help-23

[vagrant ice]

It's based on this

[ember finch]

The first way that i know of to solve this is by factorizing

but are there any other ways?

[vagrant ice]

Yeah so find the roots of r^2 + 3r + 2 are

[ember finch]

right hand side?

[vagrant ice]

Yep

[ember finch]

Ok that should be -2 and -1

but what do i do with those numbers

[vagrant ice]

e^(rx)

[ember finch]

is it always e^(rx)? When is it something else?

[vagrant ice]

The reasoning is if you sub $y = e^{rx}$, you get $r^2 e^{rx} + 3r e^{rx} + 2e^{rx} = 0$

[flat frigate]

south

[vagrant ice]

And you can divide through by e^(rx)

[vagrant ice]

When you don't have a linear DE

Like if you have xy'' + 3y' + 2y or something

[ember finch]

iirc you just have to divide that by x

[vagrant ice]

Ah it's unsolvable

[ember finch]

oh

why?

[vagrant ice]

The Bernoulli DE is actually x^2 y'' + 3xy' + 2y = 0

Where the solution is x^r

[vagrant ice]

There's no why to it

It's just like why most indefinite integrals don't have a closed form (can't be written in terms of familiar functions)

[ember finch]

and how can you recognize when its unsolvable? does the exponent of x always have to descent from left to right?

[vagrant ice]

Basically when you've thrown every tool humans know to solve DEs and it doesn't work

I'm not aware of any unsolvability theorems for DEs

[ember finch]

I see

Thanks!

[vagrant ice]

https://math.stanford.edu/~conrad/papers/elemint.pdf

There's a paper on unsolvability theorems for integration

[vagrant ice]

You definitely need a graduate maths background to work on this

Peano's uniqueness theorem is something different; whether a solution can exist in a given interval

Not whether it can be written down in terms of functions we know

[ember finch]

half the text is just symbols ;-;

[vagrant ice]

Exactly

[ember finch]

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[subtle wedge]

I've never done this kind of work before and dont understand where to start.

[safe radish]

@subtle wedge Has your question been resolved?

[subtle wedge]

<@&286206848099549185>

[subtle wedge]

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[slate elbow]

so

[slate elbow]

how do u find last digit of a number 2023^2024

[lean otter]

oh

i used to know those

let me thinl

k

[upbeat ridge]

Regular SAT question lol

[lean otter]

modular arithmetic

is the way

[lean otter]

clearly

[slate elbow]

:D so

how do udo it

[lean otter]

find remainder of 2023^2024 divided by 10

so

wait no

the last digit of 2023^n depends on the last digit (3)

and exponent n

for n>= 2 the pattern of last digits of powers of 3 repeat every 4 powers

there u go

now find it

then calculate

this should be a enough nudge

[slate elbow]

yeah

so how do you reason it mathematically

[lean otter]

!status

[safe radish]

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

[slate elbow]

?

I am asking

how do you mathematically prove that it is correct

[lean otter]

i gave you the biggest nudge

[main mural]

$2023 \equiv 3 \mod 10$\
$3^1 \equiv 3 \mod 10$\
$3^2 \equiv 9 \mod 10$\
$3^3 \equiv 7 \mod 10$\
$3^4 \equiv 1 \mod 10$\
$3^5 \equiv 3 \mod 10$\

[flat frigate]

artemetra

[main mural]

just start with this

then you find that $2024 \equiv 0 \mod 4$

[flat frigate]

artemetra

[slate elbow]

yea i see

then i say so its perfect mutliplication of 4, hence the number ends with 1

[main mural]

yep

or in other words it's equivalent to 3^0 = 1

if you need more math

[slate elbow]

@main muralwhat abou tthis number 9^(7^{6} + 5^{4} + 3^{2})

[main mural]

$9^{7^{6} + 5^{4} + 3^{2}}$

[flat frigate]

artemetra

[main mural]

also last digit?

[slate elbow]

yh

[main mural]

same thing

you start out with 9

find the cycle length

and then consider $7^{6} + 5^{4} + 3^{2} \mod \textrm{cycle length}$

[flat frigate]

artemetra

[main mural]

it's actually a very nice length

this problem is even easier than the previous one

[slate elbow]

i solved it

its 1

9^k = 9 mod 10 for k odd

9^k = 1 mod 10 for k even

5^k is always odd

3^2 is odd

hence their sum is even

7^6 ends by 9

hence 7^6 + 3^2 + 5^4 is odd

@main mural correct?

so 9^k - k odd -> ends by 1

[main mural]

uhh

no not really

you have odd + odd + odd = odd

so it is an odd power

so it should end in 9

[main mural]

so ends in 9 here

[slate elbow]

yeah im stupid

i wrote it correctly but made stupid finish

[main mural]

lol

it happens

[slate elbow]

ok good

@main mural last one

[main mural]

go for it

[slate elbow]

find all prime numbers p, such p | 5^p + 1

this goes for small fermat theorem, right?

[main mural]

i think so

[slate elbow]

howM?

@raven vessel

or @main mural if u know

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[devout echo]

Simplify: $r\cdot\frac{2n!}{n!n!}=\frac{2n!+2n!}{(n-1)!(n+1)!}$

[flat frigate]

ChocolateFudge

[rose plume]

just to make sure (2n)! or 2n!

[devout echo]

Sorry

?

[devout echo]

Idk what you are saying

[main mural]

is it (2n)! or 2(n!)

[devout echo]

(2n)!

sry

[devout echo]

Simplify: $r\cdot\frac{(2n)!}{n!n!}=\frac{(2n)!+(2n)!}{(n-1)!(n+1)!}$

[flat frigate]

ChocolateFudge

[devout echo]

@rose plume

[rose plume]

you can start rewriting one n! as n(n-1)! and the other n! as $\frac{n! (n+1)}{n+1}$

[flat frigate]

svc

[devout echo]

which n?

[rose plume]

on the lefthand side

[devout echo]

(2n)! - (N!)^2

OK i got the answer thx

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[minor gale]

can someone explain this to me pls

how does it become 8-2t

i dont understand

[timber moss]

do you mean 8-12t?

[minor gale]

yes

mb for typ

[timber moss]

they just multiplied both sides by 12 to get rid of the common denominator

[minor gale]

but then how did it become 12t and 8

2/3 x 12 -t

[timber moss]

what is 12 times -t?

[minor gale]

but how 8

[timber moss]

multiply 2/3 by 12

[minor gale]

but only one 12

[timber moss]

when you multiply both sides, the multiplication contributes to all the terms, not just one

[minor gale]

ohhh

ok

thats why

i kept getting 4

thx

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[warm oyster]

If an object started moving from stillness

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[coral shoal]

Hey i got stuck here

[shadow glade]

factor into (sinx - cos x)^2 ?

[coral shoal]

Thanks i figured it out

was a dumb question

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[obsidian talon]

Any help?

[proud belfry]

What

[obsidian talon]

I need help with my geometry

...

[proud belfry]

Ok

[proud belfry]

2 trapezoids

Lets see, do you know the trapezoid formula

[obsidian talon]

I don't, can you explain it to me

I'm new at geometry

[proud belfry]

Ok so

This trapezoid

Is just a square plus a triangle

[obsidian talon]

ohh

[proud belfry]

If you draw a line when the line starts to turn

So whats the area of the square?

One side 11 other side 23/2

[obsidian talon]

length of its side square?

[proud belfry]

Its a rectangle

Actually

One side is 11 other side is 23/2

Which is 11.5

Do u see where 23/2 camefrom

[obsidian talon]

No i dont

[proud belfry]

We split the original shape into 2 equal area trapezoids

So the side length of each one of them is 23/2

We split it along the dashed line

[obsidian talon]

I'm still a little confused

[proud belfry]

Ok so the original shape

Is complicated and messy

So we split it into 2 easier to handle shapes

Right

And each of them has half the width of the original shape

[obsidian talon]

Oh thank you i got the answer

I have to go though

thx for the help

[proud belfry]

Np

[obsidian talon]

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[lean otter]

i forgot to answer..

i still dont get it tho

do we have to do it? is it necessary?

[safe radish]

@lean otter Has your question been resolved?

[stoic dune]

I mean if you were able to find the most general solution right away, then no

But if you're not a superhero, and are dumb like me, then odds are you didn't find the most general solution right away. You need to construct it. Adding solutions together can do this.

[lean otter]

hmmmm

but like if u have multiple solutions one of them is a solution already

so why extra work

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[golden pumice]

This is trig sub right

Or no

[severe pond]

is that an integral symbol?

[golden pumice]

Yeah

Lol

[severe pond]

oof

gotta work on that man

looks like you messed up an absolute value bar

anyways

is it sec^4

[golden pumice]

Yea

[severe pond]

what trig sub were you going to do this is already a trig integral

i would distribute

then rewrite tan or sec in terms of the other

split into multiple integrals

[golden pumice]

Aight I'll try that out

Thanks!

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[last wing]

how do i sketch the inverse?

with no information other than the picture

rn i know that the inverse will hit 1,0

[median vigil]

the inverse of a function looks like the function reflected about the line y = x

[last wing]

i cant imagin eit

[rose plume]

the inverse is (y,x)

think about (x,y) and (y,x) have the same distance

[last wing]

f(x) has asymptote at y = 0

so inverse of f(x) has asymptote at x = 0

[last wing]

ohh i kinda get this

[rose plume]

just throwing some examples

[last wing]

gah damn

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[rose plume]

@last wing

look purple and red is just a mirror of the purple part

[last wing]

yea

i can get it if i just do more questoins like that

thank you

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[shadow glade]

wut is this book talking about, isn't this just a big misprint? 🤔

[buoyant shadow]

it's the first sentence that's cap

[shadow glade]

ya...someone done fucked up eh?

[buoyant shadow]

if y = 0 you don't get to ignore the rest

[shadow glade]

ok, just double checking that im not here hallucinating something, but something went very wrong with this example :p

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[buoyant shadow]

@shadow gladewhat they mean is that points on the x axis outside the circle are already mentioned, but inside the circle they are still (y=0) so they are added back

so it's not 2 different issues i think

[shadow glade]

but the function is undefined at points on the x axis where -5 < x < 5 right?

ya, not two different issues, but a fuck up coming from the wrong thing they say in the first sentence?

[buoyant shadow]

usually not, but maybe it's like a different context where you can ignore the root if y=0

yes

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[molten idol]

Hi guys, how do you go from this

[molten idol]

to this

[tall bough]

What's 177/100 and 46/100

[molten idol]

yeah but

i dont understand how you can just divide them

what is the rule

behind this

i want to understand

you cant just put 100 under 177

[lavish blade]

why not

you actually can

[molten idol]

oh okay

[molten idol]

do you know why 25 and not -25 here?

so it is basically a rule

i dont get it

._.

-a/-b is the same as a/b

[lavish blade]

no sorry

i said shit

[molten idol]

oh

no worries

[lavish blade]

but as a general rule when you have one - divide by - the result is an positiv

[molten idol]

dividing two negative numbers is a positive number

[lavish blade]

yes

[molten idol]

?

that is so weird

i dont know how to imagine this in real life scenerio

but ok

[lavish blade]

It's the same thing as when you multiply a negative number by anegative

[stark stratus]

Its cause negative numbers cant really be imagined as a scenario

Or in a scenario rather

[molten idol]

thats why we just say

it's a positive number

?

[stark stratus]

Sorry what?

[molten idol]

like multiplying two negative numbers

we just say its positive

[stark stratus]

Yeah the best analogy you can find is dont dont turn around

[molten idol]

whaat

[stark stratus]

Say youre facing right

I tell you to dont dont turn around, youre still facing the same direction

[molten idol]

ah yeah and if you continue to move to the negative direction it's a plus

[stark stratus]

Yes for example i tell you to dont turn around

Then i tell you to turn around

Youre now facing left

Hence why negative x positive is negative

[molten idol]

ah okay, so dividing two negative numbers is plus and multiplying two negative numbers results in a positive number because we dont change direction

[stark stratus]

Yes

Exactly

[molten idol]

okay

Wow, thank you!

[stark stratus]

Since multiplying and dividing are 2 sides of the same coin

They usually have the same signs

[molten idol]

i see

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@molten idol Has your question been resolved?

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[reef quartz]

can someone please help me with this

[reef quartz]

sorry for not replying @split ether I had to go eat

but I get what you said about what i will be if you use modulus

[split ether]

Any progress?

[reef quartz]

not particularly

like I get the modulus part

but how do I work out the factorials

[split ether]

What will be i^{4!}

[reef quartz]

🤔

i^24

[split ether]

And 24 is what modulo 4?

[reef quartz]

6

wait shit

0

[split ether]

Right, so i^{4!} is?

[reef quartz]

1?

[split ether]

Yes

Now, what will be i^{n!} given n >= 4?

[reef quartz]

🤔 is there a pattern

[split ether]

Keep in mind that n! Is divisible by any integer starting from 1 to n

[reef quartz]

right

i'm stumped

[split ether]

What does that imply n! to be modulo 4?

Given n >= 4

[reef quartz]

oh is it 0

because its just

a multiple

of

4

[split ether]

Yes

[reef quartz]

damn 💀

[split ether]

What does that imply i^{n!} to be?

[reef quartz]

so for 96 of them it will just be 1

[split ether]

Yes, so you have i^{1!} + i^{2!} + i^{3!} + 97

There are 97 integers on the interval [4, 100], not 96

[reef quartz]

wait what

oh 4 + 96 is 100

so you count the 4

[split ether]

Right

[reef quartz]

ok so 97 -1 + i -1

95 + i

so C

[split ether]

Right

[reef quartz]

that was the correct answr

[reef quartz]

thanks a lot for the help

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[inner parrot]

is a calculator allowed or no?

[thin bridge]

consider completing the square / converting to vertex form

[lean otter]

no

[inner parrot]

the range would just be from (-infinity, k] where k is the y coordinate of the vertex

we just need to convert this equation to vertex form is all

[lean otter]

is thi applicable too all similar functions or just this

[inner parrot]

it’s only applicable to quadratic equations with a negative a value

a value is the coefficient in front of the x^2 term

since then the parabola would open downwards

[lean otter]

correct me if I'm wrong, Vertex = (h,k) = (-b/2a,-D/4a)

[lean otter]

what if we have a positive value

[inner parrot]

it would go from [k, infinity)

[lean otter]

if we have a positive value?

[inner parrot]

yep

[dusk nova]

for a parabola, the vertex is either the minimum or the maximum point

[inner parrot]

not quite sure if that’s correct

[dusk nova]

it's correct yeah

[inner parrot]

but we can totally plug in the axis of symmetry into the function

[dusk nova]

lol

[lean otter]

D is discriminant

[inner parrot]

alright then

that saves a lot of time

[lean otter]

I'm getting (-3,2)

so

(infinity,2)

I don't think its quite right

[inner parrot]

yeah

the x coordinate is correct

not the y coordinate

[lean otter]

I think I made a calculation error in D

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

I got it

-3,17

[inner parrot]

yeah that looks good

[lean otter]

how do I write infinity in khan academy

[inner parrot]

oh

maybe check the toolboxes

if not just try to type infinity

[lean otter]

wait

[lean otter]

x smaller than or equal to 17?

[inner parrot]

yep

[lean otter]

its incorrect

minor error

it should be y

[inner parrot]

oh right yeah

since we’re talking about range

mb

[lean otter]

its fine

how do I close this question?

[inner parrot]

type .close

[lean otter]

.close

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[storm ice]

what's the general idea for this?

Like as we approach infiniy the +1 isnt going to matter so sqrt(x^2) is x

then we get x * (x-x)

i was thinking about possible exponent tricks but a bit stuck there

[compact pelican]

If you factor another x out of it and write x^2 (something) that might help

[storm ice]

i feel like its the same issue though

x^2 ( (sqrt(x^2+1) / x) -1)

when i say that x^2+1 is roughly x^2

[compact pelican]

Well, x^2*(sqrt(1+1/x^2)-1)

is another way to do it

But you're right that doesn't help

[storm ice]

how did u get 1+1 o_O

[compact pelican]

I mean you can L'Hopital it in this form

Oh, sqrt(x^2 + 1) divided by x is the same as dividing by sqrt(x^2) to get sqrt(1 + 1/x^2)

[storm ice]

and if I am not allowed to take L'hopital? ;D

[compact pelican]

Your intuition is correct though. The -x^2 term will dominate the other terms

[thin bridge]

conjugates

[compact pelican]

Actually, wait, I could be wrong. The sqrt(x^2+1) also grows as x, so maybe not

OK, without L'Hopital, I'm stuck

[storm ice]

how would I do conjugates in a limit expression

[compact pelican]

I mean, you could power series

[storm ice]

wouldnt i have to multiply the conjugate to both sides o_O

[thin bridge]

multiply by (conjugate of sqrt(x^2 + 1) - x)/ (conjugate of sqrt(x^2 + 1) - x)

[storm ice]

ohhh yeah the 1 multiplication thingyjingy

got it

yep that worked

cheers!

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[lethal wagon]

Can someone tell me why that works for the denominator?

[cloud citrus]

(n+3)^4 is divided by n^4
sqrt((n-2)(n-1)) is divided by n
The denominator is divided by n^5 in total

$\frac{\sqrt{(n-2)(n-1)}}{n}=\sqrt{\frac{(n-2)(n-1)}{n^2}}$

[flat frigate]

Afi

[frozen marlin]

missed the dollar sign

oh

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[lethal wagon]

So they just "split" the division right?

[cloud citrus]

Yep

[lethal wagon]

Okay, thanks!

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[tepid flicker]

can anyone help with this?

[tepid flicker]

i wanna see if im right

[limber patio]

You're right

[limber patio]

You can also plug in the values back in the equation to check your answer

[tepid flicker]

thank you man i apperciate it God bless

🙏

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[wraith yoke]

Can someone take me step by step on how to solve this problem

[wraith yoke]

This topic is sin, cos and tan

Advanced word problems

Ok I tried another method that I think worked

I got 195 foot per minute

Can someone verify this pls

<@&286206848099549185>

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[latent gull]

My guys I am struggling like soggy biryani

[simple oak]

What part

[latent gull]

just solving I know it is solved

[simple oak]

Wait wtf is that 58’97”

Wot

[latent gull]

?

[simple oak]

It’s solved?

[latent gull]

I know it is solved but I dont how to solved by my self

yeah

my teacher solved

[simple oak]

Oh

Ok so right angle triangle

60 degree and we have a side

Use SOH CAH TOA

sinx=opposite side/hypotenuse

a for adjacent

And do some solving shit

[latent gull]

ok hold up

You know thanks anyways .close

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[lean otter]

is it possible to find the blue part , if yes then how ;-;

[hasty warren]

yes, first extend the large sector to be a triangle. Then compute the inradius of that triangle. The inradius of that triangle is the radius of the inner circle. Now draw a line from the tangent point of the outer lines to the center of the inner circle

Then draw a line from the center to the 60 angle

this forms a 30 60 90 triangle which you can compute the length from the angle to the blue region along the outer lines.

you have sufficient information now to compute this area with some more simple geometry tricks

Geometry tip: draw lines!!!

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[steel crater]

$$\int(\frac{1}{(a^2 + x^2)^\frac{1}{2}})dx$$

[flat frigate]

SkyGazer

[steel crater]

im in high school and not very familiar with integral substitution

[peak kiln]

the key is this: if you subsitute, you must also multiply by the derivative of what you subsitute

[steel crater]

?

what do i substitute it by? i tried simple u-sub it didnt work out then i tried trig sub and i got stuck with integral of sec-theta d-theta

[left gyro]

you can integrate sec theta, you can google it if necessary

[steel crater]

i am not able to understand the steps they just state a particular identity and just give the answer i need to show procedure in exam i cant do that if i cannot do it at home😕

[lean otter]

from the old step formula booklet]

[steel crater]

sorry but i do not under hyperbolic trignometry 😦

[lean otter]

me neither

i personally dont know how you integrate to get that

[steel crater]

it is the correct answer tho

[lean otter]

it might be one of those t substitutions?

[steel crater]

i need to solve for practice

[steel crater]

i tried it comes out to be integration of sec

[lean otter]

i dont think this is what sec integrates too tho

[lean otter]

you can integrate sec into both of these forms using a t substituion

let me find it i cant remember

its like (t^2)/t^2 -1 or something

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[dense saddle]

i made the fractions have common denominators which turns out to be (4x + 4a + x^2 -2x)/4x = 0 which turns out to be x^2 + 2x + 4a = 0. From here im not sure how to get it into the required form or if ive done smth wrong

[hasty warren]

thats good, lets just complete the square

x^2+2x+? completes the square

[dense saddle]

i tried to complet the square but that got me (x + 1)^2 - 1 = -4a

what do i do with the -1

[proud belfry]

add it on both sides

[dense saddle]

even so that it not the required form?

the right side is simply just q

[proud belfry]

(x+1)^2=4a+1

(x+p)^2=q

[dense saddle]

but i thought q was meant to be only in terms of a

[proud belfry]

yes

theres no x in 4a+1

its in terms of a

[dense saddle]

oh so it can contain the +1 ?

[proud belfry]

yeah

would you say y=x+1 is not in terms of x

[dense saddle]

oh yeah

thank you

can you help with the next part please

im not sure where to begin

i think i got it

[lean otter]

hi

[dense saddle]

is it a < 0.25

[lean otter]

re-arrange to find a quadratic somewhere

and then set the discriminant greater than or equal to zero

[dense saddle]

i already figured this out thanks

can you check my answer to the next part

[lean otter]

i havent solved it so idk but itll be greater than or equal to

[dense saddle]

ah ok

[lean otter]

because if theres a repeated root, there is a real root

[lean otter]

this part?

[dense saddle]

this part

[lean otter]

ill just do it over message

so u can see

moving one term over,
(x+a)/x = (-x+2)/4

multiplying up
4x + 4a = -x^2 + 2x

re-arranging
x^2 + 2x + 4a = 0

completing the square and moving across
(x + 1)^2 = -4a + 1

is that waht u got

[dense saddle]

yes

[lean otter]

ok

[lean otter]

for this part

[dense saddle]

i meant -0.25 btw

[lean otter]

x^2 + 2x + 4a = 0

take discriminant and set it greater than or equal to zero

(2)^2 - 4(1)(4a) >= 0

so 4 >= 16a

a <= 1/4

[lean otter]

i think its positive

[dense saddle]

oh yes

i had the inequality wrong way around

a <= 0.25

[lean otter]

i think so

[dense saddle]

thanks

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[fervent girder]

I need to find the perimeter of this shape

the shaded region

is not just 17+ the green curve + Pythagoreans of the given points?

I dont know what im doing wrong?

equation of the line is y= -0.781285626507x

That intersects the ellipse on the bottom right

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@fervent girder Has your question been resolved?

[lusty ridge]

i think it is

[fervent girder]

idk why it is still wrong

[lusty ridge]

what does the answer key say?

[fervent girder]

Idk have a answer key

what in the world am i doing wrong ?

[lusty ridge]

did u recheck that you calculated the area of the green part correctly?

[fervent girder]

🤷‍♂️

Do you think i might need more decimal places for the intersection points?

Could be rounding error or smthn

[lusty ridge]

maybe the answer has it rounded to 91?

cause your wokr seems correct to me i cant see anything wrong with it

[fervent girder]

91 is wrong as well

the only thing i can think of is keeping more digits for intersection points

is there a way to get more decimal places on desmos?

[lusty ridge]

not sure but i dont think that'd help cause theyve only asked for 3 siginificant figured

figures*

[fervent girder]

Yeah, idk what to do :

😭

[thin bridge]

equation of the line is y= -0.781285626507x
how are you getting that

[fervent girder]

tan -38

[lusty ridge]

that part is correct, i checked once w geogebra to make sure

that their co ordinates are correct

[lusty ridge]

did you try 90.9 ?

[fervent girder]

Yeah I did

[lusty ridge]

cause i cant see anything wrong w ur work i rechecked it

[fervent girder]

yeah i double checked it as well

[thin bridge]

what was the unrounded value from arc length

[fervent girder]

52.5000996383

[thin bridge]

try 91.0

[fervent girder]

Nope

[lusty ridge]

its probably a flaw in the answer key

[fervent girder]

Could be atp

If someone else has an input please let me know

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[tight scroll]

How to do this?

[tight scroll]

I did do like half of it but don't know if it's correct

I don't know what to do after this

<@&286206848099549185>

<@&286206848099549185>

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[burnt night]

I am using Markov Chains to determine the optimal property to maximize profits in Monopoly. I made a Probability Mass Function for the dice probability, and I think I can make a Transition Matrix soon.

[burnt night]

However, the chance and community chest cards some of them are designed to move you across the board. Let's say 20% to draw a card that moves you across the board. How would I show these probabilities in my transition matrix?

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[willow fossil]

I tried expanding but I get math error for tan 90

[earnest zinc]

ah nvm

I'm not sure if an answer exists for this integral

[willow fossil]

Yea maybe not

Is there another way

[earnest zinc]

It means both are undefined

[vagrant ice]

You can do it by using the sub u = pi - x

Your teacher has chosen an integral that's impossible to evaluate, that's on purpose

[willow fossil]

Oh ok

sin pi - u = sin u?

Im always bad with the pi - x and pi/2 - x how do you know what it becomes

[vagrant ice]

You need your unit circle

sin x is the y-coordinate or the height of a point on the unit circle

The two triangles are congruent

And this is the other one, sin(pi/2 - x) = cos x and cos(pi/2 - x) = cos x

[willow fossil]

Is there a way to remember it for all of the trig functions

[vagrant ice]

Yes, keep drawing the unit circle

[willow fossil]

Coz i know when you do pi/2 - x it becomes the co-function

[vagrant ice]

Well I guess it won't help you with the double angle identities

And the compound angle ones

But all the other ones

[vagrant ice]

Yeah, so that's where cosine comes from

co-sine

cosine and sine always go together

[willow fossil]

so cot (pi-x) = cot x

for eg

[vagrant ice]

Break it up into cos and sin

No actually

[willow fossil]

cos (pi-x)/sin (pi-x), assuming x is acute then cos negative sin positive

so neg cot x

?

[vagrant ice]

Right, exactly

Yeah so cos(pi - x) = -cos(x)
sin(pi - x) = sin(x)

Nice work actually, I'm impressed

[willow fossil]

Oh ok i see

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[devout shale]

Take the derivative

Okay so seperate

differentiation is linear

[thin bridge]

no

watch your signs

linearity of derivative operator
differentiation is linear

linearity of derivative operator
differentiation is linear

[twilit hare]

correct

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[median flare]

I dont know hiw to do this i did once but wrong according to marking scheme

[vagrant ice]

Group the two boys and four girls together like this:

(BGGGGB)(G)(G)

There are 3 objects so 3!

Now how many ways are there to arrange the boys and girls in the big group?

[median flare]

2 for boys

[vagrant ice]

Yep

[median flare]

And 6!/4! For girls?

Within the group

Just 4! Actually

[vagrant ice]

Actually it's 6P4, I think order matters here

[median flare]

That formula in that book is inteoduced later

[ionic lark]

yes ig cuz boys are not identical

[pseudo scroll]

each boy and girl should be considered unique

[median flare]

So i wanna know how to do this without it

Ye each unique

So the girls within the boys group can be arranged in different ways and it will count as seperate

[pseudo scroll]

you haven't learnt nCr and nPr yet?

[vagrant ice]

Yeah well there's a clever way to do it if we just ignore the boys for now

[median flare]

Ik the formulas and what they dk but this ques is part of the i troduction before nPr

The answer is 288

[vagrant ice]

It's just 6!

[median flare]

But boys only at ends

[vagrant ice]

I know for the girls in total

[median flare]

So is it 2x4! ?

[vagrant ice]

Wait hold on

Yeah 6P4 * 2! = 6!

[median flare]

Huh

[pseudo scroll]

should just be ||6!2!*3||?

[median flare]

6!/2! Right?

[ionic lark]

apply the formula and youll know

[median flare]

Wai wait

[ionic lark]

thats 6P4, multiply with 2!

[median flare]

But this is suppoed to be done without it i wanna get my thought process on perms correct so while doing ques ik what im doing

[vagrant ice]

It's weird cause you can have say (G)(BGGGGB)(G)

But you can ignore the brackets and there are just 6 positions, even though the first girl could go in the big group and also not go in the big group

You just ignore the boys

[median flare]

For a few ques i got really confused

[vagrant ice]

6P4 = 6 * 5 * 4 * 3

[median flare]

The method described in worked example says exactly what u did before

[vagrant ice]

So start from 6 and multiply down until you have 4 numbers

[median flare]

3 elements so 3!

[vagrant ice]

Yep

[median flare]

This u said

[vagrant ice]

I'm guessing they didn't use the notation 6P4

[median flare]

And then within the bigger one we have to find the ways to rearrange

[median flare]

No

[vagrant ice]

Ah ic

[median flare]

Where there’s 2 boys and 4 girls

So is it 2 x 4! Or 2 x 6!/2!

[median flare]

This is for the girls i think this ur supposed to multiply by 2 because the boys can switch places

[vagrant ice]

For the girls it's 6 * 5 * 4 * 3
And for the boys it's 2!

[median flare]

If i say i wanna use 4x3x2x1 then for the ones outside the group i say 6x5?

[earnest rapids]

This itself is 360 and answer is 288

[vagrant ice]

Nah I don't think it works like that

Cause if you have 6 people left

Then 6 people can go in the first spot

[vagrant ice]

How do you know the answer

[earnest rapids]

He told already

[earnest rapids]

.

[median flare]

For each exercises theres answer at the end of the book

[earnest rapids]

I think they assumed specific 4 girls take the spot hence
G, BGGGGB, G
3! for 3 groups
4! for 4 girls
And 2! for 2 boys in group of 6

3! 4! 2! = 288

Logic derived from answer of course but if we take 6c4 and take choices of ANY 4 girls, cases increase

[median flare]

If i do 2 x 3(this is for the 3 elements) and within the big group: 2(boys) x 4!(girls) then i get 288

[median flare]

What do u get for any girls anywhere

[earnest rapids]

6c4 * 288

[median flare]

Idk so i was doing for any girls earlier and i got 4320

[median flare]

I don’t understand this

@earnest rapids

[earnest rapids]

You know concept of nCr?

[median flare]

Yes combinations

Bur doesnt order matter

[earnest rapids]

Yes it does

Actually It's included in 288

In reality I did

(6C4 * 4!) * (3!) * (2!)

But as i said answer is given thinking that 4 specific girls are sitting in between they either gave wrong answer or framed question wrong

[median flare]

I did it a seperate way without form i got same answer as the value of this

Hey i need help with another hing

In this question why is the answer not simply 5!

[earnest rapids]

Op

Because of reptition of P two times

[vagrant ice]

Yes if you swap the 2 Ps, you get the same word

It's indistinguishable vs distinguishable

[median flare]

Yes

Thats just 1

Permutation

That looks the same then right?

So then 5!-1

Oh wait is it that for each person ermutation there is 1 repetition

[ionic lark]

nope

[median flare]

Ye so this right?

That means half the results are same as other half so divide by 2

[ionic lark]

yes exactly

[vagrant ice]

Yepppp

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[median flare]

.reopen

[safe radish]

✅

[median flare]

Same goes for other nums right?

Like for 3 there will be 3 repetitions

No wait 6

The factorial of the number of repeated element

[ionic lark]

Yes

We don't use 3 for 3 repetitions but we use 3!

[median flare]

.close

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[median flare]

Ight thanks

[median flare]

3! Means 6

Total reps

[ionic lark]

Yes

Ik

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[hardy flower]

I feel like such an idiot for not understanding this but why is $(2sin(x) + 2cos(x))^2 \ne 4sin^2(x) + 4cos^2(x) + 4sin(x)cos(x)$

[flat frigate]

KJ

[hardy flower]

ohhh it's 8cos(x)sin(x)

.clone

.close

[safe radish]

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[safe radish]

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[steel forge]

can anybody help me with this?

[steel forge]

using gamma or beta function

i really find it difficult to solve

[dusk nova]

lol what

is that sin(x^2)?

[steel forge]

yes

[dusk nova]

that's not solvable

that's fresnel's integral

[steel forge]

why

i know, i saw it on yt but it was a long vid

but my prof assigned this as our quiz on beta and gamma function

is it really unsolvable?

[dusk nova]

i mean for the bound it's solvable

iirc, you had $\int_{-\infty}^{\infty} \sin (x^2) = \sqrt{\frac{\pi}{2}}$

[flat frigate]

nyxie9151

[dusk nova]

lmao and since sin(x^2) is even so you have your answer for 0 to inf

[steel forge]

yeah, definitely a 0

but i dont know how to solve this using beta func

how about xcosx^3?

[limpid holly]

where can I learn more about these functions and stuff, do you guys know of any good book?

[steel forge]

bound from 0 to infinity

[steel forge]

what functions

[limpid holly]

betta gamma

all dat

[dusk nova]

https://math.stackexchange.com/questions/1848018/find-int-0-infty-sin-x2-dx

he uses laplace and beta function

[steel forge]

im looking for those books too

[steel forge]

might use this

[limpid holly]

oh so the branch is called complex analysis?

[steel forge]

bec our next topic will be laplace

[dusk nova]

nah they use complex analysis for evaluating this thing

[limpid holly]

uh alright

[steel forge]

i also have another question hihi

not this one

my answer is 8pi

is it correct?

this one's easy. but idk, maybe this long weekend create a doubt with my solutions

[eternal carbon]

n

[limpid holly]

n

[steel forge]

pi?

[eternal carbon]

y

[dusk nova]

,w integral of 1/sqrt((x-1)(3-x)) from 1 to 3

[flat frigate]

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Requested by nyxie9151

[steel forge]

huhuhhu

okay

[dusk nova]

bruh

[steel forge]

hahahha sorry sorry

we're solving it using beta and gamma function, i saw it on website that the answer is pi but i doubt it haha

or maybe i was too confident with my wrong answer

[dusk nova]

wolfram is mostly right

also i have no idea how to do that with beta and gamma functions

[steel forge]

i used mathway

[steel forge]

oh, that one's easy

[dusk nova]

did you also have to do xcos(x^3) with beta and gamma function?

i just bash power series

[steel forge]

yes

[safe radish]

@steel forge Has your question been resolved?

[steel forge]

can anybody help me with this. i just want to check my solution using beta function

[sacred pollen]

答案是我其實不知道，祝你好運

[kindred wraith]

Il est ennuyeux

[safe radish]

@steel forge Has your question been resolved?

[safe radish]

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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>.
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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.

[safe otter]

helpp

[safe otter]

the question is to solve that arithmetic equation

i thought about doing 2 cases where n is even and uneven

but idk what to do after

[proud belfry]

like, find all solutions?

[safe otter]

yeah

[proud belfry]

bc there are infinitely many solutions to this

like find m in terms of n?

[safe otter]

i think the only solution is 0

[proud belfry]

why do you think this

[safe otter]

i mean (0,0)

[proud belfry]

why

given n, we can always find m=sqrt(n(n+1)(n+2))

[upper rivet]

i think they are supposed to be natural numbers

n and m

[proud belfry]

ohh

so we need to find "is n(n+1)(n+2) ever a perfect square"

[safe otter]

yes exactly

and i have the answer here and it says (0,0) is the only solution but with no proof

so i just need to prove that is the only one

[lusty ridge]

This only has solutions for when m=0

[lusty ridge]

Well, just to make your life easier I would substitute t=n+1 into the eq, leading us to
(t-1)(t+1)(t)=m² for some positive integer m

Here, t must be a perfect square too. Can you try and tell me why?

[safe otter]

this means (t^(2)-1)t=m^2 but idk why t must be a perfect square

[lusty ridge]

Well, any 2 successive numbers are going to be relatively co prime

So t and t-1 are coprime, t and t+1 are co prime

Ie they share no common factor. But as they all multiply out to give a perfect square, so t has to be raised to an even power

[hasty warren]

i live and die for this question

[lusty ridge]

Else it wouldnt multiply to a perfect square

[hasty warren]

this is a mod 4 perfect square q

what are the choices for a perfect square mod 4

[safe otter]

i think this must be due to a language barrier as i don't study maths in english because i have no idea what you are saying im sorry

[safe otter]

huh

[hasty warren]

Are you familiar with modular arithmetic

[safe otter]

yes but i am not at the part in the exercices where it is needed so i think this question can be solved without using it

[lusty ridge]

Yeah it can be solved without it

[hasty warren]

oh okay

😦

[lusty ridge]

So as I said, any 2 consecutive positive integers must be co prime

Meaning they don't have any common factors other than 1

[safe otter]

yes but i dont understand what it has to do with perfect squares