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Clarence

  • one year ago

Riemann Sum...

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  1. clarence
    • one year ago
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  2. clarence
    • one year ago
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    I got as far as figuring out delta x to be 2/n

  3. clarence
    • one year ago
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    And what I preceived the second one to be \[(\frac{ 12i ^{2} }{ n ^{2} } + \frac{ 40i }{ n }) (\frac{ 2 }{ n })\]

  4. anonymous
    • one year ago
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    ah its been so long since ive done this. can you jog my memory when it says the general formulae of \[x _{i}\]

  5. clarence
    • one year ago
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    You mean the part where \[x _{i}=0+i \Delta x\]?

  6. anonymous
    • one year ago
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    yeah is that what the formulae is for that bit? ah i should know this cause its so fundemental to calc but i only vaguely remember haha

  7. anonymous
    • one year ago
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    perhaps @IrishBoy123 can help he's a bit of a god

  8. clarence
    • one year ago
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    Aha, I think so. There was a previous question regarding Riemann Sum but tackling it from the left and the question is a bit different. I could potentially upload that and tell you what I got for if you'd prefer it.

  9. anonymous
    • one year ago
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    perhaps that could help but since we have n divisions, then our partitions are 0, \[\Delta x, 2\Delta x, ... , (n-2) \Delta x, (n-1) \Delta x, \]

  10. anonymous
    • one year ago
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    2

  11. anonymous
    • one year ago
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    hmm that won't help

  12. anonymous
    • one year ago
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    for right hand side, \[\Delta x \left[ f( a + \Delta x ) + f(a + 2 \Delta x)+\cdots+f(b) \right].\]

  13. anonymous
    • one year ago
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    where a=0 b=2

  14. anonymous
    • one year ago
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    that is your approx function if you do the right end method

  15. anonymous
    • one year ago
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    it looks like they are leading you the way step by step, i'm just unsure what it means by the general formulae of x(i)

  16. anonymous
    • one year ago
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    is this like a web this where you enter an answer and you can see if its right or wrong?

  17. clarence
    • one year ago
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    It isn't a web thing, more just like an extended question that the professor gave at the end of lecture. I've been trying to follow a link that my friend shared with me on Facebook. It seems pretty legit despite being a yahoo answer source, but even then as I say that it was hard for me to get my head around the answer: https://answers.yahoo.com/question/index?qid=20110910015823AARrDF4

  18. anonymous
    • one year ago
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    sweet, i have something to work with now

  19. clarence
    • one year ago
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    Hope that it'd be more useful to you than it was to me! :P

  20. IrishBoy123
    • one year ago
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    https://gyazo.com/ce50525553b52e9dbd904fb7970fc13a i'm with you on this

  21. IrishBoy123
    • one year ago
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    if that helps :p

  22. anonymous
    • one year ago
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    ok so x(i) is defined as \[x _{i}=i \Delta x\]

  23. IrishBoy123
    • one year ago
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    yes

  24. IrishBoy123
    • one year ago
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    \[x_i = i \frac{2}{n}\]

  25. anonymous
    • one year ago
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    you know \[\Delta x= \frac{ 2 }{ n }\]

  26. anonymous
    • one year ago
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    yep!^^

  27. anonymous
    • one year ago
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    follow @Clarence

  28. IrishBoy123
    • one year ago
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    i'm pretty rusty on this too so careful with the symbology but i worked it through and got 48 both ways for the integral at least that means we are on track

  29. anonymous
    • one year ago
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    haha

  30. clarence
    • one year ago
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    I'm understanding what you two are saying so far so that's good I suppose. :p

  31. anonymous
    • one year ago
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    now since you know {x _{i}\]

  32. anonymous
    • one year ago
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    \[x _{i}\]

  33. anonymous
    • one year ago
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    sub that into the x variables in the function 3x^2+20x

  34. IrishBoy123
    • one year ago
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    @chris00 yes!

  35. anonymous
    • one year ago
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    so you get \[3x _{i}^2+20x _{i}\]

  36. anonymous
    • one year ago
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    can you do this long hand @Clarence

  37. anonymous
    • one year ago
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    its bringing back memories now ahaha

  38. anonymous
    • one year ago
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    you should get \[\frac{ 12i^2 }{ n^2 }+\frac{ 60i }{ n }\]

  39. clarence
    • one year ago
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    I think I got 40 rather than 60 for that bit...

  40. IrishBoy123
    • one year ago
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    \[f(x_i) = 3 (\frac{2i}{n})^2 + 20 (\frac{2i}{n})\]

  41. anonymous
    • one year ago
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    hahah opps

  42. IrishBoy123
    • one year ago
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    let's agree it before we move on :-)

  43. anonymous
    • one year ago
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    i can math

  44. anonymous
    • one year ago
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    :P

  45. clarence
    • one year ago
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    Aha, we all make these mistakes :)

  46. IrishBoy123
    • one year ago
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    so we are here, right?? \[\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }\]

  47. anonymous
    • one year ago
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    yeah so \[f(x _{i})=\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }\]

  48. IrishBoy123
    • one year ago
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    yes, i forgot the \(f(x_i)\) cool next bit

  49. anonymous
    • one year ago
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    \[f(x _{i}) \Delta x=(\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }) \Delta x\]

  50. anonymous
    • one year ago
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    and you know what delta x is right clarence?

  51. IrishBoy123
    • one year ago
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    exellent!

  52. clarence
    • one year ago
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    Yup, so just input that into the equation to get 2/n at the right end

  53. IrishBoy123
    • one year ago
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    and the answer is???!!!!

  54. anonymous
    • one year ago
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    simply sub 2/n into delta x to get: \[f(x _{i}) \Delta x=(\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n })(\frac{ 2 }{ n })\]

  55. anonymous
    • one year ago
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    which is what you got!

  56. clarence
    • one year ago
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    Before I got stuck, yes ;)

  57. anonymous
    • one year ago
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    now we approximate the area

  58. IrishBoy123
    • one year ago
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    my go!! \[f(x _{i}) \Delta x=\frac{ 24i^2 }{ n^3 }+\frac{ 80i }{ n^2 }\]

  59. IrishBoy123
    • one year ago
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    do we agree that?!

  60. anonymous
    • one year ago
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    yep!

  61. IrishBoy123
    • one year ago
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    cool, now for the real fun bit!

  62. anonymous
    • one year ago
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    \[A _{app}\approx \sum_{i=0}^{n}(f(x _{i}) \Delta x\]

  63. anonymous
    • one year ago
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    i think thats by definition...

  64. anonymous
    • one year ago
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    |dw:1441541710476:dw| somthing like that. this is not your graph but it just shows that the area approximation of one rectangle is Delta x multiplied by the function governing the rectangle

  65. IrishBoy123
    • one year ago
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    yes, we approximate the area by adding up all the little elements \[\Sigma_{i = 1}^{\infty} f(x _{i}) \Delta x=\Sigma_{i = 1}^{\infty} \left[ \frac{ 24i^2 }{ n^3 }+ \frac{ 80i }{ n^2 } \right]\] \[=\Sigma_{i = 1}^{\infty} \frac{ 24i^2 }{ n^3 }+\Sigma_{i = 1}^{\infty} \frac{ 80i }{ n^2 }\]

  66. IrishBoy123
    • one year ago
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    and now we use these https://gyazo.com/776b4188ce5e7548a6cad4c61eda1c3a

  67. IrishBoy123
    • one year ago
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    \[=\frac{24}{n^3}\Sigma_{i = 1}^{\infty} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{\infty} i\]

  68. anonymous
    • one year ago
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    beat me too it haha

  69. anonymous
    • one year ago
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    should be hunky dory from there @Clarence

  70. IrishBoy123
    • one year ago
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    i am latexing like a crazy man here :-))

  71. IrishBoy123
    • one year ago
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    @Clarence take your time we can stop here for a while if that suits!

  72. clarence
    • one year ago
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    Never thought that maths could look so pretty... But now I don't even know what to do with the information that I have!

  73. anonymous
    • one year ago
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    you are given critical information in the latter part of the question

  74. anonymous
    • one year ago
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    \[\sum_{i}^{n}i=\frac{ n(n+1) }{ 2 }\]

  75. anonymous
    • one year ago
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    and \[\sum_{i}^{n}i^2=\frac{ n(n+1)(2n+1) }{ 6 }\]

  76. anonymous
    • one year ago
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    can we not just sub these corresponding summations into our question?

  77. IrishBoy123
    • one year ago
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    yes we can!

  78. anonymous
    • one year ago
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    where @IrishBoy123 mentioned that \[A=\frac{24}{n^3}\Sigma_{i = 1}^{\infty} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{\infty} i\]

  79. anonymous
    • one year ago
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    or not to be confused with the infinity clarence, \[A=\frac{24}{n^3}\Sigma_{i = 1}^{n} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{n} i\]

  80. anonymous
    • one year ago
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    see the relation?

  81. clarence
    • one year ago
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    Aha, yes, I can see that now! Sorry, totally forgot about the original question, I'm not the brightest freshman about if you couldn't already tell :p

  82. IrishBoy123
    • one year ago
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    @Clarence if you're doing this stuff, you *are* very bright this is amazing maths as we shall see at the end!!

  83. anonymous
    • one year ago
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    so fundamental to calculus.

  84. anonymous
    • one year ago
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    can you solve the area now clarence?

  85. clarence
    • one year ago
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    Maybe... :p

  86. anonymous
    • one year ago
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    its just simple substitution

  87. IrishBoy123
    • one year ago
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    this is a pig latex-wise shall i just do it? it gets easier when we lose all the terms

  88. anonymous
    • one year ago
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    go for it!

  89. IrishBoy123
    • one year ago
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    ok give me a few minutes to copy and paste it all together

  90. IrishBoy123
    • one year ago
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    \[A=\frac{24}{n^3} \frac{ n(n+1)(2n+1) }{ 6 }+\frac{ 80 }{ n^2 } \ \frac{ n(n+1) }{ 2 }\] this good?

  91. clarence
    • one year ago
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    That's what I got

  92. IrishBoy123
    • one year ago
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    excellent so we now have a formula that approximates the area under the curve. we choose n as the number of elements and we plug it in and find the area let's simplify it remembering we want to set \(n = \infty\) to get the perfect answer

  93. IrishBoy123
    • one year ago
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    \[A=\frac{24}{n^3} \frac{ 2n^3 + 3n^2 +n}{ 6 }+\frac{ 80 }{ n^2 } \ \frac{ n^2 +n }{ 2 }\]

  94. IrishBoy123
    • one year ago
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    everyone agree?

  95. anonymous
    • one year ago
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    yep

  96. clarence
    • one year ago
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    Yeah, I got up to that, so far :p

  97. IrishBoy123
    • one year ago
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    \[A=4 (2 + \frac{3}{n} +\frac{1}{n^2})+40 (1 +\frac{1}{n} )\]

  98. anonymous
    • one year ago
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    good so far

  99. anonymous
    • one year ago
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    @Clarence what happens now if n approaches infinity?

  100. IrishBoy123
    • one year ago
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    some more simplifying \[A= 48 + \frac{52}{n} +\frac{4}{n^2} \]

  101. anonymous
    • one year ago
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    haha more simplifying, love it

  102. anonymous
    • one year ago
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    spelling out the answer haah

  103. IrishBoy123
    • one year ago
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    yep! totally unnecesary of course as we're setting n to infinity :-((

  104. clarence
    • one year ago
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    Aha, maths sure loves their infinities :p

  105. IrishBoy123
    • one year ago
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    \[\int_{0}^{2} \ 3x^2 + 20x \ dx\]

  106. IrishBoy123
    • one year ago
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    if that's not 48, we are all doomed is it?

  107. IrishBoy123
    • one year ago
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    please let it be so!!

  108. clarence
    • one year ago
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    I got 48 for that..

  109. anonymous
    • one year ago
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    sweet as

  110. anonymous
    • one year ago
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    what a journey

  111. anonymous
    • one year ago
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    :P

  112. IrishBoy123
    • one year ago
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    beautiful beautiful stuff thank you both for involving me, really enjoyed it @Clarence are you good to go with this now??

  113. anonymous
    • one year ago
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    such a pleasure

  114. clarence
    • one year ago
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    Aha, I think so :)

  115. IrishBoy123
    • one year ago
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    if not keep the thread open @chris00 and i will get a notification if there is more to do

  116. clarence
    • one year ago
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    Awesome, thanks! :)

  117. amistre64
    • one year ago
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    it appears to me that you ran thru the LEFT side rundown of this

  118. amistre64
    • one year ago
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    on an interval from a to b, the left side setup is: \[x_i=a+i\frac{b-a}{n}~:~0\le i<n\] this hits all the high points starting at the left end and moving in 'i'ncrements of the partition widths. in this case a=0 your inputs worked out. the right side setup is either an adjustment of the interval of i: 0 < i <= n this way we simply hit the right side of the partitions.

  119. amistre64
    • one year ago
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    another alternative to the right side setup is simply to replace the a+ with b-

  120. amistre64
    • one year ago
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    hmmm, ok, it was a hard read thru for these tired old eyes, but it does look like you implemented i going from 1 to n, so my observation is moot :) good job yall

  121. IrishBoy123
    • one year ago
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    its a really long thread but we used \(x_i = i \frac{2}{n}\) (and \(\Delta x= \frac{ 2 }{ n }\)) and ran it from i=1. so for n = i = 1 we are up at the right hand end.

  122. IrishBoy123
    • one year ago
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    and we were spoon-fed because the question was so well laid out....:-))

  123. amistre64
    • one year ago
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    if a had been say 3 (not 0),and b was 5 (not 2), our interval width is still 5-3=2 and we divide that into n partitions.\[\Delta x=\frac2n\] then the value of x for the ith iterations would have been: \[x_i=3+i\frac2n\] and \[f(x_i)=f(3+i\frac2n)\] just wanting to make sure that we remember that xi is not in general the same as: i /\x

  124. IrishBoy123
    • one year ago
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    gotcha. that particular *snag* was missing from our problem here. we had \[f(x_i)=f(0+i\frac2n) = f(i\frac2n)\] thank you @amistre.

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