help-10

[drowsy nest]

hold on let me load up the identity

[restive sapphire]

kk

[drowsy nest]

i cant find the identity 😦

[restive sapphire]

answers 1/2

[drowsy nest]

ya

[restive sapphire]

bro me when i forgor sqrt(b) = b^1/2

[fierce lagoon]

It's just power rule

[restive sapphire]

ye!

damn bruh i got 1 more

me when calculator

@drowsy nest tyvm for the help

.close

[obtuse pebble]

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[alpine plaza]

need help with some grade 9 maths

[alpine plaza]

might seem a lil stupid to some of you older ones but

[merry peak]

ask away

[alpine plaza]

don't understand fractional indices

[merry peak]

what do you not understand about them

[alpine plaza]

ok so 9 to the power of 1/2 is = 2 root to the square of 9 =3

lemme get a photo probably

Here is a photo

...

wrote this down during class
I don't go to the best school teacher doesn't help just makes us go through a powerpoint and sits on her phone or laptop

The second line under fractional indices

9 to the power of 1/2 and so on

@merry peak

ping me if you reply

or whoever replys

[obtuse pebble]

@alpine plaza Has your question been resolved?

[obtuse pebble]

@alpine plaza Has your question been resolved?

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@alpine plaza Has your question been resolved?

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[shadow estuary]

can someone pls help me evaluate sin(75) with a sum identity

[shadow estuary]

i tried it and i got sqrt6 + sqrt 2 / 2

but thats wrong

[violet sentinel]

notice that $75 = 30 + 45$. So then do $sin(30 + 45) = sin(30)cos(45) + cos(30)sin(45)$

[warm shale]

MellowDramaLlama

[shadow estuary]

is it fine if i did sin(45+30) instead

[violet sentinel]

yeah same thing

[shadow estuary]

hm

oooooo

[violet sentinel]

I think you might've goofed up the angles somewhere

[shadow estuary]

yeah i think i goofed up on a denominator

[violet sentinel]

you should get $\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)$

[shadow estuary]

um

as the final answer?

[violet sentinel]

no it needs to be simplified

[shadow estuary]

o

ok

[warm shale]

MellowDramaLlama

[shadow estuary]

sqrt 6+ sqrt2 / 4

[violet sentinel]

yep perfect!

[shadow estuary]

wow tysm

[violet sentinel]

you can also write it as $\frac{\sqrt{2}(1 + \sqrt{3})}{4}$ but both are correct

[shadow estuary]

i forgot to multiply the denominators when i did sin*cos

ok

thx

[warm shale]

MellowDramaLlama

[violet sentinel]

yep np 🙂

[obtuse pebble]

@shadow estuary Has your question been resolved?

[obtuse pebble]

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[timid silo]

what am I supposed to do here?

[timid silo]

do I just subsititute for 2?

I dont understand how im supposed to use the given formula to solve this

differentiate

[timid silo]

what

y = f(x)

slope = gradient = tangent

gradient/slope of y at x=2 = value of dy/dx at x = 2

.

im kinda lost

do u k wat the q is asking u to find

yes the slope

what is slope?

do i estimate it here or find it here

y2-y2/x2-x1

[timid silo]

this only works for?

a linear function

yes

is the equaiton in the q a linear function?

but we make space between the points so small its basically a linear function

uh no

or u can do that

im confused

how am i supposed to solve for it without estimating it

hav u learned differentiation?

it seems no is the answer

hence do one thing

let a = 2

let h = a very small value larger than a

and then use the formula ig

kinda ridiculous but ig shd be

[timid silo]

larger than a?

shouldnt it be as close to 0 as possible?

we want the gap to be as close as to 0

not the value

shud be close to a

so it would be like a = 2

h = 2.00001

but h is the size of the gap

thats just making the gap bigger

i want to get another say in this

<@&286206848099549185>

.close

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[shadow estuary]

someone help me do this pls

[brisk matrix]

write this in terms of sine and cosine

[shadow estuary]

ok

[shadow estuary]

now what

[brisk matrix]

do you know your double angle formulas?

[shadow estuary]

yes

[brisk matrix]

then join the fractions and use your double angle formulas

[shadow estuary]

ok i will try

thx

@brisk matrix

i rlly need ur help

with one small part

[brisk matrix]

what is it

[shadow estuary]

what does sin^2x+sin^2x simplify to

o wait

[brisk matrix]

2sin^2x

[shadow estuary]

really?

omg

sorry

that was a dumb question

i got tanx as the solution

pretty sure its right

thx for helping

.close

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[mint locust]

Did i do t(x) correctly?

I used the formula
y = f(c) + f’(c)(x-c)
y = f’(c)(x-c) + f(c)

I changed it to this but the equation still marked wrong

<@&286206848099549185>

[plain stag]

you should evaluate f(5)

@mint locust

[obtuse pebble]

@mint locust Has your question been resolved?

[mint locust]

Alright

[obtuse pebble]

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[obtuse pebble]

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[storm wagon]

im not sure how to start this

how do i start parameterizing?

Vector calclus^^^^

[brisk matrix]

if x = e^y, let y = t and find what x(t) would be

[storm wagon]

lnx = t?

y = t

[brisk matrix]

yeah, but more simply x=e^t

oh doesn't really help with the integral does it

instead let y = ln(t)

[storm wagon]

so set x = t

[brisk matrix]

yep

[storm wagon]

okay so how would I find the bounds, im confused about this arc thing

[brisk matrix]

if x = t, you're going from t = 1 to t=e^8

look at the endpoints they gave you for the arc

[storm wagon]

so i have to use the equation of arc length??

[brisk matrix]

yes

[storm wagon]

ohh so you can use either bounds of x and y to solve for the bounds of t

[brisk matrix]

yes

[storm wagon]

sqrt(t^2 + (ln(t)^2) doesnt look fun

[brisk matrix]

that's not the arc length

[storm wagon]

oh yeah i need teh derivative

\sqrt{1^2+\left(\frac{1}{t}\right)^2}

how doyou show equations..

mathbot

[brisk matrix]

$\sqrt{1+\frac{1}{t^2}}$

[warm shale]

maximo

[brisk matrix]

your integral is then $\int_1^{e^8}t^2\sqrt{1+\frac{1}{t^2}},dt$

[storm wagon]

$\sqrt{1^2+\left(\frac{1}{t}\right)^2}$

[warm shale]

maximo

Whunder

[storm wagon]

okay thanks, that wansn't that bad just need to know how to deal with paramatrics...

that isnt a fun integral to do either

[brisk matrix]

it's actually quite simple

the integrand is the same as $t\sqrt{1+t^2}$

[warm shale]

maximo

[brisk matrix]

it's a simple u sub

[storm wagon]

u = t^2?

[brisk matrix]

u = 1+t^2

[storm wagon]

you altered the square root?

[brisk matrix]

wdym

[storm wagon]

how did you end up with that integrand

[brisk matrix]

$t^2\sqrt{1+\frac{1}{t^2}}$?

[warm shale]

maximo

[storm wagon]

wouldn't this require ingration by parts or something

[brisk matrix]

no

[storm wagon]

cuz subbing 1/t^2 would make 1/t^3

[brisk matrix]

$t^2\sqrt{1+\frac{1}{t^2}}=t\sqrt{t^2+t^2\cdot\frac{1}{t^2}}$

[warm shale]

maximo

[brisk matrix]

which is just t * sqrt(1+t^2)

let u = 1 + t^2

[storm wagon]

oh you brought a "t" back inside

well i just symbolabed that intregral and its wrong apparently

[brisk matrix]

what did you write as the answer

[storm wagon]

$\frac{\left(e^{16}+1\right)^{\frac{3}{2}}-2\sqrt{2}}{3}$

[warm shale]

Whunder

[storm wagon]

okay i noticed that the original equation has a dx, and no DS

[brisk matrix]

ah true

then dx/dt = 1 -> dx = dt

[storm wagon]

its a partial derivative?

[brisk matrix]

which one

[storm wagon]

so you just integrate x?

[brisk matrix]

no

so the dx makes it much easier but let's finish up with the parametrization

we now have x = t and y = ln(t)

then the integrand becomes te^ln(t) = t^2

and dx/dt = d/dt (t) = 1

so then dx = dt

so we just get the integral from 1 to e^8 of t^2 dt

did you follow that?

[storm wagon]

so dx goes away and becomes 1

[brisk matrix]

no

$\frac{dx}{dt}=\frac{d}{dt}(t) = 1\implies dx=dt$

[warm shale]

maximo

[brisk matrix]

so your integral becomes $\int_1 ^{e^8}t^2, dt$

[warm shale]

maximo

[storm wagon]

yeah i sort of followed along. basically solves for dt which is equal to dx

[brisk matrix]

yes

[storm wagon]

and allows you to sub

[brisk matrix]

it just allows us to take the integral wrt to t instead of finding t as a function of x

(which is actually just t = x)

so

if we just went with the x's from the start, we'd get the same thing

notice that x = e^y -> y = ln(x)

so the integrand is truly just x * e^ln(x) = x^2

the whole parametrization spiel was because i though it was ds instead of dx

[storm wagon]

okay i hope i willl be able to apply these to the rest of the questions tmr, this session was insightful.

didn't need to parameterize but it worked anyways

.close

[obtuse pebble]

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[storm wagon]

bed time haha

[obtuse pebble]

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[smoky slate]

30% of 80% plus 25% of 20%

[lusty cedar]

You can just assume total no. Of PPL in the department is 100

[kindred swift]

k

[lusty cedar]

Or ig this

[smoky slate]

You might get fractional people lol

[kindred swift]

ok thanks im done with that one

[kindred swift]

LOL

[kindred swift]

this one now pls

[smoky slate]

Set variables and equations

[kindred swift]

wdym

nah this isnt my homework

this is my elder brothers

i dont understand any of this

im tryna do it for him

@smoky slate

[smoky slate]

I won't do your brother's homework

[kindred swift]

alr

ty anyway

.close

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[royal basin]

@kindred swift next time get your brother to post here himself.

[kindred swift]

hes doing his SHL exam and im tryna help him

[ruby path]

Swedish Hockey League

[obtuse pebble]

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[pastel edge]

Hi, does anyone know about saddle points and stable games?

[pastel edge]

Like here how do I know if this game is stable or not

[obtuse pebble]

@pastel edge Has your question been resolved?

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@pastel edge Has your question been resolved?

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[small granite]

hey, i know this channel is supposed to answer math questions but may i post a c++ question

[drifting wraith]

some c++ questions are math but not all

[small granite]

it’s about consecutive numbers

[drifting wraith]

you can't post a non math c++ question

[small granite]

but consecutive numbers counts as math right

so basically i had to write a code that checks if 2 numbers, x and y, are consecutive

my idea was
int x,y;
cin >> x >> y;
if ((x>y && x=y+1))||(y>x && y=x+1)) {
cout << “yes”;
}
else {
cout << “no”;
}

[drifting wraith]

yeah that shuold work, you have to use == for equality

[timid silo]

Can't you just do (y == x + 1 || x == y + 1)?

I mean, why are you testing whether x > y or y > x?

[drifting wraith]

numbers can overflow into negative

[timid silo]

Oh, true.

[drifting wraith]

the test deals with it

[small granite]

#include<iostream>
using namespace std;
int main()
{
int x,y;
cin >> x >> y;
if ((x>y && x==y+1)||(y>x && y==x+1)) {
cout << "yes";
}

return 0;
}

you were right i got some errors

i needed to use == and delete one bracket from the if

thanks for the help guys

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[obtuse igloo]

How do I solve
$\lim{x\to{1}} x^\frac{1}{x-1}$
Without using l'hospital rule?

[warm shale]

Nex ex Umbra

[sage geode]

$L = \lim_{x\to{1}}x^\frac{1}{x-1} \$
$\ln{L} = \lim_{x\to{1}}\frac{\ln{x}}{x - 1}$

[warm shale]

A Lonely Bean

[sage geode]

Does the fraction on the right remind of you of anything?

[obtuse igloo]

Ah and then factor the x?

[sage geode]

Not really

[obtuse igloo]

Idk, I just see 0/0 and then I jump to using l'hop 😅

[sage geode]

$\frac{\ln{x}}{x - 1} = \frac{\ln{x} - 0}{x - 1} = \frac{\ln{x} - \ln{1}}{x - 1}$

[warm shale]

A Lonely Bean

[sage geode]

Does it look familiar now?

[obtuse igloo]

Not really, sorry Idk what I'm supposed to see

[sage geode]

That's the derivative of lnx at 1

There's actually another way of doing this but I can't seem to remember it so what comes to my mind is this lol

[obtuse igloo]

Ah sorry, I'm also trying to solve it without derivatives

[sage geode]

Ah

Okay one moment let me remember

But you can use the limit definition of e, right?

[obtuse igloo]

Yeah

There's supposed to be a method with substitution that gets the same answer

[sage geode]

Let u = x - 1

Then you have (1 + u)^1/u as u approaches 0

By definition this is equal to e

Or you could substitute u = 1/(x - 1) to get (1 + 1/u)^u as u approaches infinity

The same thing

[obtuse igloo]

Hmm Idk if that would help them, but thank you :)

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[obtuse pebble]

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[olive thorn]

how do i find the integer values of x in this equation:
6−x<4−2x≤28+2x

[obtuse pebble]

@olive thorn Has your question been resolved?

[obtuse pebble]

@olive thorn Has your question been resolved?

[obtuse pebble]

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[thin raven]

What impact should unit conversion have on the relationship between standard deviation and variance?

For example, when the units are kilograms, the sd is 3.156 kg , and the variance is 9.96 kg which is the sd^2.

However, when we convert this to grams, the sd is 3,156 g (the equivalent of 3.156 kg) but the variance is 9,963,829 g (the equivalent of 9,963.829 kg).

[royal basin]

the variance is 9.96 kg^2 actually.

[thin raven]

ok

but would it be in grams?

what would it be in grams*

[royal basin]

9,963,829 g^2

and there's no contradiction there

1 kg^2 = 1000^2 g^2

though units of squared mass have somewhat dubious physical significance, i admit

[thin raven]

so for instance,

if the standard deviation is 2 hour
if u were to convert this to variance in minutes --> would it be 240 minutes or 14400 minutes

[royal basin]

neither

it would be 4 hr^2, or 14400 min^2

[thin raven]

ah

[royal basin]

the units for variance are the SQUARE of those for the original variable and its mean and sd

[thin raven]

thanks

.close

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[supple epoch]

Why is this true?

If G is isomorphic to H, then G is abelian if and only if H is abelian.

[vapid hamlet]

Uhh, because G is isomorphic to H, they are practically indistinguishable as groups

[warm canopy]

Yeah just prove it yourself

[vapid hamlet]

Try to show it yourself, it's not hard

[supple epoch]

damn alr

[warm canopy]

Send a sum in G to a sum in H by the isomorphism

Show the sum in H commutes

[obtuse pebble]

@supple epoch Has your question been resolved?

[elfin burrow]

i think it uses the fact that the inverse of an isomorphism is an isomorphism

[obtuse pebble]

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[toxic hollow]

What does F^m mean

A Field with m Elements?

[vapid hamlet]

The Cartesian product F x ... x F m times

An element of it is a tuple (a_1, ..., a_m) with a_i in F for all i

[toxic hollow]

Oh I see thanks

I’m quite confused here

Why are they choosing two linear independent subsets of S denoted as T and T’? Why can’t they just compare S and T’

[elfin burrow]

they are choosing a linearly independent set T such that all other linearly independent sets T' have size less than or equal to that of T

[toxic hollow]

Could T be S here?

[elfin burrow]

no, because S need not be linearly independent

S is just a finite spanning set

they are showing that the linearly independent subset of S that has maximum size is a linearly independent spanning set (hence a basis)

[obtuse pebble]

@toxic hollow Has your question been resolved?

[toxic hollow]

So the proposition is claiming there is a maximum L.I subset of S which means that in any finite spanning set there’s a L.I spanning set?

Essentially what I mean is that any spanning set Z must have a maximum finite L.I subset which actually causes the set Z to be spanning?

[obtuse pebble]

@toxic hollow Has your question been resolved?

[elfin burrow]

the proposition is claiming that every finite spanning set contains a linearly independent spanning set, and that specific linearly independent spanning set happens to be the linearly independent subset of maximum size

they prove that every finite spanning set contains a linearly independent spanning set by presenting one, namely the linearly independent set with max cardinality

[obtuse pebble]

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[silver plover]

what can i do next?

[knotty crow]

u can integrate it nicely without a sub

[elfin burrow]

you're supposed to substitute x = tan(theta), not x = sec(theta)

[silver plover]

ah crap

mb

[silver plover]

the question asks to use x = tantehta

[knotty crow]

ah ok

[silver plover]

lemme correct my mistakes

and ill repost

[elfin burrow]

before your last step, use the identity $1 + \tan^2 \theta = \sec^2 \theta$

[warm shale]

tushar

[elfin burrow]

follows from the identity for sin and cos

[silver plover]

so i get tan^2 / sec^2

[ruby path]

You can do this with Feynman's Technique

If you know what that is

[silver plover]

never done that before

[ruby path]

Differentiating under the integral by introducing a parameter

[ruby path]

Alright then

[silver plover]

the questions asks anyway to use sub x = tantheta

[ruby path]

Ah okay

Continue then

[knotty crow]

which is sin^2

[ruby path]

woah you actually put down the limit every time

respect +

[knotty crow]

or rewrite it in terms of cos(2 theta) if it's easier to integrate

[silver plover]

wait so tan^2 / sec ^2 = tan^2cos^2?

[knotty crow]

yes

[ruby path]

yeah the reciprocal of cos is sec

[silver plover]

got this so far

but i assume i gotta change the limits right?

to match the sub

i forgot to

so can i just do it now?

before i integrate

[elfin burrow]

yes

[silver plover]

this correct?

[elfin burrow]

your first line here is wrong

[silver plover]

which first line

[elfin burrow]

the first line in your picture

you have the identity wrong

also you must take the limit after evaluating the integral, not before

[silver plover]

is the limits just arctan b and 0?

[elfin burrow]

the bounds, yes

the integrand is wrong though

the identity is: $\cos 2\theta = 1 - 2\sin^2 \theta$

[warm shale]

tushar

[silver plover]

^^

[elfin burrow]

i got the same thing

,w integral 0 to infinity of x^2 / (1+x^2)^2

[warm shale]

Results provided by WolframAlpha

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Requested by tushar#0942

[elfin burrow]

nice

[silver plover]

noice

hows the working out?

[elfin burrow]

yeah same thing

checked it

[silver plover]

cool

cheers for help lad

[ruby path]

It's so beautiful

Pi shows up in the area under polynomial functions

[silver plover]

maths is beautiful

i find integration so cool

especially subsitiution

[ruby path]

Yup

All my friends hate integrals

[silver plover]

like u can substitute something in for something else and u get to the same answer

[ruby path]

Like bruh why

[silver plover]

when i started it i hated out

[ruby path]

Actually same

[silver plover]

icl its taken me months to get my head round it and get better at it

[ruby path]

I was very close to hating calculus in general

But someone showed me the derivative of x^x^x^x^x^x^x... and it really interested me

One video by one YouTuber potentially picked my major in college

[silver plover]

dang

[ruby path]

BPRP btw

[silver plover]

ofc

the lgend

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[obtuse pebble]

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[fierce cedar]

Help please

[fierce cedar]

We have this equation where a, b, c and x are all positive real numbers. We search for all possible x values that satifies this equations.

[tall tusk]

only x=0 works

trivial

[obtuse pebble]

@fierce cedar Has your question been resolved?

[obtuse pebble]

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[uneven night]

How do i proof this

[uneven night]

So basically I should prove it by integral estimation?

[spiral knot]

start from :
1<=k<=40

then try to reach the 1/(k)^2

[uneven night]

How do you mean

[spiral knot]

we know that $1<=k<=40$

[uneven night]

Yes

[warm shale]

Mehdi_Moulati

[uneven night]

So I should take integral of 1 over sqrt x with the bonds 1 to 40?

[spiral knot]

so 1 <= sqrt(k)<= sqrt(40)

[uneven night]

I haven’t solved this type of problem before so I am basically no clue, but I think I can start by integrating 1/sqrt (k) with the range 1 to 40. Am I thinking right?

[gloomy valve]

isnt k up to 400?

[spiral knot]

you need to know 1/sqrt(k) between what's

[gloomy valve]

you can show the upper bound by integration

but not the lower bound

[spiral knot]

so you should start from k

[tall tusk]

surely that was a mis ping @tardy epoch srry

[spiral knot]

just follow me

we know that k between 1 and 40 right ?

[uneven night]

I will follow u

Yes

[spiral knot]

so $1<=\sqrt{k}<= \sqrt{40}$

[uneven night]

But do u mean 1/sqrt k or just k

[warm shale]

Mehdi_Moulati

[spiral knot]

square root

[uneven night]

Yea I agree

[spiral knot]

then $\frac{1}{\sqrt{40}}<=\frac{1}{\sqrt{k}}<= 1$

[fierce cedar]

.reopen

[uneven night]

What do u mean

I can’t see anything

[spiral knot]

TeXit bot will translate it

[uneven night]

Oh okay

[spiral knot]

but it's to slow

[tardy epoch]

tex bot doesn't like space between 1 and $

[spiral knot]

thanks

[tall tusk]

am i forgiven

😭

[tardy epoch]

i was never mad at you?

[tall tusk]

oh ok

[uneven night]

So is that the answer

[warm shale]

Mehdi_Moulati

[spiral knot]

from here will do the sum of each of them

sum of 1 from 1 to 40 is 1* 40

[uneven night]

Do u mean some integral

[spiral knot]

$\sum_1^{40} \frac{1}{\sqrt{40}}<=\sum_{k=1}^{40} \frac{1}{\sqrt{k}}<= \sum_1^{40} 1$

[uneven night]

Oh okay take the sum of each thing

[spiral knot]

yeah

[uneven night]

And now just solve it right

Then I am done

[spiral knot]

right

[uneven night]

Thank you so much @spiral knot

[warm shale]

Mehdi_Moulati

[gloomy valve]

am i blind or didnt youssef want to bound
$$\sum_{k=1}^{400} \frac{1}{\sqrt{k}}$$

[uneven night]

Why did you multiply everything with 40?

[warm shale]

nvx

[uneven night]

Yea it was too 400

[elfin burrow]

isn't the upper limit for the sum 400?

lol

[uneven night]

Well it’s

[spiral knot]

it's must be 40

not 400

[gloomy valve]

no its 400

[gloomy valve]

up to 40 the sum is 11.something

[uneven night]

The range is between 35 and 40

No it’s 400 in sigma notation

This is from the booj

And the question is: (show with estimating integral that)

[spiral knot]

ooh

[gloomy valve]

<= 40 is easy, just calculate the integral from 0 to 400 of 1/sqrt(x)

[uneven night]

But I think u thought right

[gloomy valve]

which is always larger than the sum

[uneven night]

Wait how do u mean

[gloomy valve]

$\sum_{k=1}^{400} \frac{1}{\sqrt{k}} \leq \int_0^{400} \frac{1}{\sqrt{x}} \dd{x}$

[warm shale]

nvx

[uneven night]

Oh just like that?

[gloomy valve]

yes

[uneven night]

But what happens with 35

And 40

[gloomy valve]

still thinking about the lower bound

the integral is exactly 40 btw

[uneven night]

The lower bound should be changed to B

Where B approcheaa 0+

[gloomy valve]

yes

[uneven night]

Because this is a generalized integral right

Improper integral**

[gloomy valve]

right

[uneven night]

Wait let me test if this work

So

I think this is right @gloomy valve

[gloomy valve]

yes

[uneven night]

Thank u so much @gloomy valve

[elfin burrow]

that does not give the value of the series

just an upper bound

[gloomy valve]

we want lower & upper bounds only

[uneven night]

So we want also 35?

[elfin burrow]

i think you can do better than 35

a lower bound is:

,w integral 0 to 400 1/sqrt(x+1)

[warm shale]

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Requested by tushar#0942

[elfin burrow]

similar to your upper bound idea

[uneven night]

So I add 1 inside the sqrt

But why did u just add 1 inside the square root why not 2

Is it because k is equal to 1 @elfin burrow ?

[elfin burrow]

the blue boxes are the sum, so it's bounded between the integral of the function in red and the integral of the function in black

[uneven night]

Oh okay that’s makes sense

But that x+1 is hard to integrate right?

[elfin burrow]

not at all

you just calculated an identical integral

u = x + 1

[uneven night]

Oh ofc

So I just do that and then I’m done

[elfin burrow]

yeah

2(sqrt(401) - 1) = 2sqrt(401) - 2 > 2sqrt(400) - 2 = 38 > 35

neat visual argument, not sure how it can be argued formally though

[uneven night]

So

It’s 4

Okay I think I got, big thanks to @elfin burrow

[elfin burrow]

others as well

nice question

[uneven night]

Thank you, ofc thank you everyone!!

.close

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[uneven night]

Guys sorry to interrupt u, but I struggled a lot with this question also, so I thought I would take the integral of 1/sqrt(x)(x+1) with the bonds 0 to infinity. And then change 0 to B where B approchaes 0+?

[tardy epoch]

not sure why taking the integral would help here?

unless you know how to bound the difference between the series and the integral

,w integral from 1 to inf 1/(sqrt(x)(x+1))

[warm shale]

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Requested by riemann#2529

[tardy epoch]

oh that's convenient. maybe that is all you have to do then

[uneven night]

Why didn’t u take the bonds zero to infinity?

[tardy epoch]

k=1

[uneven night]

Is it because k =1

Oh but in the previous problem we took 0

But k was equal 1

[tardy epoch]

depends on the problem usually

and depends what you're trying to show

[uneven night]

Hmmm

So wouldn’t that work if we took 1 to 400 in this problem

It would right

Then u just changing 1 to a B where b is approaching 0+

[tardy epoch]

sure? you can redo the proof if you want

[uneven night]

I mean if we take 1 to 400 we would get improper integral

So we would have to do the change

And then we would definitely get same answer

Anyway thank you @tardy epoch!

.close

[obtuse pebble]

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[elfin burrow]

,w integral from 0 to inf 1/(sqrt(x)(x+1))

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by tushar#0942

[uneven night]

.reopen

[obtuse pebble]

✅

[uneven night]

Why u take zero to infinity

[elfin burrow]

oops

[uneven night]

Wait what

[elfin burrow]

oh nevermind

wrong problem

[uneven night]

Hahahaha

No problem

But I have a question

Why did u take integral of 0 to 400

Why don’t u take 1 to 400

[elfin burrow]

because the first blue box extends to 0

so to obtain an overestimate, i integrated the function in black from 0

but you can just use riemann's method here

first term is 1/2

[uneven night]

How did u know

[elfin burrow]

all the terms from n = 2 to infinity you can bound above by this integral

so an upper bound is 1/2 + pi/2 = (pi + 1) / 2

[elfin burrow]

just substitute k = 1 into the sum

[uneven night]

Oh sry I looked at the wrong problem

Yes so if I plug in 1

I get 1/2

That’s right

I see

[elfin burrow]

finally

[warm shale]

tushar

[uneven night]

I didn’t understand why u added 1 inside the sqrt @elfin burrow

Can you please explain it to me

[elfin burrow]

to obtain an underestimate for the blue boxes

i shifted the function to the left

so that the area under the curve is always less than the area of the blue boxes

that way the integral will be an underestimate

[uneven night]

And because k =1?

[elfin burrow]

the boxes have width = 1 and height = f(x) = 1/sqrt(x) and area = 1/sqrt(x), so they are just the terms of your sum

[uneven night]

Oh okay

I think I get it

Thanks @elfin burrow .close

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[obtuse pebble]

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[stark galleon]

Hello, in my lecture of verifying trigonometric identities the professor showed an example of a step that I didn't understand.

cosx-cosx(sin^2x)
cosx(1-sin^2x)

I don't understand how this factoring works...

A-AB = A(1-B)

Any help?

[rigid yarrow]

Which part are you confused on?

[stark galleon]

Does it have a name?

I see the leap now....

If I reverse the identity

A(1-B) = A-AB

[rigid yarrow]

I was about to say try expanding it out to see if that helps?

[stable iron]

$\text{cos}x - \text{cos}x(\text{sin}^2x)=\text{cos}x\cdot 1 - \text{cos}x \cdot (\text{sin}^2x) = \text{cos}x(1-\text{sin}^2x)$

[warm shale]

ta

[stark galleon]

I see, I don't know why I couldn't immediately grasp it but now it makes sense, thank you

🙂

.close

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[worn yacht]

What trigonometric identities do I need for trigonometric substitution in integrals and how do I do it?

[dark mango]

for trig sub? just sin^2 + cos^2 = 1 and tan^1 + 1 = sec^2

[worn yacht]

So how do I do it?

@dark mango

[dark mango]

do u have an example question

[worn yacht]

$\frac{\pi}{\sqrt{8-2x^2}}$

[warm shale]

kangaroo rat

[dark mango]

ok notice we have a^2 - x^2

[worn yacht]

Yes

[dark mango]

what trig identity does that remind you of

think about rearranging

[worn yacht]

Sin x+cos x=1

[dark mango]

yes so rearrange that and we get

1 - cos(theta) = sin(theta)

so lets make x = asin(theta)

[worn yacht]

Why???

[dark mango]

ok so

fundamentally you see that square root?

[worn yacht]

Ye

[dark mango]

its gonna be really hard to intergrate that expression because of the square root

[worn yacht]

Yes

[dark mango]

so the idea of trig sub is to use trig identities to remove that square root

lets think more broadly.

so assume we have

$\sqrt{8-2x^2}$

[worn yacht]

Other slash

[dark mango]

ty

[worn yacht]

Np

[dark mango]

wait wrong expression lol

[worn yacht]

Lol

[dark mango]

$\sqrt{a^2 - x^2}$

[warm shale]

cantprogram

[dark mango]

ok assume we let $x = asin(\theta)$

[worn yacht]

\theya

\Theta

[dark mango]

god damn ur a latex pro

[worn yacht]

Lol

Wrong slash

[warm shale]

cantprogram

[worn yacht]

There

[dark mango]

ok lemme rewrite everything

[worn yacht]

K

[dark mango]

if we are integrating $\sqrt{a^2 - x^2}$

[warm shale]

cantprogram

[dark mango]

we need to get rid of the square root so to do this lets assume we let $x = asin(\theta)$

[warm shale]

cantprogram

[dark mango]

if we plug in $x = asin(\theta)$ we get

[warm shale]

cantprogram

[dark mango]

$\sqrt{a^2 - (asin(\theta))^2}$

[warm shale]

cantprogram

[dark mango]

or

$\sqrt{a^2 - a^2sin^2(\theta)}$

[warm shale]

cantprogram

[dark mango]

and we can factor out $a^2$

[warm shale]

cantprogram

[dark mango]

$\sqrt{a^2(1-sin^2(\theta))}$

[warm shale]

cantprogram

[dark mango]

and notice the trig identity

sooo

we can rewrite this as

$\sqrt{a^2cos^2(\theta)}$

[warm shale]

cantprogram

[dark mango]

which is just simply

$acos(\theta)$

[warm shale]

cantprogram

[dark mango]

and see we removed the square root

[worn yacht]

Why $x=asin(\theta)$

[warm shale]

kangaroo rat

[dark mango]

and made the integration wayyyy easier

[worn yacht]

Ye

[dark mango]

because when we plug in asin(theta) we can factor out the a^2 and be left with a trig identity

[worn yacht]

But what trig identity did u use

[dark mango]

we are manipulating our substitution to remove the squareroot

[dark mango]

depends on the intergral

if its a^2 - x^2 well you can use sin^2 + cos^2 = 1

if its x^2 - a^2 we can use tan^2 + 1 = sec^2

etc

[dark mango]

actually this should be the absolute value of acos(theta) but we can remove absolute value bec of the domain of cosine

just a note

[worn yacht]

So how does sin^2+cos^2=1 make u want to make x=asin(theta )

[dark mango]

because

see we have a^2 - x^2

so thats super similar to

1 - sin^2 = cos^2

[worn yacht]

Ohhhhhhhhhh

[dark mango]

(works either way)

and if we make x = sin then we will just be left with a cosine

[worn yacht]

I’m getting it

[dark mango]

yeah its honestly just practice

[worn yacht]

Wdym?

[dark mango]

but fundamentally its a technique typically used to get rid of the square root

[worn yacht]

Anything else?

[dark mango]

if we make x= sin then after the manipulation we will just have cos^2 and not a^2 - cos^2

[worn yacht]

Yes

[dark mango]

you need to replace dx with d theta

[worn yacht]

Ok

Thanks

[dark mango]

no worries

[worn yacht]

Can I dm you later if I want help with a problem, or ping you in server?

[dark mango]

Yeah I might not respond for a while though

depends on time of day

[worn yacht]

Thanks

That’s fine

[dark mango]

lots of ppl will help u with this if u just make a thread

good luck !!

[worn yacht]

Bye

.close

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[modern dragon]

im looking for help to determine the probability of getting n unique digits after selecting m digits

[modern dragon]

likle one example would be

probability of getting

8 unique digits

out of 10 total digits selected (0-9)

[wild bough]

well getting 8 unique results would just be the combination (10/10 x 9/10 x ... 2/10) wouldn't it? since the first "draw" would restrict the choices of the second number to only have nine possible positive choices out of 10. next number would only have 8 out of 10 and so on

i might be missing something in your question tho.

[modern dragon]

oh ok maybe digits isnt a good choice

uhh

lets say 5 digits

and 3 unique

wait

[modern dragon]

for this you are saying we do 10!/2!/10^8 * 8/10 * 8/10?

i understand that part

but

i think that is incorrect

[wild bough]

yes? well to be more precise $\frac{10!}{10^8 \cdot 2(!)}$

[warm shale]

Winus

[modern dragon]

wait so we are not mutlipling by

8/10?

for the final 2 digits to be overlapping

[wild bough]

no no need that is fulfilled in the fact that we can choose 10 diffrent numbers for the first number than 9 and so on. we basically cheack all the ways we can choose 8 unique numbers out of 10

[modern dragon]

8 unique numbers with 2 repeats

so like

[wild bough]

then we divide it but the total amount of ways to choose 8 numbers out of 10 which is 10^8 cuz we have to choose 8 times where we have 10 choices

[modern dragon]

yes

[wild bough]

ohh wait i get what you are asking now

[modern dragon]

so like the answer should be around 0.13

[wild bough]

i thought it was the simpler version where you ask what is the probability that you choose only unique choices not where you allow a certain amount of repetes

[modern dragon]

mhm im a bit confused i tried

quite alot of things but im doing something wrong and im not sure where

i know the combiniations is

(10!/3! + 10!/2!/2!)

right?

but then i dont know the probability

OH wait

mmm nvm i still dont have it

becuase theres one scenario where your repeated digit is the same

and then another scenario where they are different

^ this is for the 2 repeat scenario

[wild bough]

i wish i had the answer ontop of my head but statistics are ridiculous confusing

[modern dragon]

yaaaaaaaa... i used to know this when i was in my probability class but

now its slipping away

[wild bough]

found this, might be something tho they choose from 0 to k rather than only 0 to 9

https://math.stackexchange.com/questions/1087950/calculating-probability-of-getting-m-unique-numbers-when-choosing-n-times-fr

the formula in there seems pretty promising

[modern dragon]

yes the formula does seem to work

however i do not understand TT

seems to be inclusion exclusion of some sort

[wild bough]

yeah if i understand it correct the sum basically asks what's the probability to get n unique in n choices. then it asks for n unique in n+1 choices. continuing like this until you reach the desired m ways. alternating adding and removing to avoid counting any ways multiple times. but then again half of the statistics field is just pure magic anyways

[obtuse pebble]

@modern dragon Has your question been resolved?

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[coral marsh]

how many relations are reflexive?

[drifting wraith]

they gave the solution then asked the question

funny, new

[coral marsh]

?

[twin sapphire]

not exactly

[coral marsh]

for reflexive its when the matrix diagonals are all 1s right?

[twin sapphire]

yeah

[drifting wraith]

not the diagonals, just the main one

[twin sapphire]

so how many possibilities for such matrices?

[coral marsh]

when i = j?

[twin sapphire]

for each non diagonal entry you have two possibilities

[coral marsh]

n * 2 choose 1

[twin sapphire]

you have n²-n of those entries

so 2 possibilities ffor the first

2 for the second

and so on

n²-n times

i already said too much probably

[coral marsh]

for each index you have 2 choose 1 possibilities

[twin sapphire]

yeah

so 2 possibilities

2 choose 1=2

[coral marsh]

why do we consider non main diagonal entries

when looking for reflexive relations

[twin sapphire]

because the main diagonal ones

are set

we dont count the possibilities

theres only 1

all 1s on the main diagonal

[drifting wraith]

the rest are free to switch about

[coral marsh]

isn't a matrix only reflexive when the main diagonal is all 1s

[drifting wraith]

of course

[coral marsh]

why would status of non main diagonals matter

[drifting wraith]

so we're not allowed to touch those, because it will stop working if we do

it doesn't

[coral marsh]

i thought we were looking ways main diagonal can be 1

think im misunderstanding question

[drifting wraith]

we're not exactly looking for a ratio of two numbers

yes, we're looking for what you said

"a way" is the entire matrix

every one of n×n numbers is known, that's 1 way

[twin sapphire]

consider your whole matrix, we know there are 1s on the diagonal, now you have many choices for all the other coefficients (those not on the main diagonal) we are counting the possibilities for them

[coral marsh]

n^2 - n * (2 choose 1)

[twin sapphire]

why?

[coral marsh]

everything besides the diagonal how many ways can they be different

[drifting wraith]

that's not how things work

in life

[coral marsh]

if the diagonal can only be reflexive when they are all 1, then that leaves n^2 - n entries to choose 0 or 1?

[twin sapphire]

yeah

right

lets simplify

[drifting wraith]

if you can go by train or by car to Paris or to Cape Town tomorrow or today

you have 3 choices, 2 choose 1

total is not 6 choices, but 8

[twin sapphire]

if you have a list of 3 numbers that can be either 0 or 1

how many cases

[coral marsh]

2^3?

[twin sapphire]

yeah

so now if you have a list

of n²-n

numbers

[coral marsh]

is it 2^(n^2 - n)

[twin sapphire]

yeah

[coral marsh]

thanks both of you

i appreciate help

.close

[obtuse pebble]

Channel closed

Closed by @coral marsh

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[obtuse pebble]

Please don't occupy multiple help channels.

[placid sigil]

HELP\

HELP

HELP

PLS

somenody

[untold flax]

why so urgent

[placid sigil]

cause

this assignment

due

IN GOD DANG 10 MIN

AAAH

[untold flax]

what's ecb

<ECB

[loud dirge]

Ah okay, if ACB is 90 and DCB is 126, DCA = 126-90 = 36. Since DCE is 90, ACE = 90-36

Did I make a mistake bo

[untold flax]

bro you're not meant to just give the answer

no

[small granite]

pretty direct lmao

[placid sigil]

thanks

[untold flax]

lmao

[loud dirge]

Wait what did I do

[small granite]

yeah its correct

[placid sigil]

wait what

yay

ty

[small granite]

i think

[placid sigil]

how u solve it

[loud dirge]

90-36?

[untold flax]

that's why you don't just give the answer

give them a hint

and then let them come to it

[loud dirge]

Oh okay

Sorry

[untold flax]

all good

[placid sigil]

thanks anyways

.close

[obtuse pebble]

Channel closed

Closed by @placid sigil

Use .reopen if this was a mistake.

[small granite]

ecb = dcb - dce and ace = dce - ecb

[obtuse pebble]

Available help channel!

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[simple grail]

I really need help i dont understand how to do this at all

[stark galleon]

I would say try to understand the definitions of the terms at least. For example, is JM perpendicular to KL? We don't know that because we don't have the lengths of JK and JL, we only know that M evenly divides KL, so JM is (maybe) not a perpendicular bisector of KL...

Here is an explanation for altitude and medians: https://byjus.com/maths/altitude-median-triangle/

[simple grail]

i understand the definitions

[stark galleon]

then you must understand how to match the definitions the the triangle given, for example an altitude is a line that starts from one angle must intersect the opposite length of the triangle at 90 degrees...

I would also say that unless you are explicitly told, don't assume lengths or angles by observation. For example, the 2nd image looks like a right triangle, but it doesn't have the red square as in the 3rd, so I would not make that assumption.

[simple grail]

.close

[obtuse pebble]

Channel closed

Closed by @simple grail

Use .reopen if this was a mistake.

[obtuse pebble]

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Send your question here to claim the channel.

Remember:
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• 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.

[twin ferry]

Lim x -> a x-a /( lnx - lna)

[knotty crow]

l'Hospital allowed?

[twin ferry]

yes

the question is l hospital question

but I forgot how to do it on the test

so gonna ask and see if I did it right

but I have a feeling I am wrong because the answer I got was 0

[knotty crow]

take derivative of both the numerator and the denominator