Mimi_x3
  • Mimi_x3
how did they determine
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Mimi_x3
  • Mimi_x3
https://gyazo.com/32b7234d6e962e0ad9be42a08972b285
Mimi_x3
  • Mimi_x3
b_0 b_1 b_2 b_k
misty1212
  • misty1212
HI you are thinking tooo hard

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

misty1212
  • misty1212
any polynomial can be written as a power series, most of the coefficients are zero
anonymous
  • anonymous
If you want to be more systematic, you can plug that power series representation in to the differential equation and then match coefficients.
Mimi_x3
  • Mimi_x3
anonymous
  • anonymous
i would follow what @Jemurray3 said
dan815
  • dan815
watchu u mean its just what they say
dan815
  • 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
Mimi_x3
  • Mimi_x3
i mean how did they get it
Mimi_x3
  • Mimi_x3
solution pls!
Mimi_x3
  • Mimi_x3
like do i plug in the whole power series to get it
Mimi_x3
  • Mimi_x3
this is what i get: x = 0; b_0(0-1)^0 = 2 b_0 1 = 2 b_ 0 = 2 ..
dan815
  • dan815
http://prntscr.com/8o6l00
dan815
  • dan815
they just found a simple solution with (x-1) power series
dan815
  • dan815
so they didnt bother with the rest
dan815
  • dan815
but normally if u cant see that u would plug in both of the power series and solve for the coefficients
dan815
  • dan815
you know the summation has to equal x^2, so just math the coefficients
anonymous
  • 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: \[1 + 2(x-1) + (x-1)^2 \]
dan815
  • dan815
|dw:1444088025451:dw|
dan815
  • dan815
does it make sense now?
dan815
  • dan815
u can get expreesion for ak
anonymous
  • anonymous
Oops - I think I misread the intent of the question.
dan815
  • dan815
|dw:1444088417372:dw|
dan815
  • dan815
and the rest of the qeuations equal 0
dan815
  • 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
dan815
  • dan815
solve for a0,a1,a2,a3
dan815
  • dan815
|dw:1444102770646:dw|
Mimi_x3
  • Mimi_x3
I'm finding B_0
dan815
  • dan815
what do u mean finding B_0 that is given
dan815
  • dan815
do u mean u are finding Ao
dan815
  • dan815
|dw:1444102878452:dw|
Mimi_x3
  • Mimi_x3
b_0 is given frmo the solution...but i don't how they got it.
Mimi_x3
  • Mimi_x3
THIS IS TH QUESTION https://gyazo.com/e098e1ac11c87668eda64d1a3a316ee2
dan815
  • dan815
|dw:1444103319383:dw|
dan815
  • dan815
|dw:1444103402756:dw|
dan815
  • dan815
dan815
  • dan815
@oldrin.bataku
anonymous
  • 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 \(k\ge 3\)
anonymous
  • anonymous
you can determine that by many different means, including the Taylor coefficient formula: $$b_n=\frac{f^{(n)}(1)}{n!}$$ so $$f(x)=x^2\implies f(1)=1\\f'(x)=2x\implies f'(1)=2\\f''(x)=2\implies f''(1)=2\\f^{(3)}(x)=0\implies f^{(3)}(1)=0\\f^{(n)}(x)=0\implies f^{(n)}(1)=0\text{ for }n\ge 3$$
anonymous
  • anonymous
or just by expanding, like I did above: $$(x-1)^2=x^2-2x+1\\\implies x^2=(x-1)^2+2x-1\\2(x-1)=2x-2\\\implies x^2=(x-1)^2+2(x-1)+1$$
Mimi_x3
  • 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)
Mimi_x3
  • Mimi_x3
and y'(0) = 1 = a_1
anonymous
  • 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 \(a_1=1\)
anonymous
  • 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 \(x=0\)
Mimi_x3
  • Mimi_x3
thanks! you're awesome!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.