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JenniferSmart1

  • 3 years ago

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

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  1. JenniferSmart1
    • 3 years ago
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    I know I'm a disappointment :(

  2. TuringTest
    • 3 years ago
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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

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

  4. hartnn
    • 3 years ago
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    num = x^(2m)

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

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

  7. TuringTest
    • 3 years ago
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    yeah, that's it @hart

  8. JenniferSmart1
    • 3 years ago
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  9. JenniferSmart1
    • 3 years ago
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    #9

  10. JenniferSmart1
    • 3 years ago
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    how and why?

  11. hartnn
    • 3 years ago
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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!)

  12. hartnn
    • 3 years ago
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    got that ^ ?

  13. JenniferSmart1
    • 3 years ago
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    still thinking

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

  15. hartnn
    • 3 years ago
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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....

  16. JenniferSmart1
    • 3 years ago
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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

  17. hartnn
    • 3 years ago
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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 ?

  18. JenniferSmart1
    • 3 years ago
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    I didn't quite get that part...

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

  20. JenniferSmart1
    • 3 years ago
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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.... =(

  21. JenniferSmart1
    • 3 years ago
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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

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

  23. TuringTest
    • 3 years ago
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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.

  24. JenniferSmart1
    • 3 years ago
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    ok

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

  26. TuringTest
    • 3 years ago
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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.

  27. TuringTest
    • 3 years ago
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    *odds I mean

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

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

  30. TuringTest
    • 3 years ago
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    ohhhhhhhhhhhhhh

  31. JenniferSmart1
    • 3 years ago
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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?

  32. JenniferSmart1
    • 3 years ago
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    I understand the hint.... :(

  33. JenniferSmart1
    • 3 years ago
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    I don't understand I meant

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

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

  36. hartnn
    • 3 years ago
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    i try to bring factorial form wherever possible.

  37. hartnn
    • 3 years ago
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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.

  38. JenniferSmart1
    • 3 years ago
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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?

  39. JenniferSmart1
    • 3 years ago
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    and yes this part makes sense

  40. hartnn
    • 3 years ago
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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 ?

  41. JenniferSmart1
    • 3 years ago
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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?

  42. JenniferSmart1
    • 3 years ago
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    nevermind

  43. hartnn
    • 3 years ago
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    ohh..

  44. hartnn
    • 3 years ago
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    start from m=0

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

  46. JenniferSmart1
    • 3 years ago
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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

  47. JenniferSmart1
    • 3 years ago
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    *my not me

  48. JenniferSmart1
    • 3 years ago
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    soo many typos sorry

  49. JenniferSmart1
    • 3 years ago
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    2m+1

  50. JenniferSmart1
    • 3 years ago
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    LOL

  51. JenniferSmart1
    • 3 years ago
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    <---frustrated koala :S

  52. hartnn
    • 3 years ago
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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....

  53. JenniferSmart1
    • 3 years ago
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    oh so 2m+1 would be right!

  54. hartnn
    • 3 years ago
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    yes.

  55. JenniferSmart1
    • 3 years ago
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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?

  56. JenniferSmart1
    • 3 years ago
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    sorry koalas are fairly slow...I'm still thinking

  57. hartnn
    • 3 years ago
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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.

  58. JenniferSmart1
    • 3 years ago
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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

  59. JenniferSmart1
    • 3 years ago
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    I'm just very visual

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

  61. JenniferSmart1
    • 3 years ago
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    thanks @hartnn

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

  63. hartnn
    • 3 years ago
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    welcome ^_^

  64. JenniferSmart1
    • 3 years ago
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    no I think this is enough. Ok thanks

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

  66. JenniferSmart1
    • 3 years ago
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    for both a_0 and a_1

  67. hartnn
    • 3 years ago
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    thats for a0 only...

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

  69. JenniferSmart1
    • 3 years ago
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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

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

  71. JenniferSmart1
    • 3 years ago
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    ohhhhhhh....durrrr sorry. Ok I got now haha

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