Author Topic: Silly Questions Thread (Read 33147 times) Tweet Share

Aqualim · « **Reply #165 on:** April 17, 2010, 08:57:47 pm »

Quote from: Blakhitman on April 17, 2010, 08:09:16 pm

You posted it before .

$\begin{bmatrix} {x}'\\ {y}' \end{bmatrix}=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\begin{bmatrix} {x}\\ {y} \end{bmatrix}$

you get $y={x}'$ and $x={y}'$ then sub in (note it's the swapping x and y business) to $log_e{x}$ and you get $y=e^{x}$

Then do same for f(x) lol

How so?
$\begin{bmatrix} x'\\y' \end{bmatrix}=\begin{bmatrix} 0 & 1\\ \frac{1}{2} &0 \end{bmatrix}\times \begin{bmatrix} x\\y \end{bmatrix}+\begin{bmatrix} 0\\-\frac{1}{2} \end{bmatrix}= x'=x~and~y'=\frac{1}{2}y-\frac{1}{2}$

the.watchman · « **Reply #166 on:** April 17, 2010, 09:06:19 pm »

Nope, this matrix eqn gives $x'=y$ and $y'=\frac{x}{2}-\frac{1}{2}$

Aqualim · « **Reply #167 on:** April 17, 2010, 09:45:28 pm »

ahh ok, just picked up where I went wrong. Thank you

Aqualim · « **Reply #168 on:** April 20, 2010, 06:17:17 pm »

Can't seem to get this stupid literal equation.

Solve for x;
$ax+b=\frac{x+b}{a}$

m@tty · « **Reply #169 on:** April 20, 2010, 06:24:48 pm »

How is that an integral?? o.0

Anyway, $ax+b=\frac{x+b}{a}$

$\implies x(a^2-1)=b(1-a)$

$\implies x= \frac{b(1-a)}{(a+1)(a-1)}, a \ne \pm 1$

$\implies x= -\frac{b}{a+1} , a \ne \pm 1$

superflya · « **Reply #170 on:** April 20, 2010, 06:30:21 pm »

lol i think u misinterpreted what he wrote, it says literal

or he may have modified it..

Blakhitman · « **Reply #171 on:** April 20, 2010, 06:34:13 pm »

Quote from: superflya on April 20, 2010, 06:30:21 pm

lol i think u misinterpreted what he wrote, it says literal or he may have modified it..

He initially typed integral!

Aqualim · « **Reply #172 on:** April 20, 2010, 06:34:49 pm »

Yeah made a typo

Thanks for the help though

superflya · « **Reply #173 on:** April 20, 2010, 06:36:35 pm »

Quote from: Blakhitman on April 20, 2010, 06:34:13 pm

Quote from: superflya on April 20, 2010, 06:30:21 pm
lol i think u misinterpreted what he wrote, it says literal or he may have modified it..

He initially typed integral!

ahh thought so

kamil9876 · « **Reply #174 on:** April 20, 2010, 06:41:27 pm »

Quote from: m@tty on April 20, 2010, 06:24:48 pm

How is that an integral?? o.0

Anyway, $ax+b=\frac{x+b}{a}$

$\implies x(a^2-1)=b(1-a)$

$\implies x= \frac{b(1-a)}{(a+1)(a-1)}, a \ne \pm 1$

$\implies x= -\frac{b}{a+1} , a \ne \pm 1$

What happens when $a=\pm 1$ ? $a=0$ ?

m@tty · « **Reply #175 on:** April 20, 2010, 06:45:56 pm »

Quote from: kamil9876 on April 20, 2010, 06:41:27 pm

What happens when $a=\pm 1$ ? $a=0$ ?

I meant it as " $a \ne -1, 1$ ". I remember seeing textbooks use the combined $a \ne \pm 1$ , to mean the same thing, so I assumed it is mathematically correct??

the.watchman · « **Reply #176 on:** April 20, 2010, 06:50:54 pm »

I think you are 'supposed' to show what solutions there are for $a = \pm 1$

And yes, kamil is right, you forgot about $a=0$ from the original eqn

m@tty · « **Reply #177 on:** April 20, 2010, 07:03:09 pm »

Ahh, I must have misinterpreted his post.

But what? You are meant to show that

$-x+b=-x-b$ ??

And $x+b=x+b$ ? :|

Yeah, I did forget about $a=0$ :buck2: ...

Aqualim · « **Reply #178 on:** April 20, 2010, 07:06:02 pm »

Quote from: m@tty on April 20, 2010, 06:24:48 pm

$\implies x= \frac{b(1-a)}{(a+1)(a-1)}, a \ne \pm 1$

$\implies x= -\frac{b}{a+1} , a \ne \pm 1$

Also quick question, what happened here? I don't see how you could cancel out (1-a) and (a-1) since they are both different?

the.watchman · « **Reply #179 on:** April 20, 2010, 07:07:20 pm »

Quote from: m@tty on April 20, 2010, 07:03:09 pm

Ahh, I must have misinterpreted his post.

But what? You are meant to show that

$-x+b=-x-b$ ??

And $x+b=x+b$ ? :|

Yeah, I did forget about $a=0$ :buck2: ...

I think you should show what solutions you can have for x with 'a' as these values
Is it no solutions? Or infinite solutions?

Quote from: Aqualim on April 20, 2010, 07:06:02 pm

Quote from: m@tty on April 20, 2010, 06:24:48 pm
$\implies x= \frac{b(1-a)}{(a+1)(a-1)}, a \ne \pm 1$

$\implies x= -\frac{b}{a+1} , a \ne \pm 1$

Also quick question, what happened here? I don't see how you could cancel out (1-a) and (a-1) since they are both different?

$(1-a)=-(a-1)$ , hence why the (a-1) can be cancelled (note the added negative sign

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