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MathSofiya Group Title

I'm trying to find a pattern \[a_{n+2}=\frac{a_n}{(n+2)(n+1)}\] \[n=0:a_2=\frac{a_0}{2!}\] \[n=2:a_4=\frac{a_0}{4!}\] but that pattern changes because the next term is not \[\frac{a_0}{6!}\] \[n=4:a_6=\frac{a_0}{2\cdot 30}=\]

  • 2 years ago
  • 2 years ago

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  1. Algebraic! Group Title
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    ?? \[a _{4} = \frac{a _{0} }{ 4! } \]

    • 2 years ago
  2. TuringTest Group Title
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    this is the one we did yesterday? wasn't it\[a_{n+2}=-a_n\frac{1-n}{(n+1)(n+2)}\]?

    • 2 years ago
  3. MathSofiya Group Title
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    no this is new problem \[y''=y\]

    • 2 years ago
  4. TuringTest Group Title
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    oooooooh

    • 2 years ago
  5. MathSofiya Group Title
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    I'll write out what I have so far

    • 2 years ago
  6. TuringTest Group Title
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    I'll be able to able to help in about 15-20 min

    • 2 years ago
  7. MathSofiya Group Title
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    sure

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

    • 2 years ago
  9. nipunmalhotra93 Group Title
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    @MathSofiya how exactly is the pattern changing?

    • 2 years ago
  10. TuringTest Group Title
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    let's establish that we have the right pattern first... I feel like you may have missed something continue plz @MathSofiya so far so good

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

    • 2 years ago
  12. TuringTest Group Title
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    so yeah, right pattern

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

    • 2 years ago
  14. TuringTest Group Title
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    \[\sum_{n=0}^\infty x^n\left[(n+2)(n+1)a_{n+2}-a_n\right]=0\]

    • 2 years ago
  15. MathSofiya Group Title
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    oops

    • 2 years ago
  16. TuringTest Group Title
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    \[a_{n+2}=\frac{a_n}{(n+1)(n+2)}\]so you were right, just plug in each n and write out \(everything\) explicitly, even multiplication by 1

    • 2 years ago
  17. MathSofiya Group Title
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    ok

    • 2 years ago
  18. MathSofiya Group Title
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    \[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}\] \[n=4:a_{4+2}=a_6=\frac {a_4}{30}\] \[n=5:a_{5+2}=a_7=\frac {a_5}{42}\]

    • 2 years ago
  19. MathSofiya Group Title
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    so we established the relationship between \[a_2=\frac{a_0}{2}\] but how do I continue this to the next line? how do I change a_2 to a_0?

    • 2 years ago
  20. MathSofiya Group Title
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    ooohh no need to because the next line is going to be zero

    • 2 years ago
  21. TuringTest Group Title
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    you are not writing it out explicitly enough to see the pattern reduce everything to terms of either a_0 or a_1 the two constants we can represent no more simply

    • 2 years ago
  22. TuringTest Group Title
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    now is the next line zero?

    • 2 years ago
  23. TuringTest Group Title
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    how*

    • 2 years ago
  24. MathSofiya Group Title
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    well it was yesterday LOL ....Ok I'll write everything out

    • 2 years ago
  25. TuringTest Group Title
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    \[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]write everything this way, with each number on the bottom specified

    • 2 years ago
  26. TuringTest Group Title
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    yesterday we had n-1 in the top, that's why

    • 2 years ago
  27. TuringTest Group Title
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    Haha I'm exercising while you crunch the numbers :P

    • 2 years ago
  28. MathSofiya Group Title
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    \[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{2\cdot 3}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{2\cdot2\cdot3}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5}\] \[n=4:a_{4+2}=a_6=\frac {a_4}{30}=\frac{a_0}{2\cdot2\cdot2\cdot3\cdot5}\] \[n=5:a_{5+2}=a_7=\frac {a_5}{42}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5\cdot42}\]

    • 2 years ago
  29. MathSofiya Group Title
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    n=4 is wrong oh well...but I see a pattern now. How do you workout and do math?

    • 2 years ago
  30. TuringTest Group Title
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    push ups, then run over and check your work real quick :)

    • 2 years ago
  31. TuringTest Group Title
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    n=2 is wrong

    • 2 years ago
  32. TuringTest Group Title
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    12 is not 2x3

    • 2 years ago
  33. MathSofiya Group Title
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    haha! That's what I call commitment!

    • 2 years ago
  34. MathSofiya Group Title
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    give me a second to do the problem on paper. It's kinda tedious to type it...one second

    • 2 years ago
  35. TuringTest Group Title
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    \[a_n=\frac{a_n}{(n+1)(n+2)}\]\[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]\[n=2:a_4=\frac {a_2}{4\cdot3}=\frac{a_0}{4\cdot3\cdot2\cdot1}\]\[n=3:a_5=\frac {a_3}{5\cdot4}=\frac{a_1}{5\cdot4\cdot3\cdot2}\]

    • 2 years ago
  36. TuringTest Group Title
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    write out each number in the denom without simplifying every time

    • 2 years ago
  37. TuringTest Group Title
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    I cannot stress that enough to find patterns

    • 2 years ago
  38. MathSofiya Group Title
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    \[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{3\cdot 2}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{4\cdot3\cdot2\cdot1}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{5\cdot4\cdot3\cdot2\cdot1}\] Ok I'm finally convinced...had to calculate like 5 times that the denominator is actually 120

    • 2 years ago
  39. TuringTest Group Title
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    good, but more importantly what is/are the general pattern(s)?

    • 2 years ago
  40. TuringTest Group Title
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    again, there are two; one for even n and one for odd n

    • 2 years ago
  41. MathSofiya Group Title
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    sorry internet is super slow. I'm gonna restart my computer brb

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

    • 2 years ago
  43. TuringTest Group Title
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    I'm getting in the shower soon... so what are the two constants that every other constant \(a_n\) can be written in terms of ?

    • 2 years ago
  44. MathSofiya Group Title
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    I wanna say \[(n+2)!\] and then sub in n=2k

    • 2 years ago
  45. MathSofiya Group Title
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    wait...you go shower, and let me work on this

    • 2 years ago
  46. TuringTest Group Title
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    no, and that would only be evens, and needs a base constant like \(a_0\) write it as\[a_n=?\]and remember this time we do have odd terms

    • 2 years ago
  47. MathSofiya Group Title
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    allow me to decipher it...

    • 2 years ago
  48. MathSofiya Group Title
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    The denominator for both even and odds is the following...still working on the numerator \[\frac{?}{(n+2)!}\]

    • 2 years ago
  49. TuringTest Group Title
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    look at the difference between constants with an even and odd subscript what is the main difference you see>?

    • 2 years ago
  50. MathSofiya Group Title
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    one will always be zero and the other will always be 1. I have come to that conclusion, but now I have to find a way to write it mathematically...I'm gonna cheat and look at yesterdays work

    • 2 years ago
  51. TuringTest Group Title
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    no wait, you can do it... let me give another hint...

    • 2 years ago
  52. MathSofiya Group Title
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    ok

    • 2 years ago
  53. TuringTest Group Title
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    odd are reducible to being in terms of the constant a_1 evens are reducible to being in terms of a_0 and for the denominator for any constant \(a_n\) what is it? look at the denom for a_2, a_3, a_4, what are they in terms of n ?

    • 2 years ago
  54. TuringTest Group Title
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    don't get confused with the n=(whatever) thing I mean what is the relation between the n in the subscript \(a_n\) and the number in the denominator?

    • 2 years ago
  55. MathSofiya Group Title
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    hold on I didn't write anything yet

    • 2 years ago
  56. MathSofiya Group Title
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    sorry It just took me 3 mins to load this page

    • 2 years ago
  57. TuringTest Group Title
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    no rush, still haven't showered yet

    • 2 years ago
  58. MathSofiya Group Title
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    When the numerator is \[a_0\] the denominator is even When the numerator is \[a_1\] the denominator is odd

    • 2 years ago
  59. MathSofiya Group Title
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    by odd I mean 5! 3! and so on

    • 2 years ago
  60. TuringTest Group Title
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    exactly, and all even numbers can be written how?

    • 2 years ago
  61. MathSofiya Group Title
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    2k

    • 2 years ago
  62. TuringTest Group Title
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    and odd?

    • 2 years ago
  63. MathSofiya Group Title
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    n=2k

    • 2 years ago
  64. MathSofiya Group Title
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    n=2k+1

    • 2 years ago
  65. MathSofiya Group Title
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    \[a_{2k}\] and \[a_{2k+1}\]

    • 2 years ago
  66. TuringTest Group Title
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    so then what is our formulas? we need one for all \(a_{2k}\) and another for all \(a_2k+1}\)

    • 2 years ago
  67. TuringTest Group Title
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    right, we need a formula for each one.. what are they ? that's probably the hardest part of the problem in my opinion is finding the pattern

    • 2 years ago
  68. MathSofiya Group Title
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    I would like to put a_{2k} in the numerator and (n+2) in the numerator, but that wouldn't be right. So I have write it in one formula to account for both the even and odd?

    • 2 years ago
  69. MathSofiya Group Title
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    I can't write two separate formulas correct? I've gotta write one?

    • 2 years ago
  70. TuringTest Group Title
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    everything should be written in terms of either a_0 and k, or a_1 and k n is now out of the poicture completely

    • 2 years ago
  71. TuringTest Group Title
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    picture*

    • 2 years ago
  72. MathSofiya Group Title
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    gtg...I know this is the last part, and shouldn't take too long, but I gtg to the gym. i'll be back at 7:30 8pm

    • 2 years ago
  73. MathSofiya Group Title
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    ok bye

    • 2 years ago
  74. TuringTest Group Title
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    cool, health makes one think more clearly I believe see ya

    • 2 years ago
  75. MathSofiya Group Title
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    LOL I hope soo! see ya

    • 2 years ago
  76. MathSofiya Group Title
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    where is Alan Turing?

    • 2 years ago
  77. MathSofiya Group Title
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    I'm reviewing the previous problems we did...and there might be some errors. I not convinced with the way that the final answer was written in k's

    • 2 years ago
  78. TuringTest Group Title
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    Allow me to restore your faith, though there are other manners of making the notation I'm sure... Paul to the rescue again: http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx Notice that since the constants become different depending on even-odd subscripts for many answers we divide the subscripts in the recursion formula for them into terms of 2k and 2k+1. This particular problem you are doing is almost identical to example 1 one this page, with the exception of the minus between y'' and y exchanged for a plus (reading this example you will probably discover the pattern I was asking you to look for). Notice that it is in fact essential that we change the terms from n to k, since we can get no single summation formula in terms of n since the coefficients depend on whether n is odd n. This is not true for all problems with series solutions, but when we need to distinguish between even and odd n the only way to do this is to put them as two separate series in terms of k. Read example 1 carefully to understand the intricacies of the process, and the why we put n in terms of k. Sorry I never returned last night, I was teaching political philosophy. Look forward to finishing the problem with you today if possible. Tag me when you are ready to continue :)

    • 2 years ago
  79. MathSofiya Group Title
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    I'll be back in like 10 mins, gotta start laundry

    • 2 years ago
  80. TuringTest Group Title
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    right-o

    • 2 years ago
  81. MathSofiya Group Title
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    I can't find my laundry card! Ok so I was trying to rewrite the first problem, and got to the part where 1. all even n#'s =0, or all odd "a" subscripts=0 2. We've gotten a pattern that we called \[a_n=\frac{a_0}{2^n(n!)}\] but if we call them "n" for now wouldn't that conflict because Let's say for n=5 \[\frac{a_0}{2^3(3!)}\] which is not true (I'm referring to y'=xy, the first problem)

    • 2 years ago
  82. MathSofiya Group Title
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    Well it's what we called the pattern initially before writing it in terms of k's

    • 2 years ago
  83. MathSofiya Group Title
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    ok found the laundry card brb

    • 2 years ago
  84. TuringTest Group Title
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    The pattern is not in terms of n like that, if it were it would be\[a_n=\frac{a_0}{2^{n/2}(\frac n2!)}\]for even n subscripts of a, and\[a_n=0\] for all odd n subscripts of a. I think a major problem here is that we are getting confused with the n in the \(n=...\) that we are using to plug in to find the particular constant \(a_n\), and the n in the \(a_n\) in the constant. Those two n's are not equal! What exactly is your argument that in the last problem you are referring too we do no not have that\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]??

    • 2 years ago
  85. MathSofiya Group Title
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    Ok that makes sense. I thought the n's were the same. So the n's are just a variable to hold place until we change the argument to k's I guess. wait, what do you mean by the last sentence? so the n=5 statement above is not accurate?

    • 2 years ago
  86. TuringTest Group Title
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    yes it is accurate, plugging in n=5 into the recursion formula\[a_{n+1}=\frac{a_{n-1}}{n+1}\]gives\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]the \(n=...\) is just what we use to get an expression relating two other constants \(a_{n+ p}\)and \(a_{n+ q}\) where p and q are some number (in this case above p=-1 and q=1) so the n we plug in to test is \(not\) the same as the n in the subscripts, because the subscripts in our recursion relation are usually not just \(a_n\) (the same n as we are plugging in for the test) but really \(a_{n+p}\)

    • 2 years ago
  87. TuringTest Group Title
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    We want our expression for each constant in terms of its \(subscript\), not the \(n=...\) we use to test the relation, hence we divide the subscripts into even and odd to make the pattern more clear. Once we notice that by doing that we can get meaningful expressions for each constant a in terms of the subscript (however we write) that is what we use...

    • 2 years ago
  88. TuringTest Group Title
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    in this case by dividing the subscripts into those of the form 2k and those of the form 2k+1 we get a new way to write each constant in terms of k

    • 2 years ago
  89. TuringTest Group Title
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    ...and whatever base constant we can't determine, like \(a_0\) which we do not know since we are not given the initial conditions

    • 2 years ago
  90. MathSofiya Group Title
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    ok that makes more sense now. Let's see if I can apply it to this problem now.

    • 2 years ago
  91. TuringTest Group Title
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    go for it, I'm gonna get a few empanadas for breakfast :) (sortof like Mexican hot-pockets :) brb

    • 2 years ago
  92. MathSofiya Group Title
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    sounds good

    • 2 years ago
  93. MathSofiya Group Title
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    Let's see here.... \[a_2=\frac{a_0}{2!}\] \[a_3=\frac{a_1}{3!}\] \[a_4=\frac{a_0}{4!}\] \[a_5=\frac{a_1}{5!}\] \[a_2k=\frac{a_0}{k!}\] my way of saying if the subscript is even...

    • 2 years ago
  94. MathSofiya Group Title
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    \[a_{2k+1}=\frac{a_1}{k!}\]

    • 2 years ago
  95. TuringTest Group Title
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    is it? what is\(a_3\) ?

    • 2 years ago
  96. MathSofiya Group Title
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    oh my 2k+1=3 2k=2 k=1

    • 2 years ago
  97. MathSofiya Group Title
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    yeah I'm wrong

    • 2 years ago
  98. MathSofiya Group Title
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    3k?

    • 2 years ago
  99. MathSofiya Group Title
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    your patients with me is truly astonishing. I would have strangled the student.

    • 2 years ago
  100. TuringTest Group Title
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    haha, but I haven't done this in a long time either, so it's a good review for me too.

    • 2 years ago
  101. MathSofiya Group Title
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    *patience

    • 2 years ago
  102. TuringTest Group Title
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    \[a_n=\frac{a_n}{(n+1)(n+2)}\]\[n=0:a_2=\frac {a_0}{2}=\frac{a_0}{2!}\]\[n=1:a_3=\frac {a_1}{6}=\frac{a_1}{3\cdot 2}=\frac{a_1}{3!}\]\[n=2:a_4=\frac {a_2}{12}=\frac{a_0}{4\cdot3\cdot2\cdot1}=\frac{a_0}{4!}\]\[n=3:a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{5\cdot4\cdot3\cdot2\cdot1}=\frac{a_1}{5!}\]now hopefully you have already noticed that the denominator on each \(a_n\) is \(n!\)\[a_n=\frac{a_0}{n!}\]if n is even, and\[a_n=\frac{a_1}{n!}\]if n is odd given that, simply rewrite the subscripts as 2k or 2k+1 depending on whether they are even or odd, and then sub in for n in the general expressions

    • 2 years ago
  103. MathSofiya Group Title
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    yep, I honestly saw that relationship all along, I just didn't know how to express that in k's \[a_{2k}=\frac{a_0}{2k}\] \[a_{2k+1}=\frac{a_0}{2k+1}\] sorry I kinda feel silly now

    • 2 years ago
  104. MathSofiya Group Title
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    way easy

    • 2 years ago
  105. TuringTest Group Title
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    you forgot the factorials ;)

    • 2 years ago
  106. MathSofiya Group Title
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    !!!!!!!!!!!! there

    • 2 years ago
  107. MathSofiya Group Title
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    LOL ok let me rewrite them

    • 2 years ago
  108. TuringTest Group Title
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    and you forgot that for odd subscripts it is a_1 in the numerator

    • 2 years ago
  109. MathSofiya Group Title
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    \[a_{2k}=\frac{a_0}{(2k)!}\] \a_{2k+1}=\frac{a_1}{2k+1}[\] I didn't forgot...I just checking if you were paying attention :P

    • 2 years ago
  110. MathSofiya Group Title
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    \[a_{2k+1}=\frac{a_1}{2k+1}\]

    • 2 years ago
  111. MathSofiya Group Title
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    oh my

    • 2 years ago
  112. TuringTest Group Title
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    haha :P

    • 2 years ago
  113. MathSofiya Group Title
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    \[a_{2k+1}=\frac{a_1}{(2k+1)!}\]

    • 2 years ago
  114. MathSofiya Group Title
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    happy?

    • 2 years ago
  115. MathSofiya Group Title
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    oh wait there is more...

    • 2 years ago
  116. TuringTest Group Title
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    Much better :) I wanna make a subtle point about the possible values of k we are talking about, but I will save it till the end so as not to confuse you.

    • 2 years ago
  117. MathSofiya Group Title
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    \[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!} \] \[y=\sum_{n=0}^\infty a_nx^n=a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!} \]

    • 2 years ago
  118. TuringTest Group Title
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    you've got the right idea, but we need to use superposition to get the full answer

    • 2 years ago
  119. TuringTest Group Title
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    if we accept only one answer or the other we are excluding either all the even or odd exponents, so how can we fix that?

    • 2 years ago
  120. MathSofiya Group Title
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    put them into one formula somehow

    • 2 years ago
  121. MathSofiya Group Title
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    oh y and c_1 and so on and so forth...hold on

    • 2 years ago
  122. TuringTest Group Title
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    superposition...

    • 2 years ago
  123. MathSofiya Group Title
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    adding them?

    • 2 years ago
  124. TuringTest Group Title
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    yep :) c1=a0 c2=a1

    • 2 years ago
  125. MathSofiya Group Title
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    \[c_1\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+c_2\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\] huh?

    • 2 years ago
  126. MathSofiya Group Title
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    gosh the sums probably shouldn't be there...

    • 2 years ago
  127. TuringTest Group Title
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    no I was just saying that the c1 you are used to is called a0 in this case, so...

    • 2 years ago
  128. TuringTest Group Title
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    \[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\]

    • 2 years ago
  129. TuringTest Group Title
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    wait, why =0 ?

    • 2 years ago
  130. TuringTest Group Title
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    \[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}\]

    • 2 years ago
  131. MathSofiya Group Title
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    I just looked at the example in the book Uhm Ohhh,,, but they don't have to be written in c_1 and c_2 right?

    • 2 years ago
  132. MathSofiya Group Title
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    can I put my laundry in the dryer?

    • 2 years ago
  133. MathSofiya Group Title
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    brb

    • 2 years ago
  134. MathSofiya Group Title
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    2 mins

    • 2 years ago
  135. TuringTest Group Title
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    nah, constants are constants, you can call them whatever you want but I just wanted to keep our notation consistent, and since they are equivalent to c1 and c2 (constants that depend on the initial conditions) and yes, you have my permission to do your laundry :P

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

    • 2 years ago
  137. MathSofiya Group Title
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    yep makes sense

    • 2 years ago
  138. TuringTest Group Title
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    so review what we did, look again at example one one here: http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx and tell me if you have any questions

    • 2 years ago
  139. TuringTest Group Title
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    example 1 on*

    • 2 years ago
  140. MathSofiya Group Title
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    I read it, I don't think I have any questions. 22 hrs later I finally understand it (hopefully I'll be able to recreate it)

    • 2 years ago
  141. TuringTest Group Title
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    awesome! there is, as I said, a subtle point to be made about the possible values of k (as paul mentions) but since it really doesn't affect this problem I'll leave it for you to read, or we'll cross that bridge when we come to it :)

    • 2 years ago
  142. MathSofiya Group Title
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    sounds good...I'll do 2 problems on my own from beginning to end and we'll see if I make error or still struggle on some points here and there...

    • 2 years ago
  143. TuringTest Group Title
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    I do encourage you to read the whole page I linked you to though, to see some of the differences encountered in other problems. Good luck!

    • 2 years ago
  144. MathSofiya Group Title
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    I will. Thanks! quick question, did you make your own empanadas?

    • 2 years ago
  145. TuringTest Group Title
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    nope, bought 'em at the corner I only cook hamburgers and pasta

    • 2 years ago
  146. MathSofiya Group Title
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    what? you've been in Mexica for a little while now! Gotta learn to cook like the locals!

    • 2 years ago
  147. TuringTest Group Title
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    well I make quesadillas, I guess that's a start lol

    • 2 years ago
  148. MathSofiya Group Title
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    yep, work your way up from here :P

    • 2 years ago
  149. TuringTest Group Title
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    :)

    • 2 years ago
  150. mahmit2012 Group Title
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    |dw:1348252365896:dw|

    • 2 years ago
  151. mahmit2012 Group Title
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    |dw:1348252603116:dw|

    • 2 years ago
  152. mahmit2012 Group Title
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    |dw:1348252694415:dw|

    • 2 years ago
  153. mahmit2012 Group Title
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    |dw:1348252778070:dw|

    • 2 years ago
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