help-23

[wide hamlet]

im also

curious

[lapis vault]

Same

[compact ferry]

there and there

i see what you want to do but its not quite right

[queen parcel]

Your mistake lies here

[lean otter]

was i meant to half 6/5

[queen parcel]

Be careful of what you're doing, show a couple more steps to see exactly what's going on

[lean otter]

wait no u dont half

could u tell me where there i made the mistake pls

[upbeat swan]

You have an extra square

[compact ferry]

x shouldnt be squared inside the brackets

and the corrective term is wrong

[upbeat swan]

Yes

[compact ferry]

it should be -1/4

because (1/2)^2 = 1/4

[wide hamlet]

so where is the -

from

where is minus from

[lean otter]

OH YEAH

thanks alot

ok ill see how that goes

i make the lcm /20 right

guys now what

oh

19/20

@compact ferry

[safe radish]

@lean otter Has your question been resolved?

[verbal oriole]

@lean otter

react mate

[lean otter]

sfe

[flat frigate]

Mehdi_Moulati

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

@shut inlet how did it go from -5+24/20 to 19/12

[shut inlet]

i made a mistake , 20 instead of 12

[lean otter]

but the answer is 19/14

isnt it

or 12

[shut inlet]

$5x²+5x +6 = 5(x+\frac{1}{2})² + \frac{19}{4}$

[flat frigate]

Mehdi_Moulati

[lean otter]

oh yeah that was the answer

how tho

[flat frigate]

Mehdi_Moulati

[queen parcel]

@lean otter ^^

(Just in case you haven't seen it)

[safe radish]

@lean otter Has your question been resolved?

[wide hamlet]

ruhan

are u gonna

close this

@lean otter

[lean otter]

where does 30 come from
. 5

[queen parcel]

Think about what changes from top to the bottom

[wide hamlet]

the top times by 5

on each

[queen parcel]

Yes, but that idea won't work for all fractions

It's + 1 and - 1 pretty much

[wide hamlet]

we r gonna fail

exams

well

i am

imma start revising friday

have to go cousins house

tomorrow

@lean otter

😭

but i only have to revise science and maths

[lean otter]

I have to go somewhere tmrw too

[wide hamlet]

bro

its so annoying

[lean otter]

ima revise in their house

[wide hamlet]

i dont wanna go

no way im revising there

its gonna be chaos

[lean otter]

I'm not sure

[wide hamlet]

in every room

1/4 goes to 5/4

[lean otter]

if u X 5 won't everything be timsed by 5

[wide hamlet]

6/5 goes to 30/5

[lean otter]

ohhh it was all timsed by 5

I see now

[queen parcel]

Ye - it's all distributed

[lean otter]

I'm gna need way more practice on this

thanks for the help

do you suggest any apps

to work out completing the square

[queen parcel]

Just Google "completing the square practice problems"

[lean otter]

they don't show working out unfortunately

[lean otter]

I tend to forget this step and find the LCM and add immediately, any tips?

[queen parcel]

In that case you can look it up on YouTube

Just Google what you need, it's there for you

[lean otter]

alright thanks

thanks @queen parcel @shut inlet

ima close this

[queen parcel]

You can find the least common multiple of the denominators and add them before distributing

One is not better than the other

[lean otter]

oh

so it leads to same result right

distribution before or after

[queen parcel]

Correct

Again, you're not changing anything about what the expression is equal to

[lean otter]

alright thanks

.close

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[rugged dragon]

Can someone please help me with this kinematic question?

[plucky elk]

,rotate

[safe radish]

@rugged dragon Has your question been resolved?

[rugged dragon]

I don’t understand anything 😅

Like where to begin at all

Or how to tackle questions like this

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[queen parcel]

Why is this highlighted portion negative?

[queen parcel]

Does it have to do with F(r) vs F(r + dr)?

Given that dr is really small, can we not just say that the force for that line segment dr is a constant F(r)?

[lapis shadow]

does the force point along the direction of dr

if it does then it is F*dr

[queen parcel]

It’s just a vector field, we aren’t given any information as to the direction of F

Regardless of what direction it points, whether it’s toward or opposite dr will be captured by the dot product, so I’m unsure of why the - is there to begin with

Any opposite-pointing vectors will have a negative sign introduced via the dot product itself

[lapis shadow]

maybe coz it says work done "against" a force

they do mention that the direction of the integral along the curve must be specified

and if the direction is reversed u multiply again by -1

[queen parcel]

That’s for a general line integral, I’m pretty sure

I’m not sure it’s work done against a force

Regardless, it doesn’t impact the rest of the section

But thank you!

.close

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[hard canopy]

I got 15/32 by finding the probability of X, and the probability of Y and then multiplying X and Y together, however I did not get the answer. What did I do wrong?

[quasi bison]

confusing setup and confusing question

the table seems to show the fraction of novels sold to each gender per novel type...

and you seem to be asked for two numbers, yet the answer options only list one...?

i'd skip this problem and seek clarification as to what actually needs to be done @hard canopy

[hard canopy]

I’m pretty sure they are asking the probability of X and Y together, like the intersection of X and Y

Is that better?

[quasi bison]

so the probability that a randomly chosen novel was both a teenage novel sold to a man and a children's novel sold to a woman?

that sounds like zero to me.

as i understand it, the entries in the table correspond to disjoint categories of books sold...

[hard canopy]

yeah, you can’t be male and female at the same time

[quasi bison]

more importantly a novel cannot be both children's and teenage

[hard canopy]

Yeha I’ll just close this channel since the question is confusing, I’ll ask my teacher, thanks

.close

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[tawdry wing]

How do you do this:

[tawdry wing]

proof that 5^(n) +3 is divisible by 4 for all positive integers

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[lean otter]

whered i mess up

[lean otter]

this is the answer

^

[empty gyro]

Line 3

[lean otter]

is line 3 including the question

or line 3 of the workng

oh

what did i do wrong

ohh was it the halving

[empty gyro]

Expand your squared term and see if it's right

[lean otter]

alright

im not that good at expanding leme try

uhh i feel like im expanding wrong already

what do i do with the square

okay i got that

now i gotta minus

acc ima just redo it all 😭

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[bold aurora]

I intuitively know it’s C. How do I show it?

[upbeat swan]

you mean your friend told you the answer but didn't want to tell you the explanation

[compact ferry]

no its quite obvious intuitively that its C

[bold aurora]

No, A and B are obviously wrong. D has periodic functions which have no reason to converge

[upbeat swan]

ok

[timid pasture]

How is it intuitive that it is C ?

[compact ferry]

because its not A B or D

[bold aurora]

Check my previous message. It’s the process of elimination 😄

[timid pasture]

bruh

[bold aurora]

Positive sequence implies lim a_n = 0

[compact ferry]

and for C the only way its divergent is if a_n would approach a vertical asymptote at 0

[bold aurora]

Not sure if that’s any use

[compact ferry]

but tan(x) has no vertical asymptote at 0

[compact ferry]

probably is

[bold aurora]

Positive sequence approaching 0 implies a_n is also decreasing

[compact ferry]

doesnt have to be decreasing

[bold aurora]

Yep, nvm. It can be oscillating down too

[lean otter]

like for A, just take the sequence
1 1 1/2 1/4 1/8 1/16 ... and it contradicts, for B take the same sequence and it doesnt even converge and for D i think the same sequence works too, it gets really close to zero, so cos gets close to 1 and sin close to 0 and it doesnt converge

[bold aurora]

I’m kinda looking for something more rigorous

[compact ferry]

do you need to directly show C is correct>

[lean otter]

yeah, but for anyone wondering why its intuitive

[compact ferry]

because it seems showing A B and D are false is easier

[lean otter]

i think showing C is more interesting

[compact ferry]

for sure

[bold aurora]

I technically don’t but I honestly don’t feel good about that. What if a b and d werent so obv wrong?

[compact ferry]

true its good practice

[lean otter]

i think you can do it with comparison test

[compact ferry]

is there a nice iff test

[bold aurora]

tan is continuous, i think I could also show tan(a_n) is diminishing

[lean otter]

compare tan(a_n) with 2*a_n

since tan(x) < 2x for small enough x (i think?)

[bold aurora]

tanx < x for small enough x even

[lean otter]

dont think so

[compact ferry]

for negative x yes

[bold aurora]

Isn’t the maclaurin for tanx = x + o(x^3)?

[lean otter]

,w series tan(x)

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by rbit ✨#4860

[bold aurora]

For positive x too look 😛

[compact ferry]

thats 2x

[bold aurora]

,w plot x vs tanx

[flat frigate]

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Requested by Learath2#2281

[timid pasture]

I have a question, what if one of the terms of a_n is pi/2 won't that result into non convergence of tan(a_n)?

[lean otter]

it would make the whole series undefined

[timid pasture]

cause tan(pi/2) diverges

[compact ferry]

x/cos(x) is larger than tan(x) if that helps

for pos x

[compact ferry]

not defined at pi/2

[bold aurora]

Hm, so anyway. 2a_n does indeed also converge, and tan(a_n) is definitely < 2a_n

I think comparison test does work

Eh, at the very least I’m convinced. Thanks @compact ferry && @lean otter

.close

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[mossy pilot]

how do i do this

ping me if you know pls

part a

if a is the real part and it equals 1/2 how can a/a = 1/2
a/a should just be one

[mossy pilot]

Given that the real part of z is 1/2, find*

[cosmic grove]

The real part of z is what

[mossy pilot]

1/2

@cosmic grove

[cosmic grove]

Ok, put z in the form x+iy

[mossy pilot]

so :
a + ib = (a+2i)/(a-i) ?

[cosmic grove]

nah

[mossy pilot]

oh wdym

[cosmic grove]

Put it in the form that show us the real part and the imaginary part of z

x+iy

or a+ib, whatever its just variable

[mossy pilot]

oh

so like separate the fraction

?

a/a - 2i/i

[cosmic grove]

bro just destroyed algebra

[mossy pilot]

fax

help me pls

[cosmic grove]

$z = \frac{a+2i}{a-i}$

[flat frigate]

Ｈｅｒｅｌｓ

[cosmic grove]

we have on the denominator so we need to rationalize it

[mossy pilot]

oh i see

so time is by a + i

[cosmic grove]

$z = \frac{(a+2i)(a+i)}{a^2 + 1}$

[flat frigate]

Ｈｅｒｅｌｓ

[mossy pilot]

ohh wait

lemme do it now

i got root 5 for a

[static patio]

well u can just multiply z by (a+i) /(a+i), separate the i's and comparr it to 1/2

[cosmic grove]

$z = \frac{a^2 + 3ai - 2}{a^2 + 1} = \frac{a^2 - 2}{a^2 +1} + i \frac{3a}{a^2 +1}$

[flat frigate]

Ｈｅｒｅｌｓ

[static patio]

yeah that

[cosmic grove]

$\frac{a^2 -2}{a^2 +1} = \frac{1}{2}$

[flat frigate]

Ｈｅｒｅｌｓ

[cosmic grove]

$2a^2 - 4 = a^2 +1$

[flat frigate]

Ｈｅｒｅｌｓ

[cosmic grove]

a^2 = 5

hes right

[static patio]

yeah

[cosmic grove]

a = sqrt5

but not only sqrt5

[static patio]

and since a>0 u pick the positive ones

so its only sqrt(5)

[cosmic grove]

ah they said a>0

[static patio]

yeah

[mossy pilot]

Thank you guys for helping me understand🙏❤️

[static patio]

and the argument is atan2(im(z), re(z))

[cosmic grove]

nobody ask for it

[mossy pilot]

Part b

But yh thanks💯

[static patio]

np

[mossy pilot]

Ima close now have a good day/ evening🙏

.close

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[slow fern]

How union of two vector subspaces is a subspace of the vector space, iff one of the subspaces contain the other?

[marsh walrus]

can you be more clear about the assumptions

[sly field]

u want proof

[slow fern]

The proofs aren't given here. I don't understand these facts.

[slow fern]

Yes

[sly field]

https://math.stackexchange.com/questions/2325502/prove-that-the-union-of-two-subspaces-of-v-is-a-subspace-of-v-if-and-only-if

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[versed thorn]

how is 288 and 342 obtained?

[versed thorn]

How is 288 and 342 obtained as solutions

arcsin(sin324)

which gave me -36

and then i did 180 - -36

to give me 216

then i did 216 + 3

oh wait#

i got 288, 162 and 108

how do i get 324

[winged cloak]

2x = 324 + 360k or 2x = 180 - 324 + 360k

when you solve for x

[versed thorn]

yup got it

cheers mate

[winged cloak]

you'll get x = 162 + 180k or x = 108 + 180k

when you work with k = 0 and then k = 1 you'll get the 4 solutions

[versed thorn]

didnt work with k

i did

324 + 360

then that was 684

684/2

gave me 342

[winged cloak]

yeah that can work as well

[versed thorn]

.close

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[last imp]

how can I study the parity of a function?

[cosmic grove]

by calculating f(-x) ?

[last imp]

graphically madam

not by calculation

[clear blade]

check whether it is symmetric about the x-axis and/or the origin

[winged cloak]

if the graph is symmetrical according to the origin coordinate, your function is odd
if the graph is symmetrical according to the y axis, your function is even

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[cosmic dune]

Anybody going to use this channel?

[cosmic dune]

Micos?

[zealous hawk]

I going to move

[cosmic dune]

Well its mine now

Do u still want it

Hello ?

Hmm

Anyways

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[prime cobalt]

hi!

[prime cobalt]

how would I do part (iii)

[hearty egret]

its too difficult

[safe radish]

@prime cobalt Has your question been resolved?

[prime cobalt]

ohh it’s difficult for me too 😭

[hearty egret]

is it a college exercise ?

[prime cobalt]

yes preparation paper to prepare for A levels

not college yet

entering university next year

[hearty egret]

good luck

[prime cobalt]

ty!

[toxic stratus]

this is legendres formula if you want to look it up

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[prime cobalt]

ty!

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[prime cobalt]

managed to get it :D

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[frosty fox]

sup

[frosty fox]

@quasi bison you’ve helped me yesterday so I have pinged you to see if you’re able to help me with this today

first part

[shut inlet]

just calculate the limit @frosty fox

1-2/(k+3) tend to 1 when k tend to infinity

so :
lim a_k when k tend to inf is ln(1) which is 0

[vast obsidian]

*Using that log is continuous

[frosty fox]

thx

.close

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[frosty fox]

.

what

[frosty fox]

nice

ok posting my question

[frank glen]

[frosty fox]

alright

[peak estuary]

polynomial division, then quadratic formula

[frosty fox]

how exactly?

[prime cobalt]

hi!

[frosty fox]

sup

[peak estuary]

divide the polynomial by (x+5/2)

you'll get a quadratic

[frosty fox]

hm

[prime cobalt]

since you know 2z+5 is a factor you can use 8z^3 +125 = (2z+5)(Az^2 +Bz + C)

C is 25 here

[static shore]

You can multiply z by a third root of unity to get another solution

[prime cobalt]

then you can compare coefficient of z^2 maybe

[frosty fox]

uh

[static shore]

And the square of said root of unity

[frosty fox]

whats a unity?

[static shore]

A unity is 1

[peak estuary]

"roots of unity" means the solutions to x^n=1

[frosty fox]

i didnt learn that method

[frosty fox]

also whats this

[static shore]

Polynomial long division

[peak estuary]

well a bit of a nonstandard way to write it down tho

[frosty fox]

how you know C is 25??

[peak estuary]

multiply out the RHS

compare coefficients

[frosty fox]

oh dang

[peak estuary]

if we want another few overkill methods, you can also view 8z^3+125 as a geometric sum

[frosty fox]

I compare coefficient of z^2?

[frosty fox]

uh lmao nahh im good

[prime cobalt]

oh haha sorry!

[prime cobalt]

yes

[frosty fox]

how?

[prime cobalt]

2B + 20=0

B=-10

[frosty fox]

oh ok nvm

i see what you did

[prime cobalt]

yay!

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[keen mural]

really quick question: how do I factor (10.sqrt(2))^2?

[mortal pawn]

?

[frank glen]

Well it's already factored

[keen mural]

I mean solve

sorry

[mortal pawn]

(10sqrt(2))^2

Do you want to simplify it

[frank glen]

Expand then

[keen mural]

yes

[frank glen]

Remember

[mortal pawn]

You can distribute squaring if it’s multiplication

[frank glen]

$(ab)^c = a^c b^c$

[mortal pawn]

exactly

[flat frigate]

VulcanOne

[keen mural]

ohh so it's 200?

[mortal pawn]

He

Yes

[keen mural]

alright, it didn't matter anyways because I got lost in the problem and realized I didn't have to figure that hypotenuse to solve it

regardless, thank you

.close

[safe radish]

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[keen mural]

.reopen

[safe radish]

✅

[keen mural]

here's the problem solved, by the way

I think it's correct

I had to figure out the value for the shaded areas, in case it wasn't clear

[mortal pawn]

That’s just half the original thing

[mortal pawn]

Wait are you sure

Nvm

You got it

But yeah the shaded region is half the entire thing

[keen mural]

don't get misled by my drawing, the length of BS is 10.sqrt(2)cm

[mortal pawn]

Oh

Ok

I’m dumb

Wait a second

I think you got it

I’m brain shutting

[keen mural]

don't brain shut for me, it's not worth it

but yeah I solved it again and it gave me the same result

so I think I got it right

now I can close this

.close

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[vague prism]

why is m+1 choose 2 the same as m(m+1) / 2

[plucky elk]

Use the formula for binomial coefficient

[mental lodge]

U choose first item from m+1

Then second from only m

So m*(m+1)

But u can choose second firstly, then first secondly

So u divide it all on 2

Cmon

Classy handshake problem

Google handshake problem

[mortal pawn]

Don’t say cmon

Not everyone is as gifted as you

[vague prism]

i dont really understand tbh

[mortal pawn]

There’s two ways you can think about this

first of all the formula for n choose 2

[vague prism]

Thats the way i learned it so far

[mortal pawn]

n!/((n-2)!(2)!)

Right

[mental lodge]

I can explain it for u but tommorow

[mortal pawn]

Think about n!/(n-2)!

[vague prism]

yea ik u wanted to go to sleep thats why i asked in pub

[mortal pawn]

Everything below (n-2) cancels out

[mental lodge]

Thats formally good way but imo wrong way to think

[mortal pawn]

Because it both in the numerator and denominator

[mental lodge]

Yeah but

[mortal pawn]

I said there were two ways before

This is the first way

[vague prism]

Understandable so far

[mortal pawn]

And no it’s not the wrong way to think

[mortal pawn]

So then you just left with n(n-1)/2!

Which is just n(n-1)/2

[vague prism]

wait i need to reproduce that step

[mortal pawn]

It’s not wrong to think this way

[mortal pawn]

So right now you have n!/((n-2)!(2)!)

[vague prism]

So when having

n!

(n-2)!

it will become

n(n-1)

  2!

?

[mortal pawn]

Which is n(n-1)(n-2)(n-3)…/((n-2)(n-3)…(2!))

[mortal pawn]

No

It will just become n(n-1)

I forgot to include the 2!

That’s also there

[vague prism]

i forgot to do the faculty after 2

[mortal pawn]

Factorial

[vague prism]

english problems

hard to translate some words

[mental lodge]

It is cause it kills developing special kind of intuition destinated to task bkit probablity

[mortal pawn]

No

[mortal pawn]

It’s cool I just didn’t want you thinking it was faculty 😂

[vague prism]

😄

ok so to recap this

thats what i tried to reproduce here

but i get 1 as denominator due to shorten. how do you get the 2! ?

@mortal pawn

[mortal pawn]

In the definition of choose

It’s n!/(n-2)!2!

That’s where the two comes from

[vague prism]

oh ok so i just forgot the 2! from your start term

yea ok i get what you mean

[mortal pawn]

Now if you want to

You can think of it this way

N choose 2 is like choosing 2 items from a set of n things

With no preference for order

n(n-1) is the number of ways choosing 2 items with order mattering

How many ways can you order two things

Only 2 ways

[vague prism]

e.g. if i have 2 thins ( A and B) i can only sort it AB or BA ?

you mean that ?

[mortal pawn]

Yep

[vague prism]

👍

[mortal pawn]

So if I only want one of them

I divide by two

[vague prism]

so i need to devide n(n-1) / 2! by 2 ?

[mortal pawn]

Just n(n-1)

Read what I said

[vague prism]

i got one question to the stuff above

[mortal pawn]

Yeah

[mental lodge]

It kills

[vague prism]

Why is N Choose 2 no preference for order
but
N(N-1) does have a preference

[mortal pawn]

N choose 2 just is that’s the definition of choose

[vague prism]

ok ok

[mortal pawn]

N(n-1) is like

First I’m going to choose 1 of n items, I have n choices

Then I will choose 1 of the remaining n-1 items, I have n-1 choices

The thing is

That you can choose the saim pair twice

For example I can choose A first and B second or B first and A second

[vague prism]

ok and to connect this to your previous explanation, if i just want one of n items, i will divide my possibilities n(n-1) by 2 right?

oh damn i finally understood it

after writing it all down and put it together it made sense

big thanks @mortal pawn

.close

[safe radish]

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Channel closed

Closed due to the original message being deleted

[mortal pawn]

.close

Np

[safe radish]

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[cloud elm]

is this for all n greater or equal to 1

[mortal sandal]

yeah

[split ether]

Yes

[cloud elm]

how do i prove n=1

[mortal sandal]

plug it in

[cloud elm]

does 0/12 prove it as being able to divide by 12

[mortal sandal]

definition of divisibility

[cloud elm]

ye im confused by these notations

[mortal sandal]

a is divisible by b if there exists an integer q such that a=bq

[cloud elm]

oh i understand

.close

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[wind fog]

\begin{tikzpicture}
\draw[ blue, very thick] (-4,-4) -- (4,-4);
\draw[ blue, very thick] (-2,4) -- (-4,-4);
\draw[ blue, very thick] (-2,4) -- (6,4);
\draw[ blue, very thick] (6,4) -- (4,-4);
\draw[ blue, very thick] (4,-4) -- (2,4);
\draw[ blue, very thick] (-2,4) -- (0,-4);
\draw[ blue, very thick] (-3,0) -- (6,4);
\draw[ blue, very thick] (-4,-4) -- (5,0);
\node[ scale = 0.8pt] (a) at (-4,-4.5){A};
\node[ scale = 0.8pt] (a) at (0,-4.5){E};
\node[ scale = 0.8pt] (a) at (4,-4.5){B};
\node[ scale = 0.8pt] (a) at (5.5,0){F};
\node[ scale = 0.8pt] (a) at (6,4.5){C};
\node[ scale = 0.8pt] (a) at (-2,4.5){D};
\node[ scale = 0.8pt] (a) at (2,4.5){G};
\node[ scale = 0.8pt] (a) at (-3.5,0){H};
\node[ scale = 0.8pt] (a) at (-1,0.6){S};
\node[ scale = 0.8pt] (a) at (1.5,1.4){R};
\node[ scale = 0.8pt] (a) at (-0.2,-1.4){P};
\node[ scale = 0.8pt] (a) at (3,-0.8){Q};
\end{tikzpicture}

[flat frigate]

Anik

[wind fog]

In ABCD parallelogram E, F, G, and H are the midpoints of
AB, BC, CD, and DA respectively. Area of ABCD is 30 sq.
unit. Find the area of PQRS.

[safe radish]

@wind fog Has your question been resolved?

[safe radish]

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[prime cobalt]

hi!

[prime cobalt]

this might be a dumb question haha but for part (a) would you use horizontal line test?

[solar hazel]

probably not if it’s asking you to show it

[compact ferry]

strictly increasing/decreasing is enough

[prime cobalt]

oh okay ty!

[solar hazel]

probably worth noting that’s only enough to show it’s injective

well actually that might be all it expects cus it doesn’t mention anything about a codomain in the question

[toxic stratus]

inverting a function without a codomain

[solar hazel]

[clear blade]

well the codomain has to be its image for it to have an inverse, right?

[safe radish]

@prime cobalt Has your question been resolved?

[solar hazel]

it doesn’t say that’s the case but it’s whatevs

i just meant like

if i am asked to show something has an inverse

[solar hazel]

i would think that’s something i need to show

[toxic stratus]

showing bijection without a codomain

[safe radish]

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[wind fog]

How to find the $dx$ for any $d\theta$ from a equation of a circle? How to do that with an irregular shaped closed curve like eclipse, parabola, hyperbola etc.

[flat frigate]

Anik

[safe radish]

@wind fog Has your question been resolved?

[stoic dune]

dx and dθ don't make sense in any this context.

[vestal sedge]

guessing a tiny change in the x position and the angle value theta but still 0 sense

[arctic locust]

Perhaps if you write $ x = r\cos(\theta)$ you can find smthg

[safe radish]

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[dark ginkgo]

Is the formula for the number of self similar objects in a sierpinsky pyramid 4^n?

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@dark ginkgo Has your question been resolved?

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[prisma urchin]

How to solve this problem 18

[safe radish]

@prisma urchin Has your question been resolved?

[stray socket]

What got blacked out

You can use this rule

[marsh walrus]

can i ask from here, what the steps are

sorry @prisma urchin

[prisma urchin]

I did word blacked

The word is "is"

Probably

I don't know how to translate it

Fist log given and 2nd find out maybe hhe

[stray socket]

What steps

It's just one of the identities you learn, or at least I learned

I haven't gotten around proving it, but it probably ain't hard to prove

It just becomes dln(c)/(bln(a))

[safe radish]

@prisma urchin Has your question been resolved?

[frank glen]

Change of base will prove it lol

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[tame oxide]

what does it mean by the "the nature of" in this question

[tidal frost]

the nature of the question is whether the answer is going to be real, irrational, or complex im pretty sure

[hardy lion]

yes its the count of real zeros

[tame oxide]

oh

right that makes sense

thanks

[tidal frost]

thank you

[tame oxide]

.close

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[half flame]

Just doing through some lectures at the moment and my LA really needs some work, can someone help me understand the orthonormal basis, I have examples and just would like to verify before I go ahead with this example of orthogonal projection in R3

[junior smelt]

The bit in red?

[half flame]

Yes!

[junior smelt]

Taking the dot product and it’s generated norm, you can see that e1.e2=0

[half flame]

First off can I ask you some questions about the orthonomal basis?

[junior smelt]

And that the norm of each element is 1

[junior smelt]

[half flame]

hehe okay so

As you know my understanding of LA is weak and my understanding of the basis stuff is even worse but in this chapter I was introduced to this definition of "orthonormal basis"

Here is the definition I have been provided

First off, the inner product of the uniterary vector (I think that's what they call it) what is it?

I am a bit confused on the inner product of vector spaces in general since we have used the inner product for other stuff such as polynomials and I haven't see wnat the <> actually encases

However, looking online apparently you're supposed to just dot it ?

[junior smelt]

Not sure what you mean there, could you elaborate please?

[junior smelt]

Well there are many different inner products, the dot product being one of them

[half flame]

So I am basically a bit confused on the definition, what is < > I guess

Looking at the definitions in my workbook, an inner product on a vector space is a function that takes each ordered pair of elements V to a real number

bit doozy looking at that

[junior smelt]

But you can define an inner product over a vector space in many different ways, they just have three properties they need to satisfy

[half flame]

the main get away is just, an inner product takes some ting to a real number denoted by <u, v>

in the photo i provided before, it just says <ei, ej} = little delta

or is that epislon, i havent really been taught the greek symbols yet

like they havent really defined what the operation is right?

They just say that if the result is little delta which is either (1 or 0) then it is a orthonormal basis?

Maybe my understand of vector space and inner product is weak as well

[junior smelt]

Well yep, the inner product is just a map from your vector space to the scalar field you’re based on

Any map on a vector space that satisfies those is an inner product (though some looking weirder than others)

[half flame]

god damn

[junior smelt]

(When you’re in the reals, the complex conjugation has no effect btw)

[half flame]

LAMOOOO WHAT IS THATTTT

[junior smelt]

But basically, given an inner product

[half flame]

They have missed out so much of the detail LMAOOO

give me a second

[junior smelt]

You can talk about orthogonality and orthonormality

[half flame]

Perhaps you could elaborate on definition 4.1

I'm assuming the two cdots are operations?

Namely addition and some sort of multiplication?

[junior smelt]

Basically those dots are placeholders really for elements of the vector space

[half flame]

Side note

[junior smelt]

Takes two vectors and spits out a scalar

[half flame]

That definition stuff has nothing to do with like fields in general right

Like the groups and fields

[junior smelt]

Well I guess, your vector space is defined over a field, but other than that…? In this case we only really care about the real numbers and complex numbers (ignore any other field for now!)

[half flame]

Okay I concur

Okay so the inner product space just has to have those three hold

I'm assuming 4.1 is some sort of expansion property

4.2 is maybe additive properties

I have no clue waht 4.3 is but maybe that has something to do with the cauchy inequality? nvm

[junior smelt]

Yep, linear in your first component

(Though note that some people choose to be linear in the second component)

[junior smelt]

Conjugate symmetry

(Over the reals, that’s just symmetry as the conjugate of a real number is just itself)

[half flame]

I see I see

[junior smelt]

The inner product of an element and itself is a nonnegative real number, and only zero when it’s the same element

[junior smelt]

That property then helps with defining a norm from that

[half flame]

These are the "identities" provided from my textbook

I am assuming some of these are just an over simplification of your versions

To make it more digestable for students?

[junior smelt]

In particular, (I2) and (I3) make up my (4.1), (I1) is my (4.2) (but for real numbers), and (I4) and (I5) make up my (4.3)

[half flame]

I see

[junior smelt]

Generally I think people find it easier when you break it down like yours did

[half flame]

haha yeah

[junior smelt]

For me I’m more used to just jumping in and doing the linear combinations right away, but that’s with the time I’ve been doing this I guess

[half flame]

So just to clarify with the "naming" of these identities

(1): Inverse? (2): Expansion? (3): Scalar multiplcation? (4): idk? (5): idk

Like is there a document or resource online to search for, for the names

After doing some maths I think it's starting to get important to learn them since I have been seeing these rules since algebra

Although i think they are called axioms not rules

[junior smelt]

1 is symmetry, 2 and 3 linearity, forgot the name of the last two

Let me check that for you

[junior smelt]

Positive definiteness

Ah, to-may-toe, to-mah-toh 🍅

Same things, different names 🤷‍♀️

[half flame]

haha yeah doesnt matter

was just curious i guess

Anyways why were we talking about this again

[junior smelt]

To introduce orthogonality?

[half flame]

oh yeah inner product spaces

and orthogonality

se

[junior smelt]

Yep, so with an inner product defined, you have a norm automatically generated from that inner product, so you can talk about distances

• you can then define orthogonality with respect to that inner product

[half flame]

wait actually

It is important to really get the basics of like

vector spaces etc to go further

ah nvm ill just read it later

I think my main question was like what <> was and now I think you've cleared up the fact that <> just denotes inner product

Now, it's obvious that we have certain axioms etc but lets go back to the definition

[junior smelt]

Yep, that’s pretty much it, some inner product you’d have defined

[half flame]

I am basically confused because <ei, ej> = little delta

If the inner product of the ei and ej is equal to either 0 or 1 then it is a orthnormal basis... but like

what is <ei, ej> actually? Does it have an operation or? What are we doing two these two vectors

[junior smelt]

“The inner product between any two vectors [in that set] is zero if they’re different, or 1 if they’re the same”

[half flame]

oh i should of read the definition properly

[junior smelt]

Just the inner product between e_i and e_j

[half flame]

But what is the inner product? From definition its a function which satisifes the 3 properties?

But what is the function doing I guess

[junior smelt]

Basically that, to be fair I’m not the greatest at explaining the exact concept of the inner product (just some of the properties it has tbh)

[half flame]

Hahah that is fine

It's just

[junior smelt]

Like e.g. the fact you get a norm out of it and the orthogonality

[half flame]

Here is a definition I found online and nowhere does the definition from my book mention the dot product of two vectors any where

but from what I got from this definition is that

given a set with vectors u1, u2, ..., un

The dot product of any of the vectors (if theyre different as you mentioned) should give 0 while if the dot product of any of the vectors within the set where they are the same should give 1

Correct me if I am wrong

[junior smelt]

That’s for orthonormality

[half flame]

Oh shit we want the orthonormal basis

i am trolling

LOL

[junior smelt]

(For orthogonality you just need the inner product of any two different elements to be zero)

[half flame]

Maybe we consider an example

[junior smelt]

Oh yeah this is the Gram Schmidt stuff you’re doing now isn’t it

[half flame]

LMAOOOO YEAHHH

I want to nail this shit down to the bone

[junior smelt]

Just remembered you said

[half flame]

If we consider a set {(1, 0, 0), (0, 1, 0), (0, 0, 1)}

Well this was given in the lecture among other examples, its said that this set is an orthogonal basis

How should we go about showing it

[junior smelt]

Matter of fact, it’s orthonormal

You have the definition of the dot product somewhere right?

(Just cause I’m on mobile atm so a tiny bit harder for me to fetch stuff)

[half flame]

Yeah um

I dont want to pull it up but the dot product of two vectors is like

well in this case

(1, 0, 0) . (0, 1, 0) = 0

(1x0) + (0x1) + (0x0) = 0

Now I just "assumed" that the dot product is commutative (I think is how you say it)

So I just need to consider the following combinations

[half flame]

(1, 0, 0) . (0, 1, 0) = 0
(1, 0, 0) . (0, 0, 1) = 0
(0, 1, 0) . (0, 0, 1) = 0

I should only need to consider these three to show that this set is indeed, orthonormal

[junior smelt]

(+ it’s easy to show symmetry of the dot product)

[junior smelt]

That’s orthogonality

[half flame]

whattt

[junior smelt]

For the “normal”-ness, check the norm of each vector

Take the dot product of each of them and itself, if you get me?

[junior smelt]

Remember, orthogonality is just that the inner product of different things is zero

Orthonormality requires both orthogonality and also that the norm of each thing is 1

You’ve done the orthogonality bit so far, if you see me?

[half flame]

to check the normal ness, check the norm of each vector

um give me sec

uhhhhh

like the sqrt(v . v) stuff?

ohhhhh

||v|| = sqrt(<v,v>)

this??/

oops

[junior smelt]

Yep that 😂

Those || are how you make spoilers

[half flame]

uh

but he norm of 0 is just 0

[junior smelt]

$\norm{v} = \sqrt{v\cdot v}$

[half flame]

ohhh

[flat frigate]

chartbit

[half flame]

the norm of the inner product

if the inner product is 0 then the norm is just sqrt(0) which is still 0 so then it still follows with the definition

need to reread everything you said with orthonormal and orthogonality later as well lol

[junior smelt]

Well, the dot product of e.g. (1,0,0).(1,0,0) is just 1, no?

Similar for the others

[half flame]

oh so besides the fact I need to check for the combinations of others

i need to check 3 combinations where the vector is the dot product of itself to check if the norm of the dot product is 1 as well?

[junior smelt]

Yep, you need to check each of the ones with themselves (but that’s easy once you spot it in this case)

[half flame]

I see I see

So to check the orthogonal set of vectors (orthonormal) we can just consider the combinations where we get all the vectors and dot them together then find the norm, they all must be equal to 0 in this case, the second case is find all the combinations where we dot the vector itself and find the norm, they must be equal to 1

If they do not satisfy these two conditions, the set is not orthonormal...

[junior smelt]

Yep, if you fail for one of them, you’re not orthonormal

For orthogonality, you just need the first part

[half flame]

Perfect

Glad I understand that now

[junior smelt]

Another thing is that any orthogonal set of non-zero vectors is automatically linearly independent

[junior smelt]

(In particular, an orthonormal set is linearly independent)

But that’s for another time I guess

[half flame]

I'm going to skip the decomposition and orthgonal project theorem at the moment and do this example

I mean the proof for it

[junior smelt]

Yep, do the rest later

[half flame]

So I am assuming to use this orthogonal projection theorem, we would need the set to be an orthonormal basis

[junior smelt]

Hmmm, let me check that, I don’t think you do, unless I’m mixed up?

[half flame]

uhhh i am not sure either LOL

[junior smelt]

Actually I think you do, at least from my quick browsing haha

Pain to try and look through notes on mobile when they go through so much

[half flame]

Yeah so i think the main thing first is that

I would like to check the set if it is indeed an orthonormal basis

But we are given a span not a set

[junior smelt]

Well, it’s the span of a set

[half flame]

So can just do what we did before

[junior smelt]

Take the vectors of that set and work with those

[half flame]

Check for orthogonality first then we check the norm

sec

[junior smelt]

(Which here, orthogonality is easy!)

Heyyy @shut inlet

Can you believe I haven’t gone to bed yet

[half flame]

oops sorry

[shut inlet]

WHAT
I just woke up 2 hours ago @junior smelt
😆

[half flame]

why is orthogonality easy here

i am assuming its got something to do with the zeros

[junior smelt]

Yep basically that

[half flame]

Like

(0, 0, 1)

any second vector

at the third entry if it is 0 then it will be 0 anyways

or (1, 0, 1)

as long as the second vector is just (0, x, 0) then it is easy to compute ?

[junior smelt]

Basically, that’s the idea, you have (?, 0, ?) and (0, ?, 0)

[half flame]

so we keep saying orthogonality

[junior smelt]

Take the dot product, you see what happens haha

[half flame]

Is that the same as orthogonal?

[junior smelt]

Yep yep

[half flame]

Ah okay I have checked it is indeed an orthogonal basis

thats it for today

projection and gramm shitmz tomorrow

Goodnight and thank you @junior smelt !

.close

[safe radish]

Channel closed

Closed by @half flame

Use .reopen if this was a mistake.

[junior smelt]

Best of luck! 💪

[safe radish]

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[solemn kestrel]

i know it's either b or c

[frosty fox]

it’s d??

[solid vector]

so

the only way

well

the best way

is put ur pen on a point and rotate it yourslef

[solemn kestrel]

either b or d*

[solemn kestrel]

yeah and i rotated it all the way until b would be on the top and it seems like it would be either b or d

[solid vector]

so its

it cant be B

bc if you rotate a positive shape on a graph 180

it becomes fully negative

lemme draw it

for you

@solemn kestrel do you know

rotational rules

[solemn kestrel]

it's d

[solid vector]

yes

do another one

and show me

[solemn kestrel]

ok

this on is a lil different since we count the distance from the point but idk which point

[solid vector]

so you count it from A1

ghd the shape on the right

[solemn kestrel]

wdym a1

[solid vector]

is being translated

[solemn kestrel]

yes

[solid vector]

see the shape on the left

is labelled A2, B2,C2

[solemn kestrel]

oh you mean like A'

[solid vector]

yeah

[solemn kestrel]

i thought we count the pre image

[solid vector]

in an exam it would be A1 and !2

[solemn kestrel]

to the image

[solid vector]

wait one is being translated

does it say

[solemn kestrel]

?

[solid vector]

which one is the orignal shape

[solemn kestrel]

the one on the right

[solid vector]

so yes 2 to the left

is correct]

[solemn kestrel]

ah ok

nice

[solid vector]

give another

preferably rotational

[digital zephyr]

I’m p sure this is actually 6 to the left

[solemn kestrel]

oh?

[digital zephyr]

You wanna check how to move A to A’

[solid vector]

oh yeah

no hes right

im being dumb

[digital zephyr]

Moving only 2 to the left just takes A to D’

[solemn kestrel]

ohhhh

[solid vector]

no its 6 to the left not 2

yeah mbmb

wasnt looking properly

[solemn kestrel]

i forgot thats hgow you do it

does this count as rotational

[solid vector]

no

wait

am i being dumb

[tall bough]

It's not rotational

[solid vector]

yeah thought so

[solemn kestrel]

im sure it's either c or d

[solid vector]

@solemn kestrel whats changed about it

you're right

[solemn kestrel]

not much really

[tall bough]

Why would it be vertical?

[solid vector]

but has it gone up or down

or just sideways

[solemn kestrel]

the translation is sideways

[solid vector]

and is vertical or horizontal sideways

[solemn kestrel]

horizontal

[solid vector]

so

ur answer

[solemn kestrel]

horizontal translation

i'm slow

[solid vector]

yes

correct

well done

alr give another

[solemn kestrel]

i've been stuck on this one for some time idk what to do if anybody who knows how to do it can pls help

[solid vector]

so its similarities

it says the trees are mathimatically similiar

[solemn kestrel]

i don't understand

[solid vector]

does it say they are similar

[solemn kestrel]

it just says two trees are planted next to eachother

[vestal bear]

You have idea about how shadow forms?

[solemn kestrel]

not rlly

[vestal bear]

It forms when an opaque object comes between the path of light

This much you must be knowing

The angle at which light rays fall on tree 1 should be same as on tree 2

[solemn kestrel]

the clsoest that would be is 3.6

[vestal bear]

Through the given data you can relate their height

[vestal bear]

What's with this closest

It's exact

Theoretically

[solemn kestrel]

oh