sigma notation (1 Viewer)

user18181818 · Oct 4, 2023

can someone please show me how to do (ii) without expanding, using sigma notation?

cossine · Oct 4, 2023

user18181818 said:
can someone please show me how to do (ii) without expanding, using sigma notation?

View attachment 40189

So you have been told to do (ii) using mathematical induction. Sigma notation is part of the question. I don't understand what trouble you are having please attempt the question first.

user18181818 · Oct 4, 2023

cossine said:
So you have been told to do (ii) using mathematical induction. Sigma notation is part of the question. I don't understand what trouble you are having please attempt the question first.

i got to the third RTP step where i said:

from there i know i have to split LHS to use the assumption of:

but i'm not sure how

cossine · Oct 4, 2023

user18181818 said:
i got to the third RTP step where i said:
View attachment 40192
from there i know i have to split LHS to use the assumption of:
View attachment 40193
but i'm not sure how

Assume true for n = k

=> sigma(j=3 -> k) (j-1)C2 = kC3

Required to prove
=> sigma(j=3->k+1) (j-1)C2 = (k+1)C3

Your notation is off. Basically n is the total number of terms and j variable is used for indexing. However you seem to have replace j with k.

Using the assumption we get

sigma(j=3->k) (j-1)C2 + kC2

= kC3 +kC2

= kC2 + KC3 (Not necessary step just rearranging)

= (k+1)C3 using part i

cossine · Oct 4, 2023

user18181818 said:
can someone please show me how to do (ii) without expanding, using sigma notation?

View attachment 40189

Basically for question what is being used is

sigma(i=1 ->n) a_i = sigma(i=1 ->(n-1) ) a_i + a_n. This result comes directly from how sigma notation is defined

SB257426 · Oct 4, 2023

Screen Shot 2023-10-04 at 4.13.03 pm.jpg

Assume true for n = k:

Screen Shot 2023-10-04 at 4.14.07 pm.jpg

Prove true for n = k+1, so we are required to prove:

Screen Shot 2023-10-04 at 4.18.32 pm.jpg

user18181818 · Oct 4, 2023

cheers guys!

SB257426 · Oct 4, 2023

user18181818 said:
cheers guys!

no worries !!! always happy to help

user18181818 · Oct 4, 2023

i have another question, i'm not sure how to prove (i) as all i know is that they're parallel (gradients are same). i also have no idea what (ii) is saying

hogzillaAnson · Oct 4, 2023

(ii) We have the line $ax + by = c \implies y = (-a/b) x +c/a$ . So the gradient of the line is $-a/b$ . Now if we look at the vector $\binom{-b}{a}$ treating $\vec{i}, \vec{j}$ as being in the positive $x, y$ directions respectively, then its gradient is rise/run, or $a/(-b) = -a/b$ which is the same gradient as the line. Therefore, it overlaps with the line.

(ii)

$\begin{align*} \text{Perpendicular distance} &= \left|\overrightarrow{OC}\right| =\left|\overrightarrow{OA} + \overrightarrow{AC}\right| \\ &= \left|\binom{m}{n} + \left(\text{projection of }\overrightarrow{AO}\text{ onto the direction vector of the line}\right) \right|\\ &= \left|\binom{m}{n} + \text{proj}_{\binom{-b}{a}} \overrightarrow{AO}\right| = \left|\binom{m}{n} + \text{proj}_{\binom{-b}{a}} \binom{-m}{-n} \right| \text{using previous part}\\ &= \left| \binom{m}{n} + \frac{\binom{-m}{-n} \cdot \binom{-b}{a}}{\binom{-b}{a} \cdot \binom{-b}{a}} \binom{-b}{a}\right| \\ &= \left| \binom{m}{n} + \frac{bm - an}{a^2 + b^2} \binom{-b}{a}\right| \\ &= \left| \frac{1}{a^2+b^2}\binom{m(a^2+b^2) - b^2m + anb}{n(a^2+b^2)+abm - a^2n}\right|\ \because \text{vector addition, combining over single denominator}\\ &= \left| \frac{1}{a^2+b^2}\binom{ma^2+\cancel{mb^2} - \cancel{mb^2} + anb}{\cancel{na^2}+nb^2+abm - \cancel{a^2n}}\right| = \left|\frac{1}{a^2+b^2}\binom{a(am+bn)}{b(am+bn)}\right|\end{align*}$

But $(m,n)$ is on the line $ax+by = c$ , therefore $am + bn = c$ . Substituting back in, we get

$\begin{align*} \left|\overrightarrow{OC}\right| &= \left|\frac{1}{a^2+b^2} \binom{ac}{bc}\right| \\ &=\frac{1}{a^2+b^2} \left|\binom{ac}{bc}\right| \text{noting that }\frac{1}{a^2+b^2} > 0\\ &= \frac{|c|}{a^2+b^2} \left|\binom{a}{b}\right| \\ &= \frac{|c|}{a^2+b^2} \sqrt{a^2+b^2} = \frac{|c|}{\sqrt{a^2+b^2}}\end{align*}$
as required.

user18181818 · Oct 4, 2023

hogzillaAnson said:
(ii) We have the line $ax + by = c \implies y = (-a/b) x +c/a$ . So the gradient of the line is $-a/b$ . Now if we look at the vector $\binom{-b}{a}$ treating $\vec{i}, \vec{j}$ as being in the positive $x, y$ directions respectively, then its gradient is rise/run, or $a/(-b) = -a/b$ which is the same gradient as the line. Therefore, it overlaps with the line.

(ii)

$\begin{align*} \text{Perpendicular distance} &= \left|\overrightarrow{OC}\right| =\left|\overrightarrow{OA} + \overrightarrow{AC}\right| \\ &= \left|\binom{m}{n} + \left(\text{projection of }\overrightarrow{AO}\text{ onto the direction vector of the line}\right) \right|\\ &= \left|\binom{m}{n} + \text{proj}_{\binom{-b}{a}} \overrightarrow{AO}\right| = \left|\binom{m}{n} + \text{proj}_{\binom{-b}{a}} \binom{-m}{-n} \right| \text{using previous part}\\ &= \left| \binom{m}{n} + \frac{\binom{-m}{-n} \cdot \binom{-b}{a}}{\binom{-b}{a} \cdot \binom{-b}{a}} \binom{-b}{a}\right| \\ &= \left| \binom{m}{n} + \frac{bm - an}{a^2 + b^2} \binom{-b}{a}\right| \\ &= \left| \frac{1}{a^2+b^2}\binom{m(a^2+b^2) - b^2m + anb}{n(a^2+b^2)+abm - a^2n}\right|\ \because \text{vector addition, combining over single denominator}\\ &= \left| \frac{1}{a^2+b^2}\binom{ma^2+\cancel{mb^2} - \cancel{mb^2} + anb}{\cancel{na^2}+nb^2+abm - \cancel{a^2n}}\right| = \left|\frac{1}{a^2+b^2}\binom{a(am+bn)}{b(am+bn)}\right|\end{align*}$

But $(m,n)$ is on the line $ax+by = c$ , therefore $am + bn = c$ . Substituting back in, we get

$\begin{align*} \left|\overrightarrow{OC}\right| &= \left|\frac{1}{a^2+b^2} \binom{ac}{bc}\right| \\ &=\frac{1}{a^2+b^2} \left|\binom{ac}{bc}\right| \text{noting that }\frac{1}{a^2+b^2} > 0\\ &= \frac{|c|}{a^2+b^2} \left|\binom{a}{b}\right| \\ &= \frac{|c|}{a^2+b^2} \sqrt{a^2+b^2} = \frac{|c|}{\sqrt{a^2+b^2}}\end{align*}$
as required.

thanks!!

SB257426 · Oct 4, 2023

hogzillaAnson said:
(ii) We have the line $ax + by = c \implies y = (-a/b) x +c/a$ . So the gradient of the line is $-a/b$ . Now if we look at the vector $\binom{-b}{a}$ treating $\vec{i}, \vec{j}$ as being in the positive $x, y$ directions respectively, then its gradient is rise/run, or $a/(-b) = -a/b$ which is the same gradient as the line. Therefore, it overlaps with the line.

(ii)

$\begin{align*} \text{Perpendicular distance} &= \left|\overrightarrow{OC}\right| =\left|\overrightarrow{OA} + \overrightarrow{AC}\right| \\ &= \left|\binom{m}{n} + \left(\text{projection of }\overrightarrow{AO}\text{ onto the direction vector of the line}\right) \right|\\ &= \left|\binom{m}{n} + \text{proj}_{\binom{-b}{a}} \overrightarrow{AO}\right| = \left|\binom{m}{n} + \text{proj}_{\binom{-b}{a}} \binom{-m}{-n} \right| \text{using previous part}\\ &= \left| \binom{m}{n} + \frac{\binom{-m}{-n} \cdot \binom{-b}{a}}{\binom{-b}{a} \cdot \binom{-b}{a}} \binom{-b}{a}\right| \\ &= \left| \binom{m}{n} + \frac{bm - an}{a^2 + b^2} \binom{-b}{a}\right| \\ &= \left| \frac{1}{a^2+b^2}\binom{m(a^2+b^2) - b^2m + anb}{n(a^2+b^2)+abm - a^2n}\right|\ \because \text{vector addition, combining over single denominator}\\ &= \left| \frac{1}{a^2+b^2}\binom{ma^2+\cancel{mb^2} - \cancel{mb^2} + anb}{\cancel{na^2}+nb^2+abm - \cancel{a^2n}}\right| = \left|\frac{1}{a^2+b^2}\binom{a(am+bn)}{b(am+bn)}\right|\end{align*}$

But $(m,n)$ is on the line $ax+by = c$ , therefore $am + bn = c$ . Substituting back in, we get

$\begin{align*} \left|\overrightarrow{OC}\right| &= \left|\frac{1}{a^2+b^2} \binom{ac}{bc}\right| \\ &=\frac{1}{a^2+b^2} \left|\binom{ac}{bc}\right| \text{noting that }\frac{1}{a^2+b^2} > 0\\ &= \frac{|c|}{a^2+b^2} \left|\binom{a}{b}\right| \\ &= \frac{|c|}{a^2+b^2} \sqrt{a^2+b^2} = \frac{|c|}{\sqrt{a^2+b^2}}\end{align*}$
as required.

Damn nice lol... much better than the way I did it

sigma notation (1 Viewer)

user18181818

Active Member

cossine

Well-Known Member

user18181818

Active Member

cossine

Well-Known Member

cossine

Well-Known Member

SB257426

Very Important User

user18181818

Active Member

SB257426

Very Important User

user18181818

Active Member

hogzillaAnson

Member

user18181818

Active Member

SB257426

Very Important User

Users Who Are Viewing This Thread (Users: 0, Guests: 1)