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Mimi_x3

  • one year ago

how did they determine

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  1. Mimi_x3
    • one year ago
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    https://gyazo.com/32b7234d6e962e0ad9be42a08972b285

  2. Mimi_x3
    • one year ago
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    b_0 b_1 b_2 b_k

  3. misty1212
    • one year ago
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    HI you are thinking tooo hard

  4. misty1212
    • one year ago
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    any polynomial can be written as a power series, most of the coefficients are zero

  5. anonymous
    • one year ago
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    If you want to be more systematic, you can plug that power series representation in to the differential equation and then match coefficients.

  6. Mimi_x3
    • one year ago
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    @dan815

  7. anonymous
    • one year ago
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    i would follow what @Jemurray3 said

  8. dan815
    • one year ago
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    watchu u mean its just what they say

  9. dan815
    • one year ago
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    1+2(x-1) + 1*(x-1)^2 + 0*(x-1)^3 + 0*(x-1)^4.... so the power series sum bk*(x-1)^k where b0=1 b1=2 b2=1 bk, k>2 is all 0

  10. Mimi_x3
    • one year ago
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    i mean how did they get it

  11. Mimi_x3
    • one year ago
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    solution pls!

  12. Mimi_x3
    • one year ago
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    like do i plug in the whole power series to get it

  13. Mimi_x3
    • one year ago
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    this is what i get: x = 0; b_0(0-1)^0 = 2 b_0 1 = 2 b_ 0 = 2 ..

  14. dan815
    • one year ago
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    http://prntscr.com/8o6l00

  15. dan815
    • one year ago
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    they just found a simple solution with (x-1) power series

  16. dan815
    • one year ago
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    so they didnt bother with the rest

  17. dan815
    • one year ago
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    but normally if u cant see that u would plug in both of the power series and solve for the coefficients

  18. dan815
    • one year ago
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    you know the summation has to equal x^2, so just math the coefficients

  19. anonymous
    • one year ago
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    Here's how you do it. If \( y = \sum_{k=0}^\infty b_k (x-1)^k \), then \[ y' = \sum_{k=1}^\infty k\cdot b_k (x-1)^{k-1} \] Notice how I replaced the bottom limit with 1, because the 0 term is just 0 now. Next, you rename the index by letting \( n = k-1 \rightarrow k = n+1 \). That means \[y' = \sum_{n=0}^\infty (n+1)\cdot b_{n+1} (x-1)^n \] Now, the left hand side of the differential equation is \(y+y'\). So we can combine those two terms together using the same index: \[y + y' = \sum_{k=0}^\infty b_k (x-1)^k + \sum_{k=0}^\infty (k+1)\cdot b_{k+1} (x-1)^k\] \[ = \sum_{k=0}^\infty \left\{ b_k + (k+1)b_{k+1}\right\}(x-1)^k \] The first few terms if you expand that out are \[(b_0 + b_1) +(b_1 + 2b_2)(x-1) + (b_2 + 3b_3)(x-1)^2 + ... \] At the same time, you have the right hand side: \[1 + 2(x-1) + (x-1)^2 \]

  20. dan815
    • one year ago
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    |dw:1444088025451:dw|

  21. dan815
    • one year ago
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    does it make sense now?

  22. dan815
    • one year ago
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    u can get expreesion for ak

  23. anonymous
    • one year ago
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    Oops - I think I misread the intent of the question.

  24. dan815
    • one year ago
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    |dw:1444088417372:dw|

  25. dan815
    • one year ago
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    and the rest of the qeuations equal 0

  26. dan815
    • one year ago
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    at least u will get 3 equations and 4 unkowns so you will get some constant of variation but all your Ak can be written as a function of 1 variable

  27. dan815
    • one year ago
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    solve for a0,a1,a2,a3

  28. dan815
    • one year ago
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    |dw:1444102770646:dw|

  29. Mimi_x3
    • one year ago
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    I'm finding B_0

  30. dan815
    • one year ago
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    what do u mean finding B_0 that is given

  31. dan815
    • one year ago
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    do u mean u are finding Ao

  32. dan815
    • one year ago
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    |dw:1444102878452:dw|

  33. Mimi_x3
    • one year ago
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    b_0 is given frmo the solution...but i don't how they got it.

  34. Mimi_x3
    • one year ago
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    THIS IS TH QUESTION https://gyazo.com/e098e1ac11c87668eda64d1a3a316ee2

  35. dan815
    • one year ago
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    |dw:1444103319383:dw|

  36. dan815
    • one year ago
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    |dw:1444103402756:dw|

  37. dan815
    • one year ago
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    @Empty

  38. dan815
    • one year ago
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    @oldrin.bataku

  39. anonymous
    • one year ago
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    $$x^2=(x-1)^2+2(x-1)+1$$ we also know $$x^2=\sum_{n=0}^\infty b_n (x-1)^n$$ so it follows \(b_0=1,b_1=2,b_2=1\) and \(b_k=0\) for \(k\ge 3\)

  40. anonymous
    • one year ago
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    you can determine that by many different means, including the Taylor coefficient formula: $$b_n=\frac{f^{(n)}(1)}{n!}$$ so $$f(x)=x^2\implies f(1)=1\\f'(x)=2x\implies f'(1)=2\\f''(x)=2\implies f''(1)=2\\f^{(3)}(x)=0\implies f^{(3)}(1)=0\\f^{(n)}(x)=0\implies f^{(n)}(1)=0\text{ for }n\ge 3$$

  41. anonymous
    • one year ago
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    or just by expanding, like I did above: $$(x-1)^2=x^2-2x+1\\\implies x^2=(x-1)^2+2x-1\\2(x-1)=2x-2\\\implies x^2=(x-1)^2+2(x-1)+1$$

  42. Mimi_x3
    • one year ago
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    ok i got it but i dont' understand how they determine boundaries it feels like magic https://gyazo.com/848f27d09ebe1758077d624b8cc6ab0e how y(0) = 0 = a_0 (all even coef)

  43. Mimi_x3
    • one year ago
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    and y'(0) = 1 = a_1

  44. anonymous
    • one year ago
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    in that problem, consider: $$y(x)=\sum_{n=0}^\infty a_n x^n\\y(x)=a_0+a_1x+a_2x^2+\dots\\y(0)=a_0+a_1\cdot 0+a_2\cdot 0+\dots=a_0$$ so if \(y(0)=0\) is given as the BC, then it follows \(a_0=0\) similarly, $$y(x)=\sum_{n=0}^\infty a_n x^n\\y'(x)=\sum_{n=0}^\infty na_n x^n\\y'(x)=a_1+2a_2 x+3a_3 x^2+\dots\\y'(0)=a_1+2a_2\cdot 0+3a_3\cdot 0+\dots=a_1$$so \(y'(0)=1\) gives us \(a_1=1\)

  45. anonymous
    • one year ago
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    setting \(x=0\) makes every term in the power series with nonzero power equal \(0\), and for convenience here we define \(x^0=1\) even at \(x=0\)

  46. Mimi_x3
    • one year ago
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    thanks! you're awesome!!

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