... I'm not beautiful, I am handsome. hahaha.... jk

  92. anonymous
    • one year ago
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    lol you're probably going to be the deciding factor in how well I do in my test next week actually, because I clearly had no idea what I was doing.

  93. anonymous
    • one year ago
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    Handsome, beautiful, the actual envisionment of God, whatever you are dude you saved my life here

  94. anonymous
    • one year ago
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    You got the method, right? don't scare me. I spent more than 1h to explain and you said you don't know what you are doing. it is a real scaring.

  95. anonymous
    • one year ago
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    I had no idea BEFORE this chat string

  96. anonymous
    • one year ago
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    Believe me, I was taking notes

  97. anonymous
    • one year ago
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    ok, good luck

  98. anonymous
    • one year ago
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    Thanks again

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