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

**leehuan** · 10 Aug 2016, 11:27 PM

$In each case use the substitution $v=\frac{dy}{dt}$ to solve$

$\\ $a) $\frac{d^2y}{dt^2}+2\frac{dy}{dt}=6\\ $b) $\frac{d^2y}{dt^2}+\omega^2 y = 0\qquad (\omega > 0)$

Part a) was quite ok

Part b)... what? How does that even work

**Paradoxica** · 10 Aug 2016, 11:46 PM

Originally Posted by leehuan

$In each case use the substitution $v=\frac{dy}{dt}$ to solve$

$\\ $a) $\frac{d^2y}{dt^2}+2\frac{dy}{dt}=6\\ $b) $\frac{d^2y}{dt^2}+\omega^2 y = 0\qquad (\omega > 0)$

Part a) was quite ok

Part b)... what? How does that even work

Once you apply the substitution, maybe you could differentiate both sides again?

Edit: nah that's useless.

**leehuan** · 14 Aug 2016, 9:54 AM

Just out of curiosity, what's an example of a first order linear ODE that's also exact?

Or does this not exist

**InteGrand** · 14 Aug 2016, 11:22 AM

$\noindent An exact ODE is of the form $I + J \frac{\mathrm{d}y}{\mathrm{d}x} = 0$, where $I$ and $J$ are functions of $x$ and $y$. If this is to be a linear ODE, the $J$ must be a function of $x$ alone, say $g(x)$ (no dependence on $y$). We would also need to make sure $I$ doesn't contain things like $y^2$ etc. For the equation to be linear, we would want the function $I$ to be of the form $\phi (x) + f(x)y$, where $\phi$ and $f$ are single variable functions. And for exactness, we would want $\frac{\partial I}{\partial y} = \frac{\partial J}{\partial x}$. So we would want $f(x) = g^\prime (x)$.$

$\noindent So we can come with many examples of ODE's that are both exact and first-order linear, as it suffices to just have $f$ be the derivative of $g$. So we can take the ODE to be things of the form $\boxed{\phi (x) + g^\prime (x)y + g(x) \frac{\mathrm{d}y}{\mathrm{d}x} = 0}$. E.g. The ODE with $\phi (x) = x^2 + 1$ and $g(x) = e^x$, with $g^\prime (x) = e^x$, is $x^2 + 1 + e^x y + e^x \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \Longleftrightarrow y^\prime + y = -e^{-x}\left(x^2 + 1\right)$. This is an example of a first-order ODE that is both exact and linear.$

**InteGrand** · 14 Aug 2016, 11:28 AM

Of course, an obvious and trivial example would be the ODE 0y + y' = 0, i.e. y' = 0.

**Flop21** · 15 Aug 2016, 5:24 PM

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

**InteGrand** · 15 Aug 2016, 5:32 PM

Originally Posted by Flop21

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** · 15 Aug 2016, 5:37 PM

Originally Posted by InteGrand

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

Oh yeah ha whoops thanks.

**Flop21** · 16 Aug 2016, 5:55 PM

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** · 16 Aug 2016, 6:21 PM

Originally Posted by Flop21

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** · 16 Aug 2016, 7:21 PM

How to integrate sec^4 x dx ?

**InteGrand** · 16 Aug 2016, 7:33 PM

Originally Posted by Flop21

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** · 17 Aug 2016, 10:44 PM

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** · 17 Aug 2016, 11:53 PM

Originally Posted by leehuan

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** · 18 Aug 2016, 12:34 AM

$\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** · 18 Aug 2016, 9:18 AM

Originally Posted by InteGrand

$\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

Sent from my iPhone using Tapatalk

**leehuan** · 18 Aug 2016, 12:39 PM

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** · 18 Aug 2016, 1:17 PM

Originally Posted by leehuan

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** · 21 Aug 2016, 5:33 PM

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

**RenegadeMx** · 21 Aug 2016, 8:01 PM

Originally Posted by Mr_Kap

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** · 21 Aug 2016, 8:33 PM

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** · 21 Aug 2016, 8:44 PM

Originally Posted by Flop21

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** · 21 Aug 2016, 8:46 PM

Originally Posted by Flop21

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** · 22 Aug 2016, 11:03 PM

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

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

**InteGrand** · 22 Aug 2016, 11:13 PM

Originally Posted by leehuan

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

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.)$

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