First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

InteGrand · Jun 7, 2016

Re: MATH1131 help thread

Flop21 said:
Think I'm just confused how to do these questions involving fractions you cannot simplify and have to do with integration by parts. Can do integration by parts if it's just f(x)*g(x).

$\noindent You can think of this one as $\int f(x)g(x)\text{ d}x$, where $f(x) =1-x$ and $g(x)= \frac{1}{\left(1+x\right)^3}$.$

Red_of_Head · Jun 7, 2016

Re: MATH1131 help thread

Flop21 said:
How do we do that?

Think I'm just confused how to do these questions involving fractions you cannot simplify and have to do with integration by parts. Can do integration by parts if it's just f(x)*g(x).

So the numerator becomes f(x) and 1/denominator becomes g(x).

To integrate 1/(1+x)^3, change it to (1+x)^-3, and then the normal integration process, changing back to the fraction for the final result.

Flop21 · Jun 7, 2016

Re: MATH1131 help thread

Find the arg(z) of z= -1-i

I'm getting the wrong answer.

InteGrand · Jun 7, 2016

Re: MATH1131 help thread

Flop21 said:
Find the arg(z) of z= -1-i

I'm getting the wrong answer.

The answer is -3pi/4. Did you get pi/4?

Flop21 · Jun 7, 2016

Re: MATH1131 help thread

InteGrand said:
The answer is -3pi/4. Did you get pi/4?

Yeah I did get pi/4.

InteGrand · Jun 7, 2016

Re: MATH1131 help thread

Flop21 said:
Yeah I did get pi/4.

I'm guessing you did inverse tan of (-1/-1).

Since the real part of the complex number is negative, we need to add or subtract pi from what you got by doing that. Since the imaginary part is negative, we subtract pi. So the answer is pi/4 – pi = -3pi/4.

It's easier to do this one graphically though. The relevant point is (-1,-1), which is in quadrant 3 and lies on the 45 degree line y = x, so the angle it makes to the positive x-axis is -3pi/4.

Flop21 · Jun 7, 2016

Re: MATH1131 help thread

InteGrand said:
I'm guessing you did inverse tan of (-1/-1).

Since the real part of the complex number is negative, we need to add or subtract pi from what you got by doing that. Since the imaginary part is negative, we subtract pi. So the answer is pi/4 – pi = -3pi/4.

It's easier to do this one graphically though. The relevant point is (-1,-1), which is in quadrant 3 and lies on the 45 degree line y = x, so the angle it makes to the positive x-axis is -3pi/4.

Ohhh okay we subtract pi. I was originally trying to add pi. Thanks

Flop21 · Jun 7, 2016

Re: MATH1131 help thread

How do I sketch this region???

{z (element symbol) C : 0<=Arg(z-i) <= pi/4}

I'm confused with the arg part.

drainbammage · Jun 7, 2016

Re: MATH1131 help thread

http://imgur.com/a/2EI0A

Can someone explain this type of question? I cant seem to understand it from the solutions.

InteGrand · Jun 7, 2016

Re: MATH1131 help thread

drainbammage said:
http://imgur.com/a/2EI0A

Can someone explain this type of question? I cant seem to understand it from the solutions.

$\noindent I'm assuming you mean the matrix one. Let $X=A^2$, so we are after $X^{-1}$. We can see that $A^6 = 64I$. Therefore, $A^4 A^2 = 64I$, as $A^4 A^2 = A^6$. So $A^4 X = 64 I \Rightarrow \left(\frac{1}{64}A^4\right)X=I$ (dividing through by $64$). This implies that $X^{-1}$ is just $\frac{1}{64}A^4$. Since we are given what $A^4$ is, we can write down $X^{-1}$.$

leehuan · Jun 7, 2016

Re: MATH1131 help thread

Flop21 said:
How do I sketch this region???

{z (element symbol) C : 0<=Arg(z-i) <= pi/4}

I'm confused with the arg part.

Because we're talking about z-i, we need to consider the point (0,1)

$$In general, $\arg{z}=\theta$ represents a $\textit{ray}$ from the point (0,0) shooting off at an angle of $\theta$, where obviously $-\pi < \theta \le \pi\\ $However, not actually including the point $(0,0)$

So for the part: arg(z-i) = 0

We draw a ray at an angle of 0 (so, just pointing to the right) from the point (0,1).

For the other part: arg(z-i) = pi/4
We draw a ray at an angle of pi/4 from the point (0,1) instead

And we put in a discontinuity at the point (0,1) because we don't include it (this is just a hole in the graph)

If you're interested, a geometric interpretation would be the lines y=1 and y=x+1, but just the top halves of the line

Because we have 0<=arg(z-i)<=pi/4, we shade all the region that's between these two lines.

The geometric interpretation would then be x>0, y>=1, y<=x+1

Flop21 · Jun 8, 2016

Re: MATH1131 help thread

InteGrand said:
$\noindent I'm assuming you mean the matrix one. Let $X=A^2$, so we are after $X^{-1}$. We can see that $A^6 = 64I$. Therefore, $A^4 A^2 = 64I$, as $A^4 A^2 = A^6$. So $A^4 X = 64 I \Rightarrow \left(\frac{1}{64}A^4\right)X=I$ (dividing through by $64$). This implies that $X^{-1}$ is just $\frac{1}{64}A^4$. Since we are given what $A^4$ is, we can write down $X^{-1}$.$

I'm having a lot of trouble with these questions, any tips or places I can go to see these done more?

Or here's another one of these questions if someone wants to show me the working that'd be awesome:

They give you A, A^2, A^4, A^8, and ask for the inverse of A^7.

A^8 is 16I also.

InteGrand · Jun 8, 2016

Re: MATH1131 help thread

Flop21 said:
I'm having a lot of trouble with these questions, any tips or places I can go to see these done more?

Or here's another one of these questions if someone wants to show me the working that'd be awesome:

They give you A, A^2, A^4, A^8, and ask for the inverse of A^7.

A^8 is 16I also.

$\noindent Since $A^8 =16I$, we have $\left(A\right)\left(A^7\right)=16I$. So dividing through by $16$, we have $\left(\frac{1}{16}A\right)\left(A^7\right)=I$. Hence the inverse of $A^7$ is $\frac{1}{16}A$, which we can write down since we're given $A$.$

Flop21 · Jun 8, 2016

Re: MATH1131 help thread

Why is this true? What happens with the "z -("part?

InteGrand · Jun 8, 2016

Re: MATH1131 help thread

Flop21 said:
Why is this true? What happens with the "z -("part?

$\noindent In general, say $w$ is a fixed complex number, then $|z-w|$ is the distance between $z$ and $w$ in the complex plane. So geometrically, if $a$ is a positive constant, all the points with $|z-w|< a$ will be correspond to the set of all points in the complex plane that are within distance $a$ of $w$, which is exactly an open disk centred at $w$ with radius $a$ (if it were a $\leq $ sign, it'd be the closed disk, since we'd include the points on the boundary, as these are exactly distance $a$ from $w$, by definition of a circle of radius $a$). In other words, everything inside the circle (technically disk) is closer than distance $a$ from $w$ (and everything outside it is farther).$

$\noindent In your example, we have $w=2+i$ and $a=3$.$

Flop21 · Jun 8, 2016

Re: MATH1131 help thread

InteGrand said:
$\noindent In general, say $w$ is a fixed complex number, then $|z-w|$ is the distance between $z$ and $w$ in the complex plane. So geometrically, if $a$ is a positive constant, all the points with $|z-w|< a$ will be correspond to the set of all points in the complex plane that are within distance $a$ of $w$, which is exactly an open disk centred at $w$ with radius $a$ (if it were a $\leq $ sign, it'd be the closed disk, since we'd include the points on the boundary, as these are exactly distance $a$ from $w$, by definition of a circle of radius $a$). In other words, everything inside the circle (technically disk) is closer than distance $a$ from $w$ (and everything outside it is farther).$

$\noindent In your example, we have $w=2+i$ and $a=3$.$

Oh interesting, I didn't know |z-w| is in general the distance between z and w in the complex plane (but yeah makes sense now).

InteGrand · Jun 8, 2016

Re: MATH1131 help thread

$\noindent Also, if we view complex numbers as being represented vectors in the plane (very common thing to do), then $z-w$ would be represented by the vector that points from $w$ to $z$. Also, $|z|$ would be the length of the vector for $z$ (which goes from the origin to the point $z$ in the complex plane). So $|z-w|$ would be the length of the vector going from $w$ to $z$, which is just the distance between $z$ and $w$ in the complex plane. So this is a way to see why $|z-w|$ is the distance between $w$ and $z$ in the complex plane.$

Flop21 · Jun 8, 2016

Re: MATH1131 help thread

Oh lord, I've got 1 whole day left to study for this exam.

I just want to pass but not sure if possible.

iforgotmyname · Jun 8, 2016

Re: MATH1131 help thread

Flop21 said:
Oh lord, I've got 1 whole day left to study for this exam.

I just want to pass but not sure if possible.

You can get lower than 50 in the finals and still pass. Just putting it out there

Flop21 · Jun 8, 2016

Re: MATH1131 help thread

iforgotmyname said:
You can get lower than 50 in the finals and still pass. Just putting it out there

No I cannot.

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