help-28

[pliant fractal]

idk what to do

[minor crater]

are you sure? this looks like part II of something

[pliant fractal]

nevermind, theres x approaching a value, i js couldnt see it on my laptop. it's all good now, thank you.

how do i close this

[spiral vigil]

.close

[full forum]

Channel closed

Closed by @spiral vigil

Use .reopen if this was a mistake.

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[sturdy vine]

To see if X and Y are independent, can I simply multiply the two diagonals and see if the two results are different?

[onyx glen]

"the two diagonals"?

[spiral vigil]

like P(0, 0) * P(1, 1) vs P(1, 0) * P(0, 1)

[sturdy vine]

yes

[onyx glen]

why would this give an indication of independence

[spiral vigil]

dunno but that's the question

[onyx glen]

and more importantly how would you even do this if either variable had 3 or more values

[sour torrent]

you can just find a 2x2 matrix where they give different result

i think he wants to know why this works

[sturdy vine]

so

$1/31/3=1/9 \ 1/101/3=1/30$

[glossy valve]

Goofy Joe

[sour torrent]

$$
\begin{array}{c|cc}
& x_1 & x_2\ \hline
y_1 & \mathbb{P}(X=x_1,;Y=y_1) & \mathbb{P}(X=x_2,;Y=y_1)\
y_2 & \mathbb{P}(X=x_1,;Y=y_2) & \mathbb{P}(X=x_2,;Y=y_2)
\end{array}
$$

[glossy valve]

alee

[sturdy vine]

not indipendent

[sour torrent]

The standard way to check whether $X$ and $Y$ are independent is to compute the marginals $p_X(x)=\sum_y p_{XY}(x,y) $ and $ p_Y(y)=\sum_x p_{XY}(x,y) $, and then verify that for every pair $(x,y)$ in the table one has $p_{XY}(x,y)=p_X(x),p_Y(y) $

[glossy valve]

alee

[sour torrent]

Quicker way
to disprove independence: for any $x_1\neq x_2$ and $y_1\neq y_2$, independence would imply $p(x_1,y_1),p(x_2,y_2)=p(x_2,y_1),p(x_1,y_2) $
So one counterexample $2\times 2$ block is enough to conclude $X$ and $Y$ are not independent

[glossy valve]

alee

[sturdy vine]

can you see if i did it right above pls

but are you sure about these things?

<@&286206848099549185>

[spiral vigil]

yea but does that counterexample exist? for such a simple case i could totally see this being a way to test

[sour torrent]

yes

[spiral vigil]

obv it doesn't scale to larger matrices

[sturdy vine]

can you see if i did it right above pls

[sour torrent]

for larger table, it is enough to find a single $2\times 2$ block for which this cross-product identity fails

[glossy valve]

alee

[sour torrent]

why?

[spiral vigil]

well mainly that you can't find the determinant of a nonsquare matrix

[sturdy vine]

$1/3\cdot1/3=1/9 ---1/10\cdot1/3=1/30$

ao not indipendent?

[sour torrent]

in exercises where the table is intentionally constructed so that X and Y are independent

the direct way is faster for largish matrices

[spiral vigil]

i mean yeah your arithmetic is fine

[glossy valve]

Goofy Joe

[sour torrent]

this is better if its not the case

[sour torrent]

i mean if we are not in this case

[sturdy vine]

😠

🙂

[spiral vigil]

alee i'm having trouble following what you're saying. Two things:

• is the determinant of a 2x2 joint pdf matrix being 0 sufficient to conclude that the two variables are independent?
• is his arithmetic correct if so

[sturdy vine]

i didnt the determinant

[spiral vigil]

you did; determinant of a 2x2 matrix like that is just the product of the diagonals like you did. You'd just have to subtract them

[sturdy vine]

what

[spiral vigil]

never mind

anyway 0.30 is not 1/3

[sturdy vine]

its 3/10

[spiral vigil]

yesh

[sturdy vine]

i know

[spiral vigil]

so.... you've written 1/3

[sturdy vine]

ops

sorry

$3/103/10=9/100---1/103/10=3/100$

[glossy valve]

Goofy Joe

[sour torrent]

A single 2×2 block that violates the cross-product identity is enough to disprove independence

But a single 2×2 block with determinant zero is not enough to conclude independence unless the whole table is 2×2

[sturdy vine]

so different->indipendent???

[sour torrent]

In general, X and Y are independent if and only if the entire joint pmf table has rank 1, i.e. it can be written as an outer product of a column and a row

[sturdy vine]

$9/100=3/100$

[glossy valve]

Goofy Joe

[sturdy vine]

no

[sour torrent]

So to prove independence in larger tables, checking rank = 1 (e.g. via Gaussian elimination) is the clean and efficient method

[spiral vigil]

if the product of diagonals are different, the variables are dependent on each other

[sturdy vine]

yes

[spiral vigil]

you can imagine a degenerate case of [ 1, 0 ] ; [ 0, 0 ]

where P(X = 0, Y = 0) = 1 and the rest are 0

those variables are clearly dependent on each other

[sturdy vine]

is there a proof?

[spiral vigil]

oh nvm that's too degenerate

[spiral vigil]

there is.... i did one with algebra on a piece of paper here but what do you know about independent events?

[sturdy vine]

Two events are independent if the occurrence of one does not change the probability of the other.

[spiral vigil]

the degenerate case is [ 0.5, 0 ] ; [ 0, 0.5 ] aka the identity matrix:
that is, P(X = 0, Y = 0) = P(X=0, Y=1) = 0.5
clearly dependent
diagonal products are 0.25 and 0

[spiral vigil]

okay, can you express that in the form of an equation?

[sturdy vine]

if they are indipendent

[spiral vigil]

if your probability matrix was $\begin{pmatrix}a & b \ c & d \end{pmatrix}$ what would it mean for them to be independent?

[sturdy vine]

$p_{X,Y}(x,y)=p_Y(x)\cdot p_X(x)$

[glossy valve]

Goofy Joe

hayliänus austrǎlis

[sturdy vine]

rouchè capelli

rho=1

[spiral vigil]

what?

[sturdy vine]

I have to see the system first if it has solutions and how many

[spiral vigil]

where did we get systems from

i was talking about independent events

[sturdy vine]

I don't remember

rho=1

maybe

[spiral vigil]

,tex
\begin{matrix}
p_{X,Y}(x,y) & y=0 & y=1 \
x=0 & a & b \
x=1 & c & d
\end{matrix}

are these events independent?

[glossy valve]

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

[sturdy vine]

I have to see if the rank rho=1

the rank of the matrix is 1

[spiral vigil]

how does that relate to your definition of independence?

Two events are independent if the occurrence of one does not change the probability of the other.

[sturdy vine]

the vectors are independent

i need to see this

[sturdy vine]

idk

[spiral vigil]

i think you're talking about linear independence of a vector space? that's not really related to independence of events

[sturdy vine]

idk

[spiral vigil]

if you want to prove that your diagonal thingo is a valid way to check for event independence, then you're going to need to know what it means for two events to be independent

you're welcome to just use the method regardless of proving it

[sturdy vine]

yes

Two events are independent if the occurrence of one does not change the probability of the other.

[spiral vigil]

okie dokie and how does that look in an equation? use the probability values I've given above.

[sturdy vine]

for example if X is toss 1st coin and also Y, toss the first coin and the result of the second toss does not depend on the first

[spiral vigil]

yes you've said that. the probabilities of each coin toss are given above. Are they independent? How would you compute that? What would it mean for the result of Y to depend on the result of X?

[sturdy vine]

but if for example X is choosing between two dice, loaded or not, and Y is tossing a coin, there is a dependency because Y depends on the chosen die.

but if for example X is choosing between two dice, loaded or not, and Y is tossing a coin, there is a dependency because Y depends on the chosen die.

[spiral vigil]

yes

[sturdy vine]

,tex
\begin{matrix}
p_{X,Y}(x,y) & y=0 & y=1 \
x=0 & HH=1/4 & HT=1/4 \
x=1 & TH=1/4 & TT=1/4
\end{matrix}

[spiral vigil]

double slashes for newline

what does this matrix mean?

[sturdy vine]

all combination of heads and tails

i need to put the numbers

[glossy valve]

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

[spiral vigil]

perhaps it would make more sense to have y = H and y = T instead of y= 0 and y = 1?

i don't understand what the value is of HH and HT

[sturdy vine]

X is the first throw and Y is the second, so the first throw is heads or tails, the same thing happens on the second.

X={0,1}={H,T}

[spiral vigil]

anyway, yes that matrix is clearly independent given that all probabilities are the same. But that's not the only way that events can be independent

[sturdy vine]

$HHTT=HTTH$

[glossy valve]

Goofy Joe

[sturdy vine]

also this

are the same ->indipendent

[spiral vigil]

why does that have anything to do with what you said earlier about "Two events are independent if the occurrence of one does not change the probability of the other."

again you're welcome to simply use this property without proving it

but if you want a proof you're going to need to be precise and refer to definitions

[sturdy vine]

but are there other methods?

[spiral vigil]

well yes, you can show that P(x = 0 | y = 0) is equal to P(x = 0)

have you worked with conditional probability before?

and if not, then how does your book normally prove events are independent?

[sturdy vine]

yes

[spiral vigil]

okay, so that's how you'd do it

you'd do that for all 4 possibilities

[sturdy vine]

$P(X=0|Y=0)=\frac{P(X=0 \cap Y=0)}{P(Y=0)}=\frac{P(X=0)P(Y=0)}{P(Y=0)}=P(X=0)$

this is valid if x and y are indipendent

[spiral vigil]

real and true

[glossy valve]

Goofy Joe

[sturdy vine]

so from the others I will get accordingly

$P(X=0),P(X=1),P(Y=0),P(Y=1)$

[glossy valve]

Goofy Joe

[sturdy vine]

is it enough for me to see that in each box there are these 4 probabilities?

no

and now?

@spiral vigil

[spiral vigil]

the exhaustive way would be to compute all of those and compare them to e.g. P(X = 1 | Y = 0)

[sturdy vine]

$P(X=0)=1/2,P(X=1)=1/2,P(Y=0)=1/2,P(Y=1)=1/2$

[glossy valve]

Goofy Joe

[sturdy vine]

P(X=1 | Y=0) = 1/2

so true for all

[spiral vigil]

sure

[sturdy vine]

but in (x=1,y=0) i have 1/4 in the table

[spiral vigil]

generally if you can show that your matrix is factorable into a product of two vectors then it's independent. that's also exactly when the matrix has rank 1, but idk if you've used those concepts yet

[spiral vigil]

yes you do

so?

[sturdy vine]

its not 1/2

[spiral vigil]

correct

so?

[sturdy vine]

I don't understand what I'm doing

[spiral vigil]

you want to show that P(X=1 | Y=0) is equal to P(X=1)

[sturdy vine]

its true

if they are indipendent

[spiral vigil]

kinda seems like you've already done that? how did you calculate this?

[sturdy vine]

ok

writing

$P(X=1|Y=0)=\frac{P(X=1 \cap Y=0)}{P(Y=0)}=\frac{P(X=1)P(Y=0)}{P(Y=0)}=\frac{1/2*1/2}{1/2}=1/2=P(X=1)$

[glossy valve]

Goofy Joe

[spiral vigil]

this is only true for independent events

[sturdy vine]

yes

two tosses of a coin are two independent events

[spiral vigil]

yes...... yes they are. you already knew that. if you're going to assume they're independent then why bother calculating anything here

[sturdy vine]

to see the probability

so I have to go by intuition to see if they are independent or not, right?

[spiral vigil]

no.... you've already written out the equations to tell you whether they are

[sturdy vine]

yes

[spiral vigil]

,tex
\begin{matrix}
p_{X,Y}(x,y) & y=0 & y=1 \
x=0 & 0.1 & 0.3 \
x=1 & 0.2 & 0.4
\end{matrix}

are these events independent?

[glossy valve]

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

[spiral vigil]

without using your diagonal thing

use conditional probability

[sturdy vine]

how

can

[spiral vigil]

well, start by finding P(X = 0)

[sturdy vine]

but

how can i calculate P(X|Y) now

i dont know if they are indipendent

[spiral vigil]

correct, we're going to find that out

$P(X = x | Y = y) = \frac{ P(X = x \cap Y = y) }{ P(Y = y) }$

[glossy valve]

hayliänus austrǎlis

[sturdy vine]

P(X=0) its 0.4

[spiral vigil]

and you have a matrix telling you all of the values for P(X = x & Y = y)

[sturdy vine]

yes

[spiral vigil]

okay, do that for all the other ones such as Y = 0

[sturdy vine]

P(Y=0)=0.3 , P(Y=1)=0.7 , P(X=1)=0.6

[spiral vigil]

okay great, now what's P(X = 0 | Y = 0)?

[sturdy vine]

i think i understand

Now I have to get that P(X=0|Y=0)=P(X=0)

[spiral vigil]

yes, that would be one point of evidence towards these events being independent

[sturdy vine]

P(X=0|Y=0)=1/10 * 3/10=3/10=0.3=P(Y=0)

[spiral vigil]

$\f1{10}\cdot\f3{10}\theq\f3{10}$

[glossy valve]

hayliänus austrǎlis

[sturdy vine]

ops

0.03

they are not indipendent

[spiral vigil]

3/100 = 0.03

but why were you comparing to P(Y = 0) in the first place?

[sturdy vine]

🙁

i need to see X=0 sorry

P(X=0)=0.4

[spiral vigil]

hang on i'm very confused in general hold up

P(X=0|Y=0)=1/10 * 3/10
where did this even come from

what are we multiplying

[sturdy vine]

P(X=0|Y=0)=1/10 * 10/4=1/4=0.25

i fix

[white badge]

1/4 = 0.4?

[sturdy vine]

🙁

[spiral vigil]

also, where did 10/4 come from?

[sturdy vine]

(1/10)/(4/10)=1/10 * 10/4 =1/4=0.25

[spiral vigil]

okay, where did 4/10 come from?

[sturdy vine]

P(X=0)=0.1+0.3=0.4=4/10

[spiral vigil]

yes, but why are we dividing by P(X = 0)?

[sturdy vine]

🙁

[sturdy vine]

but so I had done well before?

[spiral vigil]

i don't understand that question

[sturdy vine]

im fixing

[spiral vigil]

question: are you writing any of this down on a piece of paper or whiteboard or are you trying to solely keep notes via discord message?

[sturdy vine]

discord message

[white badge]

maybe you should jot it down on MS Paint or paper or something.

[sturdy vine]

now im on a paper

[white badge]

writing gives you time to process what you wrote instead of quickly hitting the Enter key.

[sturdy vine]

P(X=0|Y=0)=0.3 and P(X=0)=0.4

[spiral vigil]

it's also much faster to quickly reference stuff like P(X = 0) = 0.4

I don't know where you got P(X=0|Y=0)=0.3

[sturdy vine]

but

im using

$p_Y (0)=\sum_{x\in X} p_{x,0}(x,0)$

[spiral vigil]

nah yeah your individual probabilities are correct

the 0.4, 0.6 stuff

[glossy valve]

Goofy Joe

[white badge]

maybe you can try something like this.

[sturdy vine]

so 0.1+0.2=0.3

this is P(Y=0)

[spiral vigil]

yes like i said those are all fine

[white badge]

(values are all examples.)

[sturdy vine]

P(X=0|Y=0)=0.3 and P(X=0)=0.4 so they are not indipendent

[white badge]

this way you get to keep track exactly what value you are using for each and can hopefully immediately notice if some value doesn't make sense.

[spiral vigil]

can you please explain how you calculated P(X=0|Y=0)=0.3 ? I get something else.

[sturdy vine]

(1/10) / (3/10) = 1/3

i take a pic

[spiral vigil]

I don't understand why you keep going into fractions but that's okay

yes, it's 1/3

and 1/3 is not equal to 0.3

[sturdy vine]

its 0.3 periodic

[spiral vigil]

yes.... that's a different number than 0.3

[sturdy vine]

yes

i made a mistake

1/3 is not equal to 0.4

so not indipendent

[spiral vigil]

that's correct yes

if they did match you'd have to keep trying all the others as well

if they all match properly then you can conclude it's independent

[sturdy vine]

ah ok

clear

thanks all

.close

[full forum]

Channel closed

Closed by @sturdy vine

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gusty flicker]

i want to understand this

[sonic stratus]

I wonder if you can make the usual trick of y = log(x+y)

[gusty flicker]

i believe you can

i want to do $f(x)=\sqrt2^x-x$, then i believe this is $\int_0^{2026}f^{-1}(x)\text{ d}x$?

[glossy valve]

ijo Sani

[gusty flicker]

so it would be $(2026-0)(f^{-1}(2026)-f^{-1}(0))-\int_{f^{-1}(0)}^{f^{-1}(2026)}f(x)\text{ d}x$?

[glossy valve]

ijo Sani

[gusty flicker]

which would simplify to $2026\times20-\int_2^{22}\sqrt2^x-x\text{ d}x$

[glossy valve]

ijo Sani

[gusty flicker]

$\sqrt2^x-x=e^{x\ln\sqrt2}-x$

[glossy valve]

ijo Sani

[gusty flicker]

$$\int\sqrt2^x-x\text{ d}x=\frac{e^{x\ln\sqrt2}}{\ln\sqrt2}-\frac{x^2}2$$

$$\int\sqrt2^x-x\text{ d}x=\frac{\sqrt2^x}{\ln\sqrt2}-\frac{x^2}2$$

[glossy valve]

ijo Sani

[gusty flicker]

$$\int_2^{22}\sqrt2^x-x\text{ d}x=\frac{2048-2}{\ln\sqrt2}-\frac{22^2-2}2$$

[glossy valve]

ijo Sani

[gusty flicker]

wai wai wait, apparently it's at 4?

$f^{-1}$ is not really well-defined, and i think the current method errors

[glossy valve]

ijo Sani

[gusty flicker]

$2026\times18-\int_4^{22}\sqrt2^x-x\text{ d}x$

[glossy valve]

ijo Sani

[gusty flicker]

$$\int_4^{22}\sqrt2^x-x\text{ d}x=\frac{2048-4}{\ln\sqrt2}-\frac{22^2-4^2}2$$

[glossy valve]

ijo Sani

[gusty flicker]

$$\cdots=\frac{4088}{\ln2}-234$$

[glossy valve]

ijo Sani

[gusty flicker]

$2026\times18-\int_4^{22}\sqrt2^x-x\text{ d}x=2026\times18-\frac{4088}{\ln2}+234$

[glossy valve]

ijo Sani

[gusty flicker]

$$\cdots=\boxed{44806-\frac{4088}{\log(2)}}$$

[glossy valve]

ijo Sani

[gusty flicker]

ok lemme compile this into one thing

[whole lily]

AOPS sol
https://artofproblemsolving.com/community/c7h3756199p37013470

[gusty flicker]

i want to type this out myself

[whole lily]

no worries

[gusty flicker]

ty!

[knotty grail]

i think its meant to be 2026 x 22 here..

[gusty flicker]

nope, $f^{-1}(0)=4$

[glossy valve]

ijo Sani

[knotty grail]

isnt it 2026 x f-1(2026) - 0 x f-1(0)

[gusty flicker]

i dont think so? im not sure though

im in latex hell

Let $f(x) = \left(\sqrt2\right)^x-x$ for $4\le x\le22$, then \begin{align*}I&=\int_{0}^{2026}f^{-1}(x)\text{ d}x\&=2026f^{-1}(2026)-0f^{-1}(0)-\int_{f^{-1}(0)}^{f^{-1}(2026)}f(x)\text{ d}x\&=(2026-0)(22-4)-\int_4^{22}\left(\sqrt2\right)^x-x\text{ d}x\&=2026\cdot22-\left[\frac{\left(\sqrt2\right)^x}{\ln\sqrt2}-\frac{x^2}2\right]_4^{22}\&=44572-\left(\left(\frac{\left(\sqrt2\right)^{22}}{\ln\sqrt2}-\frac{22^2}2\right)-\left(\frac{\left(\sqrt2\right)^4}{\ln\sqrt2}-\frac{4^2}2\right)\right)\&=44572-\frac{\left(\sqrt2\right)^{22}-\left(\sqrt2\right)^4}{\ln\sqrt2}+\frac{22^2-4^2}2\&=44572-\frac{2^{11}-2^2}{\ln2/2}+\frac{468}2\&=44572-\frac{2\cdot2044}{\ln2}+234\&=\boxed{44806-\frac{4088}{\ln2}}\end{align*}

[glossy valve]

ijo Sani

[gusty flicker]

8 minutes of latex hell lmao

Let $f(x) = \left(\sqrt2\right)^x-x$ for $4\le x\le22$, then \begin{align*}I&=\int_{0}^{2026}f^{-1}(x)\text{ d}x\&=2026f^{-1}(2026)-0f^{-1}(0)-\int_{f^{-1}(0)}^{f^{-1}(2026)}f(x)\text{ d}x\&=(2026-0)(22-4)-\int_4^{22}\left(\sqrt2\right)^x-x\text{ d}x\&=2026\cdot22-\left[\frac{\left(\sqrt2\right)^x}{\ln\sqrt2}-\frac{x^2}2\right]_4^{22}\&=44572-\left(\left(\frac{\left(\sqrt2\right)^{22}}{\ln\sqrt2}-\frac{22^2}2\right)-\left(\frac{\left(\sqrt2\right)^4}{\ln\sqrt2}-\frac{4^2}2\right)\right)\&=44572-\frac{\left(\sqrt2\right)^{22}-\left(\sqrt2\right)^4}{\ln\sqrt2}+\frac{22^2-4^2}2\&=44572-\frac{2^{11}-2^2}{\left(\ln2\right)/2}+\frac{468}2\&=44572-\frac{2\cdot2044}{\ln2}+234\&=\boxed{44806-\frac{4088}{\ln2}}\end{align*}

[glossy valve]

ijo Sani

[knotty grail]

line 4 is an impostor

[gusty flicker]

i need someone to teach me though

[gusty flicker]

why

[gusty flicker]

how do i integrate $f^{-1}(x)$

[glossy valve]

ijo Sani

[knotty grail]

how did $2026 f^{-1}(2026) - 0f^{-1}(0)$ become $(2026-0)(22-4)$ and then again $2026 \cdot 22 - 0$ in the next line

[gusty flicker]

i thought by line 4 you meant line 4 of the align. whoops

Let $f(x) = \left(\sqrt2\right)^x-x$ for $4\le x\le22$, then \begin{align*}I&=\int_{0}^{2026}f^{-1}(x)\text{ d}x\&=2026f^{-1}(2026)-0f^{-1}(0)-\int_{f^{-1}(0)}^{f^{-1}(2026)}f(x)\text{ d}x\&=2026\cdot22-0\cdot4-\int_4^{22}\left(\sqrt2\right)^x-x\text{ d}x\&=2026\cdot22-\left[\frac{\left(\sqrt2\right)^x}{\ln\sqrt2}-\frac{x^2}2\right]_4^{22}\&=44572-\left(\left(\frac{\left(\sqrt2\right)^{22}}{\ln\sqrt2}-\frac{22^2}2\right)-\left(\frac{\left(\sqrt2\right)^4}{\ln\sqrt2}-\frac{4^2}2\right)\right)\&=44572-\frac{\left(\sqrt2\right)^{22}-\left(\sqrt2\right)^4}{\ln\sqrt2}+\frac{22^2-4^2}2\&=44572-\frac{2^{11}-2^2}{\left(\ln2\right)/2}+\frac{468}2\&=44572-\frac{2\cdot2044}{\ln2}+234\&=\boxed{44806-\frac{4088}{\ln2}}\end{align*}

[glossy valve]

ijo Sani

[whole lily]

looks good

[gusty flicker]

i still need help with this

nevermind im just stupid

[whole lily]

you should use y instead of f(x)

so it works exactly like a u-substitution

[gusty flicker]

hmm, right, makes sense

welp

.close

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[whole lily]

if you express it in terms of y it's simply integration by parts

[gusty flicker]

ty!

wait

.reopen

[full forum]

✅ Original question: #help-28 message

[gusty flicker]

why did $f^{-1}(0)=2$ fail?

[glossy valve]

ijo Sani

[whole lily]

because you want the substitution to be injective

[gusty flicker]

i intuitively know why but can't pinpoint

the entire portion has to be?

[whole lily]

yeah

[gusty flicker]

k

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[spice grail]

is this correct?

[slate glacier]

Looks right

[grave elm]

but your 8 looks like 6

[slate glacier]

True

The one in the denominator of the power of e

[spice grail]

thanks

is dy = 1

[onyx glen]

no, dy is dy

but $\int \frac{\dd{y}}{8+y} = \int \frac{1}{8+y} \dd{y}$ if that's what you really wanted to know

[glossy valve]

Ann

[spice grail]

shouldn't it be this?

[wheat fern]

use absolute value for the log and according to substitution you need to divide by the coefficent of the variable here its -1

[spice grail]

can you show me

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[brittle bison]

hey, how to know when to do Integration by parts or change of variable?

[terse bramble]

hi

[brittle bison]

hii

[fast peak]

experience

[brittle bison]

ik but like is there a tech ?

sorry about the poor language i'm not native

[fast peak]

well its a product and there is no composition, so ibp

[onyx glen]

you try, you fail, you try again

[terse bramble]

are you asking how to do it

[brittle bison]

no, how to recognize it, i know how to do ibp or cov

[onyx glen]

there is no silver bullet

[brittle bison]

i see that there's no tech at all, ty all

[onyx glen]

unfortunately you just have to do 100 integrals

[terse bramble]

google 100 integral challenge and have fun

[brittle bison]

ok ok thank you to you all 🙏 🙏 🙏

How can I tell if I need to apply integration by parts a second time? I do it once, but then how do I know whether the new integral also requires another round of integration by parts?

[void nova]

Again, experience

And also if you see that the degree of the polynomials you have are getting smaller and smaller

[brittle bison]

oh rightt, and i need to keep doing ibp until my polynomial is =0 ?

[void nova]

Until the degree of the polynomial is 0

[brittle bison]

yeah i minded that srry

thanks alberto

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[sturdy oriole]

hello.

[sturdy oriole]

i wanna solve it without doing trigonometric steps, could anyone help?

[full forum]

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

[grave elm]

here is a hint

make it isosceles

[pearl sparrow]

if u want it without trigo

u can use the ange properties i mean here

or u can say angle chasing

Because A D equals D C and the angles at A and at B on the two sides are equal to alpha, the configuration is symmetric about B D, so the 45 degree angle at D is split equally, giving alpha equal to 22.5

[grave elm]

oh nvm ig this doesnt help

[grave elm]

this is weird, what do you mean by symmetric about BD?

[round gust]

doesnt seem symmetric to me

[grave elm]

I checked with trig and ur alpha isnt correct

[sturdy oriole]

hmm.

really?

the correct answer is 30°.

[sturdy oriole]

fr.

[round gust]

why do you want to solve it without trig?

[pearl sparrow]

i didnt realised i mean, i was sticking to the basic symmetry but yea dude, why u legit want to solve without trig i mean it would be soo easy with trig

[full forum]

@sturdy oriole Has your question been resolved?

[grave elm]

A silly way to do it would be imitating proof of sine law by similarity

Otherwise i dont really see what to add to make it angle-chasable, but it should be possible

[boreal slate]

this was not worth it

[sturdy oriole]

🤨

interesting.

[sturdy oriole]

understood, thank you very much!

so, tan(30)=1/sqrt(3) explains, yeah?

[boreal slate]

yeah, its a unit circle/special triangle thing

[sturdy oriole]

aah, aight, appreciate your support also yours, guys.

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[spice grail]

could we separate it and then integrate?

[round gust]

no

you cant have dy/dx = 7 + 3/x * y since the right side would depend on both x and y

so you cant use g(y) dy = h(x)

[spice grail]

can someone help with showing the prodcut rule

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@spice grail Has your question been resolved?

[ivory cairn]

what do you mean showing the product rule?

[spice grail]

like how it is the product rule but i figured it out

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[brittle spade]

A square and an equilateral triangle have equal perimeters. The area of the triangle is $2\sqrt {3}$ square inches. What is the number of inches in the length of the diagonal of the square?

[glossy valve]

soloplanet

[brittle spade]

What i have done already is

realised square sidelenght is 3/4 of triangle sidelength

and i think the height is sqrt (3)

and if i half it i get 1

and the hypotenuse is 2

[neon basin]

let me ask you one question

What is the formula for the area of an equilateral triangle?

🙂

[brittle spade]

(width * heihgt) / 2?

oh so it cant be 2

wait is that correct

is there like a specific formula for an equalaterial traingle

[neon basin]

there is

it's a very special triangle

it's bound to have a very special formula

[brittle spade]

height = xsqrt(3)

if you half it

height is xsqrt(3), hypotenuse is 2x and base is x. otherwise i do not know it

[neon basin]

$A = \frac{\sqrt{3}}{4} \cdot a^2$

[glossy valve]

USS-Enterprise

[brittle spade]

what

is that

[neon basin]

Area of an equilatereal triangle, where A is the area, and a is the side length

[brittle spade]

.

what

that exists

[neon basin]

of course

so many things exist in triangles

and equilaterals let them shine

[brittle spade]

why does that make math seem like much more exciting

so i just solve for it?

and then a * 3/4

then (a * 3/4) ^2 + (a * 3/4^2 = answer?

[neon basin]

Well you are looking for the length of the square's diagonal

a * 3/4 gets you the square's side length

And how do we calculate a square's diagonal length

[brittle spade]

2a^2 = c

theres just 1 problem

[neon basin]

I notice it yes

[brittle spade]

i kinda suck at radical operations.....sry

[neon basin]

c^2

[brittle spade]

oof

i forogt

cooked i am

[neon basin]

If you have c^2

and want c

what do you do

[brittle spade]

sqrt(c)

.sqrt(c)

um

how do you simplify

[neon basin]

$c^2 = 2a^2$

[glossy valve]

USS-Enterprise

[brittle spade]

$2sqrt(3) = sqrt(3)/4 * a^2

[neon basin]

$c = \sqrt{2a^2}$

[glossy valve]

USS-Enterprise

[neon basin]

$c = \sqrt{2} \cdot a$

[glossy valve]

USS-Enterprise

[neon basin]

right

[brittle spade]

this one

no

[neon basin]

[brittle spade]

this one

sry mb gng

[neon basin]

hang on

[brittle spade]

so we just get a alon

[neon basin]

where did 2sqrt(3) come from

[brittle spade]

thats the equalateral triangle area

[neon basin]

oh right

I thought we had a already

[neon basin]

From this

[neon basin]

yes

sorry

I misread

[brittle spade]

so like

2 sqrt(3) / sqrt(3)/4 = a^2

[neon basin]

$2\sqrt{3} = \frac{\sqrt{3}}{4} \cdot a^2$

[glossy valve]

USS-Enterprise

[neon basin]

Yes, we divide by sqrt(3)/4

[brittle spade]

so then 2sqrt(3)/1 * 4 / sqrt(3)

[neon basin]

sure

[brittle spade]

so 8sqrt(3) / sqrt(3)

[neon basin]

yes

[brittle spade]

is the word i and q banned

[neon basin]

oh

[brittle spade]

lol

[neon basin]

seems

[brittle spade]

so 8 = a^2

[neon basin]

anyway yeah

I am slow

yes

[brittle spade]

so sqrt(8)

so 2sqrt(2)

[neon basin]

yes

[brittle spade]

so 2sqrt(2) * 3/4

wait

what

6sqrt(2) / 4?

pl

plz

plz

plz be right

[neon basin]

yes

[brittle spade]

LETS GO

[neon basin]

so 3sqrt(2)/2

is the square's side

[brittle spade]

so 2(3sqrt(2)/2)^2 = c^2

so 2(18 / 4) = c^2?

[neon basin]

It'd be much simpler if you solved for c first with a

$2a^2 = c^2$

[glossy valve]

USS-Enterprise

[neon basin]

And then just subtituted a in

[brittle spade]

wait

did i just do that

i substituted a in

[neon basin]

Uh yes

What I meant is

try to simplify things first

otherwise you might end up with lines-long terms and equations

[brittle spade]

how do i do that?

i dont know what to simplify

[neon basin]

This

You just put a in

[brittle spade]

oh

ok

[neon basin]

and then squared it, multiplied by 2, then took the square root

It's much easier if you work with a as a first

and solve for c

and then substitute a in

So first you'd take the square root and get $c = \sqrt{2} \cdot a$

[glossy valve]

USS-Enterprise

[brittle spade]

so 2(sqrt(3)/2)^2 = c^2

[neon basin]

And now you can just substitute a in

[brittle spade]

wait what

[neon basin]

From this

[brittle spade]

so you have 2a^2 = c^2

[neon basin]

Yes

And you solve for c

[brittle spade]

so you can simplify by sqrt(2 * a

so c = sqrt(2) * a

[neon basin]

sqrt(2*a^2) yes

not 2*a

a^2

[brittle spade]

so a sqrt(2)

[neon basin]

$\sqrt{x \cdot y} = \sqrt{x} \cdot \sqrt{y}$

[glossy valve]

USS-Enterprise

[neon basin]

yes

[brittle spade]

so sqrt(3)/2 * sqrt(2) = c

so sqrt(6) / 2 = c

[neon basin]

yes

but wait just a second

[brittle spade]

so sqrt(3) / 2?

wait can you do that

no

[neon basin]

why sqrt(3)?

[neon basin]

here

we got $a = \frac{3 \sqrt{2}}{2}$

[glossy valve]

USS-Enterprise

[brittle spade]

ye

[neon basin]

WHy then sqrt(3)/2 * sqrt(2)

It's 3*sqrt(2)/2 * sqrt(2)

[brittle spade]

oh it evolved overtime and i forgot

im not using paper

[neon basin]

If we go again, through everything quickly.

$2\sqrt{3} = \frac{\sqrt{3}}{4} \cdot a^2$

[brittle spade]

so i cant rember sry gng mb

[glossy valve]

USS-Enterprise

[neon basin]

Is what we began with

We solved for a, triangle's side length

a = $2\sqrt{2}$

[glossy valve]

USS-Enterprise

[neon basin]

Then, square's side length, say b, is $b = \frac{3}{4} \cdot a$

[glossy valve]

USS-Enterprise

[brittle spade]

k

[neon basin]

So $b = \frac{3\sqrt{2}}{2}$

[glossy valve]

USS-Enterprise

[neon basin]

And then, to get the diagonal

$c = \sqrt{2} \cdot b$

[glossy valve]

USS-Enterprise

[neon basin]

$c = \sqrt{2} \cdot \frac{3\sqrt{2}}{2} = \frac{3}{2}$

[brittle spade]

wait can we simplify this further

[neon basin]

No, 3/2 is all we can do

[brittle spade]

wait

[neon basin]

You used the wrong side here

so that is wrong

[brittle spade]

ohhhh

doesnt it become sqrt(2) * (3sqrt(2)

wait

so when we have whole number

only sqrt 2 mutliply with sqrt 2

[neon basin]

what do you mean

[brittle spade]

when we multiply we get sqrt(2) * (3sqrt(2)

[neon basin]

yes

OH

😂

[brittle spade]

so dont we mutliply 3 by sqrt(2) and sqrt(2) by sqrt(2)

[brittle spade]

this was the moment i realised

[neon basin]

Yes

[brittle spade]

i mucked up

[neon basin]

I did

[brittle spade]

oh

[neon basin]

I need to go to sleep really

It's 2AM

[brittle spade]

holy

[neon basin]

so yeah we get 3 * sqrt(2) * sqrt(2) / 2

[brittle spade]

you are actually a goated helper

[neon basin]

so 3 * 2 / 2

so 3

[neon basin]

thanks!

🙂

[brittle spade]

wait isnt it 3 * sqrt(2) and sqrt(2) * sqrt(2)

i understand everything except this

[neon basin]

No it's $\sqrt{2} \cdot \frac{3\sqrt{2}}{2}$

[glossy valve]

USS-Enterprise

[brittle spade]

oh

k

[neon basin]

sqrt(2) * b

and b is 3sqrt(2)/2

[brittle spade]

so that means 3

[neon basin]

yes

[brittle spade]

let me check

marvelous work

thanks

so i just have to remember

[neon basin]

correct?

[brittle spade]

dui

so i jsut have to remmber

[neon basin]

driving under influence?

[brittle spade]

sqrt(3) / 2 * a^2

[neon basin]

4

[brittle spade]

yes

no

[neon basin]

sqrt(3)/4 * a^2

[brittle spade]

i mean it means yes

k

thx

[neon basin]

It's nice if you move a^2 to the start

then the numbers go up by 1

a^2 * sqrt(3) / 4

[brittle spade]

oh ok

thanks for your help

[neon basin]

no problem

good luck

[brittle spade]

thx bro

[neon basin]

I go sleep

😴

[brittle spade]

gng

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[brisk warren]

I am kinda bad at combinations

I know some possible ways but there are too many to count how to do this

[rare dock]

we don't need to know anything about combinations for ts

[brisk warren]

Then how do I do this

I would guess 5 as answer

[rare dock]

if you knew how many paths of length 8 there were from A to each other point in the graph, would you be able to answer the question?

[brisk warren]

No

I have a few ideas like we need to repeat our path a few times

[rare dock]

a path of length 9 consists of a path of length 8 to either x or y followed by a move to B

[brisk warren]

I see

[rare dock]

so if you had this, you could just take how many paths of length 8 there are to x and y and add them up

[brisk warren]

But I don’t know how to do that

[rare dock]

so how do we get these? well you can look at paths of length 7

[brisk warren]

Ok

[rare dock]

that'll work but it's not the fastest way probably

since this graph is kinda nice

[brisk warren]

So how do I do it

[rare dock]

i'll label these too

[brisk warren]

Wait is the answer 12

[rare dock]

that looks too smol but i have not done the problem yet

notice that we first go to z/w, then we hang out switching from z to w for some time (could be no time at all)

[brittle steeple]

There's a cool matrix solution for these kind of things

It's probably quite hard without it

[rare dock]

then we move to the x/y column

[brisk warren]

Yes

Ok

[brittle steeple]

Oh yeah ok I see what you mean

[rare dock]

you can do some casework on that i guess

by how many moves you spend in the z/w column and how many you spend in the x/y column

[brisk warren]

I found out we need to go through x and y and z and w a combined 6 times

[rare dock]

2 of the 8 moves need to be moving out of A and to B

yea

[brisk warren]

So then I used cases to get 12

[brittle steeple]

Does path of length 9 mean 9 vertices and 8 moves?

[rare dock]

good question

[normal fog]

Is the question how many ways are there to reach B for a total of 9 paths lenght?

[brittle steeple]

Since they mentioned all the segments have length 1, i would assume it's 9 moves

[rare dock]

ok, scale up everything i said by 1

cool, a path is uniquely determined by the first point and how many moves you spend in the z/w column

so i think 14

10 vertices total, a path looks like

A, n vertices in {z,w}, 10-2-n vertices in {x,y}, B

if we fix n and then pick z or w to start, the path is unique. so 2 paths for each n. and any n in {1,2,...,7} is fine

[normal fog]

I think 32

[rare dock]

why

[brisk warren]

It’s 14

[rare dock]

great

[brisk warren]

I checked

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[nimble canyon]

How do you do this?

[nimble canyon]

I have a feeling this isn t year 11 rather year 12?

[robust slate]

Write as a geometric series

[nimble canyon]

1 + 10^1+10^2+10^3......10^2n-1

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[boreal slate]

when you tried it, what happens?

[boreal slate]

after your therefore symbol, dont forget thats S_1

[nimble canyon]

Alright but i assume from this i managed to turn this Expression into an actual AP?

[boreal slate]

you turned it into an infinite geometric progression

[nimble canyon]

Then once i did it again into a GP

[boreal slate]

actually after the first difference, its something known as an arithmetico geometric progression

since it has both an arithmetic part and a geometric part

[nimble canyon]

So the method of differrene allows me to turn any progression into either a AP or GP which allows me to solve it

[boreal slate]

method of difference usually ends with a gp

the ultimate goal with all this is to make something called a closed form

closed forms are formulas that simplify an infinite expression to a finite one

[nimble canyon]

How is this helpful?

[boreal slate]

not always infinite, but it shrinks a long sum into a shorter one

easier to compute

and 100% accurate

[nimble canyon]

So does methods of Difference also allows me to turn a series into a GP or a AP allowng for me to solve it?

[boreal slate]

it allows you to create a closed form

which allows you to compute the exact answer

in a reasonable amount of time

[nimble canyon]

So its just a method to find the answer of something

[boreal slate]

its a method to find the answer to something quickly and exactly

[nimble canyon]

Alright what type of questions do i normally use iti n?

[boreal slate]

questions where you have an infinite sum

and theres a pattern that you can exploit

[nimble canyon]

So lets say if a question is asking you to find the sum of something but the sequence isnt a AP or a GP you will use this to turn it into a closed form to find it?

[boreal slate]

possibly, but this is one of those tools that has its use cases

and its use cases are limited

[nimble canyon]

Alright

[nimble canyon]

Isnt this method also used to prove both Sum formulas of an AP and a GP

[boreal slate]

yeah you can use it to prove those

[nimble canyon]

What about the formulla of converging GP's?

[boreal slate]

yes, that falls under GP

[nimble canyon]

How would you prove n/2 (a + l) then?

You cant write out the geneal formulas of a AP since it doesnt consist of L?

Nvm

alr thx

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[lean rock]

is this a thing in braids?

 • 二 + ⤬
二 二   二
+
⤬ 二   ⤬

(二 + ⤬)•(二 + ⤬)=二² + 2二⤬ + ⤬²=二+ 二+ ⤬=⤬?

[upbeat knot]

what

[maiden vapor]

what exactly are you trying to do

braid group? braid algebra? skein theory?

glow $ds (1+sigma)^2 = 1 + 2 sigma + sigma^2$

[glossy valve]

Oléagineux Distilliànus VIVII

[maiden vapor]

assuming your 二 is the identity (i.e. doing nothing/parrarel strands) and ⤬ is a single crossing aka the generator sigma

[maiden vapor]

you can reduce this but you need a specific rule

that is the skein relation

to deal with the double twist

in the usual sense we have that the double twist is equal to (q-1)σ + q where q is a variable related to how stiff or twisted the knot is

[lean rock]

I wonder if multiplication translates also from ℤ

[maiden vapor]

uhm

[lean rock]

@maiden vapor thank you, lots of information

[lean rock]

this looks the same to what I wrote

similar

[maiden vapor]

yes but the steps after that are only valid id you specify your skein relation

[lean rock]

ok

[maiden vapor]

i suppose if you represent braids using only 2 strands then order does not matter

in other words, it does not matter which one you stack first because there are no other strands to get in the way

adding a third strand would break commutativity

[lean rock]

Okay, thanks lots!🎉

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[plain depot]

if i have 2 distinct classes and 2 features, LDA tries to project those 2 features into one line such that the variance in the classes are minimized and the difference between means are maximized right ?

[plain depot]

And i think i get the intuition behind it, but i dont understand the reasoning with 3 distinct classes and 2 features

sorry, my english is not the best

but i hope i made myself clear

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[night musk]

for part b is it valid to say that we could write a funciton f(x) such that f'(x) = (x-alpha)^kj(x) wher j(x) = r+1(g(x)) + (r-alpha)g'(x). and just say since integration reverse differentiation f(x) always has k+1. Also if you did not know that integration is the opposite of differentiation is there anything you could do?

[spiral vigil]

you don't really need integration since you're defining f(x) and then talking about its derivative...

i suspect you'll want to use the result of (a) in your argument for (b)

[night musk]

isn't that what I do though? I can prove the same way that I proved that if f(x) has r zeroes than f'(x) has r-1 zeroes but now have r+1 for f(x) and r for f'(x). Also I don't see why for part a they say at least some zeroes shouldnt always be 1 down

[spiral vigil]

shouldn't always be 1 down

[night musk]

but this isn't even at least r-1. we went from 1 zero to 2

[spiral vigil]

and 2 is, it turns out, at least 0

[night musk]

ok I see.

is what I said though about part b general?

[spiral vigil]

i can't really tell what your function is doing

but you must remember that f(x) might not be a polynomial or anything like that; it could be a sine function for instance

[night musk]

wait doesn't he say that f is a polynomial when setting up the problem?

[spiral vigil]

oh it sure does mb

[night musk]

ok so f(x) = (x-alpha)^r+1 g(x) then we have f'(x) = r+1(x-alpha)^rg(x) + (x-alpha)^{r+1} g'(x) = (x-alpha)^r(r+1(g(x)) + (x-alpha)g'(x)). Let j(x) = (r+1(g(x)) + (x-alpha)g'(x)). we have f'(x) = (x-alpha)^rj(x). j(alpha) does not equal 0. and then we can see for a polynomial of r roots comes from a polynomial of r+1 roots

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[misty pine]

@full forum could you teach me calculus like i know 0

[silk bridge]

bot is no ai
and no human can teach you entire here

[misty pine]

Oh ok

[onyx glen]

she*

[misty pine]

Thanks for clearing

[onyx glen]

the bot literally has a pronoun role 😭

[silk bridge]

fuk

[onyx glen]

#study-discussion

we can't teach you entire topics from scratch

[silk bridge]

specific questions yes

[misty pine]

Ok I’ll learn it through ChatGPT

I’m in middle school learning this pray for me✌️🥀

[onyx glen]

!noai

[full forum]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[white badge]

um... not a good idea to learn whole topics through AI/LLMs, in my humble opinion.

[onyx glen]

so like, we can't stop you, but it's a mega bad idea

[misty pine]

Fathers help?

[onyx glen]

... wait, are you over 13 or not

[misty pine]

What’s that mean

[misty pine]

I am why

[onyx glen]

"middle school" threw me off

[misty pine]

Oh

Mb

[onyx glen]

anyway, if your father can teach you calculus, then why not.

!noai just prints that factoid about AI

[white badge]

a heads-up though, please make sure your algebra is solid before heading into calculus.

[misty pine]

I’m just learning it for my career school teaching me easy things

[misty pine]

It is,expressions,equations what not monomials binomials trinomials

[white badge]

specifically you need to master functions.

[misty pine]

Oh ok I’ll see on that

Thanks for the advice

Guys why’s vectors important

It’s just left right up and down and forward and backwards

[upbeat knot]

for one, when handling more information we might end up looking at higher dimensions

[misty pine]

Oh like is 4d found yet or uhh not

[white badge]

I think that thinking of vectors as "quantities with directions" is very limiting and helps only up to a certain point.

[upbeat knot]

and for two, even vectors in 2D/3D can be closely related to complicated matrices that describe space, transformations, ....

and for three, even the notion of vectors as directions in 2D-3D space isnt as easy as you think when combined with geometry:

angles, perpendiculars, intersections, ...

not to scare you but if you have trouble with such geometry with lines, then good luck doing so with planes

so before you learn vectors / linear algebra, make sure your foundations for

ALGEBRA

and

BASIC GEOMETRY

are solid first

[misty pine]

I mean those are easy

[upbeat knot]

then good luck learning them

if you want to study it by yourself first then there are resources like khan academy and youtube online

but warning, learning more advanced stuff with shaky foundations will impede your current and future learning of that topic

(residual memory blah blah blah)

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[boreal token]

Hello what do you call a second (1 second) divided into 60 parts?
If 1 minute = 60 seconds
then 1 second = 60 what?

[zenith kernel]

Third (historically)
ChatGPT says historically (1/60)^k times one minute is called k+1 th. Like (1/60) one minute is called second, (1/60)^2 one minute is called third… and it says we don’t have a universal name for that nowadays

[boreal token]

Ok thanks, is it the most universally used divison of a second?

Say 3.23s = 3 seconds and 23 thirds?

[flint hull]

it doesnt go by 60s anymore, after that they start using the system with millisecond, microsecond, nanosecond, etc

where 1000 milliseconds = 1 second

[boreal token]

At least the 60s is not confusing compared to the 10s

[white badge]

I don't think I have ever seen a second subdivided into 60 equal parts, especially in modern times.

[flint hull]

nobody uses thirds anymore

you would just say three point two three seconds, or if you really want to be specific 3 seconds and 230 milliseconds

[white badge]

with the advent of computing, the Unix timestamp format and what not, we've pretty much abandoned the 60-part subdivision past seconds.

[boreal token]

I see, in video editing programs like Adobe Premiere Pro on default 1 second is still divided into 60 parts

[white badge]

that's because usually 60 frames are in a second.

[flint hull]

thats different, it counts frames

[white badge]

keyword: usually.

you could have any number of frames per second, but I don't know how Premiere works with that.

[boreal token]

oh I see, it's frames, it's still better for me than the 10s divison, thanks kind people

[white badge]

thing is, counting time by frames is not reliable, unless you are sure your frame rate is constant.

[flint hull]

well if youre referring to general time never tell someone 3.23s is 3 seconds and 23 frames

[boreal token]

By the way do you call this 3.23 = 3 seconds and 2 deciseconds and 3 centiseconds

[white badge]

anyone who has played video games in recent times will tell you that unless you have a decent PC and/or a frame rate cap, you will see your FPS counter vary between 30 and 300, depending on the game and your PC.

[white badge]

nope. 3 seconds, 230 milliseconds.

we go by powers of 10^3 usually.

[boreal token]

230 milliseconds sounds like a very big number

[flint hull]

thats how its meant to be

[white badge]

well it's 230 x 10^-3 seconds, which translates to 0.23 seconds. not a long time at all.

it's the same idea with quantities like grams and milligrams, isn't it?

[flint hull]

deciseconds and centiseconds arent used

[white badge]

or meters and millimeters?

with the exception being length, since centimeters are commonly used.

but you definitely never hear people say centigrams.

another exception is volume - cubic decimeters are still used in some syllabi for chemistry.

but by far centi- and deci- are not really standard prefixes to hear on a daily basis.

[boreal token]

I see, it's kinda confusing instead of just saying 3.23 seconds which is 3 seconds and 23 thirds (if people know 1 second = 60 thirds)

3 seconds and 230 milliseconds sounds big

[flint hull]

right but how many people do you know that know what 23 thirds means compared to 230 milliseconds

[white badge]

which is why the standard is s.mmm (where s is seconds, then m is milliseconds).

when you have a standard, there's common agreement on what things mean.

[flint hull]

99% of people above middle school level know what a millisecond is, but if you say 23 thirds people's first instinct would be "hmm 23 divided by 3..."

[boreal token]

for easy remembrace the 0.23 I just times it by 100 to get the milliseconds?

[flint hull]

its just how the brain works for how we were taught. it was a very different time back when they used to use thirds

[white badge]

1000*.

[flint hull]

1000, move the decimal over 3 places

[boreal token]

oh forgive me it's 1000

I really am bad at math sorry

[flint hull]

so 3.23 is equal to 3 seconds and 230 milliseconds, or 3230 milliseconds

[boreal token]

ok thank you people! Have a nice day!

[white badge]

we've all been there before, OP. it's fine, you'll get better!

[flint hull]

of course!

[white badge]

anything else, OP?

[boreal token]

Nothing more, thank you!

[white badge]

alright, have a nice one!

[boreal token]

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[cobalt atlas]

why am I wrong

[cobalt atlas]

its asking for -5pi/6

which is -rad3/2, 1/2

[robust slate]

cos(-5pi/6) = cos(5pi/6) -> second quadrant -> negative

you dropped the negative sign

[cobalt atlas]

oh

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[cobalt atlas]

wait

actually never mind I understand

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[boreal wave]

Can someone explain what is happening in here?

Thank you

[gritty rose]

!msgdel

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[bitter python]

Hi I'm currently new to algebra, I did not pay attention to school. I'm doing some self-study when I got breaks at work and etc. I think it's pretty fun. But I didn't understand Why I couldn't add the x into the 4x^2 to make it 5x. In the study material I am using they seem to be adding the x's together. Thanks for you help :)

[bitter python]

( the right picture was what I got )

[sharp vine]

The thing is you can't

[bitter python]

Oh, may you elaborate?

[sharp vine]

I mean, in general your intuition is right, 4x^2 + x isn't 5x^2

[boreal slate]

its like trying to add 400+10 and expecting 500

[bitter python]

is it because it's + x?

[sharp vine]

Ah, you're asking why

[bitter python]

Yeah haha

If I wanna learn I wanna know why I can't do it and build over that in the future

[sharp vine]

you can only add similar words (i.e. of the same degree)

like x^2 + 3x^2 etc.

[bitter python]

so only to the power of can be added together?

so x^2 cannot be added with * x ( x^2 * x )

[sharp vine]

Notice that for example if you have 10 + 2 * 10^2 you don't say it's 3 * 10^2, you can check it by hand

[bitter python]

Because 10+2 * 10^2 and 3 * 10^2 is two whole different numbers

[sharp vine]

x^2 cannot be added with x, that's true, you can only do x + x^2 = x(1 + x) if needed