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MathSofiya

  • 2 years ago

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

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

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

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

  4. TuringTest
    • 2 years ago
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    oooooooh

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

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

  7. MathSofiya
    • 2 years ago
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    sure

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

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

  10. TuringTest
    • 2 years ago
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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

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

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

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

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

  15. MathSofiya
    • 2 years ago
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    oops

  16. TuringTest
    • 2 years ago
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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

  17. MathSofiya
    • 2 years ago
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    ok

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

  19. MathSofiya
    • 2 years ago
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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?

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

  21. TuringTest
    • 2 years ago
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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

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

  23. TuringTest
    • 2 years ago
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    how*

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

  25. TuringTest
    • 2 years ago
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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

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

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

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

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

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

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

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

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

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

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

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

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

  38. MathSofiya
    • 2 years ago
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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

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

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

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

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

  43. TuringTest
    • 2 years ago
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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 ?

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

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

  46. TuringTest
    • 2 years ago
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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

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

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

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

  50. MathSofiya
    • 2 years ago
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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

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

  52. MathSofiya
    • 2 years ago
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    ok

  53. TuringTest
    • 2 years ago
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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 ?

  54. TuringTest
    • 2 years ago
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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?

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

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

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

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

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

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

  61. MathSofiya
    • 2 years ago
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    2k

  62. TuringTest
    • 2 years ago
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    and odd?

  63. MathSofiya
    • 2 years ago
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    n=2k

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

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

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

  67. TuringTest
    • 2 years ago
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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

  68. MathSofiya
    • 2 years ago
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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?

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

  70. TuringTest
    • 2 years ago
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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

  71. TuringTest
    • 2 years ago
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    picture*

  72. MathSofiya
    • 2 years ago
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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

  73. MathSofiya
    • 2 years ago
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    ok bye

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

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

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

  77. MathSofiya
    • 2 years ago
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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

  78. TuringTest
    • 2 years ago
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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 :)

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

  80. TuringTest
    • 2 years ago
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    right-o

  81. MathSofiya
    • 2 years ago
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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)

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

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

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

  85. MathSofiya
    • 2 years ago
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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?

  86. TuringTest
    • 2 years ago
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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 \(a_{n+p}\)

  87. TuringTest
    • 2 years ago
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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...

  88. TuringTest
    • 2 years ago
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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

  89. TuringTest
    • 2 years ago
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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

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

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

  92. MathSofiya
    • 2 years ago
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    sounds good

  93. MathSofiya
    • 2 years ago
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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...

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

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

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

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

  98. MathSofiya
    • 2 years ago
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    3k?

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

  100. TuringTest
    • 2 years ago
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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.

  101. MathSofiya
    • 2 years ago
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    *patience

  102. TuringTest
    • 2 years ago
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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

  103. MathSofiya
    • 2 years ago
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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

  104. MathSofiya
    • 2 years ago
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    way easy

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

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

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

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

  109. MathSofiya
    • 2 years ago
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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

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

  111. MathSofiya
    • 2 years ago
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    oh my

  112. TuringTest
    • 2 years ago
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    haha :P

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

  114. MathSofiya
    • 2 years ago
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    happy?

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

  116. TuringTest
    • 2 years ago
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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.

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

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

  119. TuringTest
    • 2 years ago
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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?

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

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

  122. TuringTest
    • 2 years ago
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    superposition...

  123. MathSofiya
    • 2 years ago
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    adding them?

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

  125. MathSofiya
    • 2 years ago
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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?

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

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

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

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

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

  131. MathSofiya
    • 2 years ago
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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?

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

  133. MathSofiya
    • 2 years ago
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    brb

  134. MathSofiya
    • 2 years ago
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    2 mins

  135. TuringTest
    • 2 years ago
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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

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

  137. MathSofiya
    • 2 years ago
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    yep makes sense

  138. TuringTest
    • 2 years ago
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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

  139. TuringTest
    • 2 years ago
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    example 1 on*

  140. MathSofiya
    • 2 years ago
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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)

  141. TuringTest
    • 2 years ago
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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 :)

  142. MathSofiya
    • 2 years ago
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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...

  143. TuringTest
    • 2 years ago
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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!

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

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

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

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

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

  149. TuringTest
    • 2 years ago
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    :)

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

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

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

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

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