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Loser66
 one year ago
On attachment
Please, help
Loser66
 one year ago
On attachment Please, help

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I would like to know from where we have "Hence \(x(t) = e^{tA}x_0\) at 1.4, 1.5

freckles
 one year ago
Best ResponseYou've already chosen the best response.0like how did they get that equation?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0It didn't indicate to x(t) before. It worked on \(e^{tA}\) and derivative of it. Suddenly, I have :" Hence x(t) =...." why?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0plug x(t) into the IVP \[\dfrac{dx}{dt} = Ax~;~x(0)=x_0\] and see if it is really a solution

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0If "plug x(t) into .." \(\dfrac{d}{dt}e^{tA}= Ae^{tA}\) then \(\dfrac{d}{dt}x(t)= Ax(t)\), right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since we have that \(\dfrac{d}{dt}e^{At}=Ae^{At}\), it follows that \(x=e^{At}\) is clearly a solution to \(\dfrac{dx}{dt}=Ax\) (this is just a restatement of the previous fact)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(x(t) = e^{tA}x_0\) right ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 That is what I asked!! How and why we have that equation?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0and @oldrin.bataku gave the answer, but I didn't get how "it follows that x = e^(At) :(

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0they are saying that function is a solution to the given IVP i was asking you to check if it is really the case by plugging x(t) into the differential eqn

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0btw, \(x_0\) is just a constant recall that if \(f(x)\) is a solution , then a constant multiple of it is also a solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in fact, since the derivative is linear, we actually have \(\dfrac{d}{dt}\left(ke^{At}\right)=kAe^{At}\) so the general solution is actually a whole parameterized family of solutions \(x(t)=kAe^{At}\) for an \(n\)by\(1\) constant matrix \(k\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, i meant \(x(t)=ke^{At}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if \(x(0)=x_0\) then \(ke^{A\cdot0}=x_0\implies kI=x_0\implies k=x_0\) so \(x(t)=x_0e^{At}=e^{At}x_0\) is the unique solution to \(\dot x=Ax\) subject to \(x(0)=x_0\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I got it. Thank you very much. I am confused with x_0. But now, I am ok with it.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0In the textbook, \(x_0\) is a specific constant for the IVP. Maybe think of it like the usual \(c_0\) if the past experience with simple ODE's helps..

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I think it's cute how you can kinda pretend these aren't vectors and matrices and just kinda like \[\frac{dx}{dt} = Ax \] "separate" \[\frac{dx}{x} = Adt\] \[ \ln x = At +c\] \[x=ke^{At}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0note the idea here is actually very general and underlies what we call flows of vector fields  \(d/dt\) is an infinitesimal generator of the evolution of \(x\) through time, i.e. it gives rise to flows

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I have another question, please explain me Ex 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which part is giving you trouble?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0recall that for a matrix \(A\) with columns \(A_1,A_2,A_3,\dots,A_n\) the image of a column vector \(u=(u_1,\dots,u_n)^T\) is \(Au=u_1A_1+\dots+u_nA_n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i.e. the \(n\)th column of the matrix is the action of the matrix on the \(n\)th basis vector \(e_n\), represented by a matrix of zeros with a \(1\) in row \(n\): $$e_1=\begin{bmatrix}1\\0\end{bmatrix},e_2=\begin{bmatrix}0\\1\end{bmatrix}$$so the matrix $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$represents the transformation that takes \(e_1\to \begin{bmatrix}a\\c\end{bmatrix}\) and \(e_2\to \begin{bmatrix}b\\d\end{bmatrix}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so to figure out the first column of the matrix representing \(e^{At}\), we just need to see how it behaves on \(\begin{bmatrix}1\\0\end{bmatrix}\), and similarly for the second column

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I got what you meant . How about 1.30?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.01.33 is for \(e^{tA}(1,0)^T\) that is the first column of \(e^{tA}\) 1.34 is what? why not \(e^{tA}(0,1)^T\) which is the second column of \(e^{tA}\) Why it calculate \(e^{tA}(1,1)^T\) ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider that if \(v\ne 0\) is an eigenvector of \(A\), with \(Av=\lambda v\), it follows that \(A^2v=\lambda^2 v\) and more generally \(A^nv=\lambda^n v\). so we have: $$e^{A}=\sum_{n=0}^\infty\frac1{n!} A^n\\e^Av=\left(\sum_{n=0}^\infty\frac1{n!} A^n\right)v=\sum_{n=0}^\infty\frac1{n!}\left(A^nv\right)=\sum_{n=0}^\infty\frac{\lambda^n}{n!}v=\left(\sum_{n=0}^\infty\frac{\lambda^n}{n!}\right)v=e^{\lambda}v$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if \(v\) is an eigenvector of \(A\) it's also an eigenvector of \(tA\) as \((tA)v=tAv=t\lambda v=(t\lambda )v\) so:$$e^{tA}=e^{t\lambda}v$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now we know the eigenvectors of \(e^{tA}\) are those of \(A\) and the eigenvalues \(\lambda\) become \(e^{t\lambda }\), which is what happened there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now, since we know the eigenvectors \(v_1,\dots,v_n\) of \(e^{tA}\) and we have a vector \(v\in\operatorname{span}\{v_1,\dots,v_n\}\) then we can decompose \(v_1=c_1v_1+\dots+c_nv_n\) and it follows $$e^{tA}v=c_1 e^{tA}v_1+\dots+c_2 e^{tA}v_n=c_1 e^{t\lambda_1}v_1+\dots+c_n e^{t\lambda_n} v_n$$. this gives us a way to determine how \(e^{tA}\) behaves on linear combinations of the eigenvectors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so to figure out how \(e^{tA}\) acts on \(e_1\begin{bmatrix}1\\0\end{bmatrix}\) to figure out its first column we can decompose \(e_1=\dfrac12v_1+\dfrac12v_2\) and \(e^{tA}e_1=\dfrac12 e^{tA}v_1+\dfrac12 e^{tA}v_2=\dfrac12e^{t\lambda_1}v_1+\dfrac12e^{t\lambda_2}v_2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since we have \(\lambda_1=1,\lambda_2=1\) and \(v_1=\begin{bmatrix}1\\1\end{bmatrix},v_2=\begin{bmatrix}1\\1\end{bmatrix}\) this says: $$e^{tA}\begin{bmatrix}1\\0\end{bmatrix}=\frac12e^t\begin{bmatrix}1\\1\end{bmatrix}+\frac12 e^{t}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}\frac12 e^t+\frac12 e^{t}\\\frac12 e^t\frac12 e^{t}\end{bmatrix}$$ is our first column of \(e^{tA}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in 1.34 they made a typo, they actually decomposed \(e_2=\frac12 v_1\frac12 v_2\) and then compute similarly to the above $$e^{tA}\begin{bmatrix}0\\1\end{bmatrix}=\frac12 e^t\begin{bmatrix}1\\1\end{bmatrix}\frac12e^{t}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}\frac12 e^t\frac12e^{t}\\\frac12e^t+\frac12e^{t}\end{bmatrix}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we've figured out the columns of \(e^{tA}\) and can write it now as $$e^{tA}=e^{tA}\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}\frac12(e^t+e^{t})&\frac12(e^te^{t})\\\frac12(e^te^{t})&\frac12(e^t+e^{t})\end{bmatrix}$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I guessed it is a typo but it tortured me a lot since I didn't understand it. Now I got it. Thank you so so so much.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Can I have another answer for another problem? The same topic, the same problem, just want to know the trick. Please.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.01.41 . Is there any way to quickly figure out \(\dfrac{i+1}{4}\) tem? and \(\dfrac{i}{2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the easy way is to compute the inner products, since the eigenvectors \(\{v_j\}\) are orthogonal we have \(\langle v_i,v_j\rangle=0\text{ where }i\ne j,\text{ or }v_i^2\text{ where }i=j\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so: $$\begin{bmatrix}1\\0\end{bmatrix}=c_1\begin{bmatrix}2\\1+i\end{bmatrix}+c_2\begin{bmatrix}2\\1i\end{bmatrix}\\\begin{bmatrix}2&1+i\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=c_1\begin{bmatrix}2&1+i\end{bmatrix}\begin{bmatrix}2\\1+i\end{bmatrix}+c_2\begin{bmatrix}2&1+i\end{bmatrix}\begin{bmatrix}2\\1i\end{bmatrix}\\2\cdot1+(1+i)\cdot0=c_1((2)^2+(1+i)^2)+c_2((2)^2+(1+i)(1i))\\2=c_1(4+1+2i1)+c_2(4+1+1)\\2=(4+2i)c_1+6c_2$$similarly we get$$\begin{bmatrix}2&1i\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=c_1\begin{bmatrix}2&1i\end{bmatrix}\begin{bmatrix}2\\1+i\end{bmatrix}+c_2\begin{bmatrix}2&1i\end{bmatrix}\begin{bmatrix}2\\1i\end{bmatrix}\\2\cdot1+(1i)\cdot0=c_1((2)^2+(1i)(1+i))+c_2((2)^2+(1i)^2)\\2=c_1(4+1+1)+c_2(4+12i1)\\2=6c_1+(42i)c_2$$now you can do substitution to solve \(c_1,c_2\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I got it. Again, thank you so much.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops they aren't orthonormal or even orthogonal here but they are linearly independent and we can still solve for unique \(c_1,c_2\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I did as usual and got a different solutions from the paper. I am checking.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=2%3D%284%2B2i%29x%2B6y%2C+2%3D6x%2B%2842i%29y

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Wow!! it takes a long time to work with :)
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