help-28

[vapid notch]

Yeah

[sturdy fern]

why theta + 60

why bnot just 60

i thought angles were same?

[bright bronze]

the argument angles are wrt real axis

[vapid notch]

argument is w.r.t the +x axis

yeah the real axis

[sturdy fern]

oh z1 okay yh cuz it dont start from the x axis

what angle does z2 have?>

[bright bronze]

some theta

you dont need any specific value

[sturdy fern]

oh yeah i was looking at angles inside the triangle but im supposed to look at the rotation

alright

but how do we link them toegtehr now

maybe since r1 is equal to r2

we can split up the z1?

since z1 would have the same components

so we would get

z2 * e^i pi/3

[vapid notch]

yeah you’re on the right track

thats equal to z1

[sturdy fern]

yh, alright for ii?

so i plug it in

and i get z2^2 ( e^i 2pi/3 + 1)

but here i get stuck

[bright bronze]

can you write (e^i 2pi/3 + 1) in the form a + bi?

[sturdy fern]

so e^i2pi/3 is -1/2 + (sqrt3/2)i

[bright bronze]

yea

[sturdy fern]

so we add one, that would just give us 1/2 which is in first quadrant. so pi/3?

[bright bronze]

yep

e^i 2pi/3 + 1 = e^i pi/3

[sturdy fern]

oh there we have the z1 term again

alright guys thank you so much

[bright bronze]

np

[vapid notch]

yw

.close

you should do that

unless you have more to ask

[sturdy fern]

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[golden shore]

Put your hands up now!

[golden shore]

NYPD!

Go -5 units towards the right

[golden shore]

It is something written on my book

Ridiculous

Ridiculous

Ridiculous

That’s unacceptable

For me

Define -5 units towards the right

That’s idiotic

That’s not right

[onyx glen]

not funny.

[golden shore]

Can we say go -5 units rightward

I cannot accept that expression

[hoary ember]

cry about it or sum :shrug:

[golden shore]

I can’t never accept it

I would rather die

[onyx glen]

english doesn't do double-negatives btw

but also you are being quite over-dramatic rn

it is a strange turn of phrase but "go -5 right" is very easy to reinterpret as "go 5 left"

[narrow path]

reasonable crashout

but what you gonna do

[golden shore]

I’m so sad at the fact that teacher might put this double negative expression on kindergarten question

[narrow path]

bit rude innit

[golden shore]

They are just kids!

[spiral elk]

bro ong im seeing red reading this wtf

[narrow path]

when i see red RUN.

[spiral elk]

🐺

🦅

[hoary ember]

when i see run, RED.

[onyx glen]

see red run i, WHEN.

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[vapid notch]

Yeah thats just 5 units to the left..

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[thick hedge]

Would like to do this using Lagrange interpolation

$\frac{x+1}{1} + \frac{(x+2)(x+1)x}{ 70}$ would be a degree 3 interpolation

[glossy valve]

What a wonderful world !

[thick hedge]

oops

[onyx glen]

would not be lagrange tho

[thick hedge]

the interpolation would be $\frac{(x+2)(x+1)x(x-1)(x-2)(x-3)}{ -5 \cdot -8 \codt -15 \cdot -20 \cdot -17}$

[glossy valve]

What a wonderful world !

the interpolation would be $\frac{(x+2)(x+1)x(x-1)(x-2)(x-3)}{ -5 \cdot -8 \codt -15 \cdot -20 \cdot -17}$
```Compilation error:```! Undefined control sequence.
<argument>  -5 \cdot -8 \codt 
                              -15 \cdot -20 \cdot -17
l.1422 ... \cdot -8 \codt -15 \cdot -20 \cdot -17}
                                                  $
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```

[onyx glen]

applying lagrange on this is gonna be kiiiinda painfu-

what?

are you sure about this

thats a 6th degree poly

[thick hedge]

yea, this gives the poly, right

[onyx glen]

m

you bungled it severely

also i will just say this

[thick hedge]

I'm not trying to show that it's a 3rd degree poly, just trying to interpolate it

[onyx glen]

lagrange interpolation BY HAND

is going to be mega ass

[thick hedge]

I know, but during my last viva I was asked to do it by hand for a data set

so not taking any chances this time

[onyx glen]

how big was the data set

[thick hedge]

Atleast 4-5 points

I think I should chose a smaller data set

do thse number work

wiat

oops

okay, this works

f(0)=2, f(1)=2.75, f(2)=4?

not that it matters here

$L_0(x) = \frac{ (x-2.75)(x-4)}{ (2-2.75) \cdot (2-4)}$
\
$L_1(x) = \frac{ (x-2)(x-4)}{(2.75-2) \cdot (2.75-4)}$
\
$L_2(x) = \frac{(x-2)(x-2.75)}{ (4-2) \cdot (4-2.75)}$

[glossy valve]

What a wonderful world !

[thick hedge]

And Then I use the fomula $\sum_{i=0}^{2} f(x_k) L_k$

[glossy valve]

What a wonderful world !

[thick hedge]

right

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[elder fiber]

.

@civic quarry

[elder fiber]

Finally

Should I show the question as well

Yeah

Wait

,rotate

[elder fiber]

,rotate

[elder fiber]

,rotate

[elder fiber]

Q18

Hello?

You there

Yess

[glossy valve]

or

[elder fiber]

Hmm okay

Thank u

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[manic cape]

150X=35 MOD 31

[manic cape]

so here i can say 150x=4mod31

what would be next step?

[narrow path]

do you mean congruent to

with the 3 lines

[void nova]

!occupied

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[lyric tusk]

since (a,31)=1 for every a is positive integer where a≠31 So
divided both side by 5 you’ll get
||30x con 7 (mod 31)|| and since
||30x con (31-1)x|| con ||-x con 7 (mod31)||
thus
||x con -7|| con ||24 (mod 31)|| as required

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[warm quiver]

'following the diagram, if the coefficients of friction are 0.4 and 0.5, what is the minimum value of applied force F to start movement of the system?'

the hanging body is 1 kg, middle is 2 kg, right is 3 kg.

by coefficients are 0.4 and 0.5, this means that the static coefficient is 0.5 and the kinetic coefficient is 0.4 (its a bit strange but this is how the questions are written, others are like this)

also assume g=10

now the intended answer is 35N, which i got but not quite. i treated the whole system roughly as one body. trying to find the point at which net force = 0 when friction is in play to find the point at which movement begins.

the hanging mass to the left is pulling on everything, with a force of mg or 10N. next, the friction forces for each of the two boxes on the surface. for the 2kg mass, 0.5 x 20 = 10N. same for the 3kg mass, 0.5 x 30 = 15N.

add it all up, i got 35 N going to the left. now the force must be 35N in the other direction to counteract start movement.

however, the force is inclined at 53 degrees. breaking up the force, i got F x cos53 for along the horizontal line. rounded this to 0.6F.

so i take this to mean 0.6F must be greater than or equal to 35N, meaning F is roughly 58.3. however, the given answer is just 35N.

did i do something wrong? or am i correct? (usually im wrong)

[vernal perch]

I think you are right

I checked seems like 35 to the right is the right answer

I would guess the question just mixed up the answer

[warm quiver]

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[dawn hinge]

Here I am trying to prove equality for the first case when f is injective. Is it sufficient to take the contrapositive of "If f is injective then equality" meaning if I prove that "Not equal then f is not injective "

[dawn hinge]

Also I'm having doubts about whether I correctly stated the contrapositive

[crude kayak]

!xy

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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.

[dawn hinge]

It's from Munkres' Topology

[dawn hinge]

This is the page, the problem is at the bottom

[onyx glen]

actually idt going for the contrapositive is a good idea at all.

orrr

no you can make it work actually it's not that bad

[dawn hinge]

What would be a better way(any general hints would be good)

assuming it's equality without injectivity then showing a contradiction?

[onyx glen]

when $S \subset T$ but $S \neq T$, it implies there exists a point in $T$ that doesn't belong to $S$.

[glossy valve]

Ann

[onyx glen]

though... actually

we have to be a bit more careful

[lime ether]

why not just do it directly

[onyx glen]

yeah i guess

do it directly and worry much less about quantifiers as i was about to write

[dawn hinge]

Alright👌

!close

[lime ether]

not quite

!done

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[dawn hinge]

Ah thank you

I'm still very new

!done

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[lime ether]

no worries

😭

[dawn hinge]

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[dawn hinge]

😆

.close

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[gritty rose]

use definition of derivative

[unborn acorn]

ok

[unborn acorn]

but what do i do with that?

if the first question is about differentiability and not derivability

[fast peak]

what do you think the word differentiability means

[unborn acorn]

i know that derivability implies differentiability but not always the otehr way?

orr

for question 1, do i only need to show that the partial derivatives exist and are continuous on C?

[gritty rose]

https://math.libretexts.org/Bookshelves/Analysis/Complex_Analysis_-_A_Visual_and_Interactive_Introduction_(Ponce_Campuzano)/02%3A_Chapter_2/2.03%3A_Complex_Differentiation

one of the first two equations

finding df is a little harder

oh not that much harder. if f(x, y) = u(x,y) + i v(x,y) then

https://complex-analysis.com/content/complex_differentiation.html

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[chilly pecan]

can you explain why a/b has to be between 1 and 2 for dimpled cardioid like $a+b\cos(\theta)$

[glossy valve]

Odd_the_Wolf

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[winter patio]

ok cus im confused w these lns

i know they base is e

so do i do

[rough tundra]

(\log_b(a)+\log_b(c)=\log_b(ac),\quad a,c>0)

[glossy valve]

PajamaMamaLlama

[winter patio]

Ln(x+4)(6-x)

(\ln_e(x+4)+\ln_e(6-x)=\ln_e(x+4)(6-x),\quad a,c>0)

can u do that

does that work

[rough tundra]

indeed it is true that (\ln(x+4)+\ln(6-x)=\ln((6-x)(x+4)))

[glossy valve]

PajamaMamaLlama

[winter patio]

but im confnused about the ln(21)

what am i doing with thta

[rough tundra]

well if that log of (6-x)(x+4) is equal to the log of 21, then (6-x)(x+4) must be equal to 21 itself, yes?

[winter patio]

so we remove the ln?

[rough tundra]

right, mathematically that is, (\ln((6-x)(x+4))=\ln(21)\implies (6-x)(x+4)=21)

[glossy valve]

PajamaMamaLlama

[rough tundra]

where that long arrow is read as "implies"

[brazen stirrup]

I sense the quadratic formula

[winter patio]

me too

ok cus

i have

-x^2-2x=24

we have to do quadratic cus we cant factor cus the x^2 is negative right/

sorry +24

not =

[rough tundra]

we can factor

since -x^2-2x+24=0 ==> x^2+2x-24=0

[winter patio]

ohhh

[rough tundra]

wait

[winter patio]

oh yeahhh why if forgot basic math

[rough tundra]

did you expand that right?

[winter patio]

did i?

[rough tundra]

oh wait

-x^2-2x+24=21 pardon

[winter patio]

OH YEAHHH

these are literally the mistakesi make on my exams

[rough tundra]

happens to everyone :) practicing usually helps with that

[winter patio]

so now i havd

x^2+2x-3

[rough tundra]

then we find two numbers that add to 2 and multiply to -3

[winter patio]

3 and -

1

[rough tundra]

indeed

[winter patio]

so we have x=-3, and 1

[rough tundra]

hang on

I believe you did expansion wrong: (6-x)(x+4)=-x^2+2x+24=21

[winter patio]

omg i did

So it’s reverse

3 and -1

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[digital mica]

hello! i need help with this problem

[digital mica]

i don't know why in step 3 the bottom part of the fraction gets canceled out

and why in the 5th step the (x+4) just dissappears

<@&286206848099549185>

[steel nest]

What’s the translated problem statement?

[digital mica]

i think: radical ecuations that get reduced to second grade

i'll send u the other part

[steel nest]

and this is a solution

[digital mica]

i do understand this one

steps 6 and 7

yes

the problem is getting solved

but i dont understand steps 3 and 5

why the denominator in step 3 and the x+4 in step 5 just get wrecked

[steel nest]

gimme a minute or two to try to parse the algebra

[digital mica]

for sure, tysm

[steel nest]

I think i understand the manipulation done here

[digital mica]

lemme see

[steel nest]

for step 3 if the denominator is not 0 then you can multiply it on each side of the equation and on the right side it will be eaten up by the 0

so implicitly it is assumed here that x is not 3,-4,or -5

[steel nest]

is this satisfactory

[digital mica]

i see

so

it can happen with any given denominator when the equation is =0?

cuz u multiply it on each side

[steel nest]

you can do this simplification when the denominator is not 0 yes, you might have to check the neglected cases x = 3,-4,-5 separatly

[digital mica]

and does this happen also in step 5??

just the inverse

[steel nest]

yep

[digital mica]

u divide each side of the equation

omg it makes so much sense

[steel nest]

you can divide if it’s not 0

[digital mica]

wait

but it is 0...

look

right here

i dont know what happened there

[steel nest]

no to get here we already assumed in step 3 that x is not -4

[digital mica]

wait wait, but why is that we assume that x cant be -4,-5 or +3??

[steel nest]

but then in step 5 there are two possible cases for this equation to be equal 0. either x=-4 it can’t because step 3, or the other expression is 0, what’s hilighted in yellow

[digital mica]

oh

i think i get it

so its like u get 2 different equations =0 for each parenthesis

[steel nest]

i mean one equation for each equal to 0

[digital mica]

yes

i thought it was one whole equation

okay i have one final question

[steel nest]

it’s another of seeing it beside cancelling the x+4

[digital mica]

about the x that cant be +3,-4 or -5

why is that?

[steel nest]

because if x equals one of these 3 values, we would divide by 0, and this operation is undefined

you just can’t divide by 0

[digital mica]

that is right

so

u cant divide by 0

so

[steel nest]

in these case some other simplification might happen if it’s important to take care of these cases

you have to solve another way using the fact that x = 3 for instance

[digital mica]

basically x cant be the number that's next to it with the opposite sign

when in the denominator

in this type of problems

=0

[steel nest]

yep

[digital mica]

alrighty, that would be all

have a nice afternoon, thank you very much

really appreciate it

❤️

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[stuck sleet]

.reopen

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[stuck sleet]

Hello, I speak Spanish. Can someone help me solve this please?

[stuck sleet]

<@&286206848099549185>

[placid oak]

does the overbar mean it's a number?

for example \overbar(abc) where a = 1, b = 2, and c = 3, would be 123?

[stuck sleet]

yes

[placid oak]

what have you tried?

[stuck sleet]

Sorry, I got the exercise wrong.

I have resolution problems with this

[placid oak]

Ok, what have you tried with this one?

[stuck sleet]

I have tried and come to the following conclusions: a is greater than c; z is greater than c; y is greater than b

Since they must all be one-digit numbers and a and c must be different from 0, the only numbers that meet this requirement would be a as 7, c as 3, and z as 4.

By replacing the values ​​the subtraction would be contradictory

[placid oak]

have you tried writing \overbar(abc) as 100 a + 10 b + c, and solving algebraically?

[stuck sleet]

no

My teacher in class taught us something like the minuend plus subtrahend plus subtraction equals two minuends.

I don't know if that is useful

[placid oak]

$$A - B = C \implies A + B + C = 2A$$

try that with your example

[glossy valve]

Shuba

[placid oak]

A, B, and C are all numbers, not the digits in a number

[stuck sleet]

look at this

estos son los unicos valores que cumplen los de 56 pero no cumplen la resta

Have you managed to solve the problem? I've started to think it's poorly stated.

[placid oak]

to be honest, I'm not sure of any smart ways to solve this, other than getting a computer to brute-force through every combination until it finds something that works.

[stuck sleet]

creo que el problema esta mal porque yo intente lo de la computadora

I think the problem is wrong because I tried the computer thing.

[placid oak]

from itertools import product

def num(*digits):
    buffer = 0
    for digit in digits:
        assert 0 <= digit <= 9, "Error: digit is not base 10."
        buffer = buffer * 10 + digit
    return buffer

def find_solution(verbose=False):

    for a,b,c,y,z in product(range(10), repeat=5):

        if verbose:
            print(f"Trying a={a}, b={b}, c={c}, y={y}, z={z}")

        if not (a > c):
            continue

        if not (a**2 + c**2 - z**2 == 56):
            continue

        if not (num(a,b,c) - num(6,y,z) == num(c,b,a)):
            continue

        return a,b,c,y,z

solution = find_solution(verbose=True)
if not solution:
    print(f"No solution found :(")
    exit()

a,b,c,y,z = solution

print(f"\nSolution Found!")
print(f"a={a}, b={b}, c={c}, y={y}, z={z}")
print("")
print(f"{a}{b}{c} - 6{y}{z} == {c}{b}{a}")
print(f"{a}^2 + {c}^2 - {z}^2 == 56")

[stuck sleet]

?

[placid oak]

it's a python program to find the solution

[stuck sleet]

do you execute it?

[placid oak]

Maybe you can write it as \overbar(abc) = 100a + 10b + c for every number, and solve algebraically

which I mean, find a simple algebraic expression, and then keep trying values until one suceeds

or not, there may be too many values to try

I'll give you a hint, c = 1

[stuck sleet]

Bro, if there is an answer, since I'm feeling unmotivated because I think there is no answer.

[placid oak]

there is an answer

[stuck sleet]

It's okay, I'll keep trying.

[stuck sleet]

To solve algebraically should I use this?

[placid oak]

I can't see where you'd use it

[placid oak]

you need the python interpreter

[stuck sleet]

Do you know how to program, bro?

Bro, do you know any way to solve it mathematically since I have to make a solution for my notebook?

[placid oak]

I'm working on that one XD

you can rewrite the equation into $$99(a-c) = 10y + z + 600$$

[glossy valve]

Shuba

[placid oak]

then, treat a - c as it's own digit.

Let $x = a - c$. Then $$99x = 10y + z + 600$$

[glossy valve]

Shuba

[placid oak]

This is as far as I've been able to reduce it.

[stuck sleet]

esto debera tener una utilidad si lo usamos?

[placid oak]

Ah, I've seen something

[stuck sleet]

This should have a use if we use it?

[placid oak]

$$99x = 600 + 10y + z = \overline{6yz}$$

[glossy valve]

Shuba

[placid oak]

So 6yz is a multiple of 99

this should easily tell you what y and z are

which should allow you to easily tell what x is

[stuck sleet]

the maximum value of x would be 1

[placid oak]

Now we know x, y, and z we will use this to find a and c knowing a - c = x.

[stuck sleet]

Sorry, nothing to do with this, my mistake

[placid oak]

there's only 2 options for a and c, only one of which will fit the second equation

hope this helps 🙂

[stuck sleet]

Okay bro, I'll keep trying and I'll give you the solution.

[placid oak]

1. Use \overline{abc} = 100a + 10b + c to expand the first equation.
2. Simplify to get 99(a-c) = \overline{6yz}
3. Recognise this means the RHS is a multiple of 99, and solve for y, z, and (a-c)
4. Write out all possible combinations of a and c using the solution for (a-c)
5. Use the second equation to determine which we need to use

[stuck sleet]

99(a-c) = 600-10y-z*

[placid oak]

99(a-c) = 600 + 10y + z*

plus

[stuck sleet]

why plus brother

I think it's because the negative sign comes before the 600+10y+z.

I should put parentheses

I have come to the conclusion that x equals 7. How could I continue developing the resolution?

@placid oak

[full forum]

@stuck sleet Has your question been resolved?

[full forum]

@stuck sleet Has your question been resolved?

[half ibex]

Where's your original question?

[stuck sleet]

wait moment pls

[stuck sleet]

.

here

We found the values ​​of the unknowns with a Python code that tested all possible numbers, but I need the mathematical solution.

[stuck sleet]

for the resolution I stayed here

[hearty sandal]

Xd

[full forum]

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@stuck sleet Has your question been resolved?

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[hoary ember]

how can you prove (a)?

[fervent owl]

if x is an 11th power mod p use FLT to deduce something about x

[hoary ember]

well by flt x^{p-1}=1 mod p

you can write it as (x^11)^k=1 mod p

[fervent owl]

so if x is not 0

an 11th power is a root of x^k - 1 = 0 mod p

so how many of them are there?

[hoary ember]

uhh what do you mean?

[fervent owl]

so let's say x is an 11th power

x = y^11

then x^k = y^(p-1) = 1 mod p

so x^k - 1 = 0 mod p

[hoary ember]

oo

is it k

cause a power-k polynomial has k solutions right

[jolly ferry]

the third one, idk if I used the right formulas, is it correct to use the liner interpolation to find r?

[hoary ember]

or is this completely unrelatsd and im dumb

[fervent owl]

well at most k

[hoary ember]

!occupied, and delete the message please

[full forum]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[fervent owl]

tbh the thing i guided you through only gives you that it has at most k solutions

which for the most part still motivates why we should pick primes that are 1 mod 11

anyway if you wanna prove that there are exactly k+1 11th powers, use the fact that we have a primitive root mod p

[hoary ember]

oh so basically the reason your finding p≡1 mod 11 is to find p with as little ranges for x^11 mod p?

[fervent owl]

yh

[hoary ember]

oo

ok then the smallest prime is p=23

[fervent owl]

yh and 23 should work i think

but anyway if that doesn't work you'd try 67 instead

anyway knowing how to choose moduli is a really useful trick to know about

[hoary ember]

a^11+11b^11+19c^11=0 mod 23 hmm

[hoary ember]

alright ill keep that in mind

[fervent owl]

another question i have in mind is

do there exist integers x,y, such that x^3 + y^4 = 2015^2015

(i think it was 2015^2015 at least)

and also is there a positive integer n such that n^7 - 77 is a fibonacci number

[hoary ember]

wtf 💔

btw (cheating with calculator) x^11=0,1,22 mod 23, is there a way to find that?

cus with that i think you can solve the question with that as x^11=0 mod 23 is when 23|x t

[fervent owl]

x^11 is a root of t^2 - 1 = 0 mod 23

so x^11 can be 0, +-1

[hoary ember]

uhh what

how can it be 0

[full forum]

@hoary ember Has your question been resolved?

[hoary ember]

oh wait

.csolved

.solved

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[blissful ledge]

△ABC is an acute triangle, P is a point on the circumcircle Γ of △ABC, and line PB and PC are tangent to Γ. ∠BPC=θ, what's the value of cos A (in terms of θ)?

[full forum]

@blissful ledge Has your question been resolved?

[tiny salmon]

connect OB, OC

OB is perpendicular with BP and so are OC and CP

therefore <BOC = 180-theta

and <A= 180-theta/2

[blissful ledge]

Oh I got it. Thanks.

close.

I hate geometry.

close.

[alpine pecan]

. close

[rose yew]

.close

[sullen hinge]

.close

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[median gust]

I know the answer is correct because I saw the answers, but is my proof right as well?

[full forum]

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[dusky girder]

HELP I DONT GET THIS

[slate violet]

which part don't you get?

[dusky girder]

after letting t=tanx

they skipped a step

[slate violet]

ah, so do you recall the integral of $\frac{1}{a^2 + t^2} \ dt$?

[glossy valve]

south

[slate violet]

a is a constant

[dusky girder]

ohhhhh

oh ok thanks

it's midnight here and my brain is missing neurones and my exam is tmr so im low-key panicking

i usually use θ

thanksss

[torn jolt]

Goodluck tomorrow

[dusky girder]

thank you

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[drowsy plover]

can anyone help me on how to differentiate a complex function

(2z^2 + i)^5

z= x +iy

[fast peak]

do you have to do it from first principles?

or are you allowed to use differentiation rules

[drowsy plover]

teacher said i cannot differentiate the way we differentiate real functions

[fast peak]

so what are you allowed to use

[drowsy plover]

first principle and differentiation rules

[onyx glen]

?!

oh like forbidden?

[fast peak]

so you are allowed differentiation rules but not the usual differentiation rules you know from real functions?

[drowsy plover]

kind of i guess

[fast peak]

which rules are you allowed to use

[drowsy plover]

yeah

[onyx glen]

dear lord that's gonna be so ass if you have to do it from first principles

[onyx glen]

seconding this

[drowsy plover]

power rules only

[fast peak]

multiply it out then

[drowsy plover]

yeah idk what to do

[fast peak]

binomial theorem

[onyx glen]

so (f+g)' = f'+g' is forbidden?

[onyx glen]

what if that is also forbidden?

[drowsy plover]

i forgot how to differentiate complex function

[fast peak]

oh dear lord pls no

the whole point is that you differentiate complex functions the same way as real functions

there is nothing special to do

[drowsy plover]

ohh thanks

[fast peak]

but you seemingly arent allowed to do that

[drowsy plover]

i asked friend of mine he said we cant do it the way we differentiate real ones

as teacher said it earlier

[hidden harbor]

you should have a more clear definition of "not like real ones". As in, which properties in particular you're allowed (or not) to use

[drowsy plover]

binomial and cr eqn will be enough

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[safe plaza]

Let's say we have a function like this:

x+2 = y

Now, is "x+2" called definition, and is "y" called output?

For context, I'm making a graph renderer and looking for appropiate names for my variables

[safe plaza]

I've also heard that "x+2" should be called "expression", but not sure if it's correct

[short tinsel]

Definition works.
The whole statement "y = x+2" is an equation, and x+2 is an expression, but these names would be generally understandable.

[safe plaza]

that seems good, thanks

[fast peak]

the whole thing is the definition of the function

at that point you might aswell just call it formula

[rapid rain]

x+2 would more likely be the called the expression/formula of the function

[safe plaza]

you mean like this?

"x+2" <- formula
"y" <- output

[fast peak]

I mean in the end you have to live with your code

I would start to question the whole setup. why do you need an output variable, its just gonna be y all the time, no?

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[thick hedge]

nvm

I'll do it later

slppey

eepy

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[torn jolt]

for finding domain and stuff , when to use wavy curve method and when to use intersection ( like writing out diff ineqalities and taking intersection)
i need a generalisation on this
( im having very bad brain fog rn , so i cant really pull things together :/ )

[alpine pecan]

its basically the same thing
if u r given many functions
u use wavy curve for the functions, then use the intersection part on them all

ull get it with more practice

[torn jolt]

alr

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[static verge]

why should discriminant be less then 0 for c

[robust slate]

You want the condition for invariant lines to fail

You showed that the condition is equivalent to that quadratic having (real) solutions

No invariant lines -> condition fails -> quadratic has no real solutions -> discriminant is negative

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[shut trellis]

how do I find that angle?

[quiet depot]

extrapolate the weight vector to form a right triangle. Sum of the angles in a triangle are 180, and the intersection between the ground and the weight vector is 90 degrees, so the angle you're looking for is 90-theta

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[swift bridge]

How are we able to combine $f(x+h) - f(x)$ into a single series?

[glossy valve]

LXDL

[swift bridge]

Like it seems we just somehow drop the limits

[gritty rose]

you can combine by the Weierstrass M-test

the terms inf f(x) and f(x+h) converge uniformly so the sum can be done term by term

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[swift bridge]

Ah okay. This doesn’t have to do with the fact that we can fix x and h and then apply limit rules?

[swift bridge]

Weiestrass M-test implies that the series converge uniformly to f(x) but then how come Inifork convergence implies we can sum term by term?

Like isn’t that still possible since both series converge?

[swift bridge]

Specifically is uniform convergence necessary?

To add term by term wise

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[plucky aurora]

Is the answer number 2?

[gloomy lance]

Perhaps

Can you say why it would be number 2?

[plucky aurora]

I put it in my calc and its the closest looking one

💀

[gloomy lance]

are you allowed to use calc at all times?

[plucky aurora]

Na

On my test its taken away

@gloomy lance can u explain how pie effects a graph

[gloomy lance]

This isn't something exclusive to pi, any other number in the place of pi/4 would change the graph

in general, a function y=f(x) will have a horizontal shift when you apply f(x+a)

f(x+a) if a is positive will shift the graph to the left, if it is negative it will shift the graph to the right

f(x+3) shifts the function's graph to the left by 3 units

[plucky aurora]

And x-3 moves the graph to the right

[gloomy lance]

exactly

[plucky aurora]

Alr so here its a fraction

Positive frac

So it goes to the left

[gloomy lance]

so if we have f(x)=tanx, then f(x+pi/4)=tan(x+pi/4) shifts the graph of tanx to the left by pi/4

[gloomy lance]

yes

[plucky aurora]

But how much is pie over 4

Idk how much to move

[gloomy lance]

pi/4 is just pi/4

[plucky aurora]

🗿

[gloomy lance]

you move pi/4

[plucky aurora]

O they all over 4

So jus move 1 to the left?

[gloomy lance]

you take the original points and subtract pi/4 from them

for example, the point at x=0 will be shifted to x=-pi/4

[plucky aurora]

Yea

So

Move 1 to the left

And das why its number 2

[gloomy lance]

yes

[plucky aurora]

Alr now what if

The negative was on the outside

Instead od x

[gloomy lance]

what do you mean

[plucky aurora]

Like negative tan x

Does it still go to the right

[gloomy lance]

-tan(x) would flip the graph

,w graph tanx

[glossy valve]

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[gloomy lance]

,w graph -tanx

[glossy valve]

Results provided by WolframAlpha

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[gloomy lance]

think of it as flipping upside down

like the graph of x²

,w graph x²

[glossy valve]

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[gloomy lance]

,w graph -x²

[glossy valve]

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[plucky aurora]

Wait upside down or side to side

[gloomy lance]

upside down

in this case it also makes sense to think of it as flipping to the other side

[plucky aurora]

So a negative infront of the tan or log or wtv

Flips it upside down

[gloomy lance]

yes

[plucky aurora]

Ok so

It would he number 1

[gloomy lance]

you sent the same picture as before

[plucky aurora]

Yea

[gloomy lance]

ah i see

[plucky aurora]

The question was jus about negative

[gloomy lance]

yes -tanx would be number 1

[plucky aurora]

In the front of tan

Alr thx

[gloomy lance]

yw

[plucky aurora]

What does this mean

@gloomy lance

[gloomy lance]

What, period?

[plucky aurora]

Yea

[gloomy lance]

The period of a function is the distance at which the function starts repeating itself basically

[plucky aurora]

Oh

[gloomy lance]

For sinx and cosx that would be 2pi, for tanx its pi

[plucky aurora]

So dis is tan

So its jus pi

Ooo

[gloomy lance]

The formula for the period of tan(a*x) would be pi/|a|

[plucky aurora]

Cuz the graph hits pi again on the next shift

[gloomy lance]

So for tan(3x), the period is pi/3

[plucky aurora]

Pi over 3

Ok so

For

Would it be 2pie over 4

[gloomy lance]

No

You divide by the absolute value of the number that is multiplying x

tan(ax+b) has period pi/|a|

Also the period isn't affected if there's a number multiplying tanx

4tan(x-pi/4) has the same period as 3tan(x-pi/4)

[plucky aurora]

What is the absolute value

[gloomy lance]

You have some catching up to do man

The absolute value of a number x, denoted |x| is defined as
|x|=x, if x≥0
|x|=-x, if x<0

I.e. , it turns the number that you input positive

If its already positive it stays positive

|3|=3, |-3|=3

[plucky aurora]

Right

[plucky aurora]

But dis isnt in thequestion

[gloomy lance]

It's in the formula for the period of a tangent function of the form c*tan(ax+b)

Which is pi/|a|

[plucky aurora]

Can u show an ex

@gloomy lance

[gloomy lance]

tan(3-x) has a period of pi/|-1| = pi

8tan(2x-1) has a period of pi/|2|=pi/2

[plucky aurora]

Can u explain how

Thos arw the answers

[gloomy lance]

Because of this

[plucky aurora]

I dont rlly get that

[gloomy lance]

It's just the formula

You should've been taught that

Have you paid attention to class

The formula for the period of sin(ax) and cos(ax) is 2pi/|a|

for tan(ax) it's pi/|a|

[violet lance]

the period of $\tan(ax+b)$ is $\dfrac{\pi}{|a|}$

[glossy valve]

r4f43l1006

[violet lance]

check it in desmos or smth

[plucky aurora]

But dis one has a 4 in the front

[violet lance]

the 4 in the front doesn't affect the x values

[plucky aurora]

Oo

[plucky aurora]

So will tan graphs always be pi

For p

And same with cot

[violet lance]

no its pi/|a|

[plucky aurora]

Pi over |a|

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[plucky aurora]

Mnfok

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[odd hill]

if they say
a ⊥ b ⊥ c
does that imply a ⊥ c
because A could be ⊥ to C
but
A could also be parallel
to C

[wide sundial]

Well perp is not a transitive relation

So they’re being ambiguous with it

Where are you seeing this?

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[odd hill]

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[torn jolt]

a bit of general question I have: could one visualise volumes of solids of revolution on desmos 3D?

[torn jolt]

i lowkey think it would be fun to look at

[umbral dome]

from the pins in #calculus

[torn jolt]

very cool, i should have looked there first

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[sharp harbor]

Need help in the last question

<@&286206848099549185>

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@sturdy valve Has your question been resolved?

[modest obsidian]

feels like something to do with vieta's formula

uhhh for any polynomial like $(x-r_1)(x-r_2)(x-r_3)\dots (x-r_n)$ its coefficient are symmetric sums of roots

[glossy valve]

parabolicinsanity

[modest obsidian]

i think...

something like that

so i think coeff of this would just be the sum of the roots... maybe i think

now you do

just brush up on vieta's formulas

i guess

ex : $(x-r_1)(x-r_2)(x-r_3)=x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_2r_3+r_3r_1)x-r_1r_2r_3$

[glossy valve]

parabolicinsanity

[modest obsidian]

no, they're symmetric sums

different things

just look into this

just keep in mind these are

uhhh

symmetric sums

so symmetric sum notation

weird that-

you have a problem on that but don't know how to do it...

😔 that's just vieta's formulae, brush up on it i guess and you'll be fine

but do notice

the first term is

(r-x)

so reverse it

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[modest obsidian]

happy mathing

you just have to find it so

there's a certain joy in being able to solve mathematical problems

find it

you'll never be able to stop after that

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[modest obsidian]

😔 opened another channel but sure make it quick

then close it

[tacit siren]

Why does it have to be done in 3 minutes?test question?

<@&268886789983436800>

[austere cove]

No academic dishonesty

Check the #rules

.close

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[modest obsidian]

weird...

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[torn jolt]

I'm trying to answer part e iii. In the previous parts, I claimed that B and C were both not compact so I attempted to use Hausdorfness as a topological property to prove B is not homeomorphic to C. I claimed C is Hausdorff as for all distinct x,y in Z, we can take the open neighbourhoods U of x and V of y to be the balls of radius 1/2 centered at x and y respectively which are disjoint. Im now struggling to show B is not Hausdorff. Ive written the definition of Hausdorffness but wasnt sure what to do then. I'm also wondering if contradiction is the correct path for this. Any help would be appreciated!

[rapid rain]

contradiction is always a bit easier when you want to prove two spaces are not homeomorphic to each other, because then supposing they are you get existence of a homeomorphism between them and you can play with it

so the idea is that B has accumulation points but C doesn't, all points of C are isolated

which is what you did

To show B is hausdorff though

B should be hausdorff shouldn't it?

for any two distinct points x and y of B

take B(x,delta) and B(y,delta) with delta = |x-y|/3 as neighborhoods of x and y respectively

A metric space is always gonna be hausdorff

[torn jolt]

Ah yeah good point. I forgot all metric spaces were Hausdorff haha. We've only covered the topological properties of Hausdorff, compactness and connectedness and since (I think) theyre both not compact, that only leaves Hausdorff and Connectednes and I'm fairly sure theyre both disconnected, so I'm not really sure how else to show they're not homeomorphic.

Wait unless they are homeomorphic

[rapid rain]

they're not

If B and C were homeomorphic, let f:B->C be a homeomorphism

{0} is an open subset of C (why?)
thus {a} is an open subset of B for some a in B

[torn jolt]

Is {0} an open subset of C as it has subspace toplogy and the interval say (-1/2,1/2) is open in standard topology so (-1/2,1/2) intersect C which is {0} is open in C?

And then {a} cant be open in B as whenever we intersect B with an open subset of R, it will include irrationals which aren't in B or something?

[rapid rain]

uh no that's not it

if you intersect B with an open subset of R

by definition of B

it's only gonna contain rationals

[torn jolt]

Ah ofc haha

[rapid rain]

the idea is indeed to look at what are the open subsets of B for its topology

just imagine how many numbers are gonna be in open subsets of B

imagine (a,b) intersected with B

and get rid of cases where it's empty

[torn jolt]

I see. Are there gonna be an infinite amount of points when (a,b) is intersected with B? Could this possibly have something to do with the closure of B being the whole interval [0,1] or something?

[rapid rain]

B is dense in [0,1] (or the closure of B in R is [0,1])

yeah

so it's gonna coincide with every single open subset of [0,1]

you can split an open subset into as many disjoint open subsets as you want (with some extra points here and there)

there's an infinity of rational numbers between a and b

[torn jolt]

Ah ok that makes sense!

I think I understand whats going on now. Tysm!

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[winter patio]

please

[winter patio]

like i juts cant with these natural logs

[gritty rose]

take log of both sides

[winter patio]

plesase can we like step by step

im so frustrated w it

[gritty rose]

,tex .log power

[glossy valve]

riemann

[winter patio]

so

[gritty rose]

what do you get after taking log of both sides

[winter patio]

dont we have to use ln

i thoughjt

cus its e

[gritty rose]

doesn't matter the base

you pick either base 2 or base e

[winter patio]

so log(x-3)=2log(x+2)

[gritty rose]

no

[winter patio]

log(e^x-3)

[gritty rose]

y is in the exponent on the left side

[winter patio]

so e is the x right

[gritty rose]

for the left side, yes

[winter patio]

is x-3 not y

[gritty rose]

x-3 is in the exponent yes

[winter patio]

so is this correvt

[gritty rose]

you're missing parentheses

[winter patio]

log(e^(x-3))

[gritty rose]

correct. that's the left side. now use this

[winter patio]

its 2log(x+2)

waity no

(x+2)log2

[gritty rose]

correct

[winter patio]

i dont kniw what to do net

like im extremel confuse

d

[gritty rose]

solve for x

[winter patio]

does log e equal 1

[gritty rose]

yes

[winter patio]

i have

(x-3)=(x+3)log2

am i ssupposed to divide by log

2

oh my god

girl can i have a new helper

like i just cant

[gritty rose]

wut

why are you being dramatic

[gritty rose]

solve for x

[winter patio]

girl i asked u to guide me im iunsure on how to do the whole thing its just frustratingatm

[gritty rose]

you're making progress

[winter patio]

im unsure on how to solbe for x with this set up, am i distributinh

[gritty rose]

yes

$a(b+c) = ab + ac$

[glossy valve]

riemann

[sturdy valve]

hes one of the best helpers ull get lol

[winter patio]

im sorry

idk whats wrong w my brain todsy

but it jusy doesnt wnt to understand for some resson

i stil dont understand

[gritty rose]

do this

[gritty rose]

to the right side

[winter patio]

so is it xLog2 +3log2

[gritty rose]

correct

[winter patio]

i literally have

x-3=xLog2 +3log2

.close

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[glossy valve]

Dhairya

[wispy lichen]

at alpha = 1 its zero so 0

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[winter patio]

im back

[winter patio]

i know this is elementary

ik how to do it but my teacher want it done w no calculator

like how is tht possible

[vestal moat]

How do you do it ? And Let’s see if you can figure out a way without a calculator.

[winter patio]

yes pls

cus

i have

2800(1+0.15)^4

for part

a

i simplified it ot

to

2800(4.60)

but thats def wrong

yesa it is

[wispy lichen]

maybe use the binomial

[tender delta]

Do you know about compound interest

[winter patio]

no i forgot it

[winter patio]

example

Where yall went

[winter patio]

Girl

I don’t think we need all this 😭

[wispy lichen]

use the squared term

[grave elm]

seems like you just did 4 * 1.15 = 4.60

but it's 1.15^4 = 1.15 * 1.15 * 1.15 * 1.15

[wispy lichen]

if its too small you can approx it to 1.60

[grave elm]

i wouldnt wanna do that without calculator

if i had to do it without calculator, 1.15 = 23 / 20 so i'd instead do that

23 * 23 = 529, if we are allowed to approximate a bit, say it's 530

530 * 530 = 280, 900

now just divide that by 20^4 = 16 * 10^4

28,09 / 16 (approximately 28 / 16 = 7 / 4 = 1.75)

1.75

[winter patio]

But isn’t that much like I don’t understand why he just doesn’t let us use a calculator

He’s so evil

[grave elm]

i honestly dont understand it either

how much time do you have for this

[winter patio]

It’s a review for my exam

[grave elm]

will you have exam on your exam

*calculators

[winter patio]

Mo

No

Not allowed

[grave elm]

how accurate does he require the stuff to be

[winter patio]

Idk I asked him am I just supposed to leave it as a formula

Waiting for his response