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MathSofiya Group Title

Use power series to solve the differential equation. y'=xy

  • 2 years ago
  • 2 years ago

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  1. MathSofiya Group Title
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    I guess the first step is this \[y'-xy=0\]

    • 2 years ago
  2. experimentX Group Title
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    hold on ... just let \[ y = \sum_{n=0}^\infty x^n \]

    • 2 years ago
  3. experimentX Group Title
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    \[ y = \sum_{n=0}^\infty a_n x^n \]

    • 2 years ago
  4. MathSofiya Group Title
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    \[y'=\sum_{n=1}^{\infty} nc_nx^{n-1}\] according to my book, same thing I guess

    • 2 years ago
  5. experimentX Group Title
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    yes yes ... \[ \sum_{n=0}^\infty n \; c_n x^{n-1} + \sum_{n=0}c_n x^{n+1} = 0\] find the recurrence relation.

    • 2 years ago
  6. MathSofiya Group Title
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    oh ok \[y''=\sum_{n=2}^{\infty} n(n-1)c_nx^{n-2}\] how do I apply these generalizations to this specific problem

    • 2 years ago
  7. experimentX Group Title
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    hah?? why do you want more trouble? you don't need to find y'' for this.

    • 2 years ago
  8. MathSofiya Group Title
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    really, I was just trying to follow this example in the book

    • 2 years ago
  9. MathSofiya Group Title
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    I meant: really?

    • 2 years ago
  10. experimentX Group Title
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    lol .. this example is of second order differential equation. you need it only for second order DE

    • 2 years ago
  11. MathSofiya Group Title
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    because there is only a y' that makes it first order?

    • 2 years ago
  12. experimentX Group Title
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    yep.

    • 2 years ago
  13. MathSofiya Group Title
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    ok so I only consider this then, right? \[y'=\sum_{n=1}^{\infty} nc_nx^{n-1} \]

    • 2 years ago
  14. experimentX Group Title
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    yes ... do you have a software called maple?

    • 2 years ago
  15. MathSofiya Group Title
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    nope

    • 2 years ago
  16. MathSofiya Group Title
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    but before solving for it, what do I sub?

    • 2 years ago
  17. MathSofiya Group Title
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    or how do I know what do sub, I think I'm just afraid of sums :S

    • 2 years ago
  18. MathSofiya Group Title
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    why do they have (n+2)(n+1) in the example?

    • 2 years ago
  19. experimentX Group Title
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    from maple i got y(x) = y(0)+(1/2)*y(0)*x^2+(1/8)*y(0)*x^4+O(x^6)

    • 2 years ago
  20. experimentX Group Title
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    Honesty I don't like power series method. here's how you do it. assume the solution has of the form \[ y = \sum_{n=0}^\infty a_n x^n \] so we are putting these vales in your DE and hunting a_n's

    • 2 years ago
  21. MathSofiya Group Title
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    I'm back...

    • 2 years ago
  22. MathSofiya Group Title
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    gtg again...

    • 2 years ago
  23. experimentX Group Title
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    after putting the value of y in your DE, you get \[ \sum_{n=0}^\infty n \; c_n x^{n-1} + \sum_{n=0}c_n x^{n+1} = 0 \\ \] since on the left side you don't have negative power of x, put this \( \sum_{n=0}^\infty n \; c_n x^{n-1} \) = \( \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} \) Also we can make \( \sum_{n=0}^\infty \; c_n x^{n+1} \) = \(\sum_{n=0}^\infty c_{n-1} x^{n} \) now we have \[ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + \sum_{n=0}^\infty c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty ((n+1) \; c_{n+1} + c_{n-1} ) x^{n} = 0 \]

    • 2 years ago
  24. experimentX Group Title
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    since you don't have power's of x on the right side, you have \[ (n+1) c_{n+1} + c_{n-1} = 0 \\ \text{ or, } c_{n+1} = {- c_{n-1} \over n+1}\] this is the recurrence relation you get put n=1, you get |dw:1348083846860:dw| this way you find the coefficients. hence you have solution.

    • 2 years ago
  25. experimentX Group Title
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    not a nice method ... purely brute. but this works for all non linear differential equations ... even if their solution is not closed, you (put it up) as special function like Airy function, Bessel function, Hermite ... etc. and C_0 is your constant of integration. also you might wanna check Frobenius method. Best of luck with these.

    • 2 years ago
  26. TuringTest Group Title
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    I think you made a slight mistake @experimentX though I'm no fan of power series myself

    • 2 years ago
  27. experimentX Group Title
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    hmm ... could you be more specific? power series is the method that is most prone to error ... I never get it correctly on first attempt (for second order)

    • 2 years ago
  28. experimentX Group Title
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    oh ... i see that. i put up + instead of -

    • 2 years ago
  29. TuringTest Group Title
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    I'm just going to go ahead and give mine a shot, it's hard for me tow see how you ended up with c_{n+1} and c_{n-1} though

    • 2 years ago
  30. TuringTest Group Title
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    yeah that

    • 2 years ago
  31. experimentX Group Title
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    EDIT:: \[ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} - \sum_{n=0}^\infty c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} - c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty ((n+1) \; c_{n+1} - c_{n-1} ) x^{n} = 0 \]

    • 2 years ago
  32. experimentX Group Title
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    EDIT:: \[ (n+1) c_{n+1} - c_{n-1} = 0 \\ \text{ or, } c_{n+1} = {c_{n-1} \over n+1} \]

    • 2 years ago
  33. experimentX Group Title
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    so the final solution is \[ y = c_0\sum_{n=0}^\infty {1 \over 2^n n!}x^{2n}\]

    • 2 years ago
  34. TuringTest Group Title
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    okay that was nice, let me see if doing it my way changes anything

    • 2 years ago
  35. experimentX Group Title
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    this is same as \[ \huge y = c_0 e^{x^2 \over 2}\]

    • 2 years ago
  36. experimentX Group Title
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    2 am here. gotta sleep ... i'll be watching this thread tomorrow morning :) gotta be careful with power series. this is quite annoying.

    • 2 years ago
  37. TuringTest Group Title
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    yes your solution seems right @experimentX sleep well, I still am gonna try to brush up myself :)

    • 2 years ago
  38. TuringTest Group Title
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    @MathSofiya I have confirmed @experimentX 's answer in a slightly different way if you would like to go over it, let me know

    • 2 years ago
  39. TuringTest Group Title
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    \[y'=yx\implies y'-xy=0\]\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\]\[y'=\sum_{n=1}^\infty a_nnx^{n-1}\](notice the index starts at n=1 for the series for y' because the first term in the series for y is a constant \(a_n\) and disappears upon taking the derivative) Our problem then becomes\[\sum_{n=1}^\infty a_nnx^{n-1}-\sum_{n=0}^\infty a_nx^{n+1}=0\]The first thing is to get all the x's to the \(n^{th}\) power with index shifts, so we will start the first series one lower at n=1 (so we add 1 to n in the summand) and the second series we will shift up by 1 (so that n shifts down by 1 in the summand). Note these shifts also change the indices of the constants \(a_n\)\[\sum_{n=0}^\infty( n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]Next we need to get the indices the same again (make n start at the same number for each series) To avoid doing another index shift that will take us back where we started, we will "strip" the n=0 term out of the first series\[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]notice all we did was plug in n=1 and explicitly write that term out in front of the rest of the series. They call that "stripping out a term" in case you did not know. So we now have\[a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]Here we now need to figure out a formula for any pattern(s) in the constants \(a_n\) First, plugging in \(n=0\) since the series is not defined we get\[n=0:a_1=0\] so now we just have\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]and since exponentials like \(x^n\) can never be zero, and the summation sign will not change the fact that what is within the brackets must be zero we have the "recurrence relation"\[(n+1)a_{n+1}-a_{n-1}=0\implies a_{n+1}={a_{n-1}\over n+1}\]so again we start plugging in values for n starting from n=1\[n=1:a_2=\frac{a_0}2\]\[n=2:a_3=\frac{a_1}3=\frac03=0\]\[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\]\[n=4:a_5=\frac{a_3}5=\frac05=0\]\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]\[~~~~~~~\vdots\]so what's the pattern for the constants? There seem to be two.

    • 2 years ago
  40. MathSofiya Group Title
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    I'm back

    • 2 years ago
  41. TuringTest Group Title
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    at a very opportune time too :)

    • 2 years ago
  42. MathSofiya Group Title
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    Sorry, women and hairstylist... I'm reading everything now

    • 2 years ago
  43. TuringTest Group Title
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    Take your time, it's not done... not sure if I should wait for you to try the next step. I will prepare the answer in the meantime anyway :)

    • 2 years ago
  44. MathSofiya Group Title
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    sounds good, It's gonna take me a while to understand it. I'm gonna type questions as I come upon them.

    • 2 years ago
  45. MathSofiya Group Title
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    \[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\] because \[\sum_{n=0}^\infty a_nx^n x \] since the exponents are added that's why it's \[x^{n+1}\] correct?

    • 2 years ago
  46. TuringTest Group Title
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    yep

    • 2 years ago
  47. MathSofiya Group Title
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    I understand what you did by changing, we start one lower at n=1, \[na_nx^{n-1} \implies (n+1)a_nx^n\] but why does that change where the sum starts from n=1 instead now from n=0 \[\sum_{n=1}^{\infty} \implies \sum_{n=0}^{\infty}\] sorry I guess I'm still struggling with the basics of summations

    • 2 years ago
  48. TuringTest Group Title
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    try it both ways, write out the first two or three terms and convince yourself how starting earlier means you need to add to the argument

    • 2 years ago
  49. TuringTest Group Title
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    for\[\sum_{n=1}^\infty na_nx^{n-1}\]what is the first term?

    • 2 years ago
  50. MathSofiya Group Title
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    \[1\]

    • 2 years ago
  51. TuringTest Group Title
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    not quite, it's \(a_n\)

    • 2 years ago
  52. MathSofiya Group Title
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    right

    • 2 years ago
  53. TuringTest Group Title
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    what about the first term of\[\sum_{n=0}^\infty(n+1)a_nx^n\]?

    • 2 years ago
  54. MathSofiya Group Title
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    \[a_n\]

    • 2 years ago
  55. MathSofiya Group Title
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    ?

    • 2 years ago
  56. MathSofiya Group Title
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    weird

    • 2 years ago
  57. TuringTest Group Title
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    see how by starting our index at n=0 we had to add 1 everywhere to keep the terms the same try it for more terms until you get the feel if you still doubt

    • 2 years ago
  58. MathSofiya Group Title
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    does this mean that we're trying to manipulate the way we write the sum without changing the value of it?

    • 2 years ago
  59. TuringTest Group Title
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    yes if we started from zero the first term would be zero, because it has n as a coefficient the next term would have a coefficient of 1, the one after of 2 etc. if we add 1 then we get the correct coefficients the whole way through because we made up for changing the starting point of the count

    • 2 years ago
  60. MathSofiya Group Title
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    ok so we change the starting point and made up for it. Why are we changing the starting points, what is our eventual goal....ahhh, we want to be able to write the two sums as one sum?

    • 2 years ago
  61. TuringTest Group Title
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    exactly :)

    • 2 years ago
  62. TuringTest Group Title
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    we can only combine the sums once we have their indices at the same starting point and as you will see, we also need the powers on x to be the same, customarily x^n....

    • 2 years ago
  63. TuringTest Group Title
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    these are not short problems :P

    • 2 years ago
  64. MathSofiya Group Title
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    indices...

    • 2 years ago
  65. MathSofiya Group Title
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    {n=..... } are those the indices? I'm kinda rusty on the terminology

    • 2 years ago
  66. TuringTest Group Title
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    yes exactly, n is the index of a sum with n=a under it, and "a" would be the starting point of the index changing the starting point of an index is called an index shift, and requires you to alter the summand (the thing under the summation sign) as I have just tried to describe.

    • 2 years ago
  67. TuringTest Group Title
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    that should say "we explicitly plugged in x=0..." *

    • 2 years ago
  68. TuringTest Group Title
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    I mean n=0 :/

    • 2 years ago
  69. MathSofiya Group Title
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    by taking an \[a_1\] out of the summation how did that change the indices? nothing else seems to have been changed. \[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]

    • 2 years ago
  70. TuringTest Group Title
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    because we wrote out the first term explitily

    • 2 years ago
  71. TuringTest Group Title
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    explicitly*

    • 2 years ago
  72. MathSofiya Group Title
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    ohh...so either way the first answer would be a_1 is the point again I guess

    • 2 years ago
  73. TuringTest Group Title
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    we don't need to start from n=0 anymore because we already wrote it out and took it out of the series, so repeating the n=0 term would add an extra term

    • 2 years ago
  74. MathSofiya Group Title
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    on either side of that equal sign I mean.

    • 2 years ago
  75. TuringTest Group Title
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    what is important right now is that we get the index to start at n=1 like the other term does

    • 2 years ago
  76. TuringTest Group Title
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    if you have a series that starts at n=0, and you want to write it starting at n=1 you can either do an index shift (which we already did to get x^n on both series) or write out the n=0 term explicitly.

    • 2 years ago
  77. MathSofiya Group Title
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    got it. I am down to this line now \[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\] and I do understand how you came to this

    • 2 years ago
  78. TuringTest Group Title
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    cool :)

    • 2 years ago
  79. MathSofiya Group Title
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    I see a pattern something like this \[\frac{a_0}{2^n(n!)}\]

    • 2 years ago
  80. MathSofiya Group Title
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    well I probably shouldn't use "n"

    • 2 years ago
  81. TuringTest Group Title
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    yes! that is exactly right for half the n

    • 2 years ago
  82. MathSofiya Group Title
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    oh so I can use "n"

    • 2 years ago
  83. TuringTest Group Title
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    well no, we need to differentiate between the two cases for n n only equals what you wrote under what condition for n ?

    • 2 years ago
  84. TuringTest Group Title
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    a_n only equals what you wrote for which n ?

    • 2 years ago
  85. MathSofiya Group Title
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    when n=0

    • 2 years ago
  86. MathSofiya Group Title
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    no no no

    • 2 years ago
  87. TuringTest Group Title
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    actually the n in your formula is outright wrong sorry but look at it another way so you can figure how to fix it... for which values of n is a_n zero?

    • 2 years ago
  88. MathSofiya Group Title
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    durrr...all the even numbers

    • 2 years ago
  89. MathSofiya Group Title
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    haha it took me while to see that :P

    • 2 years ago
  90. TuringTest Group Title
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    yes, but I screwed up the question again, sorry I must phrase more carefully

    • 2 years ago
  91. TuringTest Group Title
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    for which *subscripts on a* is a_n zero ?

    • 2 years ago
  92. MathSofiya Group Title
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    all the odd subscripts

    • 2 years ago
  93. TuringTest Group Title
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    right :) so and we can write any odd number as n=2k+1 with k=0,1,2..., right? so any term with a coefficient we can write as \(a_n=a_{2k+1}=0\) therefore our formula is only concerned with the even number subscripts since the odd ones will drop out

    • 2 years ago
  94. TuringTest Group Title
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    so we only need to think about the terms with n=2k by the way we skipped a set; plugging in n=0 to find what a_1 is (we can't do it with our formula because our series does not have a_1 defined) to back up just for a moment, what does a_1=

    • 2 years ago
  95. MathSofiya Group Title
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    0?

    • 2 years ago
  96. TuringTest Group Title
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    yes, how did you know?

    • 2 years ago
  97. MathSofiya Group Title
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    part of the pattern

    • 2 years ago
  98. TuringTest Group Title
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    that does not illustrate the proper reasoning, and I think you know that ;)

    • 2 years ago
  99. TuringTest Group Title
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    that reasoning will certainly not always work and will in fact often fail. the real reason is as follows:\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]plug in n=0 and notice something: the series is not defined at n=0, only from n=1 on, so we just wind up with the whole series not in the equation\[a_1=0\]

    • 2 years ago
  100. TuringTest Group Title
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    hence the problem we are now dealing with is really just\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]

    • 2 years ago
  101. MathSofiya Group Title
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    interesting, yes

    • 2 years ago
  102. TuringTest Group Title
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    so where were we....?

    • 2 years ago
  103. MathSofiya Group Title
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    one more thing

    • 2 years ago
  104. MathSofiya Group Title
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    so you're saying that I could drop the \[a_1\] because it is not defined at n=0 yes...ok

    • 2 years ago
  105. TuringTest Group Title
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    a_1 is defined at n=0 (a_1=0), but the series is not

    • 2 years ago
  106. MathSofiya Group Title
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    oh ok the series is not defined at n=0 and this is left \[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]

    • 2 years ago
  107. MathSofiya Group Title
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    my goodness this is exhausting!

    • 2 years ago
  108. TuringTest Group Title
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    \[a_1+\cancel{\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n}^{\text{not defined}}=0\]\[a_1=0\]leaving\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]and since \(x^n\neq0\) then the coefficients must be zero\[(n+1)a_{n+1}-a_{n-1}=0\]and we get the recurrence relation we were doing

    • 2 years ago
  109. MathSofiya Group Title
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    agreed

    • 2 years ago
  110. TuringTest Group Title
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    and yes, it is extremely exhausting I think most people who do DE's whould agree, but they handle non-constnat coefficient problems well

    • 2 years ago
  111. MathSofiya Group Title
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    so would this finally be the solution \[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\] I doubt it...seems like it's just the beginning...now it's on to solving the DE

    • 2 years ago
  112. TuringTest Group Title
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    I'm afraid we have ways to go dear, I would understand if you want a break or something... I am mainly practicing, and I have a final solution from beginning to end prepared on a word doc, but it would be better to guide you of course.

    • 2 years ago
  113. TuringTest Group Title
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    that solution above is not an explicit solution for\[ y=\sum_{n=0}^\infty a_nx^n\]which is what we want for an answer

    • 2 years ago
  114. MathSofiya Group Title
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    Wow an example they did in my book took up 2 pages. yes. I don't think I need a break yet...but when we're done with this problem I'm watching an episode of the Big Bang Theory and just laugh at all of their stupid jokes. continue.....

    • 2 years ago
  115. TuringTest Group Title
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    we need a formula for a_n in terms of the index, that's how we get our answer we are mostly there actually, I just want to make sure you understand each step

    • 2 years ago
  116. MathSofiya Group Title
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    "index"?

    • 2 years ago
  117. TuringTest Group Title
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    you will see what I mean: that the index n (which don't get attached to cuz we're gonna change the name of it in a second) is going to give an explicit formula for \(a_n\) in terms of the only constant we can't determine \(a_0\) so you need a formula for all even \(a_n\) in terms of n

    • 2 years ago
  118. TuringTest Group Title
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    you almost had it, but you wrote\[a_n=\frac{a_0}{2^nn!}\]but check the numbers and you can see that's not quite right

    • 2 years ago
  119. MathSofiya Group Title
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    what I screwed up on earlier by calling it \[\frac{a_0}{2^n(n!)}\]

    • 2 years ago
  120. MathSofiya Group Title
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    ok I'll fix it...let's see here...

    • 2 years ago
  121. TuringTest Group Title
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    hint: We are only talking about even numbers, the odd subscripts are all zero. All even numbers can be written as...?

    • 2 years ago
  122. MathSofiya Group Title
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    \[a_n= \frac{a_0}{2^{2k}(2nk!)}\] If this is right.....

    • 2 years ago
  123. TuringTest Group Title
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    almost

    • 2 years ago
  124. MathSofiya Group Title
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    woops hold your horses

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

    • 2 years ago
  126. TuringTest Group Title
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    but 2k=n.... the power on 2 is not the same as the subscript on a, check again the relation

    • 2 years ago
  127. MathSofiya Group Title
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    \[a_{2k}=\frac{a_0}{2^{2k}(2k!)}\] That just looks odd

    • 2 years ago
  128. TuringTest Group Title
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    plug in k=2 and check if that's true based on the table I made way back up there

    • 2 years ago
  129. MathSofiya Group Title
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    that gives me a_1 instead of a_0

    • 2 years ago
  130. MathSofiya Group Title
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    am I loosing it?

    • 2 years ago
  131. TuringTest Group Title
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    ? k=2 n=2k=4 check a_4

    • 2 years ago
  132. MathSofiya Group Title
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    oh I see what you mean

    • 2 years ago
  133. MathSofiya Group Title
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    \[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\] uhm? what do you want me to check? the fact that it is not zero?

    • 2 years ago
  134. MathSofiya Group Title
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    which was our goal

    • 2 years ago
  135. TuringTest Group Title
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    no, I want you to check if that fits your formula\[a_{2k}=\frac{a_0}{2^{2k}(2k)!}\]or if not, how to fix it :)

    • 2 years ago
  136. MathSofiya Group Title
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    divide it by half because it seems like double the answer so \[a_{2k}=\frac{a_0}{2^{k}(k)!}\]

    • 2 years ago
  137. TuringTest Group Title
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    yay!

    • 2 years ago
  138. MathSofiya Group Title
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    no way!

    • 2 years ago
  139. MathSofiya Group Title
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    now is that my y_p or something?

    • 2 years ago
  140. TuringTest Group Title
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    and since these are the only a_n we care about we can now write\[a_n=a_{2k}=\frac{a_0}{2^kk!}\]and no, it's much better than our y_p...

    • 2 years ago
  141. MathSofiya Group Title
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    it's my "y" final answer

    • 2 years ago
  142. TuringTest Group Title
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    basically, yes it is the key to the final answer

    • 2 years ago
  143. TuringTest Group Title
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    first notice that the above is true for \(k=0,1,2,...\) and we already made the tacit assumtion that our answer is\[y=\sum_{n=0}^\infty a_nx^n\]now since all odd will give zero terms for \(a_n\), there will be only even n in our formula so we can change all n in the series to 2k, start our series from k=0, and substitute the above formula for \(a_n\) and that will be our final answer

    • 2 years ago
  144. MathSofiya Group Title
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    \[y=\sum_{k=0}^\infty \frac{a_0}{2^kk!}x^{2k}\]

    • 2 years ago
  145. TuringTest Group Title
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    you got it :)

    • 2 years ago
  146. TuringTest Group Title
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    and just to make things crystal clear that we are in fact right...

    • 2 years ago
  147. TuringTest Group Title
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    separation of variables gives y'=yx dy/y=xdx ln y=x^2/2+C y=Ce^(x^2/2) the Tayore xpansion for e^y is\[e^y=1+\frac y{1!}+\frac{y^2}{2!}+...=\sum_{n=0}^\infty\frac{y^n}{n!}\]let \[y=\frac12x^2\]and plug that into the taylor series and you get the same answer

    • 2 years ago
  148. TuringTest Group Title
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    with \(C=a_0\) of course

    • 2 years ago
  149. TuringTest Group Title
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    experimentX was right too, but I thought I would try to practice through explanation myself :)

    • 2 years ago
  150. MathSofiya Group Title
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    Amazing! I owe you half of my paychecks when I begin working as civil engineer. Dude, you are awesome. Thanks for helping me.

    • 2 years ago
  151. MathSofiya Group Title
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    patience is a virtue...you have an abundance of that...oh yeah, sorry I forgot you don't like compliments :P

    • 2 years ago
  152. TuringTest Group Title
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    Haha, don't tempt me I could use them. And thanks as always :) For posterity's sake I wanna post my full explanation without interruption, just because I already had it prepared and it may be easier to read should you decide to review this. It will slow down the page though...

    • 2 years ago
  153. MathSofiya Group Title
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    Thank you. I'm off to watch an episode of the Big Bang Theory. Is that silly? LOL

    • 2 years ago
  154. TuringTest Group Title
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    Makes more sense than doing more series solution DE's to me! I'm off to do something unproductive too, see ya!

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

    • 2 years ago
  156. MathSofiya Group Title
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    medal?

    • 2 years ago
  157. TuringTest Group Title
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    gotta take it away from experimentX... which I'll do, he has enough sorry @experimentX ;)

    • 2 years ago
  158. TuringTest Group Title
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    \[y'=yx\implies y'-xy=0\]\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\]\[y'=\sum_{n=1}^\infty a_nnx^{n-1}\](notice the index starts at n=1 for the series for y' because the first term in the series for y is a constant \(a_n\) and disappears upon taking the derivative) Our problem then becomes\[\sum_{n=1}^\infty a_nnx^{n-1}-\sum_{n=0}^\infty a_nx^{n+1}=0\]The first thing is to get all the x's to the \(n^{th}\) power with index shifts, so we will start the first series one lower at n=1 (so we add 1 to n in the summand) and the second series we will shift up by 1 (so that n shifts down by 1 in the summand). Note these shifts also change the indices of the constants \(a_n\)\[\sum_{n=0}^\infty( n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]Next we need to get the indices the same again (make n start at the same number for each series) To avoid doing another index shift that will take us back where we started, we will "strip" the n=0 term out of the first series\[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]notice all we did was plug in n=0 and explicitly write that term out in front of the rest of the series. They call that "stripping out a term" in case you did not know. So we now have\[a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]Here we now need to figure out a formula for any pattern(s) in the constants \(a_n\) First, plugging in \(n=0\) since the series is not defined we get\[n=0:a_1=0\] so now we just have\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]and since exponentials like \(x^n\) can never be zero, and the summation sign will not change the fact that what is within the brackets must be zero we have the "recurrence relation"\[(n+1)a_{n+1}-a_{n-1}=0\implies a_{n+1}={a_{n-1}\over n+1}\]so again we start plugging in values for n starting from n=1\[n=1:a_2=\frac{a_0}2\]\[n=2:a_3=\frac{a_1}3=\frac03=0\]\[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\]\[n=4:a_5=\frac{a_3}5=\frac05=0\]\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]\[~~~~~~~\vdots\]so what's the pattern for the constants? There seem to be two; one for even subscripts and one for odd. Even numbers can be written as \(n=2k\) and odd as \(n=2k+1\) for \(k=0,1,2,...\) That means that all the \(a_n\) for even n can be written as \(a_{2k}\) and have the form\[a_{2k}=\frac{a_0}{2^kk!}\]and for all odd n we have \[a_{2k+1}=0\]therefor all those terms will disappear from the series. That lets us replace all the n's in the summand with 2k and make a new representation of the series based on that. This way we don't include the odd n terms, which are zero anyway. We can now rewrite \[a_n=a_{2k}=\frac{a_0}{2^kk!}~~~~\text{where}~~~~k=0,1,2,3...\]and our guess for y becomes\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty\frac{x^{2k}}{2^kk!}\]

    • 2 years ago
  159. experimentX Group Title
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    you guys have patience of a saint ... usually i don't write more than lines.

    • 2 years ago
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