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Mimi_x3
 one year ago
how did they determine
Mimi_x3
 one year ago
how did they determine

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misty1212
 one year ago
Best ResponseYou've already chosen the best response.0HI you are thinking tooo hard

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0any polynomial can be written as a power series, most of the coefficients are zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you want to be more systematic, you can plug that power series representation in to the differential equation and then match coefficients.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i would follow what @Jemurray3 said

dan815
 one year ago
Best ResponseYou've already chosen the best response.1watchu u mean its just what they say

dan815
 one year ago
Best ResponseYou've already chosen the best response.11+2(x1) + 1*(x1)^2 + 0*(x1)^3 + 0*(x1)^4.... so the power series sum bk*(x1)^k where b0=1 b1=2 b2=1 bk, k>2 is all 0

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0i mean how did they get it

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0like do i plug in the whole power series to get it

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0this is what i get: x = 0; b_0(01)^0 = 2 b_0 1 = 2 b_ 0 = 2 ..

dan815
 one year ago
Best ResponseYou've already chosen the best response.1they just found a simple solution with (x1) power series

dan815
 one year ago
Best ResponseYou've already chosen the best response.1so they didnt bother with the rest

dan815
 one year ago
Best ResponseYou've already chosen the best response.1but normally if u cant see that u would plug in both of the power series and solve for the coefficients

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you know the summation has to equal x^2, so just math the coefficients

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here's how you do it. If \( y = \sum_{k=0}^\infty b_k (x1)^k \), then \[ y' = \sum_{k=1}^\infty k\cdot b_k (x1)^{k1} \] Notice how I replaced the bottom limit with 1, because the 0 term is just 0 now. Next, you rename the index by letting \( n = k1 \rightarrow k = n+1 \). That means \[y' = \sum_{n=0}^\infty (n+1)\cdot b_{n+1} (x1)^n \] Now, the left hand side of the differential equation is \(y+y'\). So we can combine those two terms together using the same index: \[y + y' = \sum_{k=0}^\infty b_k (x1)^k + \sum_{k=0}^\infty (k+1)\cdot b_{k+1} (x1)^k\] \[ = \sum_{k=0}^\infty \left\{ b_k + (k+1)b_{k+1}\right\}(x1)^k \] The first few terms if you expand that out are \[(b_0 + b_1) +(b_1 + 2b_2)(x1) + (b_2 + 3b_3)(x1)^2 + ... \] At the same time, you have the right hand side: \[1 + 2(x1) + (x1)^2 \]

dan815
 one year ago
Best ResponseYou've already chosen the best response.1does it make sense now?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1u can get expreesion for ak

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops  I think I misread the intent of the question.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1and the rest of the qeuations equal 0

dan815
 one year ago
Best ResponseYou've already chosen the best response.1at 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1what do u mean finding B_0 that is given

dan815
 one year ago
Best ResponseYou've already chosen the best response.1do u mean u are finding Ao

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0b_0 is given frmo the solution...but i don't how they got it.

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0THIS IS TH QUESTION https://gyazo.com/e098e1ac11c87668eda64d1a3a316ee2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$x^2=(x1)^2+2(x1)+1$$ we also know $$x^2=\sum_{n=0}^\infty b_n (x1)^n$$ so it follows \(b_0=1,b_1=2,b_2=1\) and \(b_k=0\) for \(k\ge 3\)

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or just by expanding, like I did above: $$(x1)^2=x^22x+1\\\implies x^2=(x1)^2+2x1\\2(x1)=2x2\\\implies x^2=(x1)^2+2(x1)+1$$

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0ok 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in 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 \(a_1=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0setting \(x=0\) makes every term in the power series with nonzero power equal \(0\), and for convenience here we define \(x^0=1\) even at \(x=0\)

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0thanks! you're awesome!!
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