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MathSofiya

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 \[\frac{a_0}{6!}\] \[n=4:a_6=\frac{a_0}{2\cdot 30}=\]

  • one year ago
  • one year ago

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

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

    • one year ago
  3. MathSofiya
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    no this is new problem \[y''=y\]

    • one year ago
  4. TuringTest
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    oooooooh

    • one year ago
  5. MathSofiya
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    I'll write out what I have so far

    • one year ago
  6. TuringTest
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    I'll be able to able to help in about 15-20 min

    • one year ago
  7. MathSofiya
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    sure

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

    • one year ago
  9. nipunmalhotra93
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    @MathSofiya how exactly is the pattern changing?

    • one year ago
  10. TuringTest
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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

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

    • one year ago
  12. TuringTest
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    so yeah, right pattern

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

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

    • one year ago
  15. MathSofiya
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    oops

    • one year ago
  16. TuringTest
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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

    • one year ago
  17. MathSofiya
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    ok

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

    • one year ago
  19. MathSofiya
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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?

    • one year ago
  20. MathSofiya
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    ooohh no need to because the next line is going to be zero

    • one year ago
  21. TuringTest
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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

    • one year ago
  22. TuringTest
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    now is the next line zero?

    • one year ago
  23. TuringTest
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    how*

    • one year ago
  24. MathSofiya
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    well it was yesterday LOL ....Ok I'll write everything out

    • one year ago
  25. TuringTest
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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

    • one year ago
  26. TuringTest
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    yesterday we had n-1 in the top, that's why

    • one year ago
  27. TuringTest
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    Haha I'm exercising while you crunch the numbers :P

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

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

    • one year ago
  30. TuringTest
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    push ups, then run over and check your work real quick :)

    • one year ago
  31. TuringTest
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    n=2 is wrong

    • one year ago
  32. TuringTest
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    12 is not 2x3

    • one year ago
  33. MathSofiya
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    haha! That's what I call commitment!

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

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

    • one year ago
  36. TuringTest
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    write out each number in the denom without simplifying every time

    • one year ago
  37. TuringTest
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    I cannot stress that enough to find patterns

    • one year ago
  38. MathSofiya
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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

    • one year ago
  39. TuringTest
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    good, but more importantly what is/are the general pattern(s)?

    • one year ago
  40. TuringTest
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    again, there are two; one for even n and one for odd n

    • one year ago
  41. MathSofiya
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    sorry internet is super slow. I'm gonna restart my computer brb

    • one year ago
  42. MathSofiya
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    ok I'm back

    • one year ago
  43. TuringTest
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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 ?

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

    • one year ago
  45. MathSofiya
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    wait...you go shower, and let me work on this

    • one year ago
  46. TuringTest
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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

    • one year ago
  47. MathSofiya
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    allow me to decipher it...

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

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

    • one year ago
  50. MathSofiya
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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

    • one year ago
  51. TuringTest
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    no wait, you can do it... let me give another hint...

    • one year ago
  52. MathSofiya
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    ok

    • one year ago
  53. TuringTest
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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 ?

    • one year ago
  54. TuringTest
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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?

    • one year ago
  55. MathSofiya
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    hold on I didn't write anything yet

    • one year ago
  56. MathSofiya
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    sorry It just took me 3 mins to load this page

    • one year ago
  57. TuringTest
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    no rush, still haven't showered yet

    • one year ago
  58. MathSofiya
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    When the numerator is \[a_0\] the denominator is even When the numerator is \[a_1\] the denominator is odd

    • one year ago
  59. MathSofiya
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    by odd I mean 5! 3! and so on

    • one year ago
  60. TuringTest
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    exactly, and all even numbers can be written how?

    • one year ago
  61. MathSofiya
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    2k

    • one year ago
  62. TuringTest
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    and odd?

    • one year ago
  63. MathSofiya
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    n=2k

    • one year ago
  64. MathSofiya
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    n=2k+1

    • one year ago
  65. MathSofiya
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    \[a_{2k}\] and \[a_{2k+1}\]

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

    • one year ago
  67. TuringTest
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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

    • one year ago
  68. MathSofiya
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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?

    • one year ago
  69. MathSofiya
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    I can't write two separate formulas correct? I've gotta write one?

    • one year ago
  70. TuringTest
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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

    • one year ago
  71. TuringTest
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    picture*

    • one year ago
  72. MathSofiya
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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

    • one year ago
  73. MathSofiya
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    ok bye

    • one year ago
  74. TuringTest
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    cool, health makes one think more clearly I believe see ya

    • one year ago
  75. MathSofiya
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    LOL I hope soo! see ya

    • one year ago
  76. MathSofiya
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    where is Alan Turing?

    • one year ago
  77. MathSofiya
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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

    • one year ago
  78. TuringTest
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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 :)

    • one year ago
  79. MathSofiya
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    I'll be back in like 10 mins, gotta start laundry

    • one year ago
  80. TuringTest
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    right-o

    • one year ago
  81. MathSofiya
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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)

    • one year ago
  82. MathSofiya
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    Well it's what we called the pattern initially before writing it in terms of k's

    • one year ago
  83. MathSofiya
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    ok found the laundry card brb

    • one year ago
  84. TuringTest
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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\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]??

    • one year ago
  85. MathSofiya
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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?

    • one year ago
  86. TuringTest
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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 \(a_{n+p}\)

    • one year ago
  87. TuringTest
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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...

    • one year ago
  88. TuringTest
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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

    • one year ago
  89. TuringTest
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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

    • one year ago
  90. MathSofiya
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    ok that makes more sense now. Let's see if I can apply it to this problem now.

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

    • one year ago
  92. MathSofiya
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    sounds good

    • one year ago
  93. MathSofiya
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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...

    • one year ago
  94. MathSofiya
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    \[a_{2k+1}=\frac{a_1}{k!}\]

    • one year ago
  95. TuringTest
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    is it? what is\(a_3\) ?

    • one year ago
  96. MathSofiya
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    oh my 2k+1=3 2k=2 k=1

    • one year ago
  97. MathSofiya
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    yeah I'm wrong

    • one year ago
  98. MathSofiya
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    3k?

    • one year ago
  99. MathSofiya
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    your patients with me is truly astonishing. I would have strangled the student.

    • one year ago
  100. TuringTest
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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.

    • one year ago
  101. MathSofiya
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    *patience

    • one year ago
  102. TuringTest
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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

    • one year ago
  103. MathSofiya
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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

    • one year ago
  104. MathSofiya
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    way easy

    • one year ago
  105. TuringTest
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    you forgot the factorials ;)

    • one year ago
  106. MathSofiya
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    !!!!!!!!!!!! there

    • one year ago
  107. MathSofiya
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    LOL ok let me rewrite them

    • one year ago
  108. TuringTest
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    and you forgot that for odd subscripts it is a_1 in the numerator

    • one year ago
  109. MathSofiya
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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

    • one year ago
  110. MathSofiya
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    \[a_{2k+1}=\frac{a_1}{2k+1}\]

    • one year ago
  111. MathSofiya
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    oh my

    • one year ago
  112. TuringTest
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    haha :P

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

    • one year ago
  114. MathSofiya
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    happy?

    • one year ago
  115. MathSofiya
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    oh wait there is more...

    • one year ago
  116. TuringTest
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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.

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

    • one year ago
  118. TuringTest
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    you've got the right idea, but we need to use superposition to get the full answer

    • one year ago
  119. TuringTest
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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?

    • one year ago
  120. MathSofiya
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    put them into one formula somehow

    • one year ago
  121. MathSofiya
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    oh y and c_1 and so on and so forth...hold on

    • one year ago
  122. TuringTest
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    superposition...

    • one year ago
  123. MathSofiya
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    adding them?

    • one year ago
  124. TuringTest
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    yep :) c1=a0 c2=a1

    • one year ago
  125. MathSofiya
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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?

    • one year ago
  126. MathSofiya
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    gosh the sums probably shouldn't be there...

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

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

    • one year ago
  129. TuringTest
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    wait, why =0 ?

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

    • one year ago
  131. MathSofiya
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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?

    • one year ago
  132. MathSofiya
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    can I put my laundry in the dryer?

    • one year ago
  133. MathSofiya
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    brb

    • one year ago
  134. MathSofiya
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    2 mins

    • one year ago
  135. TuringTest
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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

    • one year ago
  136. MathSofiya
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    ok I'm back

    • one year ago
  137. MathSofiya
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    yep makes sense

    • one year ago
  138. TuringTest
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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

    • one year ago
  139. TuringTest
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    example 1 on*

    • one year ago
  140. MathSofiya
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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)

    • one year ago
  141. TuringTest
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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 :)

    • one year ago
  142. MathSofiya
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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...

    • one year ago
  143. TuringTest
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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!

    • one year ago
  144. MathSofiya
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    I will. Thanks! quick question, did you make your own empanadas?

    • one year ago
  145. TuringTest
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    nope, bought 'em at the corner I only cook hamburgers and pasta

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

    • one year ago
  147. TuringTest
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    well I make quesadillas, I guess that's a start lol

    • one year ago
  148. MathSofiya
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    yep, work your way up from here :P

    • one year ago
  149. TuringTest
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    :)

    • one year ago
  150. mahmit2012
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    |dw:1348252365896:dw|

    • one year ago
  151. mahmit2012
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    |dw:1348252603116:dw|

    • one year ago
  152. mahmit2012
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    |dw:1348252694415:dw|

    • one year ago
  153. mahmit2012
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    |dw:1348252778070:dw|

    • one year ago
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