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MathSofiya

  • 2 years ago

Use power series to solve the differential equation \[y''+xy'+y=0\] \[y=\sum_{n=0}^{\infty} a_n x^n\] \[y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\] \[y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}\] \[xy'=\sum_{n=1}^{\infty}a_nnx^n\] \[\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]

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  1. TuringTest
    • 2 years ago
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    yep, now what's first?

  2. MathSofiya
    • 2 years ago
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    changing the sums so that they all begin with n=2 changing the last sum from n=0 to n=1 would be \[a_1+\sum_{n=1}^\infty a_nx^n\]

  3. TuringTest
    • 2 years ago
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    no we don't strip out the terms yet, we just do an index shift

  4. TuringTest
    • 2 years ago
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    the first order of business is to get all the x's to be to the nth power through using index shifts

  5. TuringTest
    • 2 years ago
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    the only series that has x to some power other than n is the first one how can we change the index of the first series to make the exponent on x be n?

  6. MathSofiya
    • 2 years ago
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    you mean on \[y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\]

  7. TuringTest
    • 2 years ago
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    yes, we need to start from a different n to get x^n through an index shift

  8. MathSofiya
    • 2 years ago
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    so I'm shifting the starting point up by 2 so I have to subtract 2 from {n=2-2} \[-\sum_{n=0}^\infty n(n-1)a_nx^n\] negative to make up for the negative?

  9. MathSofiya
    • 2 years ago
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    nooooo

  10. MathSofiya
    • 2 years ago
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    hold on

  11. MathSofiya
    • 2 years ago
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    \[\sum_{n=0}^\infty n(n+1)a_nx^n\]

  12. TuringTest
    • 2 years ago
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    nice :)

  13. MathSofiya
    • 2 years ago
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    still not right

  14. MathSofiya
    • 2 years ago
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    because that would give us zero

  15. TuringTest
    • 2 years ago
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    no but closer

  16. TuringTest
    • 2 years ago
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    you didn't add 2 to every n

  17. MathSofiya
    • 2 years ago
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    \[\sum_{n=0}^\infty (n+2)(n+1)a_nx^n\]

  18. TuringTest
    • 2 years ago
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    and I do mean \(every\) n (subscripts included)

  19. TuringTest
    • 2 years ago
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    again better, but...^

  20. MathSofiya
    • 2 years ago
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    \[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n \]

  21. MathSofiya
    • 2 years ago
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    can I add them?

  22. TuringTest
    • 2 years ago
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    now that, I do believe, is correct :)

  23. MathSofiya
    • 2 years ago
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    or multiply

  24. MathSofiya
    • 2 years ago
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    or just leave as is?

  25. TuringTest
    • 2 years ago
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    add what? oh that? I wouldn't yet

  26. MathSofiya
    • 2 years ago
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    ok

  27. TuringTest
    • 2 years ago
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    yeah leave as is asd now rewrite what you have

  28. TuringTest
    • 2 years ago
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    and*

  29. TuringTest
    • 2 years ago
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    the whole expression please I'm lazy

  30. MathSofiya
    • 2 years ago
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    \[\sum_{n=0}^{\infty} (n+2)(n-1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]

  31. TuringTest
    • 2 years ago
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    \[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]okay next order of business is what you tried to do earlier; get all indices to start at the same n value the only problem child here is which term?

  32. MathSofiya
    • 2 years ago
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    the middle

  33. TuringTest
    • 2 years ago
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    so what do we do to fix it?

  34. TuringTest
    • 2 years ago
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    you have to be a bit clever here...

  35. MathSofiya
    • 2 years ago
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    \[ xa_n +\sum_{n=0}^\infty a_n(n+1)x^n \]

  36. TuringTest
    • 2 years ago
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    did you try to do an index shift?

  37. TuringTest
    • 2 years ago
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    it looks like you combined the idea of the index shift with stripping out the first term

  38. MathSofiya
    • 2 years ago
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    oh my...let's try this again

  39. TuringTest
    • 2 years ago
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    it is a bit tricky so let me give you a hint...

  40. TuringTest
    • 2 years ago
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    you cannot do another index shift. If you do you will change the exponent on x, which means we are screwed again, so we need to do something else to get this to start from n=0

  41. TuringTest
    • 2 years ago
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    we can't really strip out a term because we don't have this series defined for n<1, but ask yourself, if it was defined at n=0, what would the first term be?

  42. MathSofiya
    • 2 years ago
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    just have the addition of an a_n up front like in the last example?

  43. TuringTest
    • 2 years ago
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    no, answer my question above and you will see why...

  44. MathSofiya
    • 2 years ago
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    a 1

  45. MathSofiya
    • 2 years ago
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    times a_n

  46. TuringTest
    • 2 years ago
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    no, be more careful

  47. TuringTest
    • 2 years ago
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    what would be the first term of\[\sum_{n=0}^\infty a_n nx^n\]

  48. MathSofiya
    • 2 years ago
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    zero

  49. TuringTest
    • 2 years ago
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    right, so...

  50. TuringTest
    • 2 years ago
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    \[0+\sum_{n=1}^\infty a_n nx^n=\sum_{n=0}^\infty a_n nx^n\]so we actually didn't need to change anything; the first term we "stripped out" was a zero anyway! It makes no difference.

  51. TuringTest
    • 2 years ago
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    so we can go ahead and just change the starting point of that series without further manipulation

  52. TuringTest
    • 2 years ago
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    always watch for that, it comes in handy

  53. MathSofiya
    • 2 years ago
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    oh ok uhm.....I'll revisit the last example later, because the I'm not convinced that we needed to do a index shift (or was it stripping of a term)

  54. TuringTest
    • 2 years ago
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    first to get the x^n on all terms we did an index shift (where we add or subtract from each n in the summand) next to get all the starting points for each series the same, we strip out terms/or do the trick we just did they are different tricks, and the order in which you do them is imortant

  55. MathSofiya
    • 2 years ago
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    ok

  56. MathSofiya
    • 2 years ago
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    \[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=0}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]

  57. TuringTest
    • 2 years ago
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    it is important for you to try to understand why the index shift requires you to add/subtract each n in the summand while stripping out a term does not

  58. TuringTest
    • 2 years ago
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    good, so now make it all into one summation...

  59. MathSofiya
    • 2 years ago
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    \[\sum_{n=0}^\infty x^n[(n+2)(n+1)a_{n+2}+a_nn+a_n]=0\]

  60. TuringTest
    • 2 years ago
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    nice, so what will you set to zero to find the recurrence relation?

  61. TuringTest
    • 2 years ago
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    (I am going to type minimally now as I am eating dinner at the same time :)

  62. MathSofiya
    • 2 years ago
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    food comes first in my book... :P Ok let's see here...

  63. TuringTest
    • 2 years ago
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    I can eat and respond with "yes" or "no" pretty well, no worries :)

  64. MathSofiya
    • 2 years ago
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    \[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=-\frac{a_nn-a_n}{(n+2)(n+1)}\]

  65. MathSofiya
    • 2 years ago
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    si?

  66. TuringTest
    • 2 years ago
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    si, pero si quieras podrias factor la \(-a_n\)

  67. MathSofiya
    • 2 years ago
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    ok \[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]

  68. TuringTest
    • 2 years ago
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    and now comes the tedious part that you can do while I eat plug in n=0,1,2,3 and check what results from that recurrence relation until you notice a pattern like last time so just cranks it out; plug in n=0,n=1,n=2 and if a_2 can be written in terms of a_0 or something like last time, do it. That is how we find the pattern.

  69. TuringTest
    • 2 years ago
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    go ahead and list out the first 5 or so and let's see if we see a relation

  70. MathSofiya
    • 2 years ago
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    \[n=1:a_3=0\] \[n=2:a_4=-a_2\frac{1}{12}\] \[n=3:a_5=-a_3\frac{2}{20}\] \[n=4:a_6=-a_4\frac{3}{30}\] \[n=5:a_7=-a_5\frac{4}{42}\]

  71. TuringTest
    • 2 years ago
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    what about n=0 ? we need to start from n=0 since our series starts from n=0 this time (last time it started from n=1)

  72. MathSofiya
    • 2 years ago
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    \[n=0:a_2=-a_0\frac{-1}{2}\]

  73. MathSofiya
    • 2 years ago
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    \[n=0:a_2=a_0\frac{1}{2}\]

  74. MathSofiya
    • 2 years ago
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    whatcha eating?

  75. TuringTest
    • 2 years ago
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    torta de pollo=chicken sandwich on a french sort of bread

  76. TuringTest
    • 2 years ago
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    con chipotle

  77. MathSofiya
    • 2 years ago
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    nice

  78. MathSofiya
    • 2 years ago
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    oh yeah...the pattern

  79. TuringTest
    • 2 years ago
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    so you did well in your subs I think, but now you must try to always sub all the way back as far as you can..

  80. TuringTest
    • 2 years ago
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    yeah, the pattern write each on as some finite number of unknowns, in this case everything in terms of \(a_0\) and \(a_3\)

  81. TuringTest
    • 2 years ago
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    each one*

  82. MathSofiya
    • 2 years ago
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    in the previous example you seemed to have simply switched the \[a_{odd}\] to zero and the \[a_{even}\] to \[a_0\]

  83. MathSofiya
    • 2 years ago
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    take you're time...i'm gonna watch a few more mins of my episode

  84. TuringTest
    • 2 years ago
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    it's not different, it's the same as last time odd subscript gives zero even has a pattern reducible to a_0

  85. TuringTest
    • 2 years ago
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    okay I got a fairly simple answer, let me see if I can confirm it...

  86. TuringTest
    • 2 years ago
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    stupid wolfram is of no help :/

  87. MathSofiya
    • 2 years ago
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    \[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\] \[n=1:a_3=0\] \[n=2:a_4=-a_2\frac{1}{12}=\frac{a_0}{2(3!)}\] \[n=3:a_5=0\] \[n=4:a_6=-a_4\frac{3}{30}=\frac{a_0}{5(3!)}\] \[n=5:a_7=0\]

  88. MathSofiya
    • 2 years ago
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    how did I do?

  89. TuringTest
    • 2 years ago
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    almost, I actually realized that I made a mistake because I think I see a mistake from you so you did as well as me I'd say :D

  90. TuringTest
    • 2 years ago
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    let me try on paper again...

  91. TuringTest
    • 2 years ago
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    here I think I can demonstrate the benefit of not multiplying out terms and writing things over-explicitly to find patterns...

  92. MathSofiya
    • 2 years ago
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    ok

  93. TuringTest
    • 2 years ago
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    \[n=0:a_2=\frac{a_0}{2\cdot 1}\]\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}\]now reevaluate the pattern

  94. MathSofiya
    • 2 years ago
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    \[n=0:a_2=\frac{a_0}{2\cdot 1}=\frac{a_0}{2!}\] \[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\]

  95. TuringTest
    • 2 years ago
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    so try to do the same thing as last time name all even numbers and make a new series with index k

  96. TuringTest
    • 2 years ago
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    then represent the coefficients and plug that in for \(a_n\)

  97. MathSofiya
    • 2 years ago
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    n=2k \[a_n=a_{2k}=\frac{a_0}{k!}\]

  98. TuringTest
    • 2 years ago
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    is it over k! ?

  99. MathSofiya
    • 2 years ago
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    hhm...you know I thought about 2k!, but wouldn't that me too much? \[a_n=a_{2k}=\frac{a_0}{2k!}\]

  100. MathSofiya
    • 2 years ago
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    *be

  101. TuringTest
    • 2 years ago
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    well that is the answer I get... so unless you have a way of confirming it apart from this I say that is correct btw you should put the (2k)! with parentheses to indicate that it is not 2(k!)

  102. MathSofiya
    • 2 years ago
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    \[a_n=a_{2k}=\frac{a_0}{(2k)!}\]

  103. TuringTest
    • 2 years ago
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    yeah, the sub it into the original guess for y

  104. TuringTest
    • 2 years ago
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    then*

  105. MathSofiya
    • 2 years ago
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    \[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\]

  106. TuringTest
    • 2 years ago
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    ...and that's what I got :D I sure think it's right as it comes out quite nicely, as opposed to wolf's answer http://www.wolframalpha.com/input/?i=y''%2Bxy'%2By%3D0&t=crmtb01

  107. MathSofiya
    • 2 years ago
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    I forgot, what did you intend to study? Electrical Engineering or Mathematics?

  108. TuringTest
    • 2 years ago
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    it definitely has the e^(-x^2) think in it though just by glancing, so I am even more confident now

  109. MathSofiya
    • 2 years ago
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    why the e^?

  110. TuringTest
    • 2 years ago
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    Electrical engineering, though I admit I vacillate at times We'll seee how I feel after a year in uni

  111. TuringTest
    • 2 years ago
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    oh because the tayor expansion of e^(ax^2) always has x^(2k)/(2k)! in it (try the expansion yourself and see why, it's not that hard if you remember Taylor series)

  112. MathSofiya
    • 2 years ago
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    don't have the brain cells left to do so, but I trust your reasoning

  113. TuringTest
    • 2 years ago
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    haha, well thanks. I hope I'm right. Nos vemos!

  114. MathSofiya
    • 2 years ago
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    see ya!

  115. MathSofiya
    • 2 years ago
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    my spanish sucks as you might be able to tell

  116. TuringTest
    • 2 years ago
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    Nos vemos ~ we'll see each other

  117. MathSofiya
    • 2 years ago
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    I was right!

  118. TuringTest
    • 2 years ago
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    sweet :)

  119. MathSofiya
    • 2 years ago
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    we've made a mistake

  120. MathSofiya
    • 2 years ago
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    \[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\] is 24 not 12...so the pattern is probably incorrect too

  121. TuringTest
    • 2 years ago
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    let's see...

  122. TuringTest
    • 2 years ago
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    and why should 24 be 12 exactly?

  123. MathSofiya
    • 2 years ago
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    because we need 12 4*3 is 12

  124. MathSofiya
    • 2 years ago
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    I was doing another problem an I got as far as the recurrent relationship and it was almost identical to this one

  125. TuringTest
    • 2 years ago
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    you dropped a 2...

  126. MathSofiya
    • 2 years ago
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    where?

  127. MathSofiya
    • 2 years ago
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    \[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]

  128. MathSofiya
    • 2 years ago
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    oh I see

  129. TuringTest
    • 2 years ago
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    you wrote\[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\]\[n=2:a_4=-a_2\frac{1}{12}=\frac{a_0}{2(3!)}\]but\[a_2=\frac{a_0}2\]and\[\frac1{12}=\frac1{2(3!)}\]so 1)\[a_4=a_2\frac1{2(3!)}=\frac{a_0}2\cdot\frac1{2(3!)}=\frac{a_0}{4!}\]and that minus sign you put that minus sign does not belong

  130. MathSofiya
    • 2 years ago
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    I did not pick up on the\[ a_2=\frac{a_0}{2}\] idea...i just randomly subbed a_0 for a_2 :/

  131. MathSofiya
    • 2 years ago
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    I'll finish y''=y tomorrow...I got it except for the parts stated above

  132. TuringTest
    • 2 years ago
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    once you find a base formula in terms of \(a_0\) or whatever the lowest you can express is, milk it to find the pattern

  133. MathSofiya
    • 2 years ago
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    ok

  134. TuringTest
    • 2 years ago
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    and like I say, sometime the writing out each term, even if it's just multiplying by one, helps make the pattern obvious g'night then!

  135. MathSofiya
    • 2 years ago
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    g'night

  136. mahmit2012
    • 2 years ago
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    |dw:1348252850986:dw|

  137. mahmit2012
    • 2 years ago
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    |dw:1348252918564:dw|

  138. mahmit2012
    • 2 years ago
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    |dw:1348253019736:dw|

  139. mahmit2012
    • 2 years ago
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    |dw:1348253097421:dw|

  140. mahmit2012
    • 2 years ago
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    |dw:1348253165705:dw|

  141. MathSofiya
    • 2 years ago
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    @TuringTest would this statement about striping a term be true (generally speaking) \[0+a_1+\sum_{n=2}^\infty a_n nx^n\]

  142. TuringTest
    • 2 years ago
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    yep, exactly

  143. TuringTest
    • 2 years ago
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    oh no, you dropped an x

  144. TuringTest
    • 2 years ago
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    at n=1 the exponent on x is 1

  145. TuringTest
    • 2 years ago
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    \[\sum_{n=0}^\infty a_nnx^n=a_0(0)x^0+\sum_{n=1}^\infty a_nnx^n=0+a_1(1)x^1+\sum_{n=2}^\infty a_nx^n\]\[=a_1x+\sum_{n=2}^\infty a_nx^n\]

  146. TuringTest
    • 2 years ago
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    easier example, stripping out terms for the sum of the first n integers\[\sum_{i=1}^n i=1+\sum_{i=2}^n=1+2+\sum_{i=3}^n i=1+2+3+\sum_{i=4}^n i\]

  147. MathSofiya
    • 2 years ago
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    much better

  148. TuringTest
    • 2 years ago
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    thought so, probably should have done that a while ago, hehe :P

  149. MathSofiya
    • 2 years ago
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    Makes perfect sense now, no doubt in my mind anymore. Haha, the brain exercise was good though

  150. MathSofiya
    • 2 years ago
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    cool thanks !

  151. TuringTest
    • 2 years ago
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    Awesome, always a pleasure :D

  152. MathSofiya
    • 2 years ago
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    @TuringTest shouldn't \[a_4 \] have a minus sign? I know you said it shouldn't but \[a_2=\frac{a_0}{2}\] and now we have \[a_4=-a_2\frac{1}{12}=-\frac{a_0}{24}\]

  153. TuringTest
    • 2 years ago
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    hm... I guess you may be right

  154. MathSofiya
    • 2 years ago
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    yuuss!!!!

  155. MathSofiya
    • 2 years ago
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    ^^^^my way of saying: yes!

  156. MathSofiya
    • 2 years ago
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    so it's an alternating series? for 2k+1?

  157. MathSofiya
    • 2 years ago
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    I mean 2k

  158. TuringTest
    • 2 years ago
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    sorry I was double-checking this whole time yeah it alternates, so you know the only change we're going to make?

  159. MathSofiya
    • 2 years ago
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    let's see if I remember, can I cheat?

  160. TuringTest
    • 2 years ago
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    cheating would be having me tell you, any way you wanna figure out how to make a series change sign every other term is up to you :)

  161. MathSofiya
    • 2 years ago
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    cheating would be using a the google search engine, but let's see if I can figure it out on my own without google.com

  162. TuringTest
    • 2 years ago
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    good idea brb

  163. MathSofiya
    • 2 years ago
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    k

  164. MathSofiya
    • 2 years ago
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    argh! \[\sum_{n=0}^{\infty}x^n=0,x,x^2,etc.\] \[\sum_{n=0}^{\infty}(-1)x^n=0,-x,-x^2\]

  165. TuringTest
    • 2 years ago
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    what can we do to -1 to make it even?

  166. TuringTest
    • 2 years ago
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    sorry, positive*

  167. MathSofiya
    • 2 years ago
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    \[(-1)^n\]

  168. TuringTest
    • 2 years ago
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    yes

  169. MathSofiya
    • 2 years ago
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    AWESOME!

  170. TuringTest
    • 2 years ago
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    very awesome!!!!!!!!!!!!

  171. MathSofiya
    • 2 years ago
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    hey I got 240 for a_6

  172. TuringTest
    • 2 years ago
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    you love multiplying the numbers out, I just call it 6!

  173. MathSofiya
    • 2 years ago
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    no that's 720

  174. TuringTest
    • 2 years ago
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    haha whatever I'm not even checking if you are having a problem with the constants again give me a min and I'll brb again to help

  175. MathSofiya
    • 2 years ago
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    sounds good... sorry don't mean to be nitpicking :S

  176. TuringTest
    • 2 years ago
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    oh no, I;m glad you pointed it out, but.... must... eat

  177. MathSofiya
    • 2 years ago
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    dude go for it LOL I'll see if I can figure this pattern out

  178. MathSofiya
    • 2 years ago
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    I think this is incorrect \[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\] because \[a_{n+2}=\frac{-a_nn-1}{(n+2)(n+1)}\neq-a_n\frac{n-1}{(n+2)(n+1)}\] It should be \[-a_n\frac{(n+1)}{(n+2)(n+1)}\]

  179. MathSofiya
    • 2 years ago
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    \[-\frac{a_n}{n+2}\]

  180. MathSofiya
    • 2 years ago
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    \[a_{n+2}=-\frac{a_n}{n+2}\]

  181. TuringTest
    • 2 years ago
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    \[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=-\frac{a_nn-a_n}{(n+2)(n+1)}\]is what you wrote, let's see if that is correct...

  182. TuringTest
    • 2 years ago
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    \[(n+2)(n+1)a_{n+2}+a_n(n+1)=0\implies a_{n+2}=-a_n\frac{n+1}{(n+2)(n+1)}\]aw crap you're right again

  183. MathSofiya
    • 2 years ago
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    yuuus!!!

  184. TuringTest
    • 2 years ago
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    well, being right that we were wrong is always a little bittersweet, haha gotta do the pattern over, but at least it's only\[a_{n+2}=-\frac{a_n}{n+2}\]which should be easier I would think

  185. MathSofiya
    • 2 years ago
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    Yeah, let's see if I can figure the pattern out... (btw I did yoga today...feeling more alert, or it could be plain coincidence =)

  186. TuringTest
    • 2 years ago
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    well you're catching all my mistakes, so it seems to be working :)

  187. TuringTest
    • 2 years ago
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    the pattern seems harder to write concisely for me now... though it seems obvious

  188. MathSofiya
    • 2 years ago
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    \[a_2=-\frac{a_0}{2}=-\frac{a_0}{2!}\] \[a_3=-\frac{a_1}{3}\] \[a_4=-\frac{a_2}{4}=\frac{a_0}{3!}\] \[a_5=-\frac{a_3}{5}\] \[a_6=-\frac{a_4}{6}=-\frac{a_0}{36}\] what's withe the 36?

  189. TuringTest
    • 2 years ago
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    it's like the odds get cut out of the factorials and I'm trying to figure out how to write that same for the evens

  190. MathSofiya
    • 2 years ago
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    yeah it's quite strange, but the formula is correct, right? could we have still made an error there?

  191. TuringTest
    • 2 years ago
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    a mistake in the recursion relation? we could have... to be safe we ought to take it from the top, but I can't do that, I have to leave...

  192. MathSofiya
    • 2 years ago
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    sure I'll start from the beginning and post it in a new post...this one is getting quite long

  193. TuringTest
    • 2 years ago
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    good idea I am thinking\[a_{2k}=(-1)^k\frac{a_0}{2^kk!}\]and do not yet have an expression for the odds

  194. MathSofiya
    • 2 years ago
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    oh ok, I'll see ya trmw then

  195. TuringTest
    • 2 years ago
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    hasta luego

  196. MathSofiya
    • 2 years ago
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    hasta luego mi amigo!

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