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JenniferSmart1

how do I write this in sum form? \[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\]

  • one year ago
  • one year ago

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  1. JenniferSmart1
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    I know I'm a disappointment :(

    • one year ago
  2. TuringTest
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    no you're not, I don't have it either\[y=a_0\sum_{n=0}^\infty\frac{x^{2n}}{}\]darn I know there is a way to do the denom, but I can't think of it... I have seen final answers for power series DE's written as you have them though, so the summation notation is not necesarry

    • one year ago
  3. hartnn
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    i would have written denominator of the form 2^m m!

    • one year ago
  4. hartnn
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    num = x^(2m)

    • one year ago
  5. JenniferSmart1
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    this is the whole (series?) \[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]

    • one year ago
  6. hartnn
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    \(\large y=a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\) doesn't this work ??

    • one year ago
  7. TuringTest
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    yeah, that's it @hart

    • one year ago
  8. JenniferSmart1
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    • one year ago
  9. JenniferSmart1
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    #9

    • one year ago
  10. JenniferSmart1
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    how and why?

    • one year ago
  11. hartnn
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    2 = 2 (1) = 2^1 (1!) 2.4 = 2.2(1.2) = 2^2 (2!) 2.4.6 = 2.2.2 (1.2.3) = 2^3 (3!) in general denominator = (2^n n!)

    • one year ago
  12. hartnn
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    got that ^ ?

    • one year ago
  13. JenniferSmart1
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    still thinking

    • one year ago
  14. JenniferSmart1
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    ok from the top. so you look at this and you think..........? What is your thought process? \[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]

    • one year ago
  15. hartnn
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    my 1st thought : did u change the Q ? where does these a1's jump in..... ? anyways, i would first factor out a0 and a1....

    • one year ago
  16. JenniferSmart1
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    that's the whole sum (series?) what I wrote is just the a_1, then I added the a_0 It's from a previous problem http://openstudy.com/users/jennifersmart1#/updates/50d63d0fe4b0d6c1d541ef3a

    • one year ago
  17. hartnn
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    ok, so did u get how , after factoring a0, u get \(\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\) now your next step is to find sum for a1, right ?

    • one year ago
  18. JenniferSmart1
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    I didn't quite get that part...

    • one year ago
  19. hartnn
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    which part ? x^(2m) or 2^m or m! ??

    • one year ago
  20. JenniferSmart1
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    ok I'm back. I can make sense of the answer but how do I come up with this on my own.... =(

    • one year ago
  21. JenniferSmart1
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    \[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\] \[\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\] Let's look at this

    • one year ago
  22. JenniferSmart1
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    so as m increases ... that makes sense Problem is will I be able to do this on my own?

    • one year ago
  23. TuringTest
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    Well as you saw, I had forgotten about the 2^m for the product of only the even numbers in the factorial, but this is kind of a tricky pattern to write if you ask me. They are not all this bad. You can just reason to yourself about needing a 2 for each of the m elements in the factorial to make each factor double, and therefor even.

    • one year ago
  24. JenniferSmart1
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    ok

    • one year ago
  25. JenniferSmart1
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    ok I'll try writing the sum a_1 let me know how I did

    • one year ago
  26. TuringTest
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    I'm not sure I know that one... I actually remember trying to come up with this before, a summation representing the product of evens and could not. Maybe I'll be more lucky this time.

    • one year ago
  27. TuringTest
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    *odds I mean

    • one year ago
  28. hartnn
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    product of odds is not difficult...u'll get it turing....

    • one year ago
  29. hartnn
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    hint : 1/(1.3.5) = 2.4 /(5!)

    • one year ago
  30. TuringTest
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    ohhhhhhhhhhhhhh

    • one year ago
  31. JenniferSmart1
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    =\[a_1x^1+\frac{a_1}{3}x^3+\frac{a_1}{3\cdot5}x^5+\frac{a_1}{3\cdot5\cdot7}x^7+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9\] =\[a_1 \left(x^1+\frac{x^3}{3}+\frac{x^5}{3\cdot5}+\frac{x^7}{3\cdot5\cdot7}+\frac{x^9}{3\cdot5\cdot7\cdot9}+\frac{x^{10}}{3\cdot5\cdot7\cdot9\cdot10}\right)\] = ok let's start with the exponents....or where would I look first?

    • one year ago
  32. JenniferSmart1
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    I understand the hint.... :(

    • one year ago
  33. JenniferSmart1
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    I don't understand I meant

    • one year ago
  34. hartnn
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    1/(1.3.5) = 2.4 /(5!) = 2^2 . (1.2) / (5!) = 2^2 (2!)/(5!) got that ?

    • one year ago
  35. JenniferSmart1
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    so how ....Ohhh by having 2x4 in the numerator your eliminating the evens in the denominator!!!!

    • one year ago
  36. hartnn
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    i try to bring factorial form wherever possible.

    • one year ago
  37. hartnn
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    for 1.3.5 i needed 2.4 to get 5! so i multiplied and divided by 2.4 now from 2.4, i get 1.2=2! by factoring out 2, twice.

    • one year ago
  38. JenniferSmart1
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    \[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\] doesn't m=3 for this?

    • one year ago
  39. JenniferSmart1
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    and yes this part makes sense

    • one year ago
  40. hartnn
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    why m=3 ? i would say m=2 5 = 2m+1 so, the term will be x^(2m+1) 2^m m!/ (2m+1)! got that ?

    • one year ago
  41. JenniferSmart1
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    no I mean this is the 3rd ....whatchamacallit? "m" is from 0 to infinity ...and this would be the third term?

    • one year ago
  42. JenniferSmart1
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    nevermind

    • one year ago
  43. hartnn
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    ohh..

    • one year ago
  44. hartnn
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    start from m=0

    • one year ago
  45. hartnn
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    m=0 <-1st term m=1 <-2nd term m=2 <-3rd term

    • one year ago
  46. JenniferSmart1
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    \[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\] my point is....where does me thought process continue from here...how do I change these numbers to variables. ok so were working on the third term now. where m=2

    • one year ago
  47. JenniferSmart1
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    *my not me

    • one year ago
  48. JenniferSmart1
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    soo many typos sorry

    • one year ago
  49. JenniferSmart1
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    2m+1

    • one year ago
  50. JenniferSmart1
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    LOL

    • one year ago
  51. JenniferSmart1
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    <---frustrated koala :S

    • one year ago
  52. hartnn
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    what i do is always take 3rd or 4th term., bring it in factorial form, take lowest number =m all other numbers in terms of m....

    • one year ago
  53. JenniferSmart1
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    oh so 2m+1 would be right!

    • one year ago
  54. hartnn
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    yes.

    • one year ago
  55. JenniferSmart1
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    where would it go? \[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\] shall we put it somewhere in the numerator?

    • one year ago
  56. JenniferSmart1
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    sorry koalas are fairly slow...I'm still thinking

    • one year ago
  57. hartnn
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    all this exercise was to find the general term , which we found as x^(2m+1) 2^m m!/ (2m+1)! right ? now just put a summation sign b4 this, with limits from 0 to infinity.

    • one year ago
  58. JenniferSmart1
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    \[\frac{x^5}{3\cdot5}\] m=2 \[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\] \[\frac{x^{(2m+1)}2^mm!}{(2m+1)!}\] ok I can see it now

    • one year ago
  59. JenniferSmart1
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    I'm just very visual

    • one year ago
  60. hartnn
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    finally it will be \(\large a_1[\sum\limits _0^\infty\frac{x^{(2m+1)}2^mm!}{(2m+1)!}]\)

    • one year ago
  61. JenniferSmart1
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    thanks @hartnn

    • one year ago
  62. JenniferSmart1
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    do either of you have a similar problem somewhere? LOL I wanna try this again

    • one year ago
  63. hartnn
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    welcome ^_^

    • one year ago
  64. JenniferSmart1
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    no I think this is enough. Ok thanks

    • one year ago
  65. JenniferSmart1
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    does this make sense to anyone? \[\sum_{n=0}^{\infty}\frac{x^{2n}}{2^nn!}\]

    • one year ago
  66. JenniferSmart1
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    for both a_0 and a_1

    • one year ago
  67. hartnn
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    thats for a0 only...

    • one year ago
  68. JenniferSmart1
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    It's the solution in the back of my book for y''-xy'-y=0

    • one year ago
  69. JenniferSmart1
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    I think I wrote it out correctly, and the the sums we wrote seem right too...but somehow he took those two and made one compact solution

    • one year ago
  70. hartnn
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    u mean in the last image ? he just wrote solution of initial value problem.... with a1=0

    • one year ago
  71. JenniferSmart1
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    ohhhhhhh....durrrr sorry. Ok I got now haha

    • one year ago
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