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MathSofiya

  • 2 years ago

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

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  1. MathSofiya
    • 2 years ago
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    I guess the first step is this \[y'-xy=0\]

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

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

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

  5. experimentX
    • 2 years ago
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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.

  6. MathSofiya
    • 2 years ago
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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

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

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

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

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

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

  12. experimentX
    • 2 years ago
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    yep.

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

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

  15. MathSofiya
    • 2 years ago
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    nope

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

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

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

  19. experimentX
    • 2 years ago
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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)

  20. experimentX
    • 2 years ago
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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

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

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

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

  24. experimentX
    • 2 years ago
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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.

  25. experimentX
    • 2 years ago
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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.

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

  27. experimentX
    • 2 years ago
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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)

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

  29. TuringTest
    • 2 years ago
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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

  30. TuringTest
    • 2 years ago
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    yeah that

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

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

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

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

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

  36. experimentX
    • 2 years ago
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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.

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

  38. TuringTest
    • 2 years ago
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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

  39. TuringTest
    • 2 years ago
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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.

  40. MathSofiya
    • 2 years ago
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    I'm back

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

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

  43. TuringTest
    • 2 years ago
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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 :)

  44. MathSofiya
    • 2 years ago
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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.

  45. MathSofiya
    • 2 years ago
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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?

  46. TuringTest
    • 2 years ago
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    yep

  47. MathSofiya
    • 2 years ago
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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

  48. TuringTest
    • 2 years ago
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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

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

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

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

  52. MathSofiya
    • 2 years ago
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    right

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

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

  55. MathSofiya
    • 2 years ago
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    ?

  56. MathSofiya
    • 2 years ago
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    weird

  57. TuringTest
    • 2 years ago
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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

  58. MathSofiya
    • 2 years ago
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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?

  59. TuringTest
    • 2 years ago
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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

  60. MathSofiya
    • 2 years ago
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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?

  61. TuringTest
    • 2 years ago
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    exactly :)

  62. TuringTest
    • 2 years ago
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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....

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

  64. MathSofiya
    • 2 years ago
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    indices...

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

  66. TuringTest
    • 2 years ago
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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.

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

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

  69. MathSofiya
    • 2 years ago
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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. \[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]

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

  71. TuringTest
    • 2 years ago
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    explicitly*

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

  73. TuringTest
    • 2 years ago
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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

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

  75. TuringTest
    • 2 years ago
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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

  76. TuringTest
    • 2 years ago
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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.

  77. MathSofiya
    • 2 years ago
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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

  78. TuringTest
    • 2 years ago
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    cool :)

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

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

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

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

  83. TuringTest
    • 2 years ago
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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 ?

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

  85. MathSofiya
    • 2 years ago
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    when n=0

  86. MathSofiya
    • 2 years ago
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    no no no

  87. TuringTest
    • 2 years ago
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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?

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

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

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

  91. TuringTest
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    for which *subscripts on a* is a_n zero ?

  92. MathSofiya
    • 2 years ago
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    all the odd subscripts

  93. TuringTest
    • 2 years ago
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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

  94. TuringTest
    • 2 years ago
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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=

  95. MathSofiya
    • 2 years ago
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    0?

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

  97. MathSofiya
    • 2 years ago
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    part of the pattern

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

  99. TuringTest
    • 2 years ago
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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\[a_1=0\]

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

  101. MathSofiya
    • 2 years ago
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    interesting, yes

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

  103. MathSofiya
    • 2 years ago
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    one more thing

  104. MathSofiya
    • 2 years ago
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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

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

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

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

  108. TuringTest
    • 2 years ago
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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

  109. MathSofiya
    • 2 years ago
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    agreed

  110. TuringTest
    • 2 years ago
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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

  111. MathSofiya
    • 2 years ago
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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

  112. TuringTest
    • 2 years ago
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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.

  113. TuringTest
    • 2 years ago
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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

  114. MathSofiya
    • 2 years ago
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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.....

  115. TuringTest
    • 2 years ago
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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

  116. MathSofiya
    • 2 years ago
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    "index"?

  117. TuringTest
    • 2 years ago
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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

  118. TuringTest
    • 2 years ago
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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

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

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

  121. TuringTest
    • 2 years ago
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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...?

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

  123. TuringTest
    • 2 years ago
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    almost

  124. MathSofiya
    • 2 years ago
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    woops hold your horses

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

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

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

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

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

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

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

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

  133. MathSofiya
    • 2 years ago
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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?

  134. MathSofiya
    • 2 years ago
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    which was our goal

  135. TuringTest
    • 2 years ago
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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 :)

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

  137. TuringTest
    • 2 years ago
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    yay!

  138. MathSofiya
    • 2 years ago
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    no way!

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

  140. TuringTest
    • 2 years ago
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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...

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

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

  143. TuringTest
    • 2 years ago
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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

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

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

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

  147. TuringTest
    • 2 years ago
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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

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

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

  150. MathSofiya
    • 2 years ago
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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.

  151. MathSofiya
    • 2 years ago
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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

  152. TuringTest
    • 2 years ago
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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...

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

  154. TuringTest
    • 2 years ago
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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!

  155. MathSofiya
    • 2 years ago
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    see ya!

  156. MathSofiya
    • 2 years ago
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    medal?

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

  158. TuringTest
    • 2 years ago
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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\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty\frac{x^{2k}}{2^kk!}\]

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

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