First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

Paradoxica · Sep 13, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
How did this happen. The guy goes "and here's the trick" and writes the result. How can he just look at that LHS and do what he did? What's the trick I'm missing?

Reverse Product Rule

RenegadeMx · Sep 13, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
How did this happen. The guy goes "and here's the trick" and writes the result. How can he just look at that LHS and do what he did? What's the trick I'm missing?

differentiate top wrt x and u get those terms out, once u understand that in future DE's u can get ur int factor and skip a few steps

Memphis Depay · Sep 13, 2016

Re: MATH1231/1241/1251 SOS Thread

True

leehuan · Sep 17, 2016

Re: MATH1231/1241/1251 SOS Thread

$\text{Let's say I had a set of vectors }\{ \textbf{x}_1,\textbf{x}_2,\textbf{x}_3\}\text{ which form a basis in }\mathbb R^3$

$\\ \text{Then }\forall \, \textbf{b}\in \mathbb{R}^3\quad \exists !\, t_1,t_2,t_3 \in \mathbb{R}\text{ s.t.}\\ \textbf{b}=t_1\textbf{x}_1+t_2\textbf{x}_2+t_3 \textbf{x}_3$

$\text{Now staying in }\mathbb{R}^3\text{ suppose }\{\textbf{v}_1,\textbf{v}_2,\textbf{v}_3 \}\text{ do not form a basis because:}\\ \textbf{v}_3=\textbf{v}_1+\textbf{v}_2\text{ and thus no linear independence.}$

$\text{Discard }\textbf{v}_3\text{, then }\{\textbf{v}_1,\textbf{v}_2\}\text{ also forms a basis because it's obviously linearly independent.}$

The question
Why does {v1, v2} still form a spanning set for R³ again?

InteGrand · Sep 17, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\text{Let's say I had a set of vectors }\{ \textbf{x}_1,\textbf{x}_2,\textbf{x}_3\}\text{ which form a basis in }\mathbb R^3$

$\\ \text{Then }\forall \, \textbf{b}\in \mathbb{R}^3\quad \exists !\, t_1,t_2,t_3 \in \mathbb{R}\text{ s.t.}\\ \textbf{b}=t_1\textbf{x}_1+t_2\textbf{x}_2+t_3 \textbf{x}_3$

$\text{Now staying in }\mathbb{R}^3\text{ suppose }\{\textbf{v}_1,\textbf{v}_2,\textbf{v}_3 \}\text{ do not form a basis because:}\\ \textbf{v}_3=\textbf{v}_1+\textbf{v}_2\text{ and thus no linear independence.}$

$\text{Discard }\textbf{v}_3\text{, then }\{\textbf{v}_1,\textbf{v}_2\}\text{ also forms a basis because it's obviously linearly independent.}$

The question
Why does {v1, v2} still form a spanning set for R³ again?

It doesn't. We need at least three vectors to span R³.

$\noindent (If you were willing to believe it's a basis, you must have already believed it was spanning. Remember, a basis is a \emph{linearly independent spanning set}. So it's a set that's \emph{both} spanning and linearly independent. In this case, $\left\{\vec{v}_1, \vec{v}_2\right \}$ can't be a basis as it's not spanning. Also, you said it was `obviously' linearly independent, but nothing was said to us about $\vec{v}_1$ and $\vec{v}_2$, so it'd only be linearly independent if that was an assumption of the question or something. Because remember, a set with two vectors can be linearly dependent. (A non-empty set with any number of vectors can be linearly dependent in fact -- just take one of the elements to be the zero vector.))$

leehuan · Sep 17, 2016

Re: MATH1231/1241/1251 SOS Thread

Oh my fault on that one. Should've stated to not assume v1 and v2 were scalar multiples of each other.

I've lost bits of my question then. I'll need to find an example, sec

leehuan · Sep 17, 2016

Re: MATH1231/1241/1251 SOS Thread

Never mind, sorry I know what i did. It was asking for a basis for a span, not R³

Question types like this were what I was alluding to.

Flop21 · Sep 18, 2016

Re: MATH1231/1241/1251 SOS Thread

Why can't you just multiply top and bottom by x (bringing the top x out of the integral), thus cancelling the x on the top, and then do the tan sub? Rather than what they do? (obviously you get different answers tho)

He-Mann · Sep 18, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Why can't you just multiply top and bottom by x (bringing the top x out of the integral), thus cancelling the x on the top, and then do the tan sub? Rather than what they do? (obviously you get different answers tho)

I see no useful consequence of multiplying top and bottom by x. Also, you cannot bring the x out of the integral because it's a variable and not a constant.

Multiplying top and bottom by x

$\int [(x^2) / (x(x+1)^2 + 9x)] dx$

does nothing. Their method is already efficient. Doing integration by substitution is what you want to avoid.

Flop21 · Sep 18, 2016

Re: MATH1231/1241/1251 SOS Thread

He-Mann said:
I see no useful consequence of multiplying top and bottom by x. Also, you cannot bring the x out of the integral because it's a variable and not a constant.

Multiplying top and bottom by x

$\int [(x^2) / (x(x+1)^2 + 9x)] dx$

does nothing. Their method is already efficient. Doing integration by substitution is what you want to avoid.

oh thanks

Flop21 · Sep 18, 2016

Re: MATH1231/1241/1251 SOS Thread

What's actually happening though in the second line? Why/how does he put a (x+1) -1?

Flop21 said:

Paradoxica · Sep 18, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
What's actually happening though in the second line? Why/how does he put a (x+1) -1?

Add 0, multiply by 1. That's a key principle in algebraic manipulation. The trick is knowing the exact expression for 0 and 1, which only comes from experience or intuition.

RenegadeMx · Sep 18, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
What's actually happening though in the second line? Why/how does he put a (x+1) -1?

splits up the fraction which when u integrate gives ur ln and arctan term

He-Mann · Sep 18, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
What's actually happening though in the second line? Why/how does he put a (x+1) -1?

The third line explains why he does it. How he saw this is a result of foresight gain from experience/intuition as Paradoxica described.

Here's some insights:

1) The fraction is polynomial of degree 1 divided by polynomial of degree 2. When you have this (degree of numerator is one less than degree of denominator), then you think of a logarithms.

2) The denominator is sum of two squares, think of tan.

RenegadeMx · Sep 18, 2016

Re: MATH1231/1241/1251 SOS Thread

He-Mann said:
The third line explains why he does it. How he saw this is a result of foresight gain from experience/intuition as Paradoxica described.

Here's some insights:

1) The fraction is polynomial of degree 1 divided by polynomial of degree 2. When you have this (degree of numerator is one less than degree of denominator), then you think of a logarithms.

2) The denominator is sum of two squares, think of tan.

following on from this,

pro tip for 1231, for these types of integrals will mostly be logs or the occasional case arctan

leehuan · Sep 18, 2016

Re: MATH1231/1241/1251 SOS Thread

Is there a trick to use that's not induction?

$\text{Let }a\text{ and }b\text{ be positive numbers with }a>b$

$\\ \text{Let }a_1\text{ denote their AM }\frac{a+b}{2}\\ b_1\text{ denote their geometric mean }\sqrt{ab}$

$\\ \text{Keep going so that }a_{n+1}=\frac{a_n+b_n}{2}\text{ and }b_{n+1}=\sqrt{a_nb_n}\quad \forall n\in \mathbb Z^+$

$\text{a) Show that }a_n>a_{n+1}>b_{n+1}>b_n$

Paradoxica · Sep 18, 2016

MATH1231/1241/1251 SOS Thread

leehuan said:
Is there a trick to use that's not induction?

$\text{Let }a\text{ and }b\text{ be positive numbers with }a>b$

$\\ \text{Let }a_1\text{ denote their AM }\frac{a+b}{2}\\ b_1\text{ denote their geometric mean }\sqrt{ab}$

$\\ \text{Keep going so that }a_{n+1}=\frac{a_n+b_n}{2}\text{ and }b_{n+1}=\sqrt{a_nb_n}\quad \forall n\in \mathbb Z^+$

$\text{a) Show that }a_n>a_{n+1}>b_{n+1}>b_n$

1. The arithmetic mean of any two different real numbers strictly lies between them.
2. The geometric mean of any two different positive numbers strictly lies between them.
3. Use AM-GM for n=2.

Flop21 · Sep 19, 2016

Re: MATH1231/1241/1251 SOS Thread

dy^2/dt^2 + 4y = 0

how do i find the general solution of this equation?

I'm used to it in the form dy^2/dt^2 + dy/dt + 4y = 0.

InteGrand · Sep 19, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
dy^2/dt^2 + 4y = 0

how do i find the general solution of this equation?

I'm used to it in the form dy^2/dt^2 + dy/dt + 4y = 0.

$\noindent The ODE is $y'' = -4y$, which as we know from the Simple Harmonic Motion topic of HSC 3U maths has general solution $y = A\cos (2t) + B\sin (2t)$, for some arbitrary constants $A,B$.$

InteGrand · Sep 19, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
dy^2/dt^2 + 4y = 0

how do i find the general solution of this equation?

I'm used to it in the form dy^2/dt^2 + dy/dt + 4y = 0.

$\noindent If you want to do it using the characteristic equation rather than by inspection as done above, note that the characteristic equation is $\lambda^2 + 4 = 0$ (because the ODE has no $y'$ term, so there is no $\lambda$ term in the characteristic equation. Another way to think of it is that the ODE is $y'' + 0y' + 4y =0$, so the characteristic equation is $\lambda^2 + 0\lambda + 4=0$, which is just $\lambda^2 + 4 = 0$.).$

$\noindent So the characteristic roots are $\pm 2i$. The general solution can be written down from this.$

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