help-23

[hardy vale]

exp(x) = e^x

so that exp(3) = e^3 and exp(59.5) = e^59.5

Converts the sum inside into a multiplier

means that like

exp(3+3+3) = e^(3+3+3) = e^3 * e^3 * e^3

and you'll notice the sum (3+3+3) turned into a multiplier (e^3 * e^3 * e^3)

[maiden kernel]

Ist just a Converter?

[hardy vale]

hmm

the conversion is sort of like

a side effect of how exponents work

like,

notably exp(3+3+3) itself is a number

that is not the same as 3+3+3

,ask exp(3+3+3)

[severe pond]

they're just saying exponential functions have the property that f(a + b) = f(a) * f(b)

[flat frigate]

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[hardy vale]

as you can ssee

exp(3+3+3) is effectively just like

a really long way to write

e^(3+3+3) = e^9 = 8103.somethingorwhatever

its just a different number

but this feels like an example of the XY problem

https://xyproblem.info/

what is the context that youre trying to understand this in

like

surely this came from somewhere

it might be more useful for your understanding

if it was explained in context

[primal bone]

or in other words

!xy

[safe radish]

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.

[primal bone]

-# good bot

[hardy vale]

oh

i didnt know this was a thing

lmao

[tender meteor]

Mental idea: $e^x$ is a “multiplier”. If you increase the exponent by +1, you multiply the value by $e$. So adding $ a+b$ to the exponent is equivalent to multiplying by $e^a$ and then by $e^b$

[flat frigate]

alee

[maiden kernel]

I try to build a game and I think I understand the Geometrie brwonian Motion and now I try to add the Player Actions. And I Found

$S_{t+\Delta t} ;=; S_t \cdot \exp!\big( (\mu - \tfrac{1}{2}\sigma^{2}) \cdot \Delta t ;+; \sigma \cdot \sqrt{\Delta t} \cdot Z \big)$

[flat frigate]

Nicolas Matheisen

[maiden kernel]

And its new that there is a exp()🥲

[hardy vale]

ah

you can rewrite this as like

$S_{t+\Delta t} ;=; S_t \cdot \big(!e^{!(\mu - \tfrac{1}{2}\sigma^{2})} \cdot \Delta t ;+; \sigma \cdot \sqrt{\Delta t} \cdot Z \big)$

[maiden kernel]

😂I try to understand the whole Formel xD

[hardy vale]

oops

[flat frigate]

c9

[primal bone]

exp(x) = e^x, is all tbh afaik that you need to know

[hardy vale]

yeah

[primal bone]

It's just, if you've got a long expression that should be in the power, it's much easier just to write exp(massive abomination) than e to the (massive abomination but rly tiny now)

[maiden kernel]

Ahhh its just another Notation for the same okeo

Thx 🙂

.close

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[austere goblet]

please explain what you have tried, what you understand and where you are stuck

[dapper pelican]

The thing is, I know how to solve the equation

I just get kind of confused on my goal like why am I trying to achieve?

like with Answer 19

i’m stuck

[austere goblet]

by "knowing how to solve the inequality" (not an equation here btw), what do you mean? can you show your attempt at "solving" this inequality?

and by the way, your job here is to find the range of values of the variable for which the inequality holds

(if that range exists)

[still charm]

Lets not get pedantic here. The thing is that you just have to treat the inequality like an equal sign and then solve for x and then graph it.

[austere goblet]

in that case I'll leave the channel to you

[dapper pelican]

you have to graph it, but but if you divide by a negative you have to change the direction

[safe radish]

@dapper pelican Has your question been resolved?

[fast bolt]

which one yu want

[safe radish]

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[dapper pelican]

first on

[austere goblet]

.reopen

[safe radish]

✅ Original question: #help-23 message

[fast bolt]

ig there is no solution

[dapper pelican]

yes

[fast bolt]

haven't seen such question

what is the problem then

[dapper pelican]

because x cancel out

[safe radish]

@dapper pelican Has your question been resolved?

[safe radish]

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[silk cove]

w+o=90 worker

w+o/90=400

sum of salary (w+o)=36000

O average=600

[rain basin]

i would call “o” as 90-w

[kind seal]

Something seems off

@silk cove check your second equation

The salary is averaged, not the people

It should be weighted

[silk cove]

Ha maine average hi mana h usko

90/90 ni kar rha hoon

[kind seal]

What ??? English pls

[rain basin]

lol

[silk cove]

Just average not 90/90

[kind seal]

What are you trying to say dude

[rain basin]

make an equation relating w and o by writing o as 90-o

[kind seal]

I am trying to tell you that the average of salaries is weighted

Your current equations mean 90=3600

[silk cove]

I wrote sum of salary in the bracket

[kind seal]

That is not a good practice though

It creates ambiguity

It is better to not express statements mathematically like that

[silk cove]

.close

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[kind seal]

Lol

[safe radish]

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[light tiger]

Check it out if there is any mistake or something else

[light tiger]

Question 1,b

[light tiger]

Isn't anyone here?

@dusk wasp

[dusk wasp]

I am here

[light tiger]

Hehe

[light tiger]

Check it

[dusk wasp]

What

[light tiger]

Check images that I have sent above

[dusk wasp]

Already check it

[light tiger]

Good

Hello
anyone come and check

I'm confused about the number of simple space.
I wrote it but I'm not sure

[safe radish]

@light tiger Has your question been resolved?

[light tiger]

Ignored

.close

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[tall yarrow]

Please help me guys regarding this problem.

[still charm]

U = log x

[tall yarrow]

Elaborate it little

[quasi bison]

he's suggesting that you make a substitution

do you know how to do that in general yes/no

[tall yarrow]

Done but problem arises in putting limits

[queen ingot]

oh it has a vertical asymptote at x=1

[hollow dock]

And at x=0

[queen ingot]

i think you should split the integral at x=1

[tall yarrow]

Substituted it and used the property of log.
Putting Log x = t
x = e^t
dx = e^t dt

Limits must be change Now

[queen ingot]

oh after the substitution the asymptote is at u=0

the limits should be -infinity to 2 right?

[tall yarrow]

Yes

[tall yarrow]

What's ur final answer?

[queen ingot]

idk how to deal with the asymptote

don't you need cauchy principal value?

[tall yarrow]

It's alright.

[tall yarrow]

I don't know abt that

[queen ingot]

wait doesn't the area just blow up to -infinity ?

principal value won't save you here

[safe radish]

@tall yarrow Has your question been resolved?

[safe radish]

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[steady yoke]

Why does completing the square give the turning point

[still charm]

Ok lets consider a general quadratic

y = x^2

Where is the vertex

@steady yoke

[steady yoke]

the tuning point?

vertex means turning point right

[still charm]

Yea the turning point is the vertex

[steady yoke]

yeah okay

the vertex is 0

[still charm]

Yep

Why?

[steady yoke]

I’m not too sure

[still charm]

Yeah

So let me introduce a new word

"Piecewise monotonicity"

Basically?

A parabola DECREASES until you hit the vertex

Then INCREASES after

So far so good?

[steady yoke]

okay

yes

and then it increases

[still charm]

Great! And the reason 0 is the vertex of y=x^2 is because 0 is the point where it goes from decreasing to increasing

[steady yoke]

mhm

[still charm]

Now let's complete the square for a more general quadratic. I'm sure you've done this before

[steady yoke]

yes I have

[still charm]

y = ax^2 + bx + c becomes?

[steady yoke]

(ax+b/2)^2 -b/2^2 + c

[still charm]

Um... not quite

[steady yoke]

.

[still charm]

a(x+b/(2a))^2 + c - b^2/(4a)

Now let's just zoom in on the (•)^2 part

[steady yoke]

huh what?

how

[still charm]

That's called completing the square?

[steady yoke]

wait let’s use numbers isnteas of algebra

makes it easier

[still charm]

Ig...

[steady yoke]

what quadratic should we take

[still charm]

Let me come up with a really simple example for you

[steady yoke]

okay

[still charm]

x^2 + 6x + 10

= x^2 + 6x + 9 + 1

= (x+3)^2 + 1

[steady yoke]

here’s what I would do

[steady yoke]

(x+3)^2 - 9 + 10

$(x+3)^2 + 1$

[still charm]

Where are you getting -9

[steady yoke]

-(3)^2

[still charm]

Looks good but you need to add 10, not 1

[steady yoke]

ihhh plus 10

[flat frigate]

botou

[steady yoke]

yep

[still charm]

Ok, great!

Quick aside:

x^2 and x^2 + 1 and x^2 + 1000 all have the same turning point, right?

For the x value

[steady yoke]

I think Ann disagrees

[still charm]

I'm not sure why she does

Constants don't affect curvature

[austere goblet]

but they do affect the coordinates of the turning point.

so I think there's a reason to be pedantic here.

[still charm]

Yea but that's irrelevant

[peak estuary]

being imprecise to someone who is learning is rarely irrelevant

[austere goblet]

I'm not going to say anything more then.

[quasi bison]

why are you even talking about "curvature"

i mean sure the x coord of the turning point is the same for all 3 of these

the y coord however is different

i will concur that imprecision is no good when teaching somebody

[still charm]

Ok.

Missing the point entirely then

[steady yoke]

hanako could you explain it to me

[faint hornet]

I’ve heard the best way to learn something is 1 person explaining imprecisely with 6 people watching and waiting for them to make a mistake

[clever gale]

The turning point / vertex of y = (x - h)² + k is (h, k), and this form is called VERTEX form since it is written directly in terms of the vertex’s coordinates. Completing the square converts any old quadratic into this form in which you can identify the vertex/turning point easily.

[steady yoke]

I’m wondering why there’s just spectators

who disappear after pointing out mistakes?? And argue with the tutor

[opaque fern]

There's barely other open help channels, so helpers concentrate on the open few

[austere goblet]

have you seen six helpers all talking at once in the same channel? you probably have. did you enjoy it?

[austere goblet]

I can't speak for Dena and Ann, but you replied to my message specifically saying "I'm not going to say anything more then"

[steady yoke]

I wouldn’t but if one helper sees another making a mistake I’d assume they should take over

@faint hornet you free to explain it to me?

[austere goblet]

pointing out mistakes shouldn't mean shoving the OG helper aside

[steady yoke]

fair enough I apologise

[faint hornet]

yeah i can explain dw

[steady yoke]

thank you

[faint hornet]

do you know about graph transformations

[steady yoke]

yeah not covered it yet comes next after inequalities

[faint hornet]

okay np

[steady yoke]

but I know it from GCSEs

[faint hornet]

i'll explain a different way

[steady yoke]

rn im just tryna figure out why completing the square gives you the turning point

that was my original question

[faint hornet]

when you complete the square, you have a(x-b)^2 + c in some way right

[steady yoke]

yeah

guys please no

[faint hornet]

but, if you square something, it is always greater or equal to zero

the minimum value of (x-b)^2 is thus zero

which is why the turning point occurs precisely at x = b

[solar hazel]

watches patiently waiting for you to make a mistake to pounce on

[steady yoke]

yeah I’ve learnt about the roots of a function

[steady yoke]

SLEOGHA STOP IT

[faint hornet]

pounce on is crazy

anyway does that make sense

[abstract carbon]

Real

[steady yoke]

yes simmilar to the roots of a function and completing the square on it

And deciding the minimum value

[faint hornet]

what do you mean

[austere goblet]

I suppose I can make this simple if need be

oh but the simplification I have has been mentioned. nvm

[steady yoke]

Like function x = x^2 - 2x + 2

then you complete the square

and explain why f(x) is >0 for all values of x

[faint hornet]

yeah exactly like this

[steady yoke]

yeah

[faint hornet]

the minima / maxima are precisely are the turning points

[steady yoke]

ohhhh

that makes alot of a sense

it’s the same as finding the minimum of a function like that

[faint hornet]

yeah

[steady yoke]

so I basically comeolte the square and see the minimum value

[austere goblet]

min or max*

[steady yoke]

how would I do so for max

[faint hornet]

if the coefficient of x^2 is negative, she means

but it is the same in principle

[austere goblet]

so remember that we have a(x - b)^2 + c

[abstract carbon]

[steady yoke]

so I must show that it’s squared and it equals to 0

[austere goblet]

(x - b)^2 is always positive

or 0*

so if the a is positive as well, then the whole a(x - b)^2 is positive

but if the a is negative, you get a negative times a positive = a negative

[steady yoke]

but how does that show the maximum

[austere goblet]

hence a max

[steady yoke]

I don’t understand

[austere goblet]

ok let's forget the k for a moment

we have a(x - b)^2

(x - b)^2 is non-negative

so suppose a is negative then

[steady yoke]

it would make it negative

[austere goblet]

then as x grows, (x - b)^2 becomes super large

but then a is negative, so multiplying by a makes for a very large negative

[steady yoke]

yeah

[austere goblet]

so you can see here that the value of a(x - b)^2 can only go more and more negative if a is negative itself

[steady yoke]

yeah

oh so the toomost position is when it’s 0?

As anything else is only more negative

[austere goblet]

when (x - b)^2 = 0, yes

which is the same as x = b

which is the same as the minimum case

hence you don't have to memorize two separate cases

[steady yoke]

so I’m basically working out the same thing as the minimum

except becuase it’s negative and gets larger in the negatives

[austere goblet]

just recall that the sign of a dictates whether you have a min or a max

correct

[steady yoke]

okay that makes sense

thank you

[austere goblet]

and sorry @faint hornet for ramboing in

[faint hornet]

np

[steady yoke]

thank you guys!!

.close

[safe radish]

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[nimble vine]

https://tenor.com/view/good-morning-love-gif-14967101488090424627

[steady yoke]

.reopen

[safe radish]

✅ Original question: #help-23 message

[steady yoke]

no peace yet

@austere goblet

I tried it with a real question

I’m a bit confused on how the c becomes the y value

[nimble vine]

fok

[austere goblet]

!noping please.

[safe radish]

Please do not ping individual helpers unprompted.

[steady yoke]

my bad

[austere goblet]

I don't get what you mean here.

show your steps.

[steady yoke]

okay so I did it with the quadratic

$x^2 -5x + 4 = y$

[austere goblet]

okay. keep going.

[steady yoke]

I want to find the turning point so

I complete the square to find the minimum possible point this quadratic

the completed square is

$(x-\frac{5}{2})^2-\frac{9}{4}$

[flat frigate]

botou

[quasi bison]

woah woah woah hold on hold on hold on

are you sure it says =0 at the end

cause talking about a minimum pt would surely be more appropriate if you were starting with y = x^2 - 5x + 4

[austere goblet]

should be = y yeah

[steady yoke]

sorry I meanr y

I meant

[austere goblet]

oops, didn't catch that. sorry to both Ann and botou

[flat frigate]

botou

[austere goblet]

ok anyway proceed.

[steady yoke]

now the least possible value is if x = 5/2

[austere goblet]

mhm

[steady yoke]

Which would become 0 - 9/4 which is -9/4

so the minimum value is -9/4

[austere goblet]

okay? keep going

I still have not seen the question

[steady yoke]

I don’t know what to do with this information

what is the turning point now

[austere goblet]

so you know that the least possible value is when x = 5/2

that is your x-coordinate

and the minimum value of -9/4 is your y-coordinate

done

[steady yoke]

how is the y coordinate

[austere goblet]

wdym

[steady yoke]

??

why

[austere goblet]

^

[steady yoke]

ohhhhhhhhh

I said it myself

okay

[austere goblet]

LOL

[steady yoke]

Thank you

[austere goblet]

but yes

if x is the x-coordinate of the turning point

then f(x) is the y-coordinate

which makes sense, right?

[steady yoke]

Yes yes

[austere goblet]

if you have an x-coordinate and you wanna find its y-coordinate on the graph you do f(x)

[steady yoke]

becuase they complement each other

[austere goblet]

so yeah, same story here

[steady yoke]

okay okay

thank you

.close

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[safe radish]

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[gilded bison]

the question, How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
the answer is 20, simply 5 x 4 x 1
my first attempt which was wrong, was 5c5 * 5c4 * 3!

[gilded bison]

I get why the first is correct, but the second one is just a formula i use in questions and assume i'll get the right answer

why is it wrong?

[brave wolf]

Why should it be right? What were u trying to do with it?

[gilded bison]

i'm not sure, but prior i'd used this method from a solved question and it was seemingly correct

in my mind its like

(you can choose 5 letters from 5) (you can choose 4 letters from 5) (you can choose 1 letter from 1) * number of spaces to fill

[brave wolf]

U dont really choose 5 letters from 5

[gusty trench]

does the information that the number is divisible by 5 tell you anything useful about any of its digits?

[gilded bison]

OH

YOU CHOOSE 1 FROM 5

[brave wolf]

you are choosing 1 letter from 5, then 1 letter from 4 and then the last one must be 5

so 5c1 * 4c1 * 1

[gilded bison]

woah nice

and wb the x 3!

i saw that being done prior

i really should have tried understanding it

[brave wolf]

i dont see a reason for why it should be there

3! is the number of ways to rearrange the stuff

[gilded bison]

wait i'll check what question I saw it

[primal bone]

I mean, I'd consider how to prioritise what I'm picking tbh

For one, you know this number you make needs to be divisible by 5

There's only one digit there that permits you to make that number work, and only one place it can go

[brave wolf]

(it's already solved and the OP knows how btw, we are just discussing why the other solution doesnt work)

[primal bone]

Well, the multiplying by 3! just doesn't work, since that 5 is fixed in place

[gilded bison]

oh here

[gusty trench]

notably the solution is the same as 5c2 * 2! * 1

[gilded bison]

Out of 7 consonants and 4 vowels, how many words of 3c and 2v formed?

[brave wolf]

or 5p2 * 1

[gilded bison]

p? im prepping for an exam and I seem to be able to solve everything with C

[brave wolf]

and it was prolly 7c3 * 3! * 4c2 * 2!

[gilded bison]

would you say it helps to learn permutations

[brave wolf]

not really, its similar to c

if u dk it, ignore it

[brave wolf]

was it this?

[gilded bison]

it was 7c3 * 4c2 * 5!, and I assumed we just had to multiply the amount of spaces to fill

[brave wolf]

oh yeah, im stupid

[gilded bison]

alright

but then why multiply by 5?

[brave wolf]

This means that you first choose 3 consonants and 2 vowels, you dont yet know where to put them. You just choose them

That's 5 letters

[gilded bison]

right

[brave wolf]

now you have to figure out in how many ways those 5 letters can be arranged to form a wordd

[brave wolf]

and this is exactly 5!

[gilded bison]

okayw ait

[brave wolf]

As was said here, we could approach the og question similarly. We could first choose 2 digits out of the 5, thats 5c2, then arrange them to form a number, thats 5c2 * 2! and finally add the digit 5 to make it divisible by 5, that's 5c2 * 2! * 1

[gilded bison]

okay if I tried using the prev method on this question, it'd be
7c1 x 6c1 x 5c1 * 4c1 * 3c1?

[brave wolf]

how do you deal with the vowels / consonants distinction?

[gilded bison]

OHHHH

[gilded bison]

oh

um

ohr that's assuming those places are for vowels and consonants

[brave wolf]

ill have to got for few mins, if anyone is here u can take over, or ill come back in a min, sorry

[gilded bison]

so you'd still multiply it by 5! which is the possibilities for where c and v is

nono I got it tysm guys

.close

[safe radish]

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[safe radish]

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[main arrow]

Help

[main arrow]

Just completed my first single variable calculus problem, I know that the derivative measures the rate of change. But, I am not sure how the negative exponent works in the derivative. Because that part had my thinking.

[long oracle]

Hm regarding the power rule or?

Like what’s the exact problem

[main arrow]

Yes.

[hollow dock]

As in you're not sure why the derivative of 1/x is -1/x²?

[main arrow]

And the sum rule.

[long oracle]

Have u tried using the definition?

[main arrow]

Of a derivative?

[long oracle]

Yeah, I mean what’s ur prerequisite on this

Like you’re not sure in what way?

[main arrow]

Like my math level?

[long oracle]

Mhm

[main arrow]

Algebra 1.

[long oracle]

Okay, are you not sure because you’ve not done it before, or are you not sure because you don’t recall how, from ur lectures or notes?

[main arrow]

Never done it before.

[long oracle]

Are you seeking an intuitive answer or rigorous one? Etc

[main arrow]

I have taken notes of the different formulas what their purpose is and how there applied to ML. Since this is multivariable calculus for ML. (I know this isnt multi calc yet)

[long oracle]

If it’s the latter, try just using the definition for your own sake

[main arrow]

I would prefer a rigours one.

[quasi bison]

you're quite a long way from ML stuff

[main arrow]

To understand this.

Without the numbers.

[quasi bison]

$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$

[flat frigate]

Ann

[quasi bison]

this is the rigorous defn of a derivative

[hollow dock]

This is also known as first principles

[quasi bison]

but based on your claim of algebra 1 as current level, i can hazard a guess you've never seen this before either

[long oracle]

Don’t be too hasty with these things @main arrow

[main arrow]

?

[long oracle]

It’s built on top of previous stuff

So a solid understanding of limits also is important

[quasi bison]

that it is

well, i mean you can teach differentiation without that much in the way of limit-bashing

but being comfortable with limits sure does come in handy

[main arrow]

I just used that formula to find the gradient or otherwise known as the rate of change of y with respect to x. I know the general idea behind a limit. (Watched a Prof. Leonard vid on limits)

Gradient = rise/run

[quasi bison]

i mean ok

would you like to get walked through this first-principles formula for f(x) = 1/x

yes/no

[main arrow]

yes pls!

[quasi bison]

ok gimme a min

[long oracle]

If this is all the type of functions you’ll encounter, then all you need to know about the derivative is think of it as a linear operator D(f) that can do this:
D(x^n) = n x^(n-1)

I remember you helping someone with linear algebra

So you should know what a linear operator is

[main arrow]

Just to be clear, f'(x) the "' ' " notation for differentiation. So like, all its sayings is "take the derivative" right?

Do you say it as "f prime (x)?"

[hollow dock]

Yes, the derivative of f(x) with respect to x

[long oracle]

“f prime of x”

[hollow dock]

It differs from country to country

[quasi bison]

there you go

"f prime of x" is how you would pronounce the notation f'(x) yeah

[main arrow]

I have not learned linear operations yet, I have learned vector operations though.

[long oracle]

Okay

[quasi bison]

i am writing $\frac{1}{h}\paren{f(x+h)-f(x)}$ instead of $\frac{f(x+h)-f(x)}{h}$ purely for notational convenience (lets me avoid skyscraper fractions)

[flat frigate]

Ann

[quasi bison]

@main arrow have you read through this?

[main arrow]

Yes

[quasi bison]

ok

does it make sense to you or does any step need extra explanation

i numbered the steps for you in pencil in case you need to refer to a specific one

[main arrow]

Just to solidify my understanding of a limit to make sure it doesnt bite me in the butt with multi variable calc, the limit is your expected prediction as a f(x) approaches x, it can get really really really close but doesnt reach it because that wouldn't be a secant line anymore.. Right?

[quasi bison]

it's not an "expected prediction" of anything and i would strongly recommend against thinking in these terms

and certainly not "as f(x) approaches x" either

[main arrow]

Got it

[quasi bison]

here it's h approaching 0 if anything

[main arrow]

Right

[main arrow]

So what is the relationship that 1/x = x^-1 has?

[long oracle]

Hm?

[main arrow]

?

[long oracle]

I’m confused what you’re asking

[hollow dock]

1/x is the same thing as x^-1

[long oracle]

That’s just by definition

[main arrow]

Why do we bother writing x^-1 if

[main arrow]

ooh

That explains why

[long oracle]

It’s notation

[hollow dock]

Sometimes its to prevent careless mistakes

Especially when differentiating

[long oracle]

We define 1/x as x^-1

[hollow dock]

But just remember that 1/f(x) is not the same as f^-1(x)

[long oracle]

Where ^-1 here means multiplicative inverse

[main arrow]

and thats multiplying a number by a number to get 1 right?

[long oracle]

Mhm

[main arrow]

Ok ok. One more question

[hollow dock]

Fire away

[main arrow]

Since I am in algebra 1 and was never taught summation notation, I had to research the Greek syntax because it was being used to represent higher-dimensional spaces in Lin algebra. Is this the universal framework behind sigma?

I watched a 60 second video on this lol.

[long oracle]

It’s just shorthand notation for repeating summation

[main arrow]

Same mathematics problem dont have upper limit which tells me that there might be different ways to use sigma.

Some*

[quasi bison]

i wouldnt say it represents high-dimensional spaces or anything like that.

[main arrow]

I initially thought the lower limit was another way to represent a limit.

[long oracle]

So e.g. $\sum_{j=3}^5 j^2$

[quasi bison]

this sigma notation is simply a way to abbreviate additions of lots of shit

[flat frigate]

Aslan

[long oracle]

Means take the sum of j^2

[quasi bison]

nothing more nothing less

[dull radish]

it is not used to "represent higher-dimensional spaces in linear algebra"

[long oracle]

When j=3

And then j=4

[main arrow]

Hm. In the Lin algebra course they said it was used to represent n-dimensions.

[long oracle]

And then j=5

[quasi bison]

at best it represents a specialized for-loop

[main arrow]

In machine learning.

[quasi bison]

can you quote exactly what was said.

[main arrow]

Not verbatim but they mentioned something similar.

[long oracle]

Don’t overthink it

It’s just shorthand notation for summation

[quasi bison]

yeah i think you might be either misremembering it or attaching too much significance to it or both

[main arrow]

They said in lin alg and machine learning, sigma notation is used to write matrix-vector operators, sums over features/dims and neural network layers. Obviously, me being very new to lin alg thought that was awesome to think of.

[main arrow]

Oh

[long oracle]

It’s still being used as a shorthand notation for summation in that case

[quasi bison]

in linalg there's a lot of variable-length sums

sigma notation doesnt have any sacred significance beyond just

popping up a lot of the time to actually write those sums

[hollow dock]

I think you might be thinking about the elements of a matrix

[main arrow]

Using it for projection and dot products.

This is when I first used it earlier today.

[long oracle]

Okay, still shorthand notation for summation

It’s not different

[main arrow]

alright

[main arrow]

Yeah.

[hollow dock]

So like a1b1+a2b2+a3b3

For a dot product

[main arrow]

But I had a simple question before I go, because you guys have been extremely helpful and I appreciate the help

Why is linear algebra / calculus so different from algebra 1?

[long oracle]

In what sense?

What are you taught in algebra 1?

I mean if anything you’re just building upon what you’ve learnt before

[main arrow]

y = mx + b
Elimination
Substitution
functions

So far in a nutshell.

[long oracle]

Algebra 1 is like teaching urself a language/framework

To now be able to ask new questions that we start leaning about in calculus

Or linear algebra

[main arrow]

You know, after learning the basic syntax and grammar for calc and linear algebra math starts to become fun

[main arrow]

I always wondered why we "plug in numbers"

[long oracle]

U needed to learn a new language (algebra 1) in order to ask such questions in calculus or linear algebra

[main arrow]

You are right

[long oracle]

Hm?

[main arrow]

Well, my teacher taught us to "plug in numbers"

[long oracle]

Why it’s possible?

[main arrow]

and I thought it was stupid frankly.

[long oracle]

Hm why?

[main arrow]

Like purpose of it

[long oracle]

Do u have an example?

[main arrow]

In single variable calculus it felt purposeful and important to ML because it provided a derivative, the rate of change at a single point, which in ML provides the fundamental understanding of how models learn especially in neural networks and then that goes into gradients then algorithms. But, point being is, that in algebra 1 my teacher would say "F(x) = 3 plug in x + 5" and it felt tedious. Like what is my argument in this function, what is the relationship, what is the asking

I did the work - not a d*ck lol.

But I love my math teacher and all.

It really seems pointless to provide no context to a problem.

[long oracle]

I mean presumably u were asked to do so to learn the machinery of doing such things

But I understand ur point

Like why

[main arrow]

But in calculus, I learned that f'(x) = 3 would mean that the gradient is constantly 3 no matter what x is.

[long oracle]

Sure

[main arrow]

and then that can be a indication in a loss function or something

[long oracle]

The point of math is often to be abstract and general

To have a framework

For many things

[main arrow]

Thats why I love lin alg

Abstraction.

Feeling confused mainly.

[long oracle]

That’s why we don’t always give concrete examples of what a function is

It’s just a function

And you can do what you want with it

Once you know how to operate it

You attach meaning to it afterwards

But often times, what we do in math is go back a step and say okay what detail can we avoid saying but still be relevant

In a general way

[hollow dock]

Yeah, there were times in math that I didnt understand anything, so I had to feel my way through the topic and then I finally understood certain concepts, thats what makes math fun

[long oracle]

Mhm

Something that helped me, and it seems that’s what’s occurring to 9Lana aswell, is to have a “back burner problem”

A personal project of yours that you always go back to and try the new math tools you learn

[main arrow]

Ohh

So take problems you didnt fully understand and come back when you have new math tools?

[long oracle]

Mhm

It helps for motivation

[main arrow]

?

[long oracle]

And it seems you’ve got that going with the ML stuff

[main arrow]

Yes

I am trying to achieve a deep understanding in linear algebra for ML / GPU and research.

Since 95% of ML is linear algebra.

[hollow dock]

Of course, things like linear transformations are key for programming

[main arrow]

A month ago I deeply disliked the idea of mathematics. I didnt know I was going to be doing calculus at 4am on a Monday

A month later

[main arrow]

Of course

Well - thank you guys for much!! I am going to go back to completing this rest of the questions. Thanks @quasi bison for the demonstration of gradient!

[safe radish]

@main arrow Has your question been resolved?

[safe radish]

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[rich hull]

we’ve written the three numbers 4, 5, and 6. A move consists of picking two integers a and b on the satisfying a ≥ b ≥ 1 and replacing them with the two numbers a + b and b − 1. What is the maximum number of moves you can make until you run out of moves?

[safe radish]

@rich hull Has your question been resolved?

[glacial cairn]

What have you tried?

That does seem to be the max, are you supposed to prove it?

[rich hull]

yes and find a formula for any xyz

[glacial cairn]

Well, one thing you should see is that the greatest number in the triple doesn't matter after some point

[rich hull]

yes

[glacial cairn]

Not sure how you'd find a formula for any starting triple, but you should be able to find one for any triple where the greatest number is big enough

For instance, once you have 3 7 19, you cannot make a number greater than 19 with just 3 and 7 (max you can make is 7+3+2+1 = 13)

[rich hull]

how do i know this is the max number

[glacial cairn]

So all you care about is how to make as many moves as possible with just 3, 7, and some big third number X

I would just search the state tree up to states where the greatest number is big enough

That's not a clean answer but it might give you more intuition to solve the entire thing

[rich hull]

?

[rich hull]

no then u have 0,13,19) u can then make 13 other moves so total moves =26

[glacial cairn]

I said you cannot make a number greater than 19, I'm not talking about the number of moves

[glacial cairn]

What are you confused about?

[rich hull]

what is a state tree

[glacial cairn]

[a, b, c] is a state (assume it's always ordered so a <= b <= c)
from it, you can go to [a-1, a+b, c], or [a-1, b, a+c], or [a, b-1, b+c]

Those are the only three options you have for any starting state

From each of these new states, you have three more next states

The state tree is just that: the starting state is the root, and each branch is an option

Each option has sub-branches

Eventually you hit leaves, which are the states [0, 0, x]

[rugged bluff]

idk how to solve it but for a special case:

the max for [a, b, c] where c>=a+b seems to be adding a to b until a is 0, then decrementing b to 0
max total moves would be a(a+1)/2+a+b

[rich hull]

from where to look this up

@rugged bluff a 19 move sequence found

[rugged bluff]

works for 4, 4, 11

[rich hull]

wydm?

*wdym

[rugged bluff]

the closed form expression is correct for [4, 4, 11] (doesn't apply to 4 5 6 because 4+5>6)

[glacial cairn]

(that's what I was hinting at here)

[rich hull]

how do derive it?

[rich hull]

will it work for 25 30 45

[rugged bluff]

how do we take the optimal path to an optimal large-gap triple

[glacial cairn]

Idk, haven't looked into it

[rugged bluff]

we can only make a or b decrease once the criterion is satisfied

to maximise path length, we should always add a to b where possible (c is too large)

[rich hull]

thanks

i think i can figure out from your formula

.close

[safe radish]

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[rugged bluff]

please ping me when you find a solution for arbitrary triples!

[safe radish]

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[stark ether]

how to find the angle the ball will bounce of to

[stark ether]

i think i can use an axis pararell to the line of angle φ that way F in the perpendicular axis = 0 so momentum is sustained in that axis

[sharp sentinel]

yeah isn't it enough to solve this problem?

[stark ether]

and to then find the distance of the points

[safe radish]

@stark ether Has your question been resolved?

[stark ether]

is the answer 2sin(2φ)h?

[safe radish]

@stark ether Has your question been resolved?

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[spring saffron]

Hello, can someone give me a hint on this. We have a right triangle with angles 30,90,60 where we have a median from angle 30 and an altitude from the right angle. the median intersects with the altitude and divides it into x and y we have to find x/(x+y)

[quasi bison]

have you made a diagram yet?

[spring saffron]

like draw it?

[quasi bison]

yes

[spring saffron]

i found it

.close

[safe radish]

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[loud wren]

isn't this an amc question

[still vine]

LMAO

[spring saffron]

maybe it is

is it

[loud wren]

I swear I'll pull it up rn

@spring saffron cause I actually answered this one in the AOPS forums 😂

[spring saffron]

🤣

is the answer A?

[loud wren]

Yes

I did the trig bash solution btw 🥀

[spring saffron]

ok , thanks

isnt it easier with mass points?

[loud wren]

yeah

I just reverted to like my only strong suit 😭

[spring saffron]

hahaahaha

when nothing works trigonometry comes in clutch

[safe radish]

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[earnest mirage]

Hello, I fixed the proof, could someone check if it's ok please?

\begin{Proposition}
Assume $m$ and $n$ are integers. If $m \ | \ n$, then $m \ | \ (7n^3 + 13n^2 - n)$.
\end{Proposition}

\begin{proof}
Let $m$, $n$ be integers and $m \ | \ n$.
By the definition of divisibility, since $m \ | \ n$, this means that $n = mk$ for some integer $k$.
% Let $k = \frac{l}{7n^2 + 13n - 1}$, then $n = m(\frac{l}{7n^2 + 13n - 1})$, for some integer $l$.
Multiplying both sides by $7n^2 + 13n - 1$, we obtain $n(7n^2 + 13n - 1) = 7n^3 + 13n^2 - n = mk(7n^2 + 13n - 1)$.
Since $k$ and $n$ are integers, so is $k(7n^2 + 13n - 1)$.
Thus, $7n^3 + 13n^2 - n = ml$, where $l = k(7n^2 + 13n - 1)$.
Therefore, by definition of divisibility, $m \ | \ 7n^2 + 13n - 1$.
\end{proof}

[flat frigate]

Mor Bras

[rustic goblet]

this is fine, but a little clunky in its wording imo

[queen ingot]

yup looks correct to me

[rustic goblet]

after saying that n = mk, you can just say that 7n^3 + 13n^2 - n = n(7n^2 + 13n - 1) = mk(7n^2 + 13n - 1) = ml

and you're done

[earnest mirage]

Oh I see

Why that works?

I see, because n = mk

[hardy vale]

you're defining l = k(7n^2 + 13n - 1) anyway

its just a bit more succinct if you just do it in one go

[hollow dock]

Yeah, that works

[earnest mirage]

Like this?

\begin{proof}
Let $m$ and $n$ be integers and $m \ | \ n$.
By the definition of divisibility, since $m \ | \ n$, this means that $n = mk$ for some integer $k$.
Then, $7n^3 + 13n^2 - n = n(7n^2 + 13n - 1) = mk(7n^2 + 13n - 1) = ml$, where $l = k(7n^2 + 13n - 1)$.
Therefore, by definition of divisibility, $m \ | \ 7n^2 + 13n - 1$.
\end{proof}

[flat frigate]

Mor Bras

[hardy vale]

you still need the part specifying that l is in fact an integer, this proof is now missing a little piece

you maintain the same proof, just moving it a little bit so its shorter

but its completely optional, as higher! said

[earnest mirage]

\begin{proof}
Let $m$ and $n$ be integers and $m \ | \ n$.
By the definition of divisibility, since $m \ | \ n$, this means that $n = mk$ for some integer $k$.
Then, $7n^3 + 13n^2 - n = n(7n^2 + 13n - 1) = mk(7n^2 + 13n - 1) = ml$, where $l = k(7n^2 + 13n - 1)$.
Since $k$ and $n$ are integers, so is $k(7n^2 + 13n - 1)$.
Therefore, by definition of divisibility, $m \ | \ 7n^2 + 13n - 1$.
\end{proof}

[flat frigate]

Mor Bras

c9

[hardy vale]

something like that yeah

this works

👍

very cool

[earnest mirage]

Thanks!

.close

[safe radish]

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[somber quail]

can someone help

[kind seal]

Do you know the properties of a tangent

[somber quail]

yep

[kind seal]

Specifically about angles?

[somber quail]

wdym?

[kind seal]

Angle PQB = ?

[somber quail]

pqb is 90 degrees

q is 90

[kind seal]

Not really

[somber quail]

why isnt that a

property of a tangent

pq is a tangent right

so at q it will be 90

[kind seal]

Yes

Check your textbook, that is not true

[somber quail]

can you explain it to me?

[kind seal]

The 90 degree angle is formed with the radius

[somber quail]

yeah

[kind seal]

And the angle PQB is equal to the angle subtended by minor arc BQ on the circle

You agree on this?

[somber quail]

yeah

ig

[kind seal]

So whats PQB?

[somber quail]

30?

[kind seal]

Yes

[somber quail]

ahhh idk

yay

[kind seal]

Now think about AQB

Hint, AB is diameter

[somber quail]

hmm

[kind seal]

What angle does a diameter subtend?

[somber quail]

90-30

90-30 right?

[kind seal]

Not quite

Any diameter subtends 90 degrees

[somber quail]

no i mean because pqb is 30 and q is actullay 90 so I think it will subtract and give 30-90 so 60

[kind seal]

What is q?

[somber quail]

Pq is tangent

[kind seal]

AQB, AQP, PQB?

[somber quail]

so there will be 90 at q

pqb

[kind seal]

Not really

[somber quail]

ahh

[kind seal]

A tangent is perpendicular to the radius

OQP is 90

<@&268886789983436800>

[somber quail]

yeahhhhh

<@&268886789983436800>

[kind seal]

You understand this?

[somber quail]

yes yes

I just forgot about that a bit

[kind seal]

Now what is AQB

Again AB is a diameter

[somber quail]

aqb

hmm

wth

idk

ahhh

leave it

[kind seal]

Use this

[somber quail]

there was a or question i will do that

this shit too hard

[kind seal]

?

[somber quail]

I was doing like a sample paper

[kind seal]

What part do you find hard?

[somber quail]

the b one is way easier

I dont understand the whole question lol

[kind seal]

I would suggest you refer to your Euclidean Geometry Textbook for elementary results about circles

[somber quail]

Okay

tysm for the help

you are a good teacher but I just dont understand stuff that easily

.close

[safe radish]

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[slow patrol]

guys whats 10+9

[quasi timber]

Please don’t troll.

[quasi bison]

use a calculator. your phone or computer definitely has one

if you need help with a legitimate question then open your own channel.

[safe radish]

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[earnest mirage]

Hello, could someone check if this proof is ok please?

\begin{theorem}[Division algorithm]
For all integers $a$ and $m$ with $m > 0$, there exist unique integers $q$ and $r$ such that $$ a = mq + r$$ where $0 \le r < m$.
\end{theorem}

\begin{Proposition}
Assume $a$ and $b$ are positive integers. If $r$ is the nonzero remainder when $b$ is divided by $a$, then the remainder of $-b$ divided by $a$ is $a - r$.
\end{Proposition}

\begin{proof}
Let $r$ be the nonzero remainder when $b$ is divided by $a$.
So, $b = aq + r$ for some positive integer $q$.
Then, $-b = -(aq + r) = -aq - r$.
Adding $a - a$ to the right side of the equation gives
\begin{align*}
-b &= -aq - r + a - a \
&= -aq - a + a - r \
&= a(-q - 1) + a - r \
\end{align*}
Since $q$ is an integer, so is $-q - 1$.
Let $s = -q - a$ and $t = a - r$, then $-b = as + t$, which means that the remainder of $-b$ divided by $a$ is $t = a -r$.
\end{proof}

[quasi bison]

-aq - a != a(-q-a)

also you're missing the other part of the defn of remainder

[hardy vale]

i think you're kind of glossing over something that you might think is "obvious" that you have to state rather explicitly

[quasi bison]

yeah, that

[hardy vale]

i think you understand that -r is not the remainder

but like

be more explicit about why you think that

[quasi bison]

when b is divided by a, giving quotient q and remainder r, we have b = aq + r and what else?

[hardy vale]

little bit of trouble there

what do you know about a remainder

what do you know about t

[quasi bison]

@earnest mirage

[earnest mirage]

I'm writing more context

\begin{Theorem}[Division algorithm]
For all integers $a$ and $m$ with $m > 0$, there exist unique integers $q$ and $r$ such that $$ a = mq + r$$ where $0 \le r < m$
\end{Theorem}

\begin{Theorem}
Assume $a$ and $b$ are positive integers. If $r$ is the nonzero remainder when $b$ is divided by $a$, then the remainder of $-b$ divided by $a$ is $a - r$.
\end{Theorem}

\begin{proof}
Let $r$ be nonzero remainder when $b$ is divided by $a$.
So, $b = aq + r$ for some positive integer $q$.
Then, $-b = -(aq + r) = -aq - r$.
Adding $a - a$ to the right side of the equation gives
\begin{align*}
-b &= -aq - r + a - a \
&= -aq - a + a - r \
&= a(-q - 1) + a - r \
\end{align*}
Since $q$ and $a$ are integers, so is $-q - a$.
Let $s = -q - a$, then $-b = as + a - r$, which means that the remainder of $-b$ divided by $a$ is $a -r$.
\end{proof}

[slim pulsar]

oh shoot what happened

ah there we go

[hardy vale]

as a hint, note that the remainder when 37 is divided by 9 is not 10 even though 37 = 3 * 9 + 10

division algorithm is in fact real

[quasi bison]

ok so 0 ≤ r < m, you see that?

[hardy vale]

you're missing one line

[quasi bison]

you need to show that, for your theorem, the thing you claim to be the remainder (a-r) also lies in the right range.

knowing that r does.

if you do not justify this part then your proof's just incomplete

[queen ingot]

it should say = a(-q-1) + a - r
rather than = a(-q-a) + a - r

[earnest mirage]

Certainly, I made a typo factorsing

[hardy vale]

they fixed it earlier, i assume its just probably a bad copy paste from earlier

[flat frigate]

Mor Bras

[hardy vale]

gah

you're still missing just one line

you're really close

[earnest mirage]

Discord erases backslashes

[queen ingot]

yup then u need to show that 0 <= a-r < a

[earnest mirage]

I see, thank you!

[keen tulip]

for some nonnegative integer q

[queen ingot]

you're right
in fact you can just say "some integer q" since you don't need to use q >= 0 in the proof

(q may be zero, for example if u divide 3 by 4 you get 3 = 4*0 + 3 and here q = 0)

[earnest mirage]

Ok, I fixed and added some more lines.

\begin{Theorem}[Division algorithm]
For all integers $a$ and $b$ with $a > 0$, there exist unique integers $q$ and $r$ such that $$ b = aq + r$$ where $0 \le r < a$
\end{Theorem}

\begin{Theorem}
Assume $a$ and $b$ are positive integers. If $r$ is the nonzero remainder when $b$ is divided by $a$, then the remainder of $-b$ divided by $a$ is $a - r$.
\end{Theorem}

\begin{proof}
Let $r$ be nonzero remainder when $b$ is divided by $a$.
So, $b = aq + r$ for some nonnegative integer $q$.
Then, $-b = -(aq + r) = -aq - r$.
Adding $a - a$ to the right side of the equation gives
\begin{align*}
-b &= -aq - r + a - a \
&= -aq - a + a - r \
&= a(-q - 1) + a - r \
\end{align*}
Since $q$ is an integer, so is $-q - 1$.
Let $s = -q - 1$, then $-b = as + a - r$.
Since the remainder of $b$ divided by $a$ is a nonzero integer $r$ where $0 < r < a$, this means that $0 < r - r = 0 < a - r < a$.
So, $a - r$ is a positive integer.
Therefore, the remainder of $-b$ divided by $a$ is $a - r$.
\end{proof}

[hardy vale]

s = -q-1

[queen ingot]

u need 0 < r in order to say a - r < a

0 < r comes not from the remainder property (0 <= r < a) but from assumption (r nonzero)

well, it comes from both

u need both to conclude 0 < r

[hardy vale]

i think this works

just be careful

0 < r < a

[flat frigate]

Mor Bras

[earnest mirage]

It's better now?

It's stated that 0 < r < a

[hardy vale]

also be careful

because now you are saying 0 < r - r = 0

[queen ingot]

i think you want to say 0 = r-r < a-r < a

[earnest mirage]

Oh I see

[safe radish]

@earnest mirage Has your question been resolved?

[safe radish]

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[steel plume]

can anyone help please

so far ive worked out the formula for l1

[hollow dock]

So you know the equation of the line

[steel plume]

y= root 3 x + 2 root 3

[steel plume]

yea

[hollow dock]

So if the line intersects the x-axis, whats a substitution you can make?

[steel plume]

y=0?

[hollow dock]

Yeah

[steel plume]

ohhhhh

i got it now thanks

.close

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[winged flare]

if $x + \frac{1}{x} = \frac{\sqrt{5}+1}{2}$, find $x^{90} + \frac{1}{x^{90}}$

[winged flare]

tried solving for x

[vagrant ice]

you haven't tried squaring both sides?

[winged flare]

i think this has something to do with complex numbers

$x = \frac{\sqrt{5}+1 \pm \sqrt{2\sqrt{5}-10}}{4}$

[quasi bison]

idt solving for x is a good idea here necessarily

[winged flare]

it has worked a few times

[quasi bison]

better to work out $x^n + \frac{1}{x^n}$ for various values of $n$ here, with $n=90$ being the ultimate goal

[flat frigate]

Ann

[winged flare]

maybe

[lost jewel]

I mean we could keep squaring both sides

Ig

[winged flare]

biggest hurdle would be an intermediate fifth power expansion in that case

[vagrant ice]

no, there's a nice way with powers of 2

[vagrant ice]

I'd encourage you to find x^2 + 2 + 1/x^2 first

hopefully you know what happens when you square the golden ratio

[winged flare]

what exactly

[lone void]

what is the golden ratio 💀

[winged flare]

the RHS

thats the golden ratio

[vagrant ice]

okay, so this is where the golden ratio comes from

and we can let b = 1

[lone void]

oh but that is cos36 x2

[winged flare]

ok yeah..

and what about the squaring part

[vagrant ice]

if you sub in b = 1 directly into the equality...

[lone void]

maybe instead of keeping the whole RHS like that solve with cos 36?

u said complex numbers

might come in some form

[vagrant ice]

that's a confusing way

[winged flare]

$\sqrt{\cos^2(36^{\circ})-4}$ doesnt click

[lone void]

4 cos^2

[winged flare]

oh

right

[faint seal]

Just write 2*cos(theta) in polar form

[winged flare]

ok got it thanks

.close

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[winged flare]

confirming: $x = e^{i(\frac{36\pi}{180})}$

[vagrant ice]

oh yeah okay I see what kheer means lol

$z + z^* = 2 \cos \theta$ if $|z| = 1$

[flat frigate]

south

[faint seal]

Yes

[vagrant ice]

indeed okay I was thinking without that and there's a nice way without complex numbers

but given that you've studied this might as well

[pallid bone]

de moivre theorem

[winged flare]

true..

[vagrant ice]

yeah then hopefully you recognise the angle, cos(theta) = [sqrt(5) + 1]/4

if you do it's ezpz

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[winged flare]

um ⁉️

[vagrant ice]

well um

I do appreciate the full solution, just wondering if it's needed

[safe radish]

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[carmine tulip]

$x^{2001}+\left(\frac{1}{2}-x\right)^{2001}=0\
x^{2001}=-\left(\frac{1}{2}-x\right)^{2001}\
\sqrt[2001]{x^{2001}}=\sqrt[2001]{-1} \times \sqrt[2001]{\left(\frac{1}{2}-x\right)^{2001}}\
x=-\left(\frac{1}{2}-x\right)\
x=x-\frac{1}{2}\
0=-\frac{1}{2} \longrightarrow \longleftarrow$

[flat frigate]

fort craft 🇵🇸

[carmine tulip]

What is wrong with my calculation?

[hollow dock]

Was it assumed that there would be a real solution of the equation?

[carmine tulip]

I don't think so.

[hollow dock]

I think the calculations are fine, the problem came with the start

Because I dont think there's a real solution

[vagrant ice]

$(x^n)^{1/n} = x$ is only true for real numbers $x$

[flat frigate]

south

[vagrant ice]

remember that there are n complex roots of any number

so really, all you've done is shown (by contradiction) there are no real solutions

"assume there is a real x such that x^2001 + (1/2 - x)^2001 = 0"

[proud tree]

yeah and i dont think you can split the RHS into -1 and the other term

[keen tulip]

the split the root of a product step is the risky one in general

[vagrant ice]

yeah $\sqrt{ab} = \sqrt{a} \sqrt{b}$ only works for $a, b \ge 0$

[carmine tulip]

which step are you talking about?

[flat frigate]

south

[hollow dock]

Its an odd root

[vagrant ice]

I know, just casually mentioning it

[hollow dock]

So I think thats fine if we are talking real numbers

But for imaginary, its a gamble

[carmine tulip]

Okay. From all of this, I understand that the initial equation doesn't have a real root. Can we say for certain that it has complex roots?

[vagrant ice]

yes, from the fundamental theorem of algebra

the left hand side is a polynomial

[carmine tulip]

What about this? $x=x+1$

[flat frigate]

fort craft 🇵🇸

[hollow dock]

Again, no real solutions

[carmine tulip]

I don't think this one has any solutions at all. Be it real or imaginary.

[vagrant ice]

well FTA says that the LHS can't be a constant

yeah that's the condition behind FTA

'every non-constant single-variable polynomial with complex coefficients has at least one complex root'

[carmine tulip]

wdym can't be a constant?

[vagrant ice]

you need the right hand side to be 0

so okay, x = x + 1 becomes 1 = 0

but 1 is a constant, so FTA doesn't apply here

[carmine tulip]

Ok, I see. As a side point, is it possible to get a complex root?

[vagrant ice]

to this? yeah

[carmine tulip]

Wouldn't it take insane amounts of time?

whatever. It doesn't matter.

.close

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