help-38

[granite light]

if you have a plane and a line not on that plane, then direct sum is the 3d space in which they both lie

[urban copper]

okay

[granite light]

But what i'm saying is S has to be intersection of H and T^perp

$\mathbb{S} = \mathbb{H} \cap \mathbb{T}^\perp$

[solid kiln]

CyclicTree

[granite light]

because if it wasn't then S direct sum something will contain some vectors not in T^\perp

and can't be equal

[urban copper]

why

[granite light]

well imagine in 3d instead of 4d

H is some plane

T is some line not on the plane

(and not perpendicular to the plane, it turns out)

T^\perp is some other plane

we are asked to find a line S on H

such that when you sum it with some other line given by the angle bracket thing

it gives you precisely T^\perp

so S has to be a line on H

but it also has to be a line on T^\perp

because otherwise no matter what you choose your a to be, S (+) stuff won't be T^\perp

so then $\mathbb{S}$ is given by $x_1 - 2x_3 - x_4 = 0$ and $x_1 + x_2 - x_3 + x_4 = 0$

[solid kiln]

CyclicTree

[granite light]

so basically you then have to find a that will mean <(-1, 1, 1, a)> is on T^\perp, but isn't on S

(or prove that such a doesn't exist)

[urban copper]

how do I do that

[granite light]

angle bracket stuff gives a bunch of vectors:

$t \begin{pmatrix}
-1 \ 1 \ 1 \ a
\end{pmatrix}$

[solid kiln]

CyclicTree

[granite light]

(parametrized in t)

and for it to lie on T^\perp, this must be true for every t

so
-t + t - t + ta = 0

[granite light]

plugging ^ into the other one and expanding dot product

[urban copper]

wdym parametrized wait

wait

[granite light]

well the angle bracket stuff

at the bottom of original problem

is a subspace given by scalar multiples of the vector

[urban copper]

ywah

scalar multiples

[granite light]

so they are of the form ^ for some t

[urban copper]

yeah

[granite light]

so a = 1 makes sure that angle bracket stuff lies on T^\perp

now we need to show that this is not the same as S

because if it was the same as S

then going back to 3d example, we will be trying to construct the 2d plane from 2 lines that are actually the same line

or rather that angle bracket stuff is not a subspace of S

or that this vector (-1, 1, 1, 1) is linearly independent from some basis of S

if we could find a basis of S that would be sweet

i think

nah just show that (-1, 1, 1, 1) doesn't satisfy conditions required to be on S

(1)(-1) + (-2)(1) + (-1)(1) ≠ 0
-1 -2 - 1 ≠ 0 checks out

so yeah i believe a = 1 is the answer then?

[urban copper]

I didnt unserstood how did u proved that (-1,1,1,1) doesnt satisfies conditions for in S

wtf

[granite light]

it's not in H

see condition for being in H in the problem statement

[urban copper]

okay

[granite light]

i want to make it clear that my knowledge of linear algebra is lacking, so it's entirely possible that there is a more clear way to present this argument.

[urban copper]

its ok I am still trying to unserstand if this is enough proof or not

a = 1 satisfy the conditions for = 0

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@urban copper Has your question been resolved?

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[ripe valley]

what is the indicator function for the condition $x \geq 0$

[nimble stone]

no spaces at the start and end

[ripe valley]

oh rt single $

[solid kiln]

rak³en

[dusty hound]

f(x) = 0 if x < 0
f(x) = 1 if x >= 0
A step function basically

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@ripe valley Has your question been resolved?

[marble wharf]

do you want to hear the name heaviside?

or what exactly do you want to hear

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[dapper basin]

Is this correct?

[true kestrel]

what is the question?

[dapper basin]

f(x) = 1/x ; g(x)=1-2x

[true kestrel]

and we have to find?

fof(x) and fog(x)?

[dapper basin]

fof(x)

[wind cloak]

it is not correct

[dapper basin]

oh

[wind cloak]

messed up in third step from end

[dapper basin]

can u explain?

[split chasm]

what are you doing/applying here:

[dapper basin]

I think because that's how you always do it, I don't really know, I'm not that good when dealing with fractions in algebra

[split chasm]

firstly, ideally you'd make that top line longer here and/or put () around that 1/x,
like

if just rewriting in different notation, you'd still be dividing by that component in parentheses, which is 1/x

(not x/1)

if applying the inverse relation between multiplication and division,
you'd jump straight to the multiplication of the inverse

so you should end up with
$$\frac 11 \divisionsymbol \frac 1x$$
or
$$\frac 11 \times \frac x1$$

[solid kiln]

ℝαμΩℕωⅤ

[split chasm]

(i included the 1s in the denom since that seems to be what you prefer)

[dapper basin]

Wait

I just realized

Yeah

💀

[dapper basin]

This would be correct if it was 1/x right?

[dapper basin]

Hmm okay

I should solve this easily now

Thank you

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[potent sapphire]

I need help with this question, it's in my physics textbook, but I don't think any physics is required in this question! really appreciate any help!

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@potent sapphire Has your question been resolved?

[potent sapphire]

<@&286206848099549185>

[boreal wharf]

What do you need help for

[potent sapphire]

I'm not sure on how to begin this question

[boreal wharf]

Same

[potent sapphire]

so do i just continue waiting??

[boreal wharf]

Wait I’m cook

[potent sapphire]

bc I did x and x - 1 into the tan (angle) = o/a

but idk if thats right

and solved for x there

where 0 is the height of th mountain

and a is x and x - 1

[boreal wharf]

It’s wrong

[potent sapphire]

ahh 😭 thats why I need help

[tulip sapphire]

these are the unkowns

[potent sapphire]

yeah thats what I did

[tulip sapphire]

h being the height of the mountain

[potent sapphire]

but my answer didnt match the one at the back, maybe I did something wrong 😢

[tulip sapphire]

you have the following system of equations you need to solve:
$tan(12 degrees) = \frac{h}{x} \$
$tan(14 degrees) = \frac{h}{x-1}$

[solid kiln]

woomy

[tulip sapphire]

oh

send the progress youve made

maybe u just made a mistake in the calculations

[potent sapphire]

alr give me a sec

i need to transfer the photo from my ipad to laptop

give me a sec

nevermind i've found out my problem, i moved the terms in my head and did some errors

thank you for helping !!

[tulip sapphire]

np :)

[tulip sapphire]

happens to the best of us hahaha

[potent sapphire]

i always get too confident, gotta slow down

[tulip sapphire]

i dont blame ya, after thinking on how to solve a problem when you finally find out how you tend to do the rest quickly and without thinking

[potent sapphire]

that's exactly it loll

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[true kestrel]

I have got that f(x)-g(x) is strictly dec function

[gritty wyvern]

2x-ln(x^2+2x+1) = x^2

x^2-2x=-ln(x^2+2x+1)

e^-(x^2-2x) = x^2+2x+1

1/e^(x^2-2x) = x^2+2x+1

[true kestrel]

huh

[gritty wyvern]

hmm idk

im just doing algebra

[nova spire]

so as soon as you find the x such that f(x) = g(x)

you're good to go

(you don't have to look very far)

[true kestrel]

it's at 0

[nova spire]

yep

[gritty wyvern]

oh

[true kestrel]

but what if the eqn was complex to solbe

what would I have done?

[nova spire]

there's always a way with those exercises

if there's no nice form for the solution

you can always maybe use W lambert function

[true kestrel]

but after x=0, what to do abt interval?

since it's dec function

[nova spire]

you said f(x)-g(x) is decreasing

[true kestrel]

f(infinite) < f(0)

[nova spire]

and f(0)-g(0) = 0

so you have a decreasing function

you want to know when it's <= 0

and you already know when it's = 0

[true kestrel]

uh

[nova spire]

so... is it below 0 after? before?

[true kestrel]

after

[nova spire]

got your answer

[true kestrel]

oh

0H

OH

cool

idk I am messing up things by overthinking

thanks tho

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[true kestrel]

.reopen

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✅

[true kestrel]

ayo with that method

nvm

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[wraith hinge]

I) Consider a particle moving with constant velocity x(ihat)+y(jhat) along the line x=y.Dexribe v in polar coordinates.

[wraith hinge]

!status

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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wraith hinge]

4

V=x[sin(theta)+cos(theta)]rhat+x[sintheta-costheta]thetahat

Wtf

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[wraith hinge]

lcm(a,b) * gcd(a,b) = a*b

[wraith hinge]

how do i factorize a number using this

lcm(9,15) * gcd(9,15) = 9*15 = 135

i dont understand what the book is saying

its saying lcm(a,b) * gcd(a,b) = a*b is useful for prime factorization of large numbers

[violet gust]

can you provide the context?

coz I can just see that to find the gcd and lcm you actually need the original factors a and b

[wraith hinge]

here

[violet gust]

like a ss of the book you mentioned

[wraith hinge]

if m = lcm(a,b) then m must contain all prime factors occuring in the decomposition of a or of b, and each to the highest occuring power.
For larger numbers this method is unsuitable on account of the prime factorization An expedient is first to determine the gcd by euclids algorithm and then to utilize the relation lcm(a,b) * gcd(a,b) = a*b

This is what it says

[violet gust]

but to use euclids alg, dont we already need to know a and b?

which are ofc the factors of a*b, the same number we wanna factorize

[wraith hinge]

sorry

so i have two large numbers

i use euclids alg to get the gcd

[violet gust]

so you just wanna know why lcm(a,b) * gcd(a,b) = a*b?

[wraith hinge]

to find the lcm what is the method

i am wondering like how this works lcm(a,b) * gcd(a,b) = a*b

to get lcm

[violet gust]

coz you already know a and b

so a*b is easy

and gcd is from euclids alg

[wraith hinge]

then divide

[violet gust]

so we have 3 out of 4 things

[wraith hinge]

am i suppose to divide a*b by gcd

to get lcm

[violet gust]

yes

[wraith hinge]

ok wow

let me test

will u check if i was right

[violet gust]

With a name like perelman and stuff like prime factorization of large numbers, I was sure this was referencing some high level problem in cryptography and I just couldnt see it

[wraith hinge]

lol

[violet gust]

lol

[wraith hinge]

my bad

[violet gust]

yeah so the reason this works is your lcm has all the repeat factors counted only once instead of twice as they should occur in a*b

and those repeat factors are counted once again in the gcd

so gcd * lcm accounts for all the factors of a and b

making the two products equal

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@wraith hinge Has your question been resolved?

[wraith hinge]

what is meant that the lcm of a and b must contain all prime factors occuring in the decomposition of a or of b, and each to the highest occuring power

what does each to the highest occuring power mean

[violet gust]

so say a = 20

[wraith hinge]

lcm(9,15)
3^2 and 3*5

3^2 *5 = lcm

[violet gust]

then a has factors 4*5, but 4 is not prime

so we can make it 2*2*5

or 2^2 times 5

so the highest power would be 2^2

[violet gust]

yes like this. The highest power of the factor from either of the numbers

[wraith hinge]

why do they say each to the highest occuring power tho

contain all prime factors in each a and b to the highest occuring power

[violet gust]

coz 9 and 15 multiply to form 135, which i divisible by 3^3

[wraith hinge]

so either a or b whichever has the higher occuring power

[violet gust]

but 3^3 is not a factor to either

the highst power occurs is only 2

[wraith hinge]

how do you know 135 is divisable by 3^3

did u do that without a calculator

[nova spire]

135 = 9*15

= 9 * 3 * 5

27*5

[wraith hinge]

oh okay

thank you

okay thank you both

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[sly basalt]

a box has 10% defective items. one item is selecting at random. what is the probability that item is defective

[sly basalt]

total items is not given

will it be simply 1/10

[maiden hare]

Pretty much by definition, yes

[sly basalt]

what definition

is my reasoning correct?
total= x
defective= 10% of x= x/10

P= defective/total
= (x/10)/(x)
=1/10

[maiden hare]

I mean that's fine but "a box has 10% defective items" directly tells you that defective/total = 1/10

[sly basalt]

actually I have a different question, I was just trying to get my concept right for a part of question

[maiden hare]

If you have a set of things where a proportion x of them are blue things and the rest are red things, and then you pick one of these things at random, you will pick a blue thing with a probability x

[sly basalt]

so I can apply my thinking to this whole question

[maiden hare]

Try it

[sly basalt]

they use this binomial sort of thing. this thing also came to my mind, but they both probabilities were not equal so I didn't apply it. but they did

and how i thought the answer would be (10/100)²×(90/100)⁸ is incorrect

[maiden hare]

Yeah because it's not asking for something like "probability that only the first 2 are defective"

It could be the first 2, or the last 2, or the 3rd and the 6th, or...

[sly basalt]

so i have to do arrangements?

actually from arrangements, it'll come 10C2. but the way they showed in solution, using P and Q which is binomial distribution

[maiden hare]

It is 10C2

[sly basalt]

this solution is incorrect, right? but it gives right answer.. binomial cannot be applied like this

[maiden hare]

10C2 = 45

,calc 45 * (0.1)^2 * (0.9)^8

[solid kiln]

Result:

0.1937102445

[maiden hare]

It's correct

[sly basalt]

not talking about being numerically correct

[maiden hare]

... then what?

[sly basalt]

the way they used binomial

i studied that both events should have same probability

[maiden hare]

I just see 10C2 * (0.1)^2 * (0.9)^8

[sly basalt]

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[sick plinth]

Can anyone help me with a maths question

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[brazen zephyr]

Can u share the question?

[brazen zephyr]

Try to apply Pythagoras theorem to find the side BC. As u have diagonal length in terms of ratio and AB is same as DC. Area = DE×BC

[knotty locust]

stick to #help-14

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[dreamy spire]

For c(part 1) can I just roate the triangle to a right angle like this

Ans key is here

Also I'm not sure abt part c(ii) they say angle of elevation to D but isint this angle of elevation to A so er not sure what's wrong here tho this method is correct in ans key

[fleet bear]

D is vertically above A

A, B and C lie in the same plane(on the same level)

[dreamy spire]

Er yea but isint is elevation to a in that case

[fleet bear]

What makes you think that?

[dreamy spire]

Cause it's going up to a?

[fleet bear]

It can't be going up to A since they are all at the same level

Like ABC is a triangle, imagine that resting on a flat surface. D is some point above A, up in the air

So the line BC also rests on the flat surface

[dreamy spire]

I'm not sure I understand? But if it's this drawing wouldn't the angle of elevation from the horizontal line to D angle of elevation be 0?

[fleet bear]

That drawing doesn't look right

[dreamy spire]

Oh

Could you draw out the correct one for me?

I'm not rlly sure how to change my drawing here

[fleet bear]

Not sure if I have a pen or paper. Gimme a sec

[dreamy spire]

Ohh here I understand this

Sure

OHHH

[dreamy spire]

So point d here is not on flat ground but instead D is above a so it wld be angle of elevation to D?

[fleet bear]

Ah great 👍

[dreamy spire]

Ohhh

[fleet bear]

Yep!

[dreamy spire]

I thought d was on the flat ground too

Ahh I undetnow

Understand now

[dreamy spire]

Wait but this

Am I right to just shift the triangle to right angle to solve it like that for part c(I)

[fleet bear]

Yep

Gotta notice that h (the height) gives the greatest angle as well

[dreamy spire]

Ohhh no wonder

Alright I understand now 👍 tks a lot!

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[fleet bear]

You're welcome!

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[red sentinel]

How would you simplify this?

[vagrant prism]

step by step, start with everything under the power of two

recall that $a^{-b} = \frac{1}{a^b}$

[solid kiln]

artemetra

[red sentinel]

yeah it would be (1/2)^-6 right?

[frail heron]

Next, apply this

@red sentinel

[red sentinel]

i got it! thank you guys

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[grizzled flower]

how do I get whats in the second screenshot from the last step that I did in the first screenshot?

[split chasm]

try expanding out those cos(2x) and sin(2x) with double angle identity

[grizzled flower]

I'm looking at the formulas but I do not know how

since cos2x and cosx dont have the same x and same goes for sin

and also the first one has a 2 in front of it

[split chasm]

what formulas are you looking at

[cosmic meadow]

Focus on cos(2x) first and use double angle identity

[grizzled flower]

you meant these?

[split chasm]

yes

[grizzled flower]

I've never used them but I'm guessing I expant cos2x to 2cos^2 x - 1?

[cosmic meadow]

The second one

[grizzled flower]

what do I do with this cosx here?

[cosmic meadow]

No

Cos(2x) = 1-sin^2x

[grizzled flower]

I know, I expanded the left one

but I dont know what to do with the right one

[cosmic meadow]

The double angle for that part is not for cosx

Is for sin(2x)

[grizzled flower]

but I will get 2sinxcosx which still leaves me with 2 cosx-es

ill try

just a sec

[cosmic meadow]

Dont forget u have another cosx

[grizzled flower]

I'm stuck now

[cosmic meadow]

Show it

[cosmic meadow]

Do not expand 2(1-2sin^2x)sinx

Second part would be -cosx2sinxcosx

[grizzled flower]

how?

[cosmic meadow]

Sin(2x) = 2sin(x)cosx

You have -cosxsin(2x) so

-cosxsin(2x)cosx

[grizzled flower]

could you solve this because I am very confused

[cosmic meadow]

Or -cos^2x * 2sinx

[grizzled flower]

I dont know if you mean the left or right part for everything you say

[cosmic meadow]

Leta atart with the left one

[grizzled flower]

ok

[cosmic meadow]

You wrote 2(1-2sin^2x)sinx

Correct?

[grizzled flower]

yes

[cosmic meadow]

Ok now the second

U have -2sin(2x)cosx

Use double angle on sin(2x)

What do u get

[grizzled flower]

2sinxcosx

[cosmic meadow]

And now notice that sin(2x) is negative and multiplied by another cosx

[grizzled flower]

then I have -2*2sinx cosx *cosx

[cosmic meadow]

Ok i see the mistake

There is no -2

It is only -sin(2x)cosx

[grizzled flower]

how

[cosmic meadow]

Read in your writing

It is -sin(2x)cosx+1

[grizzled flower]

if you expand sin2x you should get 2sinxcosx

[cosmic meadow]

You mixed your mistake witht he previous step

Yes

And thats it

[grizzled flower]

oh so its -2sinxcosxcosx*

?*

[cosmic meadow]

-2sinxcosxcosx

Yes

[grizzled flower]

ok

[cosmic meadow]

Or -2sinxcos^2x

[grizzled flower]

yes I got that here

[cosmic meadow]

Now notice in the left side

[grizzled flower]

but idk what to do from this step forward

[cosmic meadow]

U can write 2(1-2sin^2x)sinx like 2sinx(1-2sin^2x)

[grizzled flower]

yes

[cosmic meadow]

What common factor

U see in left and right

[grizzled flower]

2sinx*

?*

[cosmic meadow]

Yes

Factor out

[grizzled flower]

?

[cosmic meadow]

Why do u write

+cos^2x at the end?

[grizzled flower]

because we got it from before

[cosmic meadow]

You just created that cos^2x out of nowhere

No

[grizzled flower]

oh sorry only + 1

[cosmic meadow]

Ok now solve inside parenthesis

Rewrite 2sin^2x as sin^2x + sin^2x

Remember thay 2sinx is multiplied by -1

[grizzled flower]

ok now Im confused again

[cosmic meadow]

Where

[grizzled flower]

for this part

[cosmic meadow]

Ok

No lets work only inside

$1 -2 \sin^2 x - \cos^2 x$

Sait

Wait a typo

[solid kiln]

Samuel

[cosmic meadow]

Now, work here

[grizzled flower]

oh so $sin^{2}x-2sin^{2}x$?

[solid kiln]

n1ck

[cosmic meadow]

That works

Now u only multiply and u finished

-2sin^3x+1 = 1-2sin^3x

[grizzled flower]

ooooooh ok

If the teacher puts this on the test I'm standing up and leaving the room as I get it in my hands

thank you for the help kind stranger : ) have a nice rest of your day

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[south snow]

are there any other way to solve this without substitution method?

[south snow]

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[vague walrus]

need some help understanding of triangle in equality

[vagrant prism]

what's your concern exactly?

[vague walrus]

when is |z1-z2|<=|z1+z2| equal

i dont think it should be possible

but it is when i

[vagrant prism]

within complex numbers?

[vague walrus]

subsitute z2 with -z2

[vagrant prism]

when z2 = 0

[vague walrus]

ye

not sure about bottom right condition

we get that one above by replacing z2 with -z2

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[rotund grove]

hi

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[faint fulcrum]

Hello, someone help me to calculate this limit pls?

[nova spire]

hi, try to focus on this first

(hint: factor the numerator)

[faint fulcrum]

ah

[faint fulcrum]

I have to factor the numerator, rationalize the denominator so that there is a factor of x-2 left in the denominator and simplify

[nova spire]

yeah, but I mean you can immediately simplify (x-2)/sqrt(x-2) without rationalizing

[faint fulcrum]

thankss

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[dreamy spire]

Errr I sorta understand part b and don't at the same time for qn 13

Oh wait nvm

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[pearl lotus]

Is there a Easier method of doing this other than taking log on both sides and then differentiating

[opal breach]

yes, you can just use chain rule + implicit differentiation

[pearl lotus]

That's even more complex than my method

[opal breach]

it's relative xD

i can do it in one step

you start with the outside functions then move in towards what is inside

for example the left hand side, you say we differentiate e^something which drops the derivative of that something besides the e, the derivative of that something is cosine what is inside of it times the derivative of what is inside which is easy to do in your head

you end up with
$(2x+2yy') \cdot \cos(x^2+y^2) \cdot e^{\sin{(x^2+y^2)}}$

[solid kiln]

Mohamed Mohsen

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@pearl lotus Has your question been resolved?

[pearl lotus]

Hmmm is there like a trick or some shortcut that can simplify the expression for us to easily take the derivative of

Cause both the methods take atleast 5-10 min to perform and in exam you can get max 2 min for a que

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[pearl lotus]

I know lol a did it with both your and mine method

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[flat flower]

The complex number z = a+ 2i has a real part a that satisfies 0 < a < 1. The number z^3 can be written as z^3 = b + ci with b, c ∈ R. What do you know about the signs of b and c?

[flat flower]

calculator isnt allowed

idk how to start

[wraith hinge]

(a+ 2i)(a+ 2i) = z^2

multiply those

[flat flower]

ok

ok i got it thx

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[slim bison]

I am having issues understanding why two events are independent if $P(A\cap B) = P(A)P(B)$. What is the reasoning behind this definition?

[solid kiln]

Pen

[ionic pendant]

if they are dependent, then the rule is P(A and B) = P(A) • P(B given A), so if P(B given A) = P(B), event A does not affect the probability of B

[slim bison]

I am still learning the English terminology, what would "P(B given A)" be? P(B | A)?

[ionic pendant]

yes

[slim bison]

Since P(B|A) describes the relative probability, if they are independent, then the relative otucome is the same as P(B). So, P(AnB) = P(B | A)P(A) from the definition of conditional probability, yields P(AnB) = P(B)P(A)?

[cold dune]

this seems to be correct, just to add what cloud said:
you can also look at it using sets, for example by throwing 2 fair coins, the set of all possible outcomes is going to be {H, T} x {H, T}, which is basically increasing the size of the set containing all possibilities from n(A) or n(B) to n(A)xn(B)
you can see how the probability also gets multiplied here

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@slim bison Has your question been resolved?

[slim bison]

Ah, so it's like the multiplicaiton principle being used - since they are independent, the possible outcomes in both events are unaffected, then P(A n B) is simply all possible combinations

I.e you chose n amount of outcomes from A and j amount of outcomes of B

[cold dune]

yea pretty much

[slim bison]

You showing me concrete examples of the sets helped me, I kept messing up A n B as "all elements in A and B" when it really means, in this case, "probability of an outcome in A and B" ( P(A n B) )

Thank you both, Cloud and Nima!

[cold dune]

haha yea I had that struggle for some time too, it's just another perspective to look at, using the formula works pretty well most times

cheers!

[slim bison]

I am for sure struggling : D

Indeed, cheers!

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[frozen goblet]

i honestly have no idea where to start with this

[wraith hinge]

do u know how to graph lnx n e^x

[frozen goblet]

yeah

[wraith hinge]

whats the issue

[frozen goblet]

i js dont get where the +1 in e^x+1 comes in

[wraith hinge]

its just moving e^x by 1 to the left

[frozen goblet]

oh

damn i feel dumb alright thank you

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[devout swallow]

can i have help with i and ii

[worldly wing]

what do you need to do here?

[plush tide]

True or false ig

[hushed stirrup]

have u tried anyting

[devout swallow]

yeah sorry true or false

for i)

i mean we know cos^x+sin^2x=1

so i dont think this is true

but thats all ive got

for ii the only thing i can think to do is turn i into e^ipi/2

[hushed stirrup]

if u can find at least one specific value for theta such that the given equality does not hold, u can say its false

so e^(i*theta) is costheta+isintheta

what if u wanted cos(i) and sin(i) to appear?

[devout swallow]

so i could put isini/cosi

[hushed stirrup]

yeah but to do that u must first know what cosi and sini are

[devout swallow]

ah ok

[hushed stirrup]

use this formla to try to find them

[devout swallow]

without putting the i in

[hushed stirrup]

that formula contains terms of the form costheta and sintheta for any theta

u can choose what u want the theta to be

[devout swallow]

well in the question it says its i

no sorry

it will be pi/2

im a bit confused

with what to do with the formula

[hushed stirrup]

so u understand that u must somehow find cosi and sini right?

[devout swallow]

yup

so i have e^ix=csox+sinx

so do i put i in for x

[hushed stirrup]

yes u can try doing that

wat does it give u?

[devout swallow]

e^i^2

=cosi+isini

e^-1

e^-1=cosi+isini

[hushed stirrup]

yes

try to find another expression involving cosi and sini using that formula

[devout swallow]

divide both sides by cosi

[hushed stirrup]

so u can solve for cosi and sini

like it was a system of eqns

[worldly wing]

you could also directly use $\tan z=\frac{e^{iz}-e^{-iz}}{2i(e^{iz}+e^{-iz})}$

[devout swallow]

awh ok

[solid kiln]

kheerii

[devout swallow]

how do i get another equation for that??

[hushed stirrup]

whts another number u can put for x, so thar u can still get terms with cosi and sini

another number that is not i

but very similsr to it

[devout swallow]

sqrt-1

[hushed stirrup]

no thats i

[devout swallow]

im confused

[hushed stirrup]

okay well try -i

[devout swallow]

ohhhhhh

e=cosi-isini

[hushed stirrup]

now its just a system of eqns

[devout swallow]

cosi=(e^-1+e)/2

isini=(e^-1-e)/2

so i tani=(1/e^2)/(1+e^2)

so it will always be a negative real number

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[dim falcon]

I really have no idea

[dim falcon]

Can someone help me please with this equation?
I don't know how to solve it

[sour mortar]

take x common

from 2x^2 + x

let 2x+1=t
x(t) = 3 sqrt(t)
=> 2x+1 =! 0
when cancelling sqrt(t)
x sqrt(2x+1)=3, solve as quadratic

[dim falcon]

where did you get 2x+1 =!0 from ?

[ashen mountain]

x(t) -3sqrt(t) =0
Sqrt(t)(xsqrt(t) -3) = 0

So
Sqrt(t) = 0
And
Xsqrt(t) - 3 = 0

Both are solutions right

[dim falcon]

ohhhhhhhhhhhh i get it knowww thank you guys so muchhh

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[fathom tapir]

Help with no.2

I got through phase 1 but phase 2 my answer doesn't line up with the answer key

Business math teacher emphasized the use of our calculator for this section

<@&286206848099549185>

[carmine mauve]

What's "semi anually"?

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@fathom tapir Has your question been resolved?

[fathom tapir]

Semi annually in this context is the the amount times it compounds per year

So 2 interest compounding per year

[verbal stream]

so how much is Monty adding every half year

[fathom tapir]

For the first 5 years monthly payments of 300

For the next 10 years it's $2000 per quarter

[verbal stream]

so how much is that per half year

[fathom tapir]

I'm not too sure what the point of that question is, for the first 5 years it's 6 per half year for the next 10 years it's 2 per half year

The first phase I figured out that the FV is 21968.43

I use that for phase 2 as my PV

Changing the parameters to the new set however my answer doesn't seem to match with the phase 2 answer which confuses me and the answer key using the formula method which we didn't learn in class

I'm hoping that someone that does know the formula method can enlighten me :)

[carmine mauve]

the interest is being added every half year, so you need to know how much is being added every half year

[fathom tapir]

Does that mean for I/Y it's no longer 8%? Would it mean a different number even tho for phase it was 8% with the same compound per half a year?

This is using the BA2+ calculator method where I just have to input

N, PV, FV, I/Y, P/Y, C/Y, PMT

[carmine mauve]

it means, you need to first work out how much is being paid in per half-year, for every half-year

*sorry, per half-year. not per month.

Let's give an example. Suppose I pay into my savings account $50 per month, and the bank pays 5% interest per year. How much do I have after 2 years?
Well, $50 per month is $600 per year.
After the first year, 5% interest is $30 extra, leaving me with $630 in the bank.
After the second year, I had $630, paid in another $600 myself, and the bank then calculates the 5% on this which is $1291.5 ($61.5 extra).

So, for you, step 1 is working out how much Morty pays into TFSA every half-year.

[fathom tapir]

N: 2*12= 24
PV:0
FV:?
I/Y:5%
C/Y =1 P/Y = 12
PMT = 50

Using the formula
M = P[r(1+r)^n]/[(1+r)^n - 1]
Where:

M = monthly payment
P = principal loan amount
r = monthly interest rate (annual interest rate divided by 12)
n = number of payments (total number of months)

After 2 years

FV = 1,257.94

The answer differs as the interest rate is distributed throughout the year rather than adding everything before multiplying the interest

[fathom tapir]

. leave

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[slim plinth]

given f : R -> R continuous and periodic

[slim plinth]

1. prove that f has a maximum and minimum

so intuitively it makes sense, but im not sure how to write a formal proof

[hallow kite]

what does extreme value theorem tell you about the maxima and minima of a continuous function on a closed interval?

[slim plinth]

what is the extreme value theorem

[solid kiln]

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Requested by janniku

[hallow kite]

and technically i feel like this is enough justification, but also if your function is periodic then it repeats itself after some fixed interval

[slim plinth]

oh weierstrass theorem?

[hallow kite]

meaning if you know it attains a max and min on a closed interval due to continuity, then the same can be concluded about the entire function since the closed interval repeats due to periodicity

[hallow kite]

i wasn't taught it to be called that way but if that's the same thing then yes

[slim plinth]

right

so to formalize this

we can assume it repeats every a

and say the interval is [b, b + a]

and then apply weierstrass

that would be enough no?

[hallow kite]

right, so you know that there's a max and min somewhere in [b, b+a]

which you'd justify by f being cont

[slim plinth]

yeah

[hallow kite]

then since f is periodic, this repeats for every adjacent interval of length a

like [b+a, b+2a]

etc

covering the entire domain

[slim plinth]

right yeah so we can say for every z in Z, [b + za, b + 2za] and apply the theorem on that

right?

wait no

[b + za, b + (z+1)a]

alright now i need to prove that f(ℝ) is a closed interval

but what does that mean?

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[wraith hinge]

The value of... is

[wraith hinge]

I tried using cos2x and cos3x, but I am not getting anywhere

w8

[vagrant marsh]

hint: sum from n=1 to 7 of npi/7 is 0

[wraith hinge]

I'll try with this

[vagrant marsh]

ok

[wraith hinge]

I can't use that

that is not part of the class

[vagrant marsh]

correction: 2npi/7

[wraith hinge]

or the topic

[amber python]

no complex numbers?

[wraith hinge]

I have learned that before, but no

[amber python]

i guess you'll have to tread the path of pain then

[wraith hinge]

this is the topic

and it should be solvable with this

[wraith hinge]

do u have any hint?

[amber python]

not really, i was thinking to do it with complex numbers but otherwise you'll probably just have to wade through the algebra

[lusty delta]

if this is all, then you're not given very many tools to work with here

[wraith hinge]

Yeah, I've learned so far these ones

[lusty delta]

ok that is better

[lusty delta]

derive formula for sin(a)cos(b)

[lusty delta]

using these

[wraith hinge]

I don't understand what you are asking me

this is the next topic

[lusty delta]

ok well good looks like you have it

very top right is what i am looking for

[wraith hinge]

yeah I know, but it's weird bc it wasn't taught

for that problem

well

[lusty delta]

i am pretty sure this is easiest way to do such problem without complex numbers

[lusty delta]

and it is derived from this image here

[wraith hinge]

I have never seen a problem in trigonometry w complex numbers

but in algebra yes

[lusty delta]

so the formula within ur reach. i would recommend using

[wraith hinge]

do I need to use the second formula of the right

[lusty delta]

the first is what will be used here

[wraith hinge]

do I need to clear cos(O)

[lusty delta]

sort of yes

try to find a way to apply that formula

[wraith hinge]

Is it -1/2?

how

how

did u realized this

that you could use that

to find this

I feel amazed

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[lusty delta]

this in particular is actually a very common trig problem

other exercises include finding cos(2pi/7)cos(4pi/7) + cos(2pi/7)cos(6pi/7) + cos(4pi/7)cos(6pi/7) and cos(2pi/7)cos(4pi/7)cos(6pi/7)

[lusty delta]

(yes. good job)

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[blazing geode]

Can someone explain those three terms

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[upbeat yoke]

stuck on 61d

[upbeat yoke]

that’s what i’ve tried so far but once i put it into the infinite geometric series thing i got a different answer than the one on the answer key

[blazing geode]

Do you want the help from a servant of Mickey Mouse

[upbeat yoke]

yes please

[upbeat yoke]

i get 15/13

but the answer key says( 3 sqrt 13)/13

[blazing geode]

For 61-a

[upbeat yoke]

yeah?

wait no

61d

[blazing geode]

Ohhh

[upbeat yoke]

yeah

i tried to do the same thing as the question above bc it seems similar? but apparently not the right thing to do

[radiant spindle]

I would put them into two different groups (like total distance traveled horizontallly and vertically) and compute their series then use the pythagorian theorem to find the distance from the origin

[upbeat yoke]

ooh okay i was thinking about that but forgot

i’ll try that rn

thanks

[blazing geode]

Put them into the xy plane

[radiant spindle]

Yea

Yea so I just did what I suggested and I got what you said

Hope it helps

[upbeat yoke]

thanks it is helping

still working on it tho lol

alr i finally got it

tysm

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[amber wolf]

can anyone help me with this question

[granite light]

convert to cartesian

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@amber wolf Has your question been resolved?

[amber wolf]

so would the answer be 6.08

[granite light]

no

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[amber wolf]

it was right

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[granite light]

am i going mad or is the correct answer 5.76268...?

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[gentle sleet]

what am I doing wrong

Im getting 0

[spiral ocean]

hmm

this is integrating over odd function

[opal breach]

the question here is, why did you assume zero is the wrong answer

[spiral ocean]

i got zero

even when switching order of integration

[opal breach]

because zero makes sense

[gentle sleet]

[opal breach]

this function is odd around the x axis and you are integrating it over an even region

[gentle sleet]

wait I dont get this

can u explain

[opal breach]

in the left of the region it's negative and in the right it's positive in a symmetric way

if the integral over half the region is J

the integral on the other half will be -J

so it would be J-J = 0

[burnt mulch]

(Just throwing this here for further reading: https://math.stackexchange.com/questions/1230999/definite-integral-of-an-odd-function-is-0-symmetric-interval)

[gentle sleet]

oh wait isnt this also one of the integral properties that they teach in calc II

shit it is

why am I so braindead

https://tenor.com/view/black-guy-whtie-and-black-cry-praying-looking-up-gif-553441503240360994

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[fresh pendant]

what does it mean that [ in the nbhd of...]

[spiral ocean]

i mean

i can substitute

${1 + 3 + 4z^2 - 3z^2 = 1}$ for a

[solid kiln]

:D

[spiral ocean]

${z^2 = -3}$

[solid kiln]

:D

[spiral ocean]

so z doesn't exist there

[hollow arrow]

It probably means the same thing as “where” or “when”

[hollow arrow]

We’re trying to understand the meaning of “in the neighbourhood of”

[spiral ocean]

it just means around

basically it asks whether limit of the point exists there or not, no?

[hollow arrow]

B) $z^2 = 0$

[solid kiln]

Randel_

[hollow arrow]

So the answer to B) is yes?

[spiral ocean]

think so

[fresh pendant]

why dont they just say, at the point of x=1,y=1

[spiral ocean]

cuz a value can exist at (1,1) but isnt continuous at (1,1) i think

[normal bronze]

yah basically the graph at a point can have multiple limits to it depending on the path

if you arrive from the x or y axis you might not get the same thing

[indigo idol]

I need some help

[fresh pendant]

!occupied

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[spiral ocean]

also

for the last question

two values of z exist

for some reason

[fresh pendant]

i know this but i still dont know what does this has to do with use nbhd

[spiral ocean]

neighborhood basically says around x=1, y=1

so is it defined around x=1, y=1?

[fresh pendant]

yes

[spiral ocean]

should it just asks u whether the limit exists there

[fresh pendant]

but we will never know if it's defined at x=1,y=1?

[spiral ocean]

we can

substituting and see whether

z can be found

[fresh pendant]

isnt that we only know it's defined in the nbhd, not the point?

[spiral ocean]

yes...

[fresh pendant]

then how can we know it's also define at the point?

[spiral ocean]

is proving it using clairau'ts theroem

allowed

ok

lets start again

${z=\pm \sqrt{\frac{1-x^3-3y^2}{4x-3y}}}$

u can rewrite z as this right?

[fresh pendant]

ye

[solid kiln]

:D

[spiral ocean]

and we want to see whether it exists around x=1, y=1

[fresh pendant]

so it's about [in the nbhd] part?

[fresh pendant]

yes i knew this

[spiral ocean]

doesnt this just mean the possible limits of z

wait

so

we have to prove that z is a function around x=1, y=1

but the nbhd must also contain x=1, y=1, no?

[fresh pendant]

yes

[spiral ocean]

since x=1, y=1 cant even define z there

it shouldnt exist there

[fresh pendant]

since it's not even defined at the point -> it' not defined in the nbhd

Is this logic correct？

[spiral ocean]

i think so

this is the graph of the function

x=1,y=1 doesnt exist there

theres no nbhd there

x=1, y=0 on the other hand is continuous

[fresh pendant]

can you link this graph

[spiral ocean]

https://c3d.libretexts.org/CalcPlot3D/index.html

[fresh pendant]

so this point is (1,1)?

[spiral ocean]

(1,1) should be empty

[fresh pendant]

ok still have other confusions, for (b) we have solved that it's defined at the point, but how can we know it's also defined in the nbhd?

[spiral ocean]

it looks like a normal polynomial so i just assumed that it must be continuous 😭

[fresh pendant]

aah, wait do the most implicit functions continuous? what if they give us a discontinuous implicit function?

[spiral ocean]

polynomials are always continuous

but i realize later

that this is not a regular polynomial

so pure luck i guess

[spiral ocean]

u can easily notive a discontinuous

using the two-path method

[fresh pendant]

ok assume it's discontinuous, and we knew it's defined at the point, then does it also defined in the nbhd? how can we determine that?

[spiral ocean]

i think of nbhd as a 3d epsilon-delta definition of limit

so, proving that the function is not continuous there

using whether two-paths, etc

[fresh pendant]

i mean we already know it's discontinuous

then it won't be defined in the nbhd even if it's defined at the point?

[spiral ocean]

if the point isnt there

there shoudlnt be nbhd around there, no?

[fresh pendant]

no i mean the point is there but the f is discontinuous

[fresh pendant]

wait you mean that Wherever there is a point, there is a neighborhood?

If a function is defined at a particular point, then it is also defined in the neighborhood of that point?

[spiral ocean]

No

Nbhd is there if the function is continuous there

[fresh pendant]

but here we assume the function is discontinuous

then even if it's defined at the point, it's not defined in the nbhd?

[spiral ocean]

The defined point is included in the nbhd

Cuz it is within the boundary

However

A point can be defined without nbhd around

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[sage anvil]

Guys I have task: solve matrix equation if u have 2 matrices A and B. and matrix equation is AX=B, solution writes A^-1 * AX = A^-1 * B, so then X = A^-1 * B, and then just find A^-1, but I Wonder how they get this solution?

[sage anvil]

How get from AX=B to ----> A^-1 * AX = A^-1 * B, so then X = A^-1 * B

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[wraith hinge]

<@&286206848099549185> thoughts.

[frozen plover]

bruh

[worldly wing]

!15m

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[wraith hinge]

mb

[worldly wing]

also whoever made this question should seriously reconsider their life choices

[wraith hinge]

WHY

[worldly wing]

is there some boundary condition to this?

[wraith hinge]

a,b,c belong to (0,infinity)

[worldly wing]

that's it?

[wraith hinge]

yes

[worldly wing]

nothing like abc=1 or something

[wraith hinge]

no

hahaha

i have this book i bought

it is full of problems this level

[elder field]

What are we summing on?

[worldly wing]

it's probably a cyclic sum

[wraith hinge]

it doesnt have it

yea

[elder field]

Wdym

[wraith hinge]

it means we are summing on a b c then b c a and so on

interchanging

[elder field]

On S3 ig

Ok

[wraith hinge]

i was thinking of squaring it out

but i think i will go mad

[elder field]

What about finding minimum and max

Find where the gradient is 0

Plug in and see the value is less than 3√3

[wraith hinge]

i dont know derivatives

[worldly wing]

you can divide by sqrt(abc)

basically homogenising the inequality