## Mimi_x3 one year ago how did they determine

1. Mimi_x3
2. Mimi_x3

b_0 b_1 b_2 b_k

3. misty1212

HI you are thinking tooo hard

4. misty1212

any polynomial can be written as a power series, most of the coefficients are zero

5. anonymous

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

@dan815

7. anonymous

i would follow what @Jemurray3 said

8. dan815

watchu u mean its just what they say

9. dan815

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

i mean how did they get it

11. Mimi_x3

solution pls!

12. Mimi_x3

like do i plug in the whole power series to get it

13. Mimi_x3

this is what i get: x = 0; b_0(0-1)^0 = 2 b_0 1 = 2 b_ 0 = 2 ..

14. dan815
15. dan815

they just found a simple solution with (x-1) power series

16. dan815

so they didnt bother with the rest

17. dan815

but normally if u cant see that u would plug in both of the power series and solve for the coefficients

18. dan815

you know the summation has to equal x^2, so just math the coefficients

19. anonymous

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

|dw:1444088025451:dw|

21. dan815

does it make sense now?

22. dan815

u can get expreesion for ak

23. anonymous

Oops - I think I misread the intent of the question.

24. dan815

|dw:1444088417372:dw|

25. dan815

and the rest of the qeuations equal 0

26. dan815

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

solve for a0,a1,a2,a3

28. dan815

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29. Mimi_x3

I'm finding B_0

30. dan815

what do u mean finding B_0 that is given

31. dan815

do u mean u are finding Ao

32. dan815

|dw:1444102878452:dw|

33. Mimi_x3

b_0 is given frmo the solution...but i don't how they got it.

34. Mimi_x3

THIS IS TH QUESTION https://gyazo.com/e098e1ac11c87668eda64d1a3a316ee2

35. dan815

|dw:1444103319383:dw|

36. dan815

|dw:1444103402756:dw|

37. dan815

@Empty

38. dan815

@oldrin.bataku

39. anonymous

$$x^2=(x-1)^2+2(x-1)+1$$ 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

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

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

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

and y'(0) = 1 = a_1

44. anonymous

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

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

thanks! you're awesome!!