help-27

[wicked rover]

but u still had x leftover after sub

[ivory salmon]

ooo okay

[wicked rover]

general rule is all x must go after taking limit

[ivory salmon]

so remove x then sub and take the limit?

[wicked rover]

well u take the limit by subbing, theyre not separate steps

if u sub correctly u get

[ivory salmon]

yea sorry thats what i meant 😭

[wicked rover]

$\frac{-\sqrt{0+16}}{1+0}$

[woven radish]

ロケットジャンプ

[wicked rover]

in short response its good to replace before simplifying 🙂

[ivory salmon]

🫡 on it captain

[gleaming fog]

wow

thhat was long

[wicked rover]

we took it slow

[ivory salmon]

so basically what im understanding about limits to inf is to factor the dominant on the top and bottom, remove any x, then take the limit?

[wicked rover]

to avoid common student errors

[ivory salmon]

yea i really appreciate that we took it slow 😭 limits are the most confusing part about calc for me

[wicked rover]

when u say remove any x, do u mean cancel common factors?

[ivory salmon]

yea

[wicked rover]

yes

[ivory salmon]

like how we cancelled the x/x

[wicked rover]

u can also cancel x^7 if its common

[ivory salmon]

x^7?

[wicked rover]

if u have to factor out x^7 on top and bottom

[gleaming fog]

just to make sure when we cancel out stuff like that it still leaves behind the hole on the graph right?

[ivory salmon]

why specifically x^7? /genq

[wicked rover]

not specifically

but there are examples where u cancel higher powers of x

[ivory salmon]

ohhh okay

[wicked rover]

yes

[ivory salmon]

i understand a lot better now, tysm!!

[wicked rover]

np

[faint gorge]

so much joy here

[ivory salmon]

.close

[devout snow]

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[neon wagon]

Hi so here you have to check if the like dimension that goes through the points is parallel to one of the coordinate dimensions

[neon wagon]

wait let me just translate

The plane passing through the points ABC is parallel to a coordinate plane

so when do i know if thats the case generally?

if one of the coordinate on the vectors is 0

then its parallel?

[faint gorge]

die Punkte liegen alle in der Ebene z = 4

[neon wagon]

😭

hi

also dings

[faint gorge]

das erkennst du daran weil, der Wert konstant bleibt

[neon wagon]

muss bei den punkten mindestens einmal eine 0 vorkommen?

also bei beiden vektoren

[faint gorge]

nicht wirklich

die können irgendwie in der eben z = 4 liegen

oder meinst du was anderes

[neon wagon]

also

wenn die ebene parallel zur koordinaten ebene ist

dann muss bei den vektoren x y oder z 0 sein oder

[faint gorge]

meinst du sowas wie 0x+0y+z=4

[neon wagon]

warte

ok doch du hast recht

[faint gorge]

achso ich versteh glaube ich

wenn du die ebenen darstellung hast meinst du?

die richtungsvektoren?

[neon wagon]

aber wie weiß ich generell ob die ebene die durch die punkte geht parallel zu den koordinatenebene ist

[faint gorge]

also angenommen du hast einen ortsvektor

Ich weiß nicht was genau Q war aber

Aber angenommen Q ist ein Ortsvektor

damit du feststellst ob es eine ebene ist die parallel zu einer koordinatenebene ist da muss zum beispiel hier die z komponente der Richtungsvektoren 0 sein

weil ansonsten wäre die ebene schief

weil dann ein Vektor sozusagen nicht mehr flach liegt

[neon wagon]

meinst du mit ortsfaktor

ein vektor der bei 0 anfängt

[faint gorge]

ein ortsvektor ist einfach irgendein vektor

sozusagen ein startpunkt aus

[neon wagon]

hm ok

aber warte

[neon wagon]

weißt du warum man hier minus rechnen muss

[faint gorge]

so ist das definiert

ich versuch mal ein Bild zu malen

QR heißt du willst von Q nach R gelangen

naja

Du fängst bei Q

[neon wagon]

ja

also R - Q

achso

ja

[faint gorge]

und gehst in die entgegengesetzte Richtung

das ist -Q

und dann gehst du entlang +R

also rechnst du (-Q)+R also R-Q

[neon wagon]

hä aber

die entgegengesetze richtung

achso

ok ich check

aber was hast du da gezeichnet?

[faint gorge]

ja wie man R-Q geometrisch interpretiert

Richtungen

Pfade

Vektoren

Intuition hier rein bringen

[neon wagon]

was hat dann die frau gezeichnet?

also

die vorherige zeichnung

[faint gorge]

das bist du der da entlang läuft

entlang -Q und dann R

schließlich zählst du es ja zsm den Weg

[neon wagon]

also wenn ich von q -q entlang gehe

und dann plus R

dann komme ich auf R

[faint gorge]

dann kommst du von Q nach R

ja

[neon wagon]

digga

kann man nicht einfach q + r machen

[faint gorge]

ja mach dann bist du aber nicht bei R

[neon wagon]

wo bin ich dann sonst hö

[faint gorge]

du verfährst dich komplett

[neon wagon]

RQ oder was

[faint gorge]

😂

das passiert wenn du einfach Q+R machst

[neon wagon]

aber woher weiß ich wann ich plus rechnen muss

und wann minus

[faint gorge]

merk dir das einfach

[neon wagon]

ist das 0 oder o

[faint gorge]

A nach B machst du B - A

[faint gorge]

O steht für den Nullvektor ja

ist die eigentliche Schreibweise

damit man einen Bezug hat

[neon wagon]

ich schreibe freitag eine kurzarbeit ich checke nichts

aber egal

also dings

was wäre aber A+B

[faint gorge]

hä?😂

dann stell doch fragen

weiter machen bringt dann auch nichts

das baut sich auf

[neon wagon]

ja

also

manchmal muss man doch auch vektoren addieren

[faint gorge]

ja

moment gibt eine sekunde

guck mal nochmal

Stell dir vor du bist bei A und willst nach B

orange da ist baustelle zum Beispiel

also hast du keine Wahl

du hast dich verlaufen und du must A zurück gehen also -A

angelangt bei O gehst du entlang B

um nach B zu kommen

jetzt addierst du den Weg (-A) + B = B - A das ist dann die Richtung um von A nach B zu kommen

[neon wagon]

ahhhhhhh

A ist der startpunkt?

[faint gorge]

Im Prinzip ist es besser zu sagen, du addierst immer die Vektoren, die Frage ist wann gehst du vorwärts, wann rückwärts

[faint gorge]

Ja

[faint gorge]

Angenommen du bist bei O und will nach B dann kannst du dich zwischen zwei Wegen entscheiden

entweder blau oder grün + orange

Also

OB = O + B sozusagen
OB = A + AB = A + (B - A) = B

funktioniert beides

in beiden fällen läufst du nur vorwärts

aber vorher warst du gezwungen rückwärts zu gehen, und da sagt man mathematisch minus

um die Gegenrichtung zu benennen

[neon wagon]

also bei unseren beispieLß

?

[faint gorge]

was mit dem

[neon wagon]

du meintest

vorher warst du gezwungen ..

also meinst du das hier

hier geht man aber doch nicht minus oder

[faint gorge]

ja hier gehst du auch von Q erst zurück zum Bezugspunkt und dann entlang R

[neon wagon]

okay danke

ok

es wird eine lange nacht

[faint gorge]

hast du mehr als vektorgeometrie

bis freitag

[neon wagon]

nee ich glaub nicht

also

weißt du noch letzte woche

diese parallelogram aufgabe

sowas halt

[faint gorge]

ja sowas halt ist vektorgeometrie

[neon wagon]

ok warte

5 minuten pause

[neon wagon]

ist das machbar

[faint gorge]

ja

ich vergesse aber immer dass du noch schule gehst haha

und wahrscheinlich anderes zeug

das machts bisschen stressiger

[neon wagon]

ok ok

also

mischprobleme???

digga

ehrenlos

egal

ich check dieses skalarprodukt nicht

[faint gorge]

wie das funktioniert?

[neon wagon]

warte

[faint gorge]

lmao

[neon wagon]

was soll dieses r sein

[faint gorge]

das ist nicht das Skalarprodukt

oh man die Terminologie is so scheiße auf deutsch

skalar ist einfach eine zahl

wie 4 mal (1,2,3) hier ist 4 die zahl

das ist doch intuitiv

einfach mit jeder komponente multiplizieren

[neon wagon]

achso

hä digga

warum muss man jetzt extra skalar sagen

[faint gorge]

Soll ich es dir erklären hahaha

In der Mathematik ist das Skalarprodukt eine Art Funktion

zwei Vektoren bilden auf einen Skalar ab

des wegen Skalarprodukt

das ist der Grund

skalar Multiplikation heißt wortwörtlich aber zahl mal vektor nicht vektor mal vektor

[neon wagon]

ok warte

[neon wagon]

für was braucht man die meistens

[faint gorge]

im Abi wirst dus brauchen oder im test

[neon wagon]

ja

aber

ich mein

so von der fragestellung her

werden die mir direkt fragen was ist das skalarprodukt oder wie

[faint gorge]

hahaha das denke ich nicht

zu 99 %

du wirst da iwas rechnen halt

ist halt schule

du verwendest das gelernte an, und tust es nicht erklären

der mathelehrer hat auch nichts davon wenn du auswendig lernst, wie das skalarprodukt funktioniert

[neon wagon]

ja

was wäre ein beispiel dafür

wie viel punkte hattest du im mathe abi

[faint gorge]

berechnen Sie das Skalarprodukt der Vektoren (1,2,3) und (3,4,5)

Welchen Winkel haben die Vektoren?

gut das kommt auch drauf an mit oder ohne taschenrechner wie die vektoren sind

[faint gorge]

genug zum Bestehen

[neon wagon]

ja

ok ok

[neon wagon]

hattest du auch Lk

[faint gorge]

denje ja kp mehr

[neon wagon]

guckmal bruder was ist das für ein scheiß

[faint gorge]

ja das ist eine perfekte übung

wo du das Skalarprodukt brauchst

weil die Norm eines Vektors ist nichts weiteres wie die Wurzel aus dem Skalarprodukt eines Vektors mit sich selber

oder einfacher

[faint gorge]

das rechnest du aus und setzt es gleich 10 und löst für b

das ist wie als wenn du eine quadratische gleichung lösen würdest

so 8. klasse

was du hier können musst also sind die Definitionen

was bedeutet nochmal Länge mathematisch

und dann folgt der Rest von alleine

[neon wagon]

ja

die länge

[faint gorge]

jetzt anwenden

[neon wagon]

ok warte ich überlege

also soll ich hö

ich versteh nicht was gefragt ist

[faint gorge]

also

ich bin mir sicher du bist gerade erschöpft

weil die aufgabenstellung ist eigentlich klar

die Länge soll gleich 10 sein von a

[faint gorge]

und das ist die Formel für die Länge

jetzt musst du a mal a rechnen, das ist nichts weiteres wie das skalarprodukt

[neon wagon]

ist der teil b

also die koordinaten von b

achso

es ist abc

und nicht x y z

oder

[faint gorge]

what

[neon wagon]

ach digga

[faint gorge]

ist doch wurst

[neon wagon]

ich bin grad so lost

da ist ja 2b^2

ok

[faint gorge]

2b^2-4 ist die erste Komponente

[neon wagon]

ahhh

also sind die anderen komponenten 0

also

9 * 0

oder

[faint gorge]

nein

lass mich kurz noch zeigen wie das skalarprodukt geht

[faint gorge]

bei 3 Komponenten einfach die Dritte mal die Dritte

[neon wagon]

aber wir haben doch nur a

also vektor a

[faint gorge]

mit sich selber

[neon wagon]

achso

[faint gorge]

du machst dann a mal a

[neon wagon]

ja deswegen meinte ich

warte

also

digga die sind so dumm

warum schreiben die -4

man denkt es ist noch ein vektor da drinne

ja genau

also 10 =

[faint gorge]

jetzt machst du weiter sonst bringt das nichts

[neon wagon]

ja

[faint gorge]

du musst den test schreiben nicht ich

[neon wagon]

ja hast recht sir

[faint gorge]

danke sie

[neon wagon]

also 4b^2+97+wurzel aus 3 ^2=10

dann einfach nach b umstellen?

[faint gorge]

[ \sqrt{ \begin{pmatrix} 2b^2-4 \ 9 \ \sqrt{3} \end{pmatrix} \dotproduct \begin{pmatrix} 2b^2-4 \ 9 \ \sqrt{3} \end{pmatrix} } = \sqrt{(2b^2-4)^2+9^2+(\sqrt{3})^2} \stackrel{!}{=} 10 ]

[faint gorge]

keine ahnung wie ich das lesen sol

[neon wagon]

ja das meinte ich

kann ich die -4^2 direkt mit der 9^2 zusammenfassen

[faint gorge]

nein...

du musst erstmal die quadrate weg bekommen

und dann vereinfachen

[faint gorge]

yep dann kannst du die terme zsm fassen und beide seiten quadrieren

und diese quartische gleichung kann man mit substitution in eine quadratische umwandeln

ja das alles baut auf von der 11.

[neon wagon]

locker

ich kann das noch

ahh warte

binomische formel

usw

[faint gorge]

ja cooked

[neon wagon]

wie

warte

ist doch richtig

[faint gorge]

bro ich geh jetzt penne

[neon wagon]

ok aber

war richtig nh

binomische formel

[faint gorge]

ja

[neon wagon]

gute nacht aber

danke

[faint gorge]

gute nacht

[devout snow]

@neon wagon Has your question been resolved?

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[uncut tiger]

can someone help me

[uncut tiger]

where do i start

.close

[devout snow]

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[oblique girder]

Hello! looking for help with quadratic equations

[midnight totem]

have any specific questions?

[oblique girder]

Yeah I would say a few, I'm just starting it now for a college algebra class but don't really understand a few steps

[midnight totem]

okay ask

[oblique girder]

Alrighty!

thanks

Alright I am being asked to solve quadratic equations by using the zero product property

The formula is a^2 + bx + c = 0

So for the first equation

I have x^2 - 8x = 0

[glossy basalt]

You have to factorize that

[oblique girder]

I'm extremely confused on factoring

yeah thats where I am extremely lost

[glossy basalt]

$xx - 8x$

[woven radish]

Luca M

[glossy basalt]

whats the common factor

[oblique girder]

x ?

[glossy basalt]

exactly

so you can take it outside

and re-write as

[oblique girder]

just to make sure im on the right page

that x * x

[glossy basalt]

$x(x-8)$

[woven radish]

Luca M

[oblique girder]

is because x is squared

[glossy basalt]

yeah. x^2 is just notation that means the same thing as x * x

[oblique girder]

I guess im also confused as in like

what I am allowed to do

so if its common amongst the equation thats what I factor ?

[glossy basalt]

yeah. Every term needs to have it

[oblique girder]

so a portion of that X^2 has to go somewhere

which is why we now have (x - 8)

[glossy basalt]

yeah

[oblique girder]

Alright im sorta there so far

so now we have x (x - 8) = 0

wait sorry

what happened to the x that was orginally 8x

[glossy basalt]

so in x^2-8x, we have 2 terms

One is x^2, and one is -8x

When we are factorising, we are taking out a common factor from all the terms

Which we saw was x

So we remove that x from every term

Leaving just x-8

And we then multiply this that is left all by the x we took out

[oblique girder]

oh so we are allowed to just "remove" them

[glossy basalt]

As long as you then multiply the x outside by what is left

[oblique girder]

I thought factoring was just sorta spreading out what was already there

oooo

thats going to be a little hard to grasp

but I sorta understand

[glossy basalt]

Yeah it is, they are both equivalent

You can show this by going back from the factorised form to the expanded form

[oblique girder]

I guess what im a little lost by

since this is the way im seeing it

I know if we multiplied that 8 by x we would have the 8x we originally had

but how does that x get in front of - 8

since if we were taking an x off of the x^2 what is it multiplied by

[glossy basalt]

Yeah your taking a common factor of x out of the x^2 which leaves behind 1 x

[oblique girder]

ah ok

understood

so now that brings us to x ( x - 8 ) = 0

now since we factored x we need to factor -8 ?

[glossy basalt]

That’s already fully factorised

Because there is no more factors that are the same between x and -8

Before x^2 and -8x both shared a factor of x which is why we could take the x out of both of them

[oblique girder]

Alright

whats the general rule for that

if values are being shared we need to get rid of them ?

[glossy basalt]

ab+ac=a(b+c)

in our case a was the x, b is also x, c is -8

[oblique girder]

this formula is relevant for what rule ?

[glossy basalt]

Yeah, this is the idea of factorising.

We want it in a form where none of the terms share a common factor

[oblique girder]

ok understood

and is this a rule to always follow with quadratic equations ?

[glossy basalt]

Yes, however this example is easier because notice that the constant +c is 0

[oblique girder]

oo

so going back to where we left off

I guess i couldn't see that

but x(x - 8) = 0

[glossy basalt]

Yeah. So now the zero product property

It pretty much says if you have two numbers a and b, if their product a*b=0, then the only way this can happen is if either a=0 or b=0

[oblique girder]

correct

so we need to find out if any of them are zero right ?

[glossy basalt]

Yeah the equation is satisfied when either of the factors equals 0

[oblique girder]

how do we proceed then

once we factor

[glossy basalt]

So if we want to find out when each factor is equal to 0, we need to solve 2 things

x=0 and (x-8)=0

[oblique girder]

ohh

[glossy basalt]

And x=0 is already solved, so all you need to do is solve x-8=0 for x

[oblique girder]

how is x = 0 already solved ?

[glossy basalt]

We want to find out what number x has to be to be equal to 0. What number is that?

[oblique girder]

0

oh lol

[glossy basalt]

Exactly

[oblique girder]

bc there is nothing in that equation

compared to like

x - 8 = 0

so i need to get rid of the 8

[glossy basalt]

Yeah, in this case you have to do a step

Yeah

[oblique girder]

so i add 8 and add 8 to the 0

[glossy basalt]

Yeah

[oblique girder]

so that would give me x = 8

[glossy basalt]

Yep, and that’s all

[oblique girder]

oh so the other solved it self

[glossy basalt]

You have your two solutions x=0 or x=8 which will make the original equation true

And you can check it by replacing the x in x^2-8x by 0 and then replacing it by 8

And both times the answer should be 0

[oblique girder]

Awesome so I got that down

Can I ask you another ?

[glossy basalt]

Yeah

[oblique girder]

2x (2x - 7) = -12 4

: , (

[glossy basalt]

What do you mean by -12 4, -12*4?

[oblique girder]

oops

2x ( 2x - 7) = -12

[glossy basalt]

So your first goal should always be to get one side of the equation to be = 0

How would you do that in this case

[devout snow]

@oblique girder Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

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[devout snow]

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• Show your work, and if possible, explain where you are stuck.
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[wintry flare]

i am confused about the domain of series

[wintry flare]

so, for example you have a series where the limit is |x-3| and you used the root test

wait no

[misty crest]

!xy

[devout snow]

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.

[wintry flare]

i dont even have an original problem

i am just confused about the domain part

i can show the example

[misty crest]

that’s what i meant by original problem

[wintry flare]

this

so we did root test

and the limit is |x-3|

because the limit is not a definite value, we need the domain the then find convergence ?

[misty crest]

use the ratio test to find the radius

[wintry flare]

are radius and domain not the same

[misty crest]

but yes youll get |x - 3| < 1

[wintry flare]

yeah

and then 2<x<4

[misty crest]

so your radius is…

[misty crest]

did you check the endpoints sir

[wintry flare]

its [2,4)

[misty crest]

so what’s the problem?

[wintry flare]

this is my question

i already have the work and answers

i am asking in general

if there is a value like x in the limit

[misty crest]

you mean it’s not a definite value because of |x-3|?

[wintry flare]

yeah

like 1 value

[misty crest]

well that’s the whole point, as per the ratio/root test you want the limit to be less than 1 so that it converges

[wintry flare]

my prof went over it today using the term domain and then mentioned radius

and i didnt understand

[misty crest]

and you find the interval so that it’s satisfied

[wintry flare]

so you set the limit <1 by default ?

and then solve by plugging the endpoints into the series

^

[misty crest]

well yea the entire goal is to find where the power series converges

[misty crest]

yes well you’ll need to check the endpoints because you’ll only find the radius

it won’t tell you if it converges at the endpoints

[wintry flare]

yeah

i know

i was just confirming

[wintry flare]

so this example converges conditionally

or is that wrong

[misty crest]

do you know what converges conditionally means?

[wintry flare]

evidentally not 😂

i know the definintion yes

[misty crest]

i mean like i think i know what you’re trying to say

[wintry flare]

yeah idk how else to word it

[misty crest]

are you trying to say it converges provided that x is in the interval [2, 4)?

[wintry flare]

it converges on a small interval

[misty crest]

that being the "condition"

[wintry flare]

yeah

[misty crest]

then yes that’s correct but that’s not how the term conditional convergence is used

[wintry flare]

okay

gotcha

so it just converges on [2,4) and thats all id say

right

[misty crest]

conditional convergence is used when a series converges but not absolutely

[misty crest]

so when the sum of the absolute value of the terms diverges but the original series converges

[wintry flare]

okie i understand

thank you

[misty crest]

you’re welcome

[wintry flare]

.close

[devout snow]

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[still zephyr]

How to show that $\mathbb{Z}$ with the usual addition and multiplication operations is not an algebraic structure of the same type?

[woven radish]

Halex

[graceful cosmos]

"An algebraic structure of the same type" doesn't parse for me. Are we missing some of the question?

[still zephyr]

the addition and multiplication operations in Z are not algebraic structures of the same type

maybe my question is not well formulated

[devout snow]

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@still zephyr Has your question been resolved?

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[neon folio]

Does this mean there are 2 inverse functions?

[trail eagle]

It's likely a result of how you would end up with two solutions when solving for the inverse of $1 + u + u^2$

[woven radish]

Azyrashacorki

[trail eagle]

Maybe not one sec

Yeah idk if the first one isn't even defined for any x I think.
The sqrt is only defined for x >= 3/4.
For 3/4 <= x < 1, the numerator is positive and the denominator is negative, so it's not defined in the log.
For x > 1, then the numerator is negative and the denominator is positive, so it's not defined in the log.

[devout snow]

@neon folio Has your question been resolved?

[cinder nova]

it's something about complex numbers,
but sometimes both of them doesn't work

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[outer rover]

@pseudo basin

[pseudo basin]

what are the p_i lmao

this problem is inconsistent with its notation

[outer rover]

p_i is replaced with r_i

[pseudo basin]

the best i can assume is that they meant r_i when they said p_i, and that in the first sentence they meant to put dependent in the first sentence of part d

but whoever wrote this problem was inexcusably sloppy

[outer rover]

based on the echelon form, if it's independent then how do I do part (d)?

[pseudo basin]

reduce it further until you solve the system

[devout snow]

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[smoky gyro]

How can I find the natural domain and range of f(z)=e^z

[pseudo basin]

for what z is the function defined?

[smoky gyro]

Complex

[pseudo basin]

is e^z defined for all z in C?

[smoky gyro]

I guess ? How should I find the range then

[pseudo basin]

wdym "guess"

do you know the definition of the complex exponential?

[smoky gyro]

yeah

[pseudo basin]

state it here

[smoky gyro]

e^ix = cosx +isinx

[pseudo basin]

that's the exponential of a pure imaginary number

[smoky gyro]

o

[pseudo basin]

are there any pairs of reals (x,y) for which e^(x+iy) is undefined or doesn't exist?

if so, what are they?

[smoky gyro]

Would it help to think about the default e^x graph

[pseudo basin]

to an extent, yes

think you're kinda overthinking it rn tho.

[smoky gyro]

I don’t know I don’t think so?

[pseudo basin]

e^(x+iy) = e^x (cos(y) + i sin(y))

do exp, sin and cos of a REAL variable ever fail to exist?

[smoky gyro]

Well e^x can be raised to any number and sin cos is defined for all real x

[pseudo basin]

so, they never throw an error.

does this now let you conclude with confidence that e^z is defined on the entire complex plane?

[smoky gyro]

So the imaginary part of sin, the “isin(y)” doesn’t change anything about its graph? Like I’m referring to the i being multiplied

[pseudo basin]

talking about a "graph" at this juncture is a little tricky.

[smoky gyro]

What would be the right terminology

[pseudo basin]

for what

[smoky gyro]

Here

[pseudo basin]

well i don't know what you are trying to say

you didn't answer my question either

[smoky gyro]

yes

[pseudo basin]

right.

so the domain is all of C.

[smoky gyro]

yes

[pseudo basin]

now the range

specifically for e^z, you might think about how writing a complex number in polar form is almost the same as writing it as a value of e^z at some other point

re^(iθ) = e^(log(r)+iθ)

[smoky gyro]

Why do we introduce a log?

[pseudo basin]

to make my point

and because r = e^log(r)

[smoky gyro]

How does this help us tho?

[pseudo basin]

what is the one complex number that doesn't have a polar form at all?

[smoky gyro]

Erm

0?

[pseudo basin]

indeed

[smoky gyro]

Well

[pseudo basin]

this is the only number not in the range of e^z

[smoky gyro]

0 +0i

[pseudo basin]

you don't need to pretend like 0 the real and 0+0i the complex are two somehow fundamentally different things unworthy of being treated as the same

[smoky gyro]

Well I figured I should use the proper notation even tho they are the same thing

[pseudo basin]

this adds no propriety

[smoky gyro]

What does e^z look like graphically I’m having trouble visualising this

Wait maybe I can

,w graph e^z

[pseudo basin]

this will only get you the real graph

[smoky gyro]

Hm

[pseudo basin]

graphing complex functions is much more difficult bc you need 4 real dimensions to do it the same way as real ones

[smoky gyro]

Oh so the only way to find the range is pretty much algebraically?

[pseudo basin]

... yes

[smoky gyro]

wow

So what’s the process of finding the range of complex functions then is it the same for every complex function?

Like with real functions you could either graph it to find the range or take it’s inverse and find the domain of that

[pseudo basin]

there is no "process"

[smoky gyro]

what

So then what are you supposed to do

If e^z has range 0 to all C

then for other complex functions if there’s no approach what do u do?

[pseudo basin]

no

its range is C \ {0}

• if there was something general to say then i would have said it already tbh

[devout snow]

@smoky gyro Has your question been resolved?

[devout snow]

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[green edge]

Hii, I don't even know how to start (c) and (d). Can someone help?

[winter torrent]

i'm not sure what "direction angles" means

but A x B is orthogonal to A and B

[green edge]

yup yup

[green edge]

I have no idea either T.T

[winter torrent]

i'd say just calculate A x B and call it good

you need it for d anyway

[green edge]

ohh is A x B not the answer for (b)

[winter torrent]

it's an answer for b

probably the easiest

maybe they mean putting it in spherical coordinates?

so you can calculate theta and phi

[green edge]

ohhh

[green edge]

what if they're asking the angles of A x B with respect to the axes

[winter torrent]

yea i guess

[green edge]

i have no idea what direction angles mean

[winter torrent]

yea neither do i

angles with the axis makes the most sense to me

[green edge]

hm should i just do that

okiidokii, ill do that thanks for the help!

but wait idk how to do that HAHAH

[winter torrent]

i mean the way i'd do it is by using a . b = |a| |b| cosø

[green edge]

ohh, so like arccos (a.b)/( |a| |b| )

[winter torrent]

yea

a . b in this case will be very easy to calculate

same with |b| if it's a unit vector

so it should be pretty quick

[green edge]

okay im sorry but i still don't have great intuition with these can u help me haha

[winter torrent]

sure

what did you get for A x B?

[green edge]

0 , 7 , -5

[winter torrent]

ok cool

0i + 7j - 5k

so we want to find the angle between that and 1i

the magnitude of the first thing is

[green edge]

oh?

[winter torrent]

,calc sqrt(7 * 7 + 5 * 5)

[green edge]

what is 1i? sorry for late replies the wifi is p bad

[woven radish]

Result:

8.6023252670426

[winter torrent]

like the unit vectors i j and k

1i is just <1, 0, 0>

if we want to find the angle between <0, 7, -5> and <1, 0, 0> we take their dot product, divide by 8.6, and take arccos

dot product between those is 0, so the angle is π/2

[green edge]

ohhhhhhhh rightt !!

[winter torrent]

can you find the angle to 1j?

that's <0, 1, 0>

[green edge]

okayy ill try

[winter torrent]

sure. remember, take the dot product between them, divide by 8.6 (that's the magnitude of A x B) and take arccos

[green edge]

hii oki so if im not mistaken

the dot product is 7?

[winter torrent]

yep

[green edge]

then i divide that by ||A x B| | then arcos it

ohh omg it makes sense? haha yay

[winter torrent]

yep exactly

you'd also divide by || <0, 1, 0> || but like that's just 1 haha

[green edge]

ohhhhhhhhh

okay okay

okayy thank u sm

i think i got it now. Thanks for your help!

how do i close the channel

[winter torrent]

type .close

[green edge]

thanks, have a nice dayy

.close

[devout snow]

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[ripe meteor]

integral of 1/root of 1+ sinx. please help 🙂

[solar goblet]

$\int \frac{\dd x}{\sqrt{1 + \sin x}}$

[solar goblet]

what have you tried?

[devout snow]

@ripe meteor Has your question been resolved?

[limpid dove]

try to find write this as a perfect square

[ripe meteor]

yeah i did that and got sinx/2 + cos x/2

then what

[limpid dove]

ok so that was the hard part

you can rewrite sint + cost entirely in terms of sin, then its a trivial integral

[devout snow]

@ripe meteor Has your question been resolved?

[devout snow]

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[vast delta]

a

[vast delta]

.close

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[vestal bay]

Could someone explain to me what I'm supposed to do? I don't understand what they want me to do.

[vestal bay]

Do I need to have it reflected across all 4 quadrants

[pseudo basin]

they want you to copy those shapes mirrored across each line

yes all 4 quadrants

[vestal bay]

Ahh okay

Thanks for the help

[sly tundra]

each line is a symmetrical reflection

[vestal bay]

Forgot to close

.close

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[shy hearth]

I am confused how my teacher got this as the answer

[olive snow]

,rccw

[olive snow]

Its thales theorem

If you have two triangles in this kind of setting

[remote sun]

you can use similarity to prove

[remote sun]

also what exactly is the Question

[boreal robin]

yea can do it by similarity

[crisp tapir]

Something about a pekka explaining math

[supple trench]

that's sth that OP has to explain clearly, say
||apply the Intercept Thorem to the following figure and state the results||

[devout snow]

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[serene horizon]

need help question 3

[serene horizon]

hello

<@&286206848099549185>

[devout snow]

@serene horizon Has your question been resolved?

[serene horizon]

<@&286206848099549185> *

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@serene horizon Has your question been resolved?

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[woven tundra]

Anyone know spanish..

[woven tundra]

I have some questions on a test

[olive snow]

You can post the problem dw

[woven tundra]

May be alot lmk if you can do them rq or help

[olive snow]

I mean its a math server

[woven tundra]

lol

[olive snow]

Im sure there is spanish server where you can put questions

[woven tundra]

yes i am waiting on a responce there

[olive snow]

Ok but in that case close this channel as long this is not math related

[devout snow]

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[slender lynx]

I was trying to derive the sum of squares equation and I am getting an answer that is not correct, but I can't find the flaw in my reasoning

[lunar harbor]

Where is (n-1) coming from? There's only n/2 pairs of terms, so this part of the sum should have n/2 terms.

also this isn't consistent with your first few terms in the sequence

(Also, all of this makes the implicit assumption that n is even)

[slender lynx]

I see

[lunar harbor]

Any questions before I dip? B/c imma go in a minute.

[slender lynx]

Would this produce the right equation anyway if those two errors you spotted were fixed?

And all the math was done correctly after that

[lunar harbor]

it should

I would make the arguments of why the patterns work out like that a bit more rigorous

unless this is just for yourself, in which case you can include however much detail you want

[slender lynx]

Its just for myself, my friend asked me to derive that equation and i tried like 3 different methods before i landed on this one

But anyway

.close

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[lunar harbor]

Oh on that note

The classic argument to derive the sum of squares formula is to consider a telescoping argument

I'll start with this: Find a cubic $p(n)$ such that $$p(n+1)-p(n)=n^2$$

[woven radish]

Civil Service Pigeon

[lunar harbor]

this is the same logic used to find a formula for the sums of higher powers too

[lunar harbor]

as finding patterns like this becomes increasingly difficult

[lunar harbor]

Additional remark: This relates to the idea of ||telescoping series||

[lunar harbor]

Maybe a P.S.? You can also ||consider the sum of (i+1)^3 - i^3|| and finagle the sum of squares out of that

[lunar harbor]

Oh more random shit: You can also use binomial coefficients and hockey stick

[slender lynx]

Oh I may have done something like this, where I thought about it in terms of stacking blocks, like I would have one block to represent 1^2, 4 blocks to represent 2^2, and then I just line up those right next to each other, and it grows in accordance with an nxnxn cube, so I thought: Okay, so this sum represents some fraction of an nxnxn cube, which means its cubic. So I just did a system of 5 equations to solve for the 4 different variables of a cubic

[lunar harbor]

Oh remark for that too

https://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html

This way you don't have to bash a system

although if you use finite diferences, the system isn't awful tbf

[lunar harbor]

Oh also more general remark: The sum of i^k from i=1 to n is a polynomial of degree (k+1)

ok I'm done yapping for now

[slender lynx]

I'm extremely new to this type of algebra, like i had to google what the summation notation was when my friend sent it to me 😭

[lunar harbor]

F

[slender lynx]

Okay .close

.close

welp

[lunar harbor]

oh it's already closed

there's just a lag

between when you use the command

and it actually closing

[slender lynx]

Darn

[lunar harbor]

as you can tell, I really took advantage of that gap lol

[slender lynx]

I guess that worked out

[devout snow]

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[limpid epoch]

If the tan of angle x is 22/5 and the triangle was dilated to be two times as big as the original, what would be the value of the tan of x for the dilated triangle? 
Would I double the 22/5 to make 44/10, or would it just simplify back to 22/5 anyways and there's no point?

[runic prawn]

yes

[trail eagle]

Scaling a triangle doesn’t change its angles

[runic prawn]

correct

[limpid epoch]

okay, I figured. Thank you so much!

[runic prawn]

also the angle doesn't change right

it's just tan of x

x is the same

[limpid epoch]

tan of x, but the triangle is dilated by a scale factor of 2

[runic prawn]

trig functions are by defintion scale invariant

[runic prawn]

yes ik

tan(x) just takes x as input

no side lengths

so yes it remains the same

[limpid epoch]

Oh, I see what you mean

[runic prawn]

yes

[limpid epoch]

THank you so much

.close

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[chilly pivot]

i'm unsure how to do this without numbers. i can't find the derivative

[trail eagle]

Well you can just compare the tangent lines at the points they are asking, no need for derivatives

Whichever one is steeper is going faster

[chilly pivot]

how do i compare them

how do i know the tangent line

like d/dx? i assume pink is x2 so derivative is x?

[trail eagle]

Maybe use a ruler and place it on the line at those points to see it better?

[chilly pivot]

pretty sure that's not what we are supposed to do

[trail eagle]

Hmm well I’m pretty sure it is

[chilly pivot]

how do i draw the tangent line?

[trail eagle]

It’s just about interpreting derivatives graphically

[trail eagle]

.

[chilly pivot]

ahh, but i'm still unsure how to draw it with the ruler, even with ur explanation

how do i know where to trace

[devout snow]

@chilly pivot Has your question been resolved?

[devout snow]

@chilly pivot Has your question been resolved?

[devout snow]

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[mellow willow]

If I have the equation W(x)=-x(x-32) does the -x mean I’m reflecting over the y axis or x axis? I know if I had g(x)=-f(x+3) I would reflect over the x axis. But if I had g(x)=f(-x+3) I would reflect over the y

[devout snow]

@mellow willow Has your question been resolved?

[haughty monolith]

y axis

y = x ---> y = -x (graph the difference)

More generally, when you put the negative sign in front of f(x), you're affecting the y value, so you're reflecting over the x axis. If you put the negative sign in front of x, you're felecting the x value, so you're reflecting over the y axis.

[mellow willow]

Even tho it’s outside the function?

[cinder nova]

it depends what the original function is
it's a reflection over the y axis of x(-x-32)
and it's a reflection over the x axis of x(x-32)

[mellow willow]

I’m trying to do horizontal compressions and stretched. This one is a horizontal compression and those are the two points, we are going from f to g. I know in the equation k must equal somthing greater than 1. Is mine correct?

[devout snow]

@mellow willow Has your question been resolved?

[mellow willow]

How do I do this table? Would the final row be the opppsite values of the h(x) row or the c row

[upper schooner]

,rccw

[mellow willow]

I desperately need help I am so confused. Why would B be affected by A vertical stretch or compress factor ?

[devout snow]

@mellow willow Has your question been resolved?

[wise scarab]

I'm sleeping soon, so I may not be able to give you an answer, but it might be helpful if you try to word it a little more clearly (e.g. exactly which equation/sentence are you confused about? what about it is confusing/hard to understand)

a lot of less esoteric questions in this server often remain unanswered simply because it's quite hard to understand what the author is asking

[devout snow]

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[snow verge]

Hello, I have a question about a statistic thing.

[snow verge]

I'm doing an experiment, where I'm collecting n samples from a (presumably normal) distribution and taking their average x

then doing the same, collecting n samples from another presumably normal distribution, and taking their average y

My prior belief is that like x and y are around 60 apart and that the two distributions have the same variance (sample variance after 5 samples from x is around 100)

How big should n be so that I'm certain I can distinguish mean(X) and mean(Y) if they really have different means?

..

My thoughts is that, at the end, I'm gonna do a two sample t-test.

But like, its expensive to collect more samples

So I want to know when I can stop collecting samples.

Like if i've collected 100 samples of x and the mean is 140 and the sample variance is 40. The MLE for the mean will have small variance so should be ble to distinguish small variations in the means of X and Y

[devout snow]

@snow verge Has your question been resolved?

[devout snow]

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[timid schooner]

hello. does log(A-B) have a direct formula?

[timid schooner]

i know that logA-logB can be written as log(A/B) but i don't think that's the case for log(A-B)?

is it not possible to just- write it as
log(A-B) = logA-logB??

[final osprey]

no

[timid schooner]

may i please know why?

[final osprey]

you should input some value of A and B and check for yourself using a calculator

[timid schooner]

okay... what if the question i have to solve has 'x' for an exponent instead...?

[remote sun]

yk e^x=c=>logc=x

[pseudo basin]

very much no

log(2-1) is not at all equal to log(2)-log(1) for instance

[timid schooner]

i see

[pseudo basin]

can you show what question you're looking at

[timid schooner]

sure

@pseudo basin here you go

[timid schooner]

wait in this case is the answer not the same?

oh sorry nvm

[abstract zodiac]

Take the lcm between e^x and e^-x

[pseudo basin]

taking logs right away won't help here

i would first multiply by 2 and in fact also by e^x on both sides

[timid schooner]

right, so e^x-e^-x = 2

and i should multiply that with e^x on both sides??

[pseudo basin]

yes

[timid schooner]

so what would that look like?

(e^x)^2 - (e^x)(e^-x) = 2e^x?

[pseudo basin]

yes but now simplify that

[timid schooner]

i'm a little clueless as to how to proceed; is there a formula i should be applying on the left hand side?

or is this where i use log?

[abstract zodiac]

Second terms cancels to become 1

[timid schooner]

i don't get it?

[abstract zodiac]

(e^-x) is basically 1/e^x

[timid schooner]

yeah

OH

okay

so after this it becomes

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

[abstract zodiac]

Do you see a potential for the formula (a - b)^2

[timid schooner]

yes?

[abstract zodiac]

We need to do that

[timid schooner]

okay

[abstract zodiac]

add -2e^x and +2 on both sides

[timid schooner]

(e^x)^2-2e^x-1=0

(e^x-1)^2=0?

[abstract zodiac]

Missing +2

It’s not the right formula yet you need +1 not -1

[timid schooner]

why do we have to add the +2

[timid schooner]

noted

[abstract zodiac]

The formula for (a-b)^2 is a^2 - 2ab + b^2

Now you’ll get (e^x - 1)^2 = 2

Square root both sides

[timid schooner]

(e^x-1) = square root of 2

[abstract zodiac]

e^x = sqrt(2) + 1

Now is the time to take log

[timid schooner]

okay

[timid schooner]

question; is it not possible for
(e^x - 1) = +sqrt(2) and -sqrt(2)?

oh nvm it seems log is undefined for negative numbers (i did not know that)

anyways, i have my answer now. thanks!! ^^

.close

[abstract zodiac]

Yea

[devout snow]

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[subtle marsh]

please help me solve this :

[devout snow]

@subtle marsh Has your question been resolved?

[subtle marsh]

how to find |b|.|c| |a|.|b| and |a|.|c|

[subtle marsh]

after that?

[soft umbra]

State the length of a and c vectors with alpha relatively

Plug into the last given info, that’s all

You’re just a step afar to the solution

You’ve conquered the difficult part, which is recognizing a, b, and c vectors are perpendicular to each others

@subtle marsh

[subtle marsh]

thanks

[soft umbra]

Don’t forget to close the channel if your doubts have been cleared

[subtle marsh]

.close

[devout snow]

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[devout snow]

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[restive river]

<@&286206848099549185> help

[devout snow]

Please don't occupy multiple help channels.

[fossil flume]

Hello guys I need help

[valid vector]

Okay, you want to try and notice a pattern again?

[dull heron]

oh sry

[valid vector]

!occupied

[devout snow]

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).

[restive river]

yes

[valid vector]

!occupied