help-49

[west iron]

you just have to show that multiplying a number by d (single digit) will never be bigger than adding a d in front of it

[fleet moss]

i think this could work yeah. your base cases are just trivial 0-9

[twilit field]

adding another digit moves the number by $\geq 10^{\gamma+1}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[west iron]

gamma is number of digits?

then you are incorrect

[twilit field]

yes

[west iron]

for example, adding a 9 in front of 37 increases the number by 900

which is less than 10^(2+1)

[twilit field]

ah, right

by $\leq 10^{\gamma +1}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[fleet moss]

why not think about multiplication than addition

[west iron]

additive difference can be used to get at the geometric difference

i would think about both operations (multiplying by d, and adding a d at the front) in terms of the additive increase they give

[fleet moss]

i mean we can also consider adding a digit to the end rather than the front

makes it considerably easier imo

[burnt flame]

wait...

hmm

[twilit field]

if we add a digit to the end , shouldn't this be right?

[fleet moss]

your thinking was right, it just didnt work because it should be gamma rather than gamma + 1. but adding to the end lets you give a much simpler bound

i think

[west iron]

oh fair enough

[fleet moss]

i have a solution in mind but im too sleepy to know whether its right

[west iron]

that works as well

[twilit field]

hmm

so adding another digit , moves the number by $\geq 10^{\gamma}$

say I have 100

I 1add another zero

ok, yeah, makes sense

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so $0 \leq g(n) \leq 10^{\gamma}$

[burnt flame]

leq

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$g(n)= n_1 n_2 n_3....n_{\gamma}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

how all these numbers are less than or equal to 9

so in the largest case I have $0 \leq 9^{\gamma} \leq 10^{\gamma}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is true $\forall \gamma \geq 1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

any any other sequence of gamma numbers is less than $9^{\gamma}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

thus we have our proof

is that it?

I feel I have to prove that the maxium product is $9^{\gamma}$ though

[grand pond]

ƒ(Why am. I here)=I don't Know

[fleet moss]

i think this part is clear enough, i dont follow how youre setting the lower bound on n though

[twilit field]

of 0?

[fleet moss]

i mean the leq 10^gamma part

[twilit field]

$g(n) = n_1 n_2 ....n_{\gamma}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

this is maximised when $n_1=n_2....=9$

[grand pond]

ƒ(Why am. I here)=I don't Know

[fleet moss]

that part makes sense eah

but what about n?

[twilit field]

n?

[twilit field]

here is the explaination

[fleet moss]

like this part

[twilit field]

I don't know how to explain it, it just makes intutive sense to me

like say there are $\gamma-1$ 9s

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

and some other arbitary digits

[fleet moss]

is this correct? in the case of adding a 1 to the end of 100, we have 901 < 1000

[twilit field]

then we have the product is $9^{\gamma-1} \alpha$ where $\alpha$ is the random number

[grand pond]

ƒ(Why am. I here)=I don't Know

[fleet moss]

i think in general you dont have to think about the total number of digits at all if youre doing it inductively. but youll need to think about multiplication rather than addition

[west iron]

man you really gotta find easier problems to solve

[twilit field]

I'm practising hard problems to get acquainted with proof writing before I start uni

[west iron]

are there easier proof problems you can find

[twilit field]

I guess? I'm just using this as a benchmark as it's from the hardest maths exam school going kids in my country write

save for olympiads

[west iron]

it's probably ideal to work on problems that you at least stand a chance of solving by yourself

[twilit field]

true

thanks

[west iron]

np

[twilit field]

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[midnight plank]

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[fleet moss]

oh i mean i think you can do this one though, at least the first part

[midnight plank]

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[twilit field]

3 is a common divisor of 6, 9, and 15., write a logical statement representing this

[twilit field]

Let the statement 3 is a divisor of 6 be P,9 be Q and 15 b %

then $P\wedge Q\wedge R$

[grand pond]

ƒ(Why am. I here)=I don't Know

[deep vine]

so like what's your question

[twilit field]

is this right?

[deep vine]

yea

[twilit field]

that's all, thanks

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[viral dagger]

$$a+x^2=2006$$
$$b+x^2=2007$$
$$c+x^2=2008$$
$$abc=3$$
find $\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$

[grand pond]

Skill_Issue

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[last slate]

@viral dagger

[viral dagger]

huh

[last slate]

did you get all of this under the same denominator ?

[viral dagger]

$\frac{a^2+b^2+c^2-ab-ac-bc}{abc}$

[grand pond]

Skill_Issue

[last slate]

yes

now

lets just consider the numerator

$a^2+b^2+c^2-ab-ac-bc$

[grand pond]

Cnidarian

[last slate]

multiply it by 2

$2(a^2+b^2+c^2-ab-ac-bc)$

[grand pond]

Cnidarian

[last slate]

You can write this as sum of sqaures of 3 terms

do you see how ?

[viral dagger]

$(a-b)^2+(a-c)^2+(b-c)^2$

[grand pond]

Skill_Issue

[viral dagger]

ooh

ok ic, ty

let $x-y=1$, find $$x^4-xy^3-x^3y-3x^2y+3xy^2+y^4$$

[grand pond]

Skill_Issue

[last slate]

Is this related to the same question ?

[viral dagger]

no

[last slate]

Okay

[viral dagger]

nvm lmao i solved it, sorry

uh lemme find a diffrent question

[last slate]

Could you give me a hint

[viral dagger]

$$x+\sqrt{xy}+y=9$$
$$x^2+xy+y^2=27$$
find $x-\sqrt{xy}+y$

[grand pond]

Skill_Issue

[viral dagger]

uh start with making it somehow related to (x-y)^4

[last slate]

Square the first equation

and the subtitute the second equation into it

Get it ?

[viral dagger]

uh ill try

[last slate]

This question is similar to

Say

[viral dagger]

$xy+x\sqrt{xy}+y\sqrt{xy}=27$

[grand pond]

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

[viral dagger]

oh

[grand pond]

Skill_Issue

[last slate]

yeah

gtg

[viral dagger]

i tried squaring the second equation, i got $$2xy-2x\sqrt{xy}-2y\sqrt{xy}+27$$

[last slate]

try looking into this example

[viral dagger]

oh cya

[grand pond]

Skill_Issue

[last slate]

you odn't have to do that!

[viral dagger]

huh

OH

ok nvm

[viral dagger]

i factor $\sqrt{xy}$ out of this right

[grand pond]

Skill_Issue

[last slate]

no...

[viral dagger]

whar

[last slate]

gtg bye

[viral dagger]

bye

it was 3 :p

.close

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[viral dagger]

if $x=\sqrt{19+8\sqrt{3}}$, find $$\frac{x^4-6x^3-2x^2+18x+23}{x^2-8x+15}$$

[grand pond]

Skill_Issue

[upper crag]

just curious what course is this

[low finch]

i wonder wtf is this

[twilit field]

olympiad maths maybe

[sly ravine]

Patience Algebra

[low finch]

wow

[wary trail]

are you wanting a hint or just how to do it

since yeah it looks like competition maths

[low finch]

method 1: change x into the roots and calculate it (very real and siemple)

[old storm]

goddamn, cant even use rational root theorem to simplify that one above, you have to actually play with it

[twilit field]

chcek if $x^2-8x+15$ is a factor of the numerator first

[grand pond]

ƒ(Why am. I here)=I don't Know

[old storm]

(x-3)(x-5) btw

[twilit field]

,w (x^4-6x^3-2x^2+18x+23)/(x^2-8x+15)

[low finch]

wat

[sly ravine]

Stap

[twilit field]

ok, that won't help

[low finch]

whats the answer

[upper crag]

best answer so far imo lol

if you call the numerator f(x) then f(x-1) is eisenstein for p=2 (so it's irreducible)

(although knowing 3, 5 are roots of the denominator and are not factors of +-1, +-23 already means there is no cancellation)

[midnight plank]

@viral dagger Has your question been resolved?

[viral dagger]

correct

[hard umbra]

just find the min poly of x and divide

why so complicated

[viral dagger]

dont think so

[viral dagger]

what

[hard umbra]

,w min poly of sqrt(19 + 8sqrt(3))

[grand pond]

Results provided by WolframAlpha

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

Requested by tearfox

[hard umbra]

that tells you already that the denom is 2

and now you just do like

,tex<polynom> \polylongdiv{x^4 - 6x^3 - 2x^2 + 18x + 23}{x^2 - 8x + 13}

[low finch]

w

[hard umbra]

which means the numerator is 10

[viral dagger]

huh

how can you just get 10

[upper crag]

how do you get this without wa

[hard umbra]

because
[ x^4 - 6x^3 - 2x^2 + 18x + 23 = (x^2 + 2x + 1)(x^2 - 8x + 13) + 10 ]

[hard umbra]

just work it out by hand

[viral dagger]

what do you do after that

[hard umbra]

x^2 - 8x + 13 = 0

,, \sqrt {19 + 8\sqrt 3} = \sqrt {4^2 + 2 \cdot 4 \cdot \sqrt 3 + 3} = 4 + \sqrt 3

[hard umbra]

(x - 4)^2 = 3

[viral dagger]

what do you do with the 10/x^2-8x+15

[hard umbra]

x^2 - 8x + 15 = x^2 - 8x + 13 + 2 = 2

[viral dagger]

wait

oh sorry

lemme work it out

[grand pond]

Skill_Issue

[viral dagger]

wait no

wait yea

is this correct?

[hard umbra]

what did you do there

[viral dagger]

oh sorry i misread something

its 5?

[hard umbra]

yes

[viral dagger]

oh

thanks

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[twilit field]

Which of the following expressions are well-formed formulas?:-

$\left(P\wedge Q\right)\left(P\vee R\right)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

how do I even interpret this

I feel it isn't a well formed formula

[dreamy lichen]

indeed

[twilit field]

thanks!

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[last slate]

Were you able to Solve it ?

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[viral dagger]

yea

[viral dagger]

.close

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[last slate]

Okay

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[twilit field]

How do natural numbers have a decimal expansion

[zealous schooner]

it just means the digits

[twilit field]

ah

[zealous schooner]

decimal expansion because we use a decimal system

like a base 10 system

[twilit field]

so I now effectively find the number of solutions to $x+y+z=10$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

and then permute them

[zealous schooner]

yeah

x,y,z>=0

[twilit field]

yeah

thanks

[zealous schooner]

except x cant be 0

[sage helm]

sure it's not meant to be 3.000000000...

[twilit field]

[sage helm]

never heard decimal expansion being used for any other context

[strong lava]

and y or z is 0 and x >= 1

that's very quickly enumerated

[twilit field]

wait, why should y or z be 0

oh

ok

[strong lava]

(ii)

[twilit field]

so if 0 appears twice there are 9 such numbers

zero appears once in 144 numbers

so the answer 153

wait

no

the solutions are $(1,9);(2,8);(3,7);(4,6);(5,5)$

this into 2

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

wait

shit

let 0 appear twice

there are 9 such numebrs

let 0 appear once

so I have $[(1,9),(2,8)(2,7),(4,6)(5,5)]$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

as possible solutions

which gives me an answer of 18

ok, but how would i count without listing it case-wise

<@&286206848099549185>

[simple field]

wdym by case-wise

[twilit field]

by listing the solutions out

[twilit field]

like these are the values (y,z)

now (z,y) is also valid

[simple field]

a three digit number ABC has the following properties:
one of A, B, C is 0, A cannot be 0 because it would then be at most a 2 digit number. so B or C is 0
let x = A, y = max(B, C) i.e. it is the non-zero number
x + y = 10 and 1 <= x <= 9, so there are 9 solutions (x, y)
for each solution (x,y) we can have either B = y, C = 0 or B = 0, C = y
so the total number of solutions is 2 * 9 = 18

which seems to be basically what you did, nothing wrong with listing out solutions to verify your calculations

especially when there arent a ton of solutions listing is worth doing, a sanity check

[twilit field]

Yeah, I get that

but what if there say 100+ solutiosn

*solutions

[simple field]

yeah then not so much haha

[twilit field]

thanks!

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[twilit field]

here do I assume digits can be repeated?

[twilit field]

nvm

we do

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[last slate]

Does my expression look correct for a quadratic? Was this just an error of not adding 6 decimal places?

[midnight plank]

@last slate Has your question been resolved?

[carmine sigil]

@last slate so the model you have is obviously wrong, because the model should have a maximum, but your a value is positive, indicating a minimum instead at some point

How did you arrive at this model?

[last slate]

I soved for a quadratic by filling in what they gave me from the table

[carmine sigil]

And your calculator gave you a positive value for the a coefficient?

[last slate]

yes

[carmine sigil]

Can you show me?

[junior flower]

[last slate]

yess this is what its showing

[carmine sigil]

That's a negative a

[last slate]

I though a was the coeffiecient

[carmine sigil]

a is the coefficient

It's appearing as negative on your calculator, but you entered it as positive in the answer box

Simple data entry error

[last slate]

Oh okay I see whta you're saying

[carmine sigil]

[last slate]

Thank you for helping!!

.close

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[last slate]

.reopen

[midnight plank]

✅

[last slate]

On part c I'm a little lost. Part d I put 350 since it the value of profit went down.

[split jetty]

you have to use the quadratic formula you got in part b in order to find the max

aka the peak of profit

[last slate]

would I set it equal to zero and solve for x?

[last slate]

You could think of it this way
There are 10 gifts of the same kind which are to be distributed among 3 children
Where each one can get any number of gifts

[split jetty]

not really

[last slate]

Sorry for budging in

[split jetty]

do you know how to do it?

[last slate]

no

[split jetty]

its fine

that function tells you the total gain of money, when x=ticket price

so to find when the function increases or decreases, you need to find the derivative

@last slate did you understand it?

[last slate]

No I'm still lost, like I don't get the formula to plug it into. Cause I would assume to plug it into quadratic then set it equal to zero, but thats wrong

[split jetty]

only if you want to find when they get 0 dollars @last slate

@last slate do you know what is a derivative?

[last slate]

Like a rate of change

[split jetty]

yea

[last slate]

would it invlove the slope theb

then

[split jetty]

for example: if I the ticket is 1 dollar more expensive, will I get more value than just not changing the price

[last slate]

yes...

[split jetty]

here in the table you can see that 300 is the highest (but it could be 1 dollar more or less)

[last slate]

ok

[split jetty]

lets say you sell tickets for 0 dollars, then you will get 0 dollars in total

[last slate]

its making more sense now

[split jetty]

but if you sell tickets for 1000 dollars, you will gain something, but very few people will buy it

[last slate]

That makes sense

[split jetty]

well there is a point in the middle where you gain even more than those 2 cases

and well that point, even if you add or subs 1 dollar, you will lose on some benefit

is it good now? @last slate

[last slate]

Yess it makes sense now. Thank you for explaining!!!🫶

[split jetty]

imagine this is the benefit (very simplified version)

you want to find that peak

[last slate]

Okay gotcha

[split jetty]

the function is: -x^2 + x + 5

so to find the peak, we have to take the derivative

and at the peak the derivative is always 0

[last slate]

Okay so could be a derivative elsewhere with a different value

[split jetty]

the black line is the derivative of that function

[last slate]

aside from the peak i mean

[last slate]

Okay I understand!

[split jetty]

idk is there anything else? @last slate

[last slate]

No, TYSMMM🫶

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[gritty hatch]

uhm...

[gritty hatch]

Ok, so, if I'm guessing correctly.

Is this impossible?

[sweet wing]

impossible how?

[gritty hatch]

I don't see how you could factor that.

[onyx cedar]

you can

[gritty hatch]

I can't even solve one of them after spending innumerable hours thinking about how.

Without a method.

Just trying to factor it raw.

[twilit field]

Quadratic formula

[gritty hatch]

And then I don't even know how this relates to graphs or what roots are.

[gritty hatch]

nope

not allowed

just raw

[sweet wing]

wdym just raw

[gritty hatch]

is there a realistic way you could look at that and just factor it through mental calculation

[shrewd wagon]

(3x-1)(x+2)

[gritty hatch]

bc thats literally what they did

[sweet wing]

why wouldnt you be allowed to use the quadratic formula

[gritty hatch]

cuz they didnt give me it in the video

and they expect you to just solve it

https://youtu.be/GuwofNeT9ok

[onyx cedar]

i jus split up the linear term, seperate them into two brackets, factor out the gce of each bracket and then cross one of them out

[gritty hatch]

see what im really asking is for the narrowing logic they used

this is what everyone seems to not get about what im asking

the only way i could see this being solved is narrowing

like seeing what combinations would end up giving you 5x and 2 while considering the pieces given

without actually calculating anything

sort of like estimation

cuz clearly this isnt possible outright unless you're superhuman

[shrewd wagon]

(3x-1)(x+2) is factored form

[onyx cedar]

so you wanna know how to factor simple and complex trinomials without having to write down calculations?

and just do it as fast as possible?

[gritty hatch]

mentally with no methods like in the video

[summer ravine]

a+b = 5
ab = -6

[gritty hatch]

in the video he seemed like he just magically got all the answers by writing out factors

[summer ravine]

which grade are you in?

@gritty hatch

[gritty hatch]

post school

[shrewd wagon]

you can see that the first term is 3

[summer ravine]

Ok

[gritty hatch]

mmm... oh thats a bother.

Im still not being perfect as I should

[shrewd wagon]

so you can write (3x- )(x- )

[gritty hatch]

that will take an adjustment

Meta Calculations

[shrewd wagon]

the third term is -2 so you can think as -1*2

[summer ravine]

list factors of ca and find a pair that satisfies b = sum of factors.

[shrewd wagon]

this only works if a = 1

[summer ravine]

c*a

[gritty hatch]

hm...

Meta-Flow

[summer ravine]

it works for a = n also.. i'll go now.
For integers it's quite easy. For fractions you can assume them to be of some sort of form.

ok

[midnight plank]

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[novel cobalt]

if x is in set of real numbers, then the limit doesnt exist right?

[hearty rune]

yeah

[novel cobalt]

.close

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[bleak pier]

sin^2(90-18)-3/4

cos^2(18)-3/4

<@&286206848099549185>

[summer ravine]

Yes dear?

,w sin^2(72)-sin^2(60)

[grand pond]

Results provided by WolframAlpha

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Requested by anshulyolo121

[last slate]

Sin ^ 72 = cos ^ 18
Cos ^18 😦 cos 36)/2+1/2

Whole eqn becomes

God damn

Ok basically

Convert sin 72 into cos 18

Then using double angle formula

The whole equation simplifies to

[bleak pier]

@dusky flint

[last slate]

(2Cos 36 -1)/4

[dusky flint]

I see

[last slate]

Ah im on my phone

Ypu guys continue

[bleak pier]

+1\2?

[last slate]

Ignore that

Convert the cos into sin

Then you know the value of sin 60

[bleak pier]

.

[last slate]

Use double angle formula

[dusky flint]

well its pretty simple

[dusky flint]

yeah

true

[dusky flint]

just convert sin2 to cos2

[last slate]

sorry guys I'm on my phone so i can't do calculations 😭

I can but it's like 1 am

[dusky flint]

it becomes cos^2(60)-cos^2(72)

[dusky flint]

same here its 2 here

but its ok i can do it

[dusky flint]

@bleak pier

then u use double angle formula to find cos(72)

therefore cos(72) = 2cos^2(36)-1

now cos(36) = phi/2 = (1+root(5))/4

[bleak pier]

sin^2(72) becomes -cos^2(72)?

[dusky flint]

yeah

1-cos^2(72)

[bleak pier]

i see

got it

[dusky flint]

yup

then ur cos(76) becomes [root(5)-1]/4

now substitute the value

and you have your answer

anything else?

[bleak pier]

where is this came from?

[dusky flint]

well this is a popular ratio we r simply made to remember it in jee but being a math enthusiast i know its derivation n its a little lengthy

[bleak pier]

ohh I see

i only know 30 45 60 90

could you tell me what are others which are helpful?

18 36 15 75?

[last slate]

15,75 ofc

Use sin(60-45) for that derivation

[bleak pier]

ohh got it

[last slate]

Or sin(45+30)

[dusky flint]

you can also find it like this

[bleak pier]

sin60 cos 45-cos60 sin45

[dusky flint]

well i have a whole mind map book for jee

[last slate]

I smell aspirant

[dusky flint]

it has some which r imp n useful

[bleak pier]

ye

[dusky flint]

wait a min imma tell u

[bleak pier]

nice

[dull yoke]

My dark side beast mode is very dangerous...

[bleak pier]

step hen

[dull yoke]

where is your dark side girlfriend

[last slate]

Like

Sin(3×18+2×18) =sin 90

Then sin addition formula

Then triple and double angle formula

[dusky flint]

@bleak pier well i cant find my book rn but i found some useful resource u could go through
https://www.geeksforgeeks.org/trigonometry-formulas-class-12/

well they have good identities and all

it has a good base

[bleak pier]

Tq so much bhai

[dusky flint]

and additionally i would recommend learning how to find those unique angles through patterns

instead of remembering it will give you a better order of thought

for example your 72 degrees could be rewritten as 2(36) = 2(2(18)) = 2(2(90/5))

you could have found it using those formulas above

[last slate]

Just remember sin 18 and cos 36 haha. Others can easily be came up with, with sin,cos(90-36,18)

[dusky flint]

sin 18 can also be derived actually

no need to remember them

[last slate]

Yea but that takes time

[dusky flint]

hmm true well i m good at quick calculations then haha

[last slate]

If I'm in an exam and start deriving sin 18, im 101 percent dead lmao

[dusky flint]

haha then where do i belong lol

i have bad memory power so i derive formulas in exams

and also create methods to solve problems without those formulas during the exams

[last slate]

damn

[dusky flint]

yeah i have a 1000x brain power in solving logical questions but 0 for memory

[last slate]

that's coool

[dusky flint]

🥲

not cool at all

i can do maths and phy

[last slate]

Yea but

[dusky flint]

but as a jee student i need chem as well

but i cant remember even a bit of it

[last slate]

you're in 12th?

[dusky flint]

[dusky flint]

yeah

[last slate]

im in 11th :]

[dusky flint]

oh nice

u prepping for anything?

[last slate]

jee

Lmfao

[dusky flint]

oh make sure to have good grasp on basics of every concept

especially organic chemistry or u gonna cry like me later

[last slate]

we haven't even started organic yet :]

[dusky flint]

oh damn

[last slate]

just mole concept thru chemical bonding for now

[dusky flint]

yeah its the first month i see

where u frm?

[last slate]

uhh country or state or what?

[dusky flint]

state/city ig

country ofc ind

[last slate]

punjab

near punjab

[dusky flint]

ohh

[last slate]

Around

Yk

[dusky flint]

i m from gujarat

[last slate]

nicee

[dusky flint]

yup

well u should ask your teacher to teach you interconversion formulas for concentration terms for mole concepts

they r really useful

for solutions ch in 12th

[last slate]

yea they taught us that

[sage olive]

yoo fellow gujarati spotted

[last slate]

Still haven't remembered

[dusky flint]

ohh i see

[last slate]

I just converted on spot

[dusky flint]

yoo kem cho

kya thi?

[dusky flint]

nicee

[sage olive]

technically america lol

but my family’s gujarati

[dusky flint]

aree haha

hu surat thi

i am oswal jain

[last slate]

bhai ye kya collab hai 😭😭

[dusky flint]

its gujarati lol

[last slate]

so anyone suffering from chronic depression

I know I will be

[dusky flint]

everyone after taking jee

i m srsly worried abt my jee percentile after knowing i get 0s on organic

and 90s and 100s on maths

and around 60-70 in phy

[last slate]

I am loving physics rn man

nlm and wpe

[dusky flint]

haha cuz its beginning

[last slate]

Oof

Sexy

[dusky flint]

i also felt the same at first

but its crying later

rotational motion

and many more things make u cry

even in 11th

[last slate]

I've heard myths about rotational motion

Can't believe round stuff caused so many problems

[dusky flint]

haha its because trigno was built

[last slate]

🚲🚲💿💿🚵‍♂️🚵‍♂️ boo

[dusky flint]

initially everything was calculated as straight lines or figures

but when wheel was made

it made everyting complicated

[last slate]

The problem started in 3000 bc

[dusky flint]

because now after you can do smtg called rectiilinear movement

so mathematicians developed trignometry

[last slate]

Nd vectors

love vectors

[dusky flint]

nah that got made a lot later

[dusky flint]

haha u gonna cry while applying it to integration and derivations for deriving formulas later

basics are easy

applying it to harder concepts is tougher

[last slate]

Like the WET or something?

[dusky flint]

WET?

[last slate]

Work energy theorem

Just did it yesterday

[dusky flint]

nah its pretty simple

[last slate]

Right 🙂

[dusky flint]

there are derivations more complex

[last slate]

I'm cooked

Medium rare

[dusky flint]

well u r technically

but lets just say if u have a good base n good analytical brain u can somehow survive

and u also need good memory power

good enough to remember around 1000 reactions of organic chem

[last slate]

interesting

Hmmmmmmmmmmmmmmmmmmm

im gonna

ok

possible

Can I friend you?

[dusky flint]

okk try then

[dusky flint]

yeah sure

[last slate]

hey i love tech n swimming as well

[dusky flint]

oh niceeee

i love them

[midnight plank]

@bleak pier Has your question been resolved?

[midnight plank]

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[terse mesa]

need help on b

i think im braindead

[small jasper]

It's just asking for (number of factors of 24 that are odd)/(number of odd numbers)

[terse mesa]

wouldnt that be 7/6

anyone

[terse mesa]

what is ur answer and how?

[small jasper]

how did you get this

[terse mesa]

sorry, i looked at all numbers in F and intersection

over all numbers in 0 and intersection

but idk what other answer it can be, ik 7/6 wont even work cause pb isnt like that

but what is the answr

and how?

[small jasper]

edited my original for clarity

tho you should also note that 13, 14, and 15 aren't in the venn diagram

[terse mesa]

oh yeah

so 2/8?

1,3/1,3,5,7,9,11,13,15

[small jasper]

simplify it but yeah

[terse mesa]

yh 1/4

luv bro

god bless

[small jasper]

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[terse mesa]

.close

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[steady vigil]

Hello, I need help with solving this Augmented Matrices. It’s my first time and I’m confused on what of the 3 Elementary Row Operations to use.

[steady vigil]

I already started partially with help from someone but I had to leave.

[fallen vigil]

What operation did you do

[steady vigil]

I did this: Add a Multiple of One Row to Another Row: Replace one row with the sum of that row and a multiple of another row.

I subtracted 2 from the 2nd row getting 0 in the bottom left hand column.

[fallen vigil]

Im assuming you tried to do 2nd row - twice the first row ? If so your calculation is off

[steady vigil]

What?

I subtracted 2 from the first row.

[fallen vigil]

Ok yeah you can’t just subtract constants

[steady vigil]

Ok,

[fallen vigil]

Like you said you replace rows with sums of one row with a multiple of another row

[steady vigil]

I’m confused about the rule though.

[fallen vigil]

You are correct that you want to get the bottom left element as 0 but you have to subtract a constant multiple of an entire row to do this

[steady vigil]

Ok, don’t you have to do one of the three Elementary Row Operations to get 0?

[fallen vigil]

Yeah

[steady vigil]

Which are: Interchange Rows: Swap the position of two rows in the matrix.
Multiply a Row by a Non-Zero Constant: Multiply all elements of a row by the same non-zero value.
Add a Multiple of One Row to Another Row: Replace one row with the sum of that row and a multiple of another row.

The only one that works is the last one.

[fallen vigil]

Yup. So you want add a constant multiple of a row to the 2nd row, since there are only 2 rows you will have to add a constant multiple of the 1st row to the 2nd row

[steady vigil]

Oh so you add that element and a coefficient from another number. Is that correct?

[fallen vigil]

Yeah and that coefficient should be constant across all elements of the row

[steady vigil]

Aren’t none of the coefficients constant since the x and y coefficients vary on each row?

Could I continue this later? I have to go.

[fallen vigil]

Yeah

[midnight plank]

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[low finch]

ok i hate this

[neat hamlet]

look at each term

look the the numerator compared to the denominator of each term

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[twilit field]

Simplify the following statements. Which variables are free and which are bound? If the statement has no free variables, say whether it is true or false.

$\left{5\notin\ \left{x\in\mathbb{R}\right|13-2x>c\right}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so firstly, I feel that both x and c are free variables

now I know this is wrong

the thing though is , intuitively, both x and c can vary, right?

[midnight plank]

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[lusty cave]

Can someone help me to solve this I dont know where to start

[shell berry]

think of the graph of the absolute value of x

[lusty cave]

hmm

[shell berry]

,tex
$
\begin{cases}
|x|=-x, x<0\
|x|=x, x>0
\end{cases}
$

[grand pond]

lILi

[lusty cave]

what intervals do I find

[shell berry]

does this look familiar

[lusty cave]

oh

ig

[shell berry]

this is a transformation of |x|

[lusty cave]

i see

]ok

[shell berry]

to flip it upside-down, turn it into -|x|

[lusty cave]

so how do i determine what Y equals

[shell berry]

they're asking for the function

so y= some transformation of |x|

[lusty cave]

😭

[shell berry]

not an exact value

do know how to transform functions?

[lusty cave]

no

[shell berry]

alright give me just a moment

Use the sliders to look at how each variable effects the function

https://www.desmos.com/calculator/imzxu0rbka

fixed

s changes the vertical scale
a changes the horizontal offset
b changes the vertical offset

[shell berry]

this function is mirrored across the x axis, so we would use -|x|. it's offset one to the right and one down, so you would end up with -|x-1|-1

,tex $y=-|x-1|-1$

[grand pond]

lILi

[shell berry]

in the case of this software, it would be y=-abs(x-1)-1

[lusty cave]

ahh ok

thx

.close

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[twilit field]

nvm

.close

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[twilit field]

.reopen

[midnight plank]

✅

[twilit field]

why Can't I solve this by equating the two functions?

so $x^3+ax+1=x^4+ax^2+1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

if they share a common root, the above will hold true

so that gives $x=-a$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

just drop a hint please

what am I missing

[last slate]

i think its wrong, cus having a common root doesnt mean the functions can be equal

[twilit field]

oo

$x^2=-a$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is only possible at a=0

[twilit field]

if they share a common root f(c)=g(c)

at that point

[twilit field]

is this it?

[hard shard]

so whats your answer

[twilit field]

a=0

so only one value

[hard shard]

are you sure?

[twilit field]

oh

a can be negative]

right

but $\infty$ isn't the answer, is it

[grand pond]

ƒ(Why am. I here)=I don't Know

[hard shard]

make sure theyre solutions to the original equation

s

[last slate]

oh yea true

mb

[twilit field]

I'm getting x^4+ax^2+1. evaluvated at x^2=-a is 1

[hard shard]

so that means?

[twilit field]

there' no such $a \in \R$

[grand pond]

ƒ(Why am. I here)=I don't Know

[hard shard]

but alas there is

[twilit field]

0

[hard shard]

think about where you may have discounted a solution

[twilit field]

hmm

x=1 is a valid solution

for f(x)=g(x)

[hard shard]

so is x=0, but its clear that you end up with 1=0 again

[twilit field]

hmm, so one thing I know is $a$ has to be negative

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

did I discount a solution by canceling out the 1s?

nah, that's silly

[fleet moss]

what about cancelling out x

[cobalt sigil]

Real roots or any roots ?

[twilit field]

real roots

[twilit field]

yeah, that's why I factorised

$x^3(1-x)+ax(1-x)=0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$(x^3+ax)(1-x)=0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[cobalt sigil]

So a = -2

[fleet moss]

mhm and then

[twilit field]

x=1

[cobalt sigil]

Thatz one possible solution

[fleet moss]

you can factor further

[cobalt sigil]

Not required you see it will be complex

[twilit field]

$x(x^2+a)(1-x)=0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

if a is negative

then$x^2=-a$ will have infinitely many solns

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

BUT subbing $x^2=-a$ back into the OG equation doesn't give us 0

[grand pond]

ƒ(Why am. I here)=I don't Know

[cobalt sigil]

So we get only one value x=1 and a = -2

[twilit field]

how did you get$a=-2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[cobalt sigil]

Putting x =1 in eqn 1 or 2

[hard shard]

try the other solutions to this

[cobalt sigil]

x³+ax+1=0 or 1+a+1=0 or a = -2

[twilit field]

ah

That should have been obvious to me

thanks

[hard shard]

should be more obvious now

[twilit field]

it is

[hard shard]

youll be quicker the next time you see it

[cobalt sigil]

,w plot y = -(1+x³)/x and y = -(1+x⁴)/x²

[twilit field]

yeah, but if I can't think of such simple solutions, i'mma fail uni once I start it

[grand pond]

Results provided by WolframAlpha

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

Requested by itzmecairo

[hard shard]

this isnt simple and uni is far easier

do not sweat it

[cobalt sigil]

uni is easy ?

[hard shard]

some of my engineering classmates did not know what a dot product was

[cobalt sigil]

Hey bro what to choose pure applied or dicrete ?

[hard shard]

pure ftw, but also do whatever makes you happy

[hard shard]

tbf it was more a case of not applying it properly

[twilit field]

thanks !

.close

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[twin ridge]

Anyone able to point me in a direction for this homework? I missed this day of class since I was out of town.

[twin ridge]

From what I can see, the functioning definition I need here is if a sequence is bounded and monotone, then it is convergent.

And then to find the limit, I need to figure out if it's increasing or decreasing, then the limit is the sup or inf

Admittedly I do not have very much experience with recursive formulas, but I would hazard a guess that the lower bound of this function is 2 and the upper bound is 8

I know a lower bound is 2, I don't know if it's the greatest lower bound though.

[rose trout]

Well knowing the bounds is enough for monotone convergence.
Can you show it's increasing or decreasing?

[twin ridge]

I know it has bounds. It looks like it's decreasing by plugging in a few indices. How do I know if 2 is an infimum though?

Normally I'd just plug in a really large number to see where it's trending but I can't quite do that with a recursive formula.

[wary thorn]

i'd just compute its limit

and then given that its monotone and bounded

it converges to its limit

[twin ridge]

How do you compute the limit of a recursive forumla?

That's the part I'm stuck on

[rose trout]

Let's say we start by knowing it converges.

[wary thorn]

then take lim of both sides

[rose trout]

Then we take the limit on the LHS

And on RHS

x_{n+1} tends to L

so does x_n

[wary thorn]

🤝

I'd use L instead of x personally

[rose trout]

Good idea

[twin ridge]

Do we know it converges?

[wary thorn]

we'll show that later, using monotone sequence theorem

but this way, you don't have to show any infimum nonsense

[twin ridge]

The infimum is defined as the limit of the sequence

What does "tends to L" mean?

[wary thorn]

yes but its easier to compute the limit than it is to prove something is the infimum

[wary thorn]

it means that the limit is L

suppose x_n -> L as n->infinity

[frank wolf]

This terminology kind of confuses me

[wary thorn]

then we say x_n tends to L

[frank wolf]

Because not every sequence has a limit

[wary thorn]

What's the point of bringing up this nonsense?

this sequence does have a limit

[frank wolf]

Yes, but the way you phrased it is a little confusing

[twin ridge]

A monotone sequence has to have a limit, no?

[frank wolf]

No

[wary thorn]

bounded

[frank wolf]

Or at least, not a finite limit

[frank wolf]

You said “first, we compute the limit, and then since the sequence converges, it converges to its limit”

[twin ridge]

Well this one's the first question on the homework so I'm assuming they're not trying to trick me quite yet

[frank wolf]

This doesn’t really make sense to me

[wary thorn]

I haven't been able to start my explanation yet

but what I mean is start by assuming it converges

[frank wolf]

Right

[wary thorn]

to
"L"

L = 1/2 (L) + 2

[frank wolf]

Mhm

[twin ridge]

I understood what you were saying, but that's cuz I'm looking at the book.

[frank wolf]

That I’m fine with

[wary thorn]

then you can justify that it does actually converge

using MST

[twin ridge]

Is the left side "xn"?

[wary thorn]

x_(n+1)=1/2 x_n + 2

lim n-> inf x_(n+1) = lim n-> inf 1/2 (x_n) + 2
(assuming the limit exists)

(and we denote whatever it converges to as L, so...)

L = 1/2 L + 2

now solve for L

[twin ridge]

Is the limits for the xn terms both L because xn+1 is just a tailed version of xn?

[wary thorn]

yes

[twin ridge]

Ok

So L=4?

Wait

How does that work

[wary thorn]

How does what work?

[twin ridge]

How is L=4?

[wary thorn]

Why do you think that is is not 4?

[twin ridge]

I plugged in 3 terms and got down to 5.

[wary thorn]

okay

did you ever get below 4?

[twin ridge]

I haven't gone any farther yet

[wary thorn]

haha

[twin ridge]

Decimals are weird.

[wary thorn]

well you wont get below 4

its monotonically decreasing to 4

well, it remains for you to show that its monotonically decreasing

[twin ridge]

So 8,6,5,4.5,4.25,4.125 etc?

[wary thorn]

perhaps it does something like that

i haven't computed anything

I just have faith in the limit you computed already

[twin ridge]

So I could say that 4.5 etc is 2*2+r where r<1

So I could just say the 1/2xn term never goes under 2?

This isn't quite relevant to the problem, just trying to justify in my head.

[wary thorn]

perhaps it'd be easier to show the sequence is bounded below by 0

I do think you're getting side-tracked

[twin ridge]

Well it's definitely bounded below by 2, since x_n can't be negative.

[wary thorn]

very well, then that works

[twin ridge]

So that proves it is bounded.

[wary thorn]

so now show its monotone decreasing

and you're done

[twin ridge]

I gotta figure out how to do that

[wary thorn]

show that x_(n+1) is less than x_n

[twin ridge]

Do I have to use induction?

Seems like a logical step, I am just not a fan of havign to do it lol

[past mantle]

what kind of math you guys doin?

[wary thorn]

Yeah maybe do induction

[twin ridge]

Analysis

[past mantle]

what kinda of math is it

[twin ridge]

It's mostly proofs based.

[wary thorn]

it shouldn't be bad here

[twin ridge]

The first half of my class was just proving why algebra works.

[wary thorn]

Tag my sir @ rbmuk if you have any more doubts, he will clear them

[twin ridge]

Why 1=1, why 5*(-1)=-5, that kind of thing.

[past mantle]

oh ok

[twin ridge]

Now I'm in the section of the class where I'm proving calculus

[past mantle]

i don't know calculus

[twin ridge]

Does anyone really? Lol

[past mantle]

i passed geometry

luckily enough

[twin ridge]

So base case is x1=8 obviously

I took Geometry last semester 🙂

[past mantle]

i took it for my junior year

[twin ridge]

I took it in high school but my college requires it for math majors again.

[wary thorn]

Show that x2 is smaller than x1

[twin ridge]

Not a bad idea though honestly.

[past mantle]

damn

[wary thorn]

Is the base case

[twin ridge]

Gotcha

x2 being 6

[past mantle]

you guys seem cool and i might have to call on you guys for math

[twin ridge]

And then I have to show that x(n+1)<xn?

To be fair, this entire discord is full of cool people.

[past mantle]

yea indeed

[twin ridge]

I haven't really met anyone that had any ill intentions in here.

[past mantle]

im just vibing for the summer before my senior year

[twin ridge]

Nice. I wish I was. I was stupid and decided to take 4 summer classes.

[past mantle]

im about to graduate early in december

[wary thorn]

now suppose that x_(i+1) < x_i for some i>1

[past mantle]

i need 24 to graduate, i have 20.25 so im not to far off

[wary thorn]

use it to show that x_(i+2) < x_(i+1)

[past mantle]

always show your work

[twin ridge]

This only works on paper assignments. Sometimes college teachers give online assignments.

[past mantle]

true

[twin ridge]

Generally a good practice though.

[past mantle]

yea

[wary thorn]

I was going to take 3, but summer classes are expensive. For whatever reason, I can take 12 and 17 credits for the same price in the fall, but the difference between 6 and 11 credits in the summer is like $2000, so I dropped one of mine for summer quarter.

[past mantle]

there to expensive

[wary thorn]

they're

too

[twin ridge]

Having a slight brain fart. If $x_{i+1}<x_i$, then that meas that $\frac{1}{2}x_{i+1}+2<\frac{1}{2}x_i$?

[grand pond]

Narutoes

[past mantle]

im so lost

[twin ridge]

I got summer FAFSA so it went from 5K to 1.3

[wary thorn]

I'd double check what x_(i+1) is equal to.

I'd just double check all of that work there 😓

you had the right idea though

[twin ridge]

I'm a senior in college, so this stuff won't quite get to you until you do start.