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MathSofiya Group Title

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

  • 2 years ago
  • 2 years ago

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  1. TuringTest Group Title
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    yep, now what's first?

    • 2 years ago
  2. MathSofiya Group Title
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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 \[a_1+\sum_{n=1}^\infty a_nx^n\]

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

    • 2 years ago
  4. TuringTest Group Title
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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

    • 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?

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

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

    • 2 years ago
  8. MathSofiya Group Title
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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?

    • 2 years ago
  9. MathSofiya Group Title
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    nooooo

    • 2 years ago
  10. MathSofiya Group Title
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    hold on

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

    • 2 years ago
  12. TuringTest Group Title
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    nice :)

    • 2 years ago
  13. MathSofiya Group Title
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    still not right

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

    • 2 years ago
  15. TuringTest Group Title
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    no but closer

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

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

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

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

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

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

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

    • 2 years ago
  23. MathSofiya Group Title
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    or multiply

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

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

    • 2 years ago
  26. MathSofiya Group Title
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    ok

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

    • 2 years ago
  28. TuringTest Group Title
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    and*

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

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

    • 2 years ago
  31. TuringTest Group Title
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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?

    • 2 years ago
  32. MathSofiya Group Title
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    the middle

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

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

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

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

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

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

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

    • 2 years ago
  40. TuringTest Group Title
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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

    • 2 years ago
  41. TuringTest Group Title
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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?

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

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

    • 2 years ago
  44. MathSofiya Group Title
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    a 1

    • 2 years ago
  45. MathSofiya Group Title
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    times a_n

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

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

    • 2 years ago
  48. MathSofiya Group Title
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    zero

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

    • 2 years ago
  50. TuringTest Group Title
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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.

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

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

    • 2 years ago
  53. MathSofiya Group Title
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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)

    • 2 years ago
  54. TuringTest Group Title
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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

    • 2 years ago
  55. MathSofiya Group Title
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    ok

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

    • 2 years ago
  57. TuringTest Group Title
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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

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

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

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

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

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

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

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

    • 2 years ago
  65. MathSofiya Group Title
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    si?

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

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

    • 2 years ago
  68. TuringTest Group Title
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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.

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

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

    • 2 years ago
  71. TuringTest Group Title
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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)

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

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

    • 2 years ago
  74. MathSofiya Group Title
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    whatcha eating?

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

    • 2 years ago
  76. TuringTest Group Title
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    con chipotle

    • 2 years ago
  77. MathSofiya Group Title
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    nice

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

    • 2 years ago
  79. TuringTest Group Title
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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..

    • 2 years ago
  80. TuringTest Group Title
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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 \(a_3\)

    • 2 years ago
  81. TuringTest Group Title
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    each one*

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

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

    • 2 years ago
  84. TuringTest Group Title
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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

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

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

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

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

    • 2 years ago
  89. TuringTest Group Title
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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

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

    • 2 years ago
  91. TuringTest Group Title
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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...

    • 2 years ago
  92. MathSofiya Group Title
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    ok

    • 2 years ago
  93. TuringTest Group Title
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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

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

    • 2 years ago
  95. TuringTest Group Title
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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

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

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

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

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

    • 2 years ago
  100. MathSofiya Group Title
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    *be

    • 2 years ago
  101. TuringTest Group Title
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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!)

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

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

    • 2 years ago
  104. TuringTest Group Title
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    then*

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

    • 2 years ago
  106. TuringTest Group Title
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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

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

    • 2 years ago
  108. TuringTest Group Title
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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

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

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

    • 2 years ago
  111. TuringTest Group Title
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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)

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

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

    • 2 years ago
  114. MathSofiya Group Title
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    see ya!

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

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

    • 2 years ago
  117. MathSofiya Group Title
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    I was right!

    • 2 years ago
  118. TuringTest Group Title
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    sweet :)

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

    • 2 years ago
  120. MathSofiya Group Title
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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

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

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

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

    • 2 years ago
  124. MathSofiya Group Title
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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

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

    • 2 years ago
  126. MathSofiya Group Title
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    where?

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

    • 2 years ago
  128. MathSofiya Group Title
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    oh I see

    • 2 years ago
  129. TuringTest Group Title
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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

    • 2 years ago
  130. MathSofiya Group Title
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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 :/

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

    • 2 years ago
  132. TuringTest Group Title
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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

    • 2 years ago
  133. MathSofiya Group Title
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    ok

    • 2 years ago
  134. TuringTest Group Title
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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!

    • 2 years ago
  135. MathSofiya Group Title
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    g'night

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

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

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

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

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

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

    • 2 years ago
  142. TuringTest Group Title
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    yep, exactly

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

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

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

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

    • 2 years ago
  147. MathSofiya Group Title
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    much better

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

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

    • 2 years ago
  150. MathSofiya Group Title
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    cool thanks !

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

    • 2 years ago
  152. MathSofiya Group Title
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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 \[a_4=-a_2\frac{1}{12}=-\frac{a_0}{24}\]

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

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

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

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

    • 2 years ago
  157. MathSofiya Group Title
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    I mean 2k

    • 2 years ago
  158. TuringTest Group Title
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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?

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

    • 2 years ago
  160. TuringTest Group Title
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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 :)

    • 2 years ago
  161. MathSofiya Group Title
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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

    • 2 years ago
  162. TuringTest Group Title
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    good idea brb

    • 2 years ago
  163. MathSofiya Group Title
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    k

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

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

    • 2 years ago
  166. TuringTest Group Title
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    sorry, positive*

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

    • 2 years ago
  168. TuringTest Group Title
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    yes

    • 2 years ago
  169. MathSofiya Group Title
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    AWESOME!

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

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

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

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

    • 2 years ago
  174. TuringTest Group Title
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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

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

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

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

    • 2 years ago
  178. MathSofiya Group Title
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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 \[-a_n\frac{(n+1)}{(n+2)(n+1)}\]

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

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

    • 2 years ago
  181. TuringTest Group Title
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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...

    • 2 years ago
  182. TuringTest Group Title
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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

    • 2 years ago
  183. MathSofiya Group Title
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    yuuus!!!

    • 2 years ago
  184. TuringTest Group Title
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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

    • 2 years ago
  185. MathSofiya Group Title
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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 =)

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

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

    • 2 years ago
  188. MathSofiya Group Title
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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?

    • 2 years ago
  189. TuringTest Group Title
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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

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

    • 2 years ago
  191. TuringTest Group Title
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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...

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

    • 2 years ago
  193. TuringTest Group Title
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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

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

    • 2 years ago
  195. TuringTest Group Title
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    hasta luego

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

    • 2 years ago
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