Recent content by syj

The textbook I have defines a Jordan matrix to be one where the 1s are below the diagonal. So the matrix given is already a Jordan matrix.

Homework Statement I have J=\begin{bmatrix} \frac{\pi}{2}&0&0\\ 1&\frac{\pi}{2}&0\\ 0&1&\frac{\pi}{2}\\ \end{bmatrix} I need to find \sin(J) \text{ and } \cos(J) \text{ and show that } \sin^{2}(J)+\cos^{2}(J)=I Homework Equations The Attempt at a Solution I have the...

ha ha , it required puzzling, because I knew that p(x)=\sum_{i=1}^{n}p(\lambda_{i})h_{i}(x) would have to be such that p(A)=2^{A} but because of a silly mistake of writing a 0 instead of a 1, I was not getting that conclusion and so thought all my work was incorrect.

After much puzzling, I believe I have the answer: 2^{A}=2E_{1}+2^{2}E_{2} where I have found E_{1}=\begin{bmatrix} 1&0&0\\1&1& \-1\\1&0&0\end{bmatrix} E_{2}=\begin{bmatrix}0&0&0\\-1&0&1\\-1&0&1\end{bmatrix} so I have 2^{A}=\begin{bmatrix}...

The textbook I am using says: f(A)=\sum_{i=0}^{t}f(\lambda_{i})E_{i} I still don't understand how to put everything together to get the answer. :(

Homework Statement Find the spectral decompostion of $$ A= \begin{matrix} 1 & 0 & 0\\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{matrix} $$ and use this to find $$ 2^{A} $$ Homework Equations The Attempt at a Solution I have found the eigenvalues to be : $$ \lambda_{1}=1 \text{...

oh, sorry what i mean is that when it integrates they have r_0, sorry, i meant the solution has r_0

\int(\frac{1}{ur}+\frac{1}{u(u-r)}) =\frac{1}{u}\int\frac{1}{r}+\int\frac{1}{u-r} =\frac{1}{u}(ln(r)-ln(u-r)) =\frac{1}{u}(ln\frac{r}{u-r}) my problem is that i don't know where to put r_0 here. The solution has r_0

ok so i get \frac{1}{r(u-r)}=\frac{A}{r}+\frac{B}{u-r} so that B-A=0 and Au=1 so i got A=\frac{1}{u}=B this gives me \frac{1}{ur}+\frac{1}{u(u-r)}

Homework Statement i need to integrate: \frac{1}{r(u-r)}dr Homework Equations u is a constant The Attempt at a Solution im not sure if i should decompose the fraction. i tried that, but it didnt seem to be of any help.

ok so i get \frac{ln(r)-ln(r-u)}{u}=t which i then manipulate (or try to at least ) to get: ln\frac{r}{r-u}=ut if this is correct then i think i can do: \frac{r}{r-u}=e^{ut} i know something is wrong cos i don't have r_0 im working through an example in jordan and smith...

please explain how i would solve by separation. thanks

Homework Statement \frac{dr}{dt}=r(u-r) Homework Equations i rewrote this as \frac{dr}{dt}-ur=-r^2 i think this is like bernoulli's equation The Attempt at a Solution so i let w=\frac{1}{r} so that r=\frac{1}{w} this gives me \frac{dr}{dt}=-1w^{-2}\frac{dw}{dt} so now...

Homework Statement u\frac{\delta u}{\delta x}+\frac{\delta u}{\delta y} =1 u|_{x=y}=\frac{x}{2} Homework Equations the characteristic equations to solve are: \frac{dx}{ds}=u \frac{dy}{ds}=1 \frac{du}{ds}=1 The Attempt at a Solution I got the following equations from the...

hey, so I am only just starting to understand method of characteristics, but I am gona try to help. The first step I believe is to get your characteristic equations \frac{dx}{ds}=x^{2} \frac{dt}{ds}=1 \frac{du}{ds}=t solving these equations gives x=x^{2}s +x_{0} t=s+t_{0}...