help-43

[kind viper]

some people have a mnemonic like ASTC or CAST or whatever so if you like that you can adopt that

(i prefer visual references to the unit circle)

i mean unit circle is like times tables for trig.

[vale perch]

Then I should memorise it right?

[compact pewter]

@vale perch Has your question been resolved?

[kind viper]

i can't tell you exactly how much you specifically "should" memorize, but yes.

[compact pewter]

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[kind hollow]

,texHello. Suppose we have two sequences, a_n and b_n. We know that they diverge as n approaches infinity. Thus for each one, $\forall M_1

[boreal girder]

fijokazż
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[kind hollow]

sorrt

sorry

suppose the two sequences diverge

i wanna prove that their addition comes out as infty

if they diverge in positive infty

and i can use the definition of the divergence of a sequence

take a new nEN and MER, s.t. a_n+b_n > M

i wrote it in a rush but does this make sense?

[severe flare]

But this is what you want to prove

[kind hollow]

Yeah but i can just add the two inequalities right?

[severe flare]

Yeah

[kind hollow]

okeoke thank you

.solved

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[azure vault]

the codomain is N

[copper loom]

ah

[azure vault]

well N_0 in that case because non negative

[copper loom]

but if i starts at 0 how can it terminate a variable

wouldn't it terminate at a number ?

[azure vault]

?

[copper loom]

or i starts at 1

[azure vault]

i starts at 1

and N can have value 0

the result will then be an empty sum

so S = 0 in that case

[copper loom]

but doesn't i go from 1,2,3,4,5,6,7,8

[azure vault]

?

[copper loom]

or is i going from P(N=1), P(N=2) ?

[copper loom]

i is our index right ?

index is integers

[azure vault]

no, i goes from 1 to N

where N is chosen at random in the set of non negative integers

and by convention, $\sum_{i=1}^0 X_i = 0$

[boreal girder]

Raphaelisius Maximus MMIII

[copper loom]

https://tenor.com/view/what-monkey-shook-shocked-surprised-gif-8153885586429536809

why don't we just have N be some number

why we making it a random variable ?

[azure vault]

because the fact that we may not know when we stop the sum is part of being compound

[copper loom]

oh so you quite literally mean random and not arbirary

[azure vault]

yeah, as in random variable

[copper loom]

shi this is actually confusing

[azure vault]

N is a discrete random variable

[copper loom]

every integer has equal prob of occuring ?

cuz it's random

[azure vault]

no

that's not possible in fact

you could have, idk, P(N=0) = 1/2, P(N=1) = 1/4, P(N=2) = 1/8,...

but you can't have P(N=k) = constant, because of probability sum laws

and P(Universe event) = 1

[copper loom]

yeah i figured if you deal with infinity this becomes a bit weird

[copper loom]

i don't get it

why do we want a weighted random number ?

[compact pewter]

@copper loom Has your question been resolved?

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[hearty lichen]

how can i solve this? I dont get the answers reasoning

[toxic cipher]

Do you know differentiation?

[hearty lichen]

yeah

i get 2x

so m=2x

[dusty ermine]

Hello, I'm new...

[trail cave]

you have two points. that defines a line

[hearty lichen]

right

[trail cave]

(x, x^2 + a^2) and (0,0)

[hearty lichen]

whats (x, x^2 + a^2) for

[trail cave]

the other point on the graph

[hearty lichen]

which one

0,0 and either or point would have the same equation?

[trail cave]

well the slope would be different but yeah pretty much, in this problem

[hearty lichen]

right okay bet

i got m=2x

easy

[trail cave]

do you know how to find the slope from two points??

[hearty lichen]

dont i diffrentiate y=x^2 + a^2?

so dy/dx = 2x

[trail cave]

that's part of it

what is the slope between the two points

[hearty lichen]

its not m=2x?

[hearty lichen]

would that not be a gradient for one of the points if we subsitutie it in

so like m=2x

and m=x?

sorry

m=-2x

im tweakin

[trail cave]

ok let's start over.
let's say the points of tangency occur when x = t
then the points are
(t, t^2 + a^2)

[hearty lichen]

i dont get whats (t, t^2 + a^2) for 😭

what point is that

[trail cave]

according to point-slope form equation for lines,
we have $y - (t^2 + a^2) = 2t(x - t)$
as you said, the derivative is 2x, so at x=t (point of tangency) the slope is 2t

[boreal girder]

haunt

[hearty lichen]

damn

thats cool

[trail cave]

(x, y=f(x)=x^2+a^2)

[hearty lichen]

yes i understand you put the original formula as a y

but ion get what point is that for

[trail cave]

the point of intersection

[hearty lichen]

between the two tangents?

[trail cave]

between the graph and one of the tangent

[hearty lichen]

OH

I see it now

right right

[trail cave]

so simplify the equation of the tangent line to get $y = 2tx - t^2 + a^2$

[boreal girder]

haunt

[hearty lichen]

what did you simplify to get that

[hearty lichen]

why is it negative

the graph is positive isnt it

[trail cave]

review point slope form

[hearty lichen]

BRO

im tweakin

yeah yeah its negative

[trail cave]

ok tell me what this equation represents

remember at x=t the point of tangency occurs

[hearty lichen]

so we have y = x^2 + a^2
so that means we have points (0,0) and (x, x^2 +a^2)
Then what u did was you took the tangent formula
y-y1=m(x-x1)
so now its y - (x^2 + a^2) = 2x(x - x0)
and the 2x is from the diffrentiated slope value

and this would give us the tangent equation for one of the tangents right?

[trail cave]

*not tangent formula
it's just a form of lines that uses a point and slope to describe it

[hearty lichen]

oh

its not a formula to find the tangent?

[trail cave]

nope

just a way to write a line using a point and slope

[trail cave]

the important distinction is that you need a separate variable to represent the point of tangency, x = t

using x again makes the problem different

[hearty lichen]

yeah yeah my fault

y - (t^2 + a^2) = 2t(x - t)

like that?

[trail cave]

yes, you have your point and slope
point (t, t^2 + a^2)
slope 2x

then simplify that

[hearty lichen]

t= +-a?

@trail cave

[trail cave]

yeah

[hearty lichen]

bruh thx alot

feels a lil too easy now i feel stupid 😭

[compact pewter]

@hearty lichen Has your question been resolved?

[compact pewter]

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[fallen frost]

In this passage, what is the author trying to convey by defining delta using triple equals, as opposed to normal equal?

[native shard]

in this context it’s like saying "defined to be"

you might also see $\coloneq$

[boreal girder]

knief

[fallen frost]

oh, ok.

Thanks

.close

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[wary mulch]

how do I simplify this?

[thorny urchin]

evaluate directly

don't overthink these things

(i'm assuming you were attempting to find some trig identity to apply to this?)

[wary mulch]

Im just wondering how my teacher got to

step 2

[viscid shard]

common angle values

[manic geode]

They just evaluated the sin and cosine functions

sin(pi/3) = sqrt(3)/2

cos(pi/3) = 1/2

[wary mulch]

what about them? d

did he memorise it?

[kind viper]

he used them

[viscid shard]

Idk if your teacher memorized them or not

but he definitely used them

[manic geode]

The most common angles you will encounter at your level are the multiples of pi/6 radians

[viscid shard]

this is like the fifth time you've expected me to read your teacher's mind.... 😅

[wary mulch]

if I evaluate them, woudnt i get a decimail apprx

[viscid shard]

exact values for those angles are known

[manic geode]

That's if you decimalise them

[viscid shard]

so why approximate? you have an exact form in radicals

unless you are told to approximate

but even if you are, normally we only approx. the final answer, not intermediate answers

otherwise rounding errors will rear their ugly heads

[wary mulch]

so how would I go about it? use my calc?

or manually

[thorny urchin]

trig functions evaluated at special angles, 30°,45°,60° (and 0°,90°)
are things you should aim to remember

[wary mulch]

u mean remember the radians at those special angles?

[viscid shard]

to aid in this, perhaps it would be prudent to remember the two special triangles that create these common angles

[thorny urchin]

the argument in radians is the angle

[kind viper]

not only that

[viscid shard]

30 and 60 degrees (pi/6 and pi/3 radians resp.) are created by the 1-2-sqrt(3) triangle

[thorny urchin]

remember the full
trig(angle) = value (or ratio)

[manic geode]

Yes you should remember both the sin and cos values of those angles both in degree and radian forms, but radians are used because they are easier to work with

[viscid shard]

and the 45-45 (pi/4 radians) angles are created by the 1-1-sqrt(2) triangle

[wary mulch]

how are radians easier to work with?

[kind viper]

you'll have to take our word for it bc at the level of trig table memorization it won't be visible.

[wary mulch]

i mean i can just remember the special angles in degrees

then convert it to radians

[thorny urchin]

you'd primarily be working in radians in calculus

[wary mulch]

in the test

[viscid shard]

well I guess you can, but in calc we are mostly working in radians anyway, so...

[thorny urchin]

so ideally you'd want to get comfortable with radians now, and avoid conversions if it isn't needed

[wary mulch]

ok got it

[kind viper]

trig is always done in radians in calculus bc it makes derivatives of trig functions nice

[wary mulch]

do i need to remember the values as well of tan cos and sin

[kind viper]

obviously yes you do. some of those values were used in your OP

[viscid shard]

if you have to choose, remember cos and sin first

[thorny urchin]

from 0→90
if you can remember either the values for sin or cos
the values for the other is essentially mirrored
and values for tan will be sin/cos

[wary mulch]

oh so in my course, radians is the preffered way, do I just replace sqrt3/2 to pi/3

[thorny urchin]

no

[wary mulch]

i meant replace sqrt3/2 to pi/3

[thorny urchin]

no

[viscid shard]

you replace sin(pi/3) with sqrt(3)/2

[thorny urchin]

the WHOLE sin(pi/3) is sqrt(3)/2

[wary mulch]

ohhh ok

so to clear my confusion, what are the things I neeed to memorise to do trig functions?

[wary mulch]

this is from the unit circle right?

[kind viper]

sure is

just presented in a tabular format

[wary mulch]

so If i know the unit circle

will I be able to evaluate trig functions?

[kind viper]

yes with a huge asterisk

[wary mulch]

whats the huge catch?

[kind viper]

it hinges on what you really mean by "knowing the unit circle"

[wary mulch]

basically I watched this video, he showed how to derive the values

https://www.youtube.com/watch?v=gdJq1QunN-o

ik how the unit circle works

I know that the II quadrant is (-,+), I quadrant is (+,+) and so on

[kind viper]

if you think yourself able to memorize that entire clusterfuck and recall any value from any part of it in the "woken up at 3 in the morning by police and put at double gunpoint" scenario with 110% perfect accuracy,

then yes

[wary mulch]

I mean Im not memorising it, I just know the values

of the first quadrant

and I just apply the sign change

I know the critical points

2 pi, pi, pi/2 etc

which I learned from the vid

is that acceptable?

[thorny urchin]

yeh, that's what many recommend

[wary mulch]

ok so thats all I need to know in order to do trig functions?

I feel like I need to know something else

[thorny urchin]

this is what you need for basic level stuff

you'll get more complicated stuff later on that require additional trig identities
e.g. compound angle

[wary mulch]

fair point

[wary mulch]

wheres sin pi/3

sin is in the second quadrant so my attention is over there

[viscid shard]

sine is in the second quadrant?

[thorny urchin]

why 2nd quadrant

[wary mulch]

A,S,T,C

first qudrant is A

for all

[thorny urchin]

sure, that tells you the signs where the trig functions are positive
but why are you looking at Q2

[wary mulch]

because it had sin

[thorny urchin]

ASTC tells you the sign of the result

you're interested in
trigfuntion(pi/3)

so first look at where that pi/3 angle is

[wary mulch]

here

sin is the the y value

so sqrt3/2

[thorny urchin]

yes.

Q2 isn't relevant here

[wary mulch]

oh ok, we only apply the sign changes

which is positive here

that makes alot of sense now

thank you very much

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[forest token]

K = 16^x + 16^-x + 4.4^x + 4/4^x

now when i use AM>=GM on the whole expression

i got K>=8

but when i use it seperately on 16^x + 16^-x and 4.4^x +4/4^x i get K>=10

which one is correct and why

[covert creek]

the second one

is correct one

[forest token]

ohk but why is the first one wrong

[covert creek]

What do you know about AM-GM?

[forest token]

we can only use it when all numbers are positive

[covert creek]

that's not all

When the equality hold in AM-GM?

[forest token]

the numbers are equal?

[covert creek]

gimme a min

Inequality of Arithmetic and Geometric Means

For any real positive numbers $x_1,x_2,x_3,\dots,x_n$ :
$$x_1 + x_2 + x_3 + \dots + x_n \ge n\sqrt[n]{x_1x_2x_3\dots x_n}$$
The equality hold if and only if $x_1 = x_2 = x_3 = \dots = x_n$ \

[forest token]

yeah

[boreal girder]

Fionna The Unemployed

[covert creek]

So apply this for the first route you did

what do you get

[forest token]

= 4 x 2

[covert creek]

No I meant when equality hold for the fisrt

[forest token]

uh im not able to find

when i put x=0 im getting 10

[covert creek]

$16^x + 16^{-x} + 4\cdot 4^x + 4\cdot 4^{-x}$

[boreal girder]

Fionna The Unemployed

[covert creek]

Apply the same thing here

We got $16^x = 16^{-x} = 4\cdot 4^x = 4\cdot 4^{-x}$

[boreal girder]

Fionna The Unemployed

[forest token]

ur telling me to equate this to 8 and find when equality holds yes?

Oh

[covert creek]

Yup

[covert creek]

this can't be true yeah?

[forest token]

yea

[covert creek]

So

8 is lower bound

But it never reach 8

It cannot reach 8

So the true minimum here is 10

Which is hold when x=0

[forest token]

ohk so after using am gm i always have to check for this condition also?

[covert creek]

Yes

Have you ever done Olympiad kind of inequality?

[forest token]

no 😔

[covert creek]

Some time people predict what possible way the equality can hold

And then use it to arrange and apply AM-GM in the right way

[covert creek]

Oh okay, just check if the equality can be achieved

[forest token]

oh alr alr

got it

thank you

.close

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[gilded lark]

20th
I just wrote the general term and am unable to simplify it

[kind viper]

$\sum_{n=0}^{\infty} \frac{4n+5}{(2n+3)^2(2n+7)^2}$

[boreal girder]

Ann

[kind viper]

looks like your series is this

[gilded lark]

No

[kind viper]

no? then what do you have

[kind viper]

,rccw

[kind viper]

oh yours starts from 1

[gilded lark]

Yea

[kind viper]

hang on tho, that (8n-1)^2 is sus

[gilded lark]

Ohk

[kind viper]

the numbers being squared in the bottom both go up by 4

not by 8

[gilded lark]

Then 4n+3

[kind viper]

oh yeah i fucked it up too

$\sum_{n=0}^{\infty} \frac{4n+5}{(4n+3)^2(4n+7)^2}$

[boreal girder]

Ann

[kind viper]

there

anyway you defo want to partial-fraction this

to set up some kind of telescope

[gilded lark]

This is just so hard

[kind viper]

it's going to be mildly crunchy

tho like you can kinda get away with a cheat

[gilded lark]

i was thinking same

[kind viper]

put $4n+5=z$ and then pfd on $\frac{z}{(z-2)^2(z+2)^2}$

should make some of that crunch sting a little less

[boreal girder]

Ann

[gilded lark]

ugh!!

[kind viper]

yeah some of it is unavoidable

[gilded lark]

@kind viper how Abt this

[kind viper]

not much to be done about that

,rccw

[kind viper]

you think you can do this directly?

[gilded lark]

I have done a lot of rough work behind 😞

[gilded lark]

And this is just to show you

@kind viper is it right??

[kind viper]

uhh well ngl i can't say without verification

but let's see

yeah checks out

i would prob keep writing n>=1 underneath the sigma tho

[gilded lark]

Uk I'm just too lazy

[kind viper]

well it's gonna become relevant quite soon.

[gilded lark]

Hey I wanted to ask you one thing😅
That might sound foolish or childish
Can I?

[kind viper]

anyway now the thing telescopes

[kind viper]

... go ahead?

[gilded lark]

What have you invented in ur PhD😅

[kind viper]

i do not have a phd

[gilded lark]

Oh i see I misunderstood it
Post graduate with PhD

Nvm

Let's go ahead

I see the answer is D if I'm right

.close

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[naive widget]

can someone help me finding the volume of this truncated pyramid?

[open girder]

What have you tried?

[naive widget]

i tried a lot of things actually

but my first question is, should i extend the truncated pyramid to a full one, calculate all of it and then remove the volume of small extended part?

[grizzled geyser]

I would do that, yes.

[naive widget]

i did it and found an answer

i also solved by an 2nd way and used the formula for finding volume of a truncated triangular pyramid

and the answers are not matching up

i double-checked all my calculations too

[grizzled geyser]

Show your calculation, and the formula.

[naive widget]

these two are formulas for finding volume of a truncated triangular pyramid

both of these finding volume formulas give us 78√11

[naive widget]

this method on the other hand

gives us 4750.√11 / 81

[grizzled geyser]

Show your calculation using that method

[compact pewter]

@naive widget Has your question been resolved?

[naive widget]

the small part is a 8x volume and the big part is a 19x volume

we find the height of all extended pyramid √1100/ √3

base area.height.1/3. 19/27 will give us the part we want

25√3 . √1100/√3 . 1/3 . 19/27

yes

i said x to that one

and x/x+12 = 4/10

[grizzled geyser]

Wait, yes. I thought it was incorrect.

[naive widget]

well that's not the problem

8V and 19V part is correct too

base area, height they're all correct too

so i have zero clue why this gives us a different result than 78√11

[grizzled geyser]

But are you sure that it's 19V and not 117V?

[naive widget]

ohh

okay okay

i'm dumb

but thank you for your help

can i ask one more?

this is the 2nd and last question, so don't worry

[covert creek]

Just... ask

[naive widget]

i was trying to solve it myself and i think i found it

so that's why i didn't ask

but fine, i don't have answer key so i'll just ask

The longer spatial diagonal of a regular hexagonal prism is inclined to the plane of the base at an angle of 45°. Find the surface area and volume of the prism if the area of ​​its base is 12√3 cm².

[viscid shard]

you said you tried it yourself, so you may want to share your working for helpers to check

[compact pewter]

@naive widget Has your question been resolved?

[naive widget]

<@&286206848099549185>

[covert creek]

are you sure the side of hexagon is 2sqrt(2)

[compact pewter]

@naive widget Has your question been resolved?

[naive widget]

yes

[naive widget]

<@&286206848099549185>

i showed all my work and showed my answer, but i'm not sure and i don't have answer key, it's just checking my steps and answer now really

[frigid furnace]

yo

[compact pewter]

@naive widget Has your question been resolved?

[frozen lodge]

tuff

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[radiant hedge]

$\text {Hello, does anyone know why the hypervolume of a 4-ball is equal to } \frac {1} {1^2}+ \frac {1} {2^2} + \frac {1} {3^2} ... \text { i.e. the basel problem for a 4-ball of radius 1?}$

[boreal girder]

On ne passe pas

[radiant hedge]

oh mb then

lol

.close

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[radiant hedge]

oh yeah oops it was 6-ball and that was pi^3

lol

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[sonic terrace]

Here, 12 is common numerator

[sonic terrace]

So why ain’t I doing
(x*4)/12 - 3x(-2)/12 = 0?

Why is that incorrect?

Hello?

[clever copper]

Solve for x?

[sonic terrace]

Blud…

Do you understand me?

[clever copper]

I need to know what the original questions wants you to do with the equation

[sonic terrace]

They want me to solve the equation

Simple as that…

[clever copper]

Ok so yes solve for x

[sonic terrace]

Okay why is this incorrect?

[clever copper]

Parentheses should go around the whole x-2 expresion which the 3 distributes across

To give (3x-6)/12

Then you can combine the numerators since they have a common denominator

Which gives (4x-(3x-6))/12=0

You know to distribute the negative at this point right?

[sonic terrace]

(x*4)/12 - 3(x-2)/12 = 0?

[clever copper]

That is a way you can start solving for x

[sonic terrace]

Okay son

[clever copper]

^^

[sonic terrace]

I know how to do it

[clever copper]

So distributing the negative gives (4x-3x-(-6))/12=0

-(-6) becomes +6

4x-3x becomess x

You now have the expression (x+6)/12=0

You know how to do the rest from here?

[sonic terrace]

Obviously

12x + 72 = 0

x =-6

Easy

6 grade math

[clever copper]

!done

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[slim citrus]

done?

oh its done

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[versed estuary]

you mean (4x-3(x-2))/12?

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[hexed ferry]

So i have the taylor series of $f$ at $x=a $ is $f(a) \cdot \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!}$

[boreal girder]

Julian

[hexed ferry]

how do i go about finding what the series converges to

i mean i suppose i want it to converge to f(x-a)

either i want to prove the series converges to f(x-a) i think or that the taylor is the function itself

maybe a hint would be good

wow i got something maybe

So i have the taylor series of $f(x-a)$ at $x=a $ is $f(0) \cdot \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!}$

[boreal girder]

Julian

[hexed ferry]

we have the quotient of their taylor series is $f(a)$

[boreal girder]

Julian

[hexed ferry]

idk if that helps

ho ho

i have

$\left(\frac{f(x)}{f(x - a)}\right)' = 0$

[boreal girder]

Julian

[ruby mason]

this is a classic one, a hint is let f and g both satisfy the criteria, consider $\tfrac{f}{g}$ and its derivative

[hexed ferry]

well i have to do the taylor series part later anyways

does it make sense to show f/g is constant

[ruby mason]

yes

[hexed ferry]

well i mean does it help with this taylor thing

[ruby mason]

oh wait i didnt see that part hmm

[hexed ferry]

i mean it should be straightforward to show its constant

fg-fg

=0

for the quotient rule

[ruby mason]

okay yeah they want you to do it the taylor series way mb

[hexed ferry]

i think they want fg-fg way but i have more problems

wait wait

that should help then

if we show the taylor series satisfies those things

we know its equivalent

big brain fr (im failing ts class)

wth

ofc im skipping steps but

derivative of taylor $f$ at $x=a $ is $f(a) \cdot \sum_{k=1}^{\infty} \frac{k(x-a)^{k-1}}{(k)!}$

is that even defined bruh

well i know it is but

i feel like i need to prove that

(-1)!

fries in the bag

shoot maybe im selling

[boreal girder]

Julian

[hexed ferry]

i figured it out with a substitution

j+1=k

so i know it equals its derivative

now idk how to show

t(0)=1

and

t(x)>0

.close

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[hexed ferry]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[hexed ferry]

ok original question i figured out but

i want to show this here is positive and equals 1 when x=0

[hexed ferry]

where f is defined as here

prove for all $a \in \mathbb{R}$, $\sum_{j=0}^{\infty} \frac{(-a)^j}{j!} > 0$

[boreal girder]

Julian

[compact pewter]

@hexed ferry Has your question been resolved?

[hexed ferry]

@timber locust

oops, sorry

<@&286206848099549185>

im thinking

we know its positive because exp(x)exp(-x)=1

and its trivially positive for a<0

so positive times negative is negative so we need its always positive

oh but how do we show

this equals 1 for x=0

.close

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[mossy cave]

I need to construct a simple graph for the list of numbers

0,1,2,2,3,4 which represent the degrees of the vertices

I want to check if this is a valid graph

[dusky nymph]

why do V3 and V4 appear twice?

[kind viper]

i mean...

[dusky nymph]

your idea is fine, but each vertex should only appear once
get rid of the V3 and V4 at the bottom, and instead draw lines from V2 to the other V3 and V4

[kind viper]

you can glue the two v3s together

[dusky nymph]

quotient graph

[mossy cave]

Whoops I don’t know why there are duplicate v’s

[mossy cave]

Like this?

[dusky nymph]

yep that works

[mossy cave]

Ok thank you

[dusky nymph]

yw

[mossy cave]

And I have another question if that’s ok . If I were to construct two non isomorphic graphs that have the same list of degrees how can I do that?

I was thinking like this?

[dusky nymph]

in general this can be tricky, since it's easy to draw two graphs that look different but are in fact isomorphic

but yep that example certainly works, they both have the same degree sequence and they're not isomoprhic because graph 1 is connected whereas graph 2 is not

[mossy cave]

Ok thank you 🙏🏻

[dusky nymph]

example of what i mean btw

two graphs that are isomorphic even though they don't look like it

[mossy cave]

Ohhh I see, that is confusing

[dusky nymph]

this one is cool

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[serene coral]

Is there a way to find the closest factors of a number? For example, with 100 we have 4 and 5 being the closest.

[kind viper]

you mean like... given n, find two integers p and q such that pq = n with p and q as close together as possible?

or wait no

[serene coral]

no no, just factors of n, p*q does not have to equal n

[kind viper]

you mean just two factors of it that are as close together as possible

i mean if n is even then there's 1 and 2

otherwise uhh

god idk, don't know of any way besides trial and error

!xy

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[serene coral]

it's a coding problem, so do I need to provide the whole context?

i feel like normal coding techniques wouldn't work

[kind viper]

yeah let's see the full context

[kind viper]

ooookay gonna need a translation from viet

[serene coral]

so the guy has n employees, 2 two bags, each with n coins. He gives the coins in first bag to the top x members equally, and does the same to the top y members with the other bag (know that x ≠ y). We need to find the minimum possible difference between the two reward amounts.

[kind viper]

hmm

i'd probably just do it iteratively

[quiet sentinel]

Well the two factors then need to be as close as possible to root n

Oh wait no

It could be like 1 and 2

[serene coral]

for even numbers, yes

for odd nums, idk

[serene coral]

n <= 10^14 💀

[kind viper]

i mean you can start with small n can't you

and i also dont think you ever need to go above sqrt(n) in your iterations

the idea is basically, yeah you're looking for two factors p, q of n and i think also p, q > 1

such that |p-q| is minimized

so what i'd do is a for loop

[serene coral]

10^7 is still very close to TLE

[quartz yoke]

i think you can turn that into a Diophantine Equation

and solve it

[serene coral]

,w diophantine equation

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[serene coral]

can you give me a spark?

[compact pewter]

@serene coral Has your question been resolved?

[serene coral]

any ideas or do I just code iteratively?

<@&286206848099549185>

update: iterative code works, i'm just overcomplicating the problem

.close

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[carmine garden]

I was thinking this would be $\int_{0}^{1/2} \int_{3/4}^{1} 4xy dydx$

[boreal girder]

wai

[carmine garden]

But I'm quite unsure of how I'd solve this using the defn of conditional prob

[kind viper]

the integral that you've written is P(X ≤ 1/2 and Y ≥ 3/4)

[carmine garden]

yes

so now I just need the marginal of Y, right?

[kind viper]

you need P(Y ≥ 3/4), whichever way you want to calculate it.

i'd do it with another integral just for that

there's a cheaty way of doing this problem but i will not say more

[carmine garden]

$\int_{0}^{1} \int_{3/4}^{1} 4xy dydx$

[boreal girder]

wai

[kind viper]

yes

\dd{} btw.

$4xy \dd{y} \dd{x}$ vs. $4xy dy dx$

[boreal girder]

Ann

[carmine garden]

what would that be?

[kind viper]

recognize that X and Y are in fact independent

each with density function 2x on [0,1]

[carmine garden]

ooh

I would like this verified too

$\int_{0}^{1/2} \int_{x}^{1} 1/2 \dd{y} \dd{x}$

This is the joint PDF

I'm having trouble with the marginal

[boreal girder]

wai

[slate sand]

Drawing the region helps

[carmine garden]

I did

It's the upper part of the square spanned by [0,2] along the axes

wait, is it just $\int_{0}^{2} \int_{0}^{y} 1/2 dx dy$?

[boreal girder]

wai

[slate sand]

Isn’t it a triangle

[slate sand]

Why is the marginal a constant here

Or is that your answer

[carmine garden]

I thought it is?

it isn't?

[slate sand]

Shouldn’t the marginal distribution of X depend on x

[carmine garden]

yes

[slate sand]

Like $f_{X|Y=y}(x)={}$ something with $x$ in it

[carmine garden]

I need the marginal of Y , so $\int_{0}^{y} 1/2 dx$

[boreal girder]

wai

frosst

[slate sand]

But this yoy claim to be the marginal of X but it doesn’t even depend on x

[carmine garden]

Yea, I misread the conditions

so I messed up the domain of x and y

.close

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[serene wharf]

whats the approach

[charred finch]

try using the perpendicular axis theorem

[compact pewter]

@serene wharf Has your question been resolved?

[winged dock]

Is the triangle isosceles

[compact pewter]

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[uncut cloud]

im trying to find all prime numbers, p, such that 2p-1 and 2p+1 are also prime. i think p can only be either 2 or 3. does this proof make sense? 2p-1, 2p, 2p+1 are 3 consecutive integers, such that one must be a multiple of 3 (3 consecutive integers implies that one must be congruent to 0 (mod 3)). if 2p is not a multiple of 3, then either one of 2p-1 and 2p+1 is a multiple of 3, which cannot be prime unless 2p-1 = 3, hence, p = 2 workers, or 2p+1 = 3, hence, p = 1 but this isn't prime. if 2p is a multiple of 3, then 2p is a multiple of 2 and 3, so p has a multiple of 3 which cannot be prime unless p =3. thanks

[fluid stag]

yup that looks good

[uncut cloud]

ty

.close

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[steel sedge]

Let ABCD be a cyclic quadrilateral
whose opposite sides are not parallel. Suppose points P, Q, R, S lie in the interiors of
segments AB, BC, CD, DA, respectively, such that
∠P DA = ∠P CB, ∠QAB = ∠QDC, ∠RBC = ∠RAD, and ∠SCD = ∠SBA.
Let AQ intersect BS at X, and DQ intersect CS at Y . Prove that lines P R and XY
are either parallel or coincide

[steel sedge]

Pretty sure this is the diagram

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@steel sedge Has your question been resolved?

[obtuse elk]

g im ngl this diagram is a lil hard to read

[compact pewter]

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[wary mulch]

why did he write dx after 4

and I also want to know what is the advantage of using u substituion

Is it worth knowing it?

[mild sky]

Yes

You can view it as reversing the chain rule, or doing a change of coordinates for the integral

[wary mulch]

why do we think this as a reverse chain rule?

[mild sky]

What you have here, is 8cos(4x) which is the derivative of 2sin(4x) (this derivative is calculated by the chain rule)
Doing u = 4x, gives you only the outer function to integrate and removes the outer derivative

[wary mulch]

ah I see, thanks

.close

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[serene wharf]

ITS EQUILATERAL

[compact pewter]

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[timid dune]

Can someone explain inclusion exclusion principle

[timid dune]

@help

@helper

<@&286206848099549185>

[drifting flax]

!15mins

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[lilac wraith]

Anyone know how to do this?

[hollow gust]

Find angles in between

[toxic cipher]

look for angles in the same segment

[lilac wraith]

idk even know what that means

[toxic cipher]

do you know the property that "angles in the same segment are equal"?

[lilac wraith]

oh ye

[toxic cipher]

use that

look for angles in the same segment as BFE

[lilac wraith]

isnt it just BOE

[toxic cipher]

BOE is at the center so it cant be used for the property

there's another angle

[lilac wraith]

BAE?

[toxic cipher]

yup

[lilac wraith]

ahh

so BAE and BFE are equal

[toxic cipher]

yes

[lilac wraith]

alr

.close

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[sick shard]

have you tried anything

[heady spruce]

yea

[sick shard]

elaborate

[heady spruce]

i was thinking about doing that

[bleak dock]

you should be instead thinking about the radii

[sick shard]

no there is a better way

[heady spruce]

ok

[sick shard]

what's the radius here?

[heady spruce]

hm

[bleak dock]

yeah, so combine this idea with the radius

you also know the height of the square is.... ?

[bleak dock]

yep!

[sick shard]

you know the value of radius, right? use pythagoras

[heady spruce]

ok

radius is 1

so

[sick shard]

yup done

[heady spruce]

let me use quadratic formula

x=3/5

so area is 9/25

[bleak dock]

yep!

you're getting better since the last time I helped you

[heady spruce]

taht was ez lol

[heady spruce]

ty

do i set equation for this?

[sick shard]

you know they both have same angle

[heady spruce]

yea

[sick shard]

and what's the formula for angle when the arc length is known

[heady spruce]

uh

360/x * circumference = arc

[sick shard]

yup works

[heady spruce]

arc is 7

[sick shard]

set up 2 equations with this for the two circles

[heady spruce]

ok

what do we do with the circumference part

[sick shard]

you know the formula of circumference of a circle

in terms of its radius

[sick shard]

what did you do exactly?

[sick shard]

i can't understand how you came here

[heady spruce]

v

[sick shard]

oh

yup

correct and similarly for the inner circle

[heady spruce]

ok

[sick shard]

then put their values in the two equations you made

[heady spruce]

360/ang * (2pi(x) + 6pi) = 7

360/ang *(2pi(x)) = 5

[sick shard]

yup yup you have two equations in two variables ang and x, you can solve and get their values

[heady spruce]

ok

x=7.5

oh wait

its ang/360

angle = 120/pi

i think

90/pi

[sick shard]

,w (2π(7.5)+6π)/(2π(7.5)) = 7/5

[boreal girder]

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[sick shard]

x is correct

what ang did you get?

@heady spruce

[heady spruce]

90/pi

[sick shard]

,w (90/(360π))(2π×7.5) = 5

[boreal girder]

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[sick shard]

consider rechecking your steps

90/π is wrong

[olive rivet]

yo i need help gettin a help channel again

[sick shard]

#help-27

,w (120/(360π))(2π×7.5) = 5

[boreal girder]

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[sick shard]

@heady spruce so ang = 120/π was correct

now you have ang and x can you find area of the sector of inner circle

[heady spruce]

oh

ok

18 3/4

[heady spruce]

done

<@&286206848099549185>

[toxic cipher]

you might be able to use the pythagorean theorem with some algebra to solve for r maybe

[heady spruce]

ok

[heady spruce]

how did u get those variables

wait nvm

so root(34)

[toxic cipher]

For r?

[heady spruce]

ty

ye

[toxic cipher]

lemme check

[heady spruce]

its crrect

[spice oasis]

√3^2+x^2 = √(8-x)^2+5^2

Then u get x
Then find r using Pythagoras theorem

[heady spruce]

ye

[spice oasis]

For this you can use trigonometry
Tan 45

To find the perpendicular then
Use Pythagoras again to find cp then pd

[heady spruce]

uh

no trig?

[spice oasis]

U mean we should not use trig

[heady spruce]

yea

[heady spruce]

i got 7root(2)

[spice oasis]

How.. what did u do

[heady spruce]

well

i actually dont know

ye its wrong prob

lets use trig ig

i cant find another way

[spice oasis]

I mean yea trig is more direct
I am getting 10root2

Let me see any other way

[heady spruce]

hmm

[spice oasis]

Yea u can actually use linear algebra
And the equation of a circle

Like u can take the radius as coordinate
-5,0 and 5,0
Find the equation of line cd
Use it to find it's intersection with circle
Latter use distance formula

[heady spruce]

oh

[heady spruce]

walk me through it

[spice oasis]

If A = -5,0
B = 5,0
Then P = -1,0

You have slope given 45
Find the equation of line

You will get that as
Y = X+1
Then use
X^2+Y^2=5

Find the two intersections using the above two equation
U will get c and d coordinate

Use distance formula

[heady spruce]

ok

distance is 3root(2)

[heady spruce]

then?

i dont think its correct

[spice oasis]

By that method u r getting 3√2?

[heady spruce]

ye

[spice oasis]

The coordinate of c is -4,-3
And d is 3,4 ig

[heady spruce]

so

so sqrt(98)?

[heady spruce]

how

wait i got it, it is sqrt(98(

help with this

<@&286206848099549185> help!

[regal knot]

how far have you gotten

[heady spruce]

nowhere

[toxic cipher]

i cant think of any other way to do this than coordinate geometry

[heady spruce]

ye same

[toxic cipher]

what have you tried?

also your teacher disallowed trigo are you sure he or she didnt disallow coordinate geometry?

[compact pewter]

@heady spruce Has your question been resolved?

[solar loom]

whats the question?

[compact pewter]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[heady spruce]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[heady spruce]

<@&286206848099549185> HELP

[heady spruce]

she lets us use cord.

[toxic cipher]

use it see what you get

[heady spruce]

idk cord lol

that the problem

[toxic cipher]

coordinate geometry?

[heady spruce]

yea

[toxic cipher]

let the center of the larger semicircle DABC be the origin (0, 0)

[heady spruce]

kk

then?

[toxic cipher]

what then will the coordinates of P and Q be?

define rightwards to be increasing x value btw

[heady spruce]

ok

7-x?

wait

[toxic cipher]

rightwards* i edited it

[heady spruce]

ok

[heady spruce]

wait what

[toxic cipher]

as you go rightwards, the value of your x-coordinate increases

[heady spruce]

ok

so its (x,0)

[heady spruce]

how do we use that?

[toxic cipher]

i'm just saying that

because i could have defined going leftwards to be x-coordinate increasing

[heady spruce]

oh

ok

[toxic cipher]

so i wanted to make it unambiguous which way is increasing x

[heady spruce]

so P would be

(x+1,0)

-x

(-x+1,0)

[toxic cipher]

Huh

you havent defined x yet

I never defined x

[heady spruce]

oh

mb

x is distance from origin to Q

so Q cord. is (x,0)

and P cord. is (-x+1,0)

[heady spruce]

i did

[toxic cipher]

you dont need to do this

the coordinates of P and Q can be found directly

[heady spruce]

oh

how

<@&286206848099549185> bro

i still need help

[cloud flint]

hi

[heady spruce]

yo

[cloud flint]

AB is the intersection of two circles

you can use center of first circle= midpoint of AB

[heady spruce]

oh

ok

[cloud flint]

then center the other circle at (7,0)

[heady spruce]

okay...

[cloud flint]

give me one sec

ill try to solve it

oh for the other circle also bound it using p and q

that gives a unique solution

[cloud flint]

wait no i cant math its (6,0) i thought 3+7+2 was 14 fsr

[heady spruce]

hmm

[cloud flint]

okay yeah use coordinates of P and Q into (x-a)^2+(y-b)^2=r^2 gives you the a

which is also the x coordinate average of A and B

[heady spruce]

wait

what

ok

[heady spruce]

wait so what is ans?

[cloud flint]

im not... supposed to tell you the answer

[heady spruce]

i got 4root(2)

[cloud flint]

i didnt get 4root(2)

i could be wrong but i got something similar

[heady spruce]

4root(6)?

wait

can you explain how you got whatever you got?

[cloud flint]

yep

[cloud flint]

uh i could yes

[heady spruce]

ok

[cloud flint]

use P and Q to find the average x value then since half of the chord is the same as the radius r you can form 12.25+Y^2=6^2+OS^2 then solve for 12.25+Y^2 then sqrt that

which gives root 24

[heady spruce]

okay

[compact pewter]

@heady spruce Has your question been resolved?

[compact pewter]

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[hexed ferry]

ive established $(f/g)' (x) = 0$

[boreal girder]

Julian

[hexed ferry]

so obviously g/f is constant

i need to be precise though

what theorem would i need to use to say f/g is constant

the reason i have no idea is cuz this is a project where we have to read ahead

oh dang mean value i thought ftc

yeah i got that part

fg-fg=0

alr

.close

[compact pewter]

Channel closed

Closed by @hexed ferry

Use .reopen if this was a mistake.

[hexed ferry]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[hexed ferry]

(1) is just $f'(x)=f(x), f(x)>0, f(0)=1$

[boreal girder]

Julian

[hexed ferry]

im not sure exactly what it means by saying consider y as fixed and x as variable