First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

Flop21 · Aug 15, 2016

Re: MATH1231/1241/1251 SOS Thread

What am I doing wrong in this integration question? Soz it's messy.

InteGrand · Aug 15, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
What am I doing wrong in this integration question? Soz it's messy.

Integral of cos(u) is just sin(u), not (1/u)*sin(u).

Flop21 · Aug 15, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Integral of cos(u) is just sin(u), not (1/u)*sin(u).

Oh yeah ha whoops thanks.

Flop21 · Aug 16, 2016

Re: MATH1231/1241/1251 SOS Thread

How do you differentiate e^x*y^2. ?

Actually it's a partial derivative with respect to y, but I don't know how to differentiate this one normally anyway.

InteGrand · Aug 16, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
How do you differentiate e^x*y^2. ?

Actually it's a partial derivative with respect to y, but I don't know how to differentiate this one normally anyway.

$\noindent The partial derivative with respect to $y$ is just $2ye^x$.$

$\noindent (Remember, $x$ is just treated as a constant.)$

$\noindent By the way, the above answer was assuming the function was $e^x y^2$. If the function was instead $e^{xy^2}$, then the answer would be $2ye^{xy^2}$. Since brackets weren't used, it's ambiguous what was meant. It's a good idea to use brackets to eliminate ambiguity when not writing in \LaTeX. Here's a good way to write $e^{xy^2}$ if not using \LaTeX:$

e^(x*(y^2)).

$\noindent If the intention was $e^{x} y^{2}$, then here's one way to write it unambiguously:$

(e^x)*(y^2).

Flop21 · Aug 16, 2016

Re: MATH1231/1241/1251 SOS Thread

How to integrate sec^4 x dx ?

InteGrand · Aug 16, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
How to integrate sec^4 x dx ?

$\noindent Note $\sec ^{4} x = \sec^{2}x \cdot \sec^{2} x = \left(\tan^{2}x + 1\right)\sec^{2} x$. Since $\sec^{2} x$ is the derivative of $\tan x$, it follows that $\int \sec^{4}x \, \mathrm{d}x = \int \left(\tan^{2} x+1\right)\sec^{2} x \, \mathrm{d}x = \frac{\tan^{3} x}{3} + \tan x + C$. (Use a substitution of $u=\tan x$ if you want to make it easier to see how that last integral was found.)$

leehuan · Aug 17, 2016

Re: MATH1231/1241/1251 SOS Thread

Sigh...

$\text{Proven already: The roots of }z^6+z^3+1=0\text{ are }e^{\pm \frac{2^k\pi i}{9}}, k=1,2,3$

$\text{By letting }y=z+\frac{1}{z}\text{ deduce that }\cos \frac{2\pi}{9}+\cos \frac{4\pi}{9}+\cos \frac{8\pi}{9}=0$

$\text{I'm not sure what to do after }y^3-y+1=0$

InteGrand · Aug 17, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Sigh...

$\text{Proven already: The roots of }z^6+z^3+1=0\text{ are }e^{\pm \frac{2^k\pi i}{9}}, k=1,2,3$

$\text{By letting }y=z+\frac{1}{z}\text{ deduce that }\cos \frac{2\pi}{9}+\cos \frac{4\pi}{9}+\cos \frac{8\pi}{9}=0$

$\text{I'm not sure what to do after }y^3-y+1=0$

$\noindent As you have already proven, the roots of $p(z):= z^{6} + z^{3} + 1 = 0$ are $z_1 = e^{i \frac{2\pi}{9}}$, $z_2 = e^{i \frac{4\pi}{9}}$, $z_{3} = e^{i \frac{8\pi}{9}}$, and the conjugates of these, $\overline{z_j}$, $j=1,2,3$.$

$\noindent Note that if $w = e^{i \theta}$, then $w + \overline{w} = 2\cos \theta$. Since the sum of roots of $p$ is $0$ (as there is no quintic term), we have $\sum _{j=1}^{3}\left(z_j + \overline{z_j}\right) = 0$, where $z_j = e^{i \frac{2^{j}\pi}{9}}$ (the L.H.S. is just the sum of the roots). Since the L.H.S. equals $\sum _{j=1}^{3} 2\cos \left(\frac{2^{j}\pi}{9}\right)$, dividing the last equation through by $2$ yields the result.$

InteGrand · Aug 17, 2016

Re: MATH1231/1241/1251 SOS Thread

$\noindent Realised I didn't do it by doing the $y = z + z^{-1}$ method. Using this method, you got it down to $y^{3} -y + 1=0$, where $z$ is a root of the equation $z^6 + z^3 + 1 = 0$. Since the roots of this equation produce exactly three values of $z+ z^{-1}$, namely $2\cos \left(\frac{2^j \pi}{9}\right)$, for $j=1,2,3$, it follows that $y_j := 2\cos \left(\frac{2^j \pi}{9}\right)$ for $j=1,2,3$ are roots of the polynomial equation $y^3 -y + 1 = 0$. Since this is a cubic equation, these are precisely the roots of this equation, and the result follows similarly using sum of roots being $0$ here.$

leehuan · Aug 18, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent Realised I didn't do it by doing the $y = z + z^{-1}$ method. Using this method, you got it down to $y^{3} -y + 1=0$, where $z$ is a root of the equation $z^6 + z^3 + 1 = 0$. Since the roots of this equation produce exactly three values of $z+ z^{-1}$, namely $2\cos \left(\frac{2^j \pi}{9}\right)$, for $j=1,2,3$, it follows that $y_j := 2\cos \left(\frac{2^j \pi}{9}\right)$ for $j=1,2,3$ are roots of the polynomial equation $y^3 -y + 1 = 0$. Since this is a cubic equation, these are precisely the roots of this equation, and the result follows similarly using sum of roots being $0$ here.$

Ah I see. Yeah I was mainly lost in why the roots are specifically 2cos(...) but then I realised |root|=1

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leehuan · Aug 18, 2016

Re: MATH1231/1241/1251 SOS Thread

Forgot how to puzzle things together today. Just part e) please

$p(z)=3z-z^3+5z^4-z^6$

$\text{Proven in d) }|3z-z^3+5z^4|<|z^6|\text{ provided }|z|>3$

$\text{Proven in c) at least one root has positive real part, and one negative real part.}$

$\text{e) Show that the roots }\alpha_j (1 \le j \le 6)\text{ satisfy }|a_j|\le 3$

InteGrand · Aug 18, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Forgot how to puzzle things together today. Just part e) please

$p(z)=3z-z^3+5z^4-z^6$

$\text{Proven in d) }|3z-z^3+5z^4|<|z^6|\text{ provided }|z|>3$

$\text{Proven in c) at least one root has positive real part, and one negative real part.}$

$\text{e) Show that the roots }\alpha_j (1 \le j \le 6)\text{ satisfy }|a_j|\le 3$

Suppose by way of contradiction there was a root "a" with |a| > 3.

Since a is a root, we have

a^6 = 3a - a^3 + 5a^4.

Taking modulus of both sides, |a^6| = |3a - a^3 + 5a^4|.

But this contradicts the result of part d).

Mr_Kap · Aug 21, 2016

Re: MATH1231/1241/1251 SOS Thread

im beyond math help. Dont even know how to integrate, or wtf a matrix is. Or how to diff.s. RIP.

RenegadeMx · Aug 21, 2016

Re: MATH1231/1241/1251 SOS Thread

Mr_Kap said:
im beyond math help. Dont even know how to integrate, or wtf a matrix is. Or how to diff.s. RIP.

well you can only blame yourself for continuing ur study methods from 1131, need to do a lot of practice in 12X1 especially for new material

Flop21 · Aug 21, 2016

Re: MATH1231/1241/1251 SOS Thread

Yeah 1131 started off soft, but 1231 is literally just continuing on so it starts off pretty hard.

How did they go from here to here:

Where did 1/3 come from?

Paradoxica · Aug 21, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Yeah 1131 started off soft, but 1231 is literally just continuing on so it starts off pretty hard.

How did they go from here to here:

Where did 1/3 come from?

Reverse Chain Rule

Drongoski · Aug 21, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Yeah 1131 started off soft, but 1231 is literally just continuing on so it starts off pretty hard.

How did they go from here to here:

Where did 1/3 come from?

$\int tan^2 x sec^2x dx = \int (tan x)^2 dtan x = \frac {1}{3} tan^3 x + C$

The integral is the very simple power integral of the form:

$\int w^2 dw = \frac {w^{2+1}}{2+1} + C = \frac {1}{3} w^3 + C$

leehuan · Aug 22, 2016

Re: MATH1231/1241/1251 SOS Thread

$\text{Prove that }n\textbf{v}=\underbrace{\textbf{v}+\textbf{v}+ \dots +\textbf{v}}_{n\text{-terms}}\text{ for all vector spaces.}$

The answers said to use induction. Is this really necessary?

InteGrand · Aug 22, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\text{Prove that }n\textbf{v}=\underbrace{\textbf{v}+\textbf{v}+ \dots +\textbf{v}}_{n\text{-terms}}\text{ for all vector spaces.}$

The answers said to use induction. Is this really necessary?

What other method do you have in mind? (Induction helps takes care of it nicely if you want to prove it straight from the axioms.)

$\noindent (The statement being asked to prove is not just true by definition like say the statement $nm = \underbrace{m+ m + \cdots +m}_{n \text{ times}}$ for positive integers $m$. The statement being asked to prove is saying that adding the vector $\mathbf{v}$ $n$ times is the same as multiplying the vector by the scalar $n$. This is proved with the help of the integer version of this, which is true by definition.)$

First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

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