help-17

[lethal heart]

Last question

But what would be the definition that to find the sup I can also increase it with a quantity that no longer depends on x, so that this quantity is valid for all x

That is, the quantity depends only on k after

[versed bane]

well i am not sure exactly what you mean. do you just want to look at a particular x value? that just gets you a sequence of real numbers. it would be helpful if you clarified what you meant by “increase it”

[lethal heart]

I'll send an example

[versed bane]

that’s just a sequence of bounded functions

like if f_k(x) = the left hand side

[lethal heart]

The first is the sequence of functions, the second is the limit function, I decreased this quantity with 1/k, so the sup tends to 0 as all the x's vary

[versed bane]

wait this is a uniformly bounded thing. what you’re describing is uniform convergence.

like this is an example of a sequence of functions that is uniformly bounded

[lethal heart]

Yes

What does it mean to do this thing

[versed bane]

by “mean” do you want an interpretation or like how it fits into the big picture?

[lethal heart]

Where it is written that if I decrease with something that no longer depends on x then the sup is that for every x

[versed bane]

the sup didn’t take x in as an input

in the first place

[lethal heart]

Wdym

[versed bane]

“the sup is that for every x” doesn’t make sense to say

sup here didn’t depend on x, it’s defined for functions and intervals

not a particular x value

[lethal heart]

Example pls

[versed bane]

yeah, take the function f(x) = x and i wanted to find the sup of the function. What input would i give the sup? not a specific x-value, but an interval like (0,1)

[lethal heart]

Sup is a real number right?

[versed bane]

yes, it takes in a function and an interval and outputs a real number

[lethal heart]

I don't remember exactly if it could also be +∞

[versed bane]

or infinity yeah

[lethal heart]

But in the first case isn't it infinite? (f(x)=x)

[versed bane]

well, it depends on what the domain of f was. but if you let the domain be all of the reals then yes the sup would be infinity

but if i say “f was defined on the interval (0,1)” then the sup of f on that interval would just be 1

but in any case, notice that the sup depended only on the function and its domain, and NOT on any particular x value

[lethal heart]

There is the sup is the largest element when all the x's vary

[versed bane]

yes, it’s the largest possible output (sort of)

sometimes you can’t quite get there, but it’s the least upper bound of that set

it still only depends on the function and its domain

[lethal heart]

Example $x^k in I=[0,1]$ the sup is $x^{k}-0$ so it would be 0?

[twin meteor]

sofficino

[lethal heart]

No wait

f(x)= 1 if x=1

[versed bane]

yeah, so the sup is the max output, which is 1

[lethal heart]

But it wouldn't be 1-1=0

?

Where am I doing wrong

[versed bane]

what wouldn’t be?

the supremum?

[lethal heart]

If x is 1 the limit function is 1 so sup=1^k-1=0

[versed bane]

i see what is happening here i think

[lethal heart]

But it doesn't converge uniformly, so my reasoning is wrong but I don't know why

[versed bane]

so the supremum takes in only one function. so for EACH k you need to get a supremum for |x^k-1|

also remember absolute value bars when you do this btw

so you are correct that the pointwise limit is f(x)=1

[lethal heart]

Isn't it always 0??

[versed bane]

oh yeah sorry, the pointwise limit is 0

it’s late and i’m tired mb

but okay the supremum of x^k for a fixed k is 1 though

[lethal heart]

No wait maybe you're right

[versed bane]

for example the supremum of x^2 is 1

the supremum of x^5 is 1

and so on

[lethal heart]

But don't you subtract the limit function?

[versed bane]

i skipped a step in subtracting the pointwise limit function because it didn’t affect us. here is the thing that we are computing in a series of steps:

first we subtract the x^k function and the limit function (and take the absolute value if necessary

THEN take the supremum of that, and let k go to infinity.

if we get 0 then the sequence converges uniformly, and if we don’t then it doesn’t

so let’s do the x^k example. we know that the pointwise limit is 0

[lethal heart]

May I know the steps for this example pls

[versed bane]

the x^k example?

we can do that yeah

[lethal heart]

But it is 0 if x≠1

[versed bane]

what is

[lethal heart]

The limit function is 0 if x lies between [0,1) and is 1 if x=1

[versed bane]

yes

[lethal heart]

0???

[versed bane]

well, we can say it’s this function if you want

[lethal heart]

Ok

[versed bane]

i was just gonna do x^k with domain (0,1) instead

so that the limit function is nicer

[lethal heart]

Okay

[versed bane]

it won’t change the final result i promise

[lethal heart]

Okay

[versed bane]

but in this case then you agree that the limit function is actually 0

[lethal heart]

Yes

It's as if it were an exponential with base <1

[versed bane]

okay, so now we’ll take x^k and subtract 0, with abs bars if necessary

for an arbitrary fixed k

[lethal heart]

Using which definition?

If there is

[versed bane]

what i mean is take the function f_k(x) = x^k and subtract the 0 function (which was the limit function), just like usual function subtraction

[lethal heart]

Ok

[versed bane]

we get back x^k again

[lethal heart]

Yes

[versed bane]

okay now we take the supremum of that function on the interval (0.1)

[lethal heart]

Isn't it 0?

[versed bane]

it’s the max of x^k

on the interval

that’s how supremum is defined

[lethal heart]

In (0,1)?

[versed bane]

yeah

[lethal heart]

For every fixed x the limit is not always 0

?

[versed bane]

no limit

just fix a k

like pick a number i

[lethal heart]

6

[versed bane]

okay

so what is the max of x^6 on (0,1)

[lethal heart]

There is no maximum

[versed bane]

yeah, but what is the least upper bound

[lethal heart]

1

[versed bane]

okay, so that’s the supremum of x^6 on (0,1)

that is what i mean and that’s the standard terminology

[lethal heart]

For every k it is always 1

[versed bane]

yep

[lethal heart]

But

Wait

[versed bane]

i can make a remark about why we are doing it this way

if you want

[lethal heart]

To get to this result, I had to necessarily see it as [0,1) otherwise I could never find it

Otherwise I would have found 1-1=0

[versed bane]

why? how are you getting 1-1?

there was no subtraction anywhere except for subtracting 0

[lethal heart]

If I saw it as [0,1]

However I understood

[versed bane]

okay, i’ll make a remark about it at the end. let’s keep going for now.

we note that sup x^k over (0,1) = 1 for any k

[lethal heart]

Yes

[versed bane]

okay now we let k go to infty

[lethal heart]

After subtracting the limit function

Which was 0

Okay

[versed bane]

so lim_k to infty of 1 is 1

[lethal heart]

Yes

[versed bane]

which is not 0, so the x^k are not uniformly convergent

[lethal heart]

Yes

[versed bane]

okay. so time for some remarks. do you know about metric spaces

?

[lethal heart]

No

[versed bane]

okay, a metric space is just a set equipped with a distance function

[lethal heart]

Yes

d(x,y)=|x-y|

[versed bane]

what were doing when we take the uniform convergence is we’re equipping the set of functions with a distance function (namely “how far apart CAN THEY POSSIBLY get”) and then determining if that sequence converges with respect to that distance function

[lethal heart]

I just saw something

[versed bane]

so in this example, it was possible to get far apart even as k went to infinity, so the distance between x^k and 0 didn’t go to 0

now, the second remark: even if you consider x^k on the closed interval [0,1], you still get that sup|f_k-f| = 1, where f is now the more complicated pointwise limit function

[lethal heart]

Wdym

[versed bane]

here

[lethal heart]

Yes

In what sense a complicated function

[versed bane]

but even still sup|f_k-f| was 1, so the distance between those functions was 1

[lethal heart]

This i mean

[versed bane]

eh maybe it’s not that much more complicated

the point of this uniform convergence stuff is to say either “yes the functions actually look like they’re approaching the limit” or “sure they’re doing it pointwise but it doesn’t really seem like the functions themselves are taking the shape of the limit function as k goes to infinity”

same with the equicontinuity, equilimited, etc stuff as well, but with different statements like that

[lethal heart]

Function sequences tend to become the limit function

[versed bane]

only sort of. if the convergence is uniform it is even better

the shape approaches the other shape quicker in a sense

[lethal heart]

Is it uniform?

[versed bane]

really think about how the shapes are changing. the stuff near 1 really didn’t want to change in the x^k example. i think some sleep will help this concept now. i am too tired to say more about it because i stayed up way too late.

[lethal heart]

And yes, because for every x the distance is less than epsilon, with a fixed epsilon

?

Anyway, thank you very much for this great help

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[next meteor]

what did he do here

in between

[winter thistle]

seems like they combined ln|x| + c into a single ln, then took the exponential of both sides

$\ln \left| \frac{y-1}{y} \right| = \ln|x| + \ln(e^c) = \ln(e^c|x|)$

[twin meteor]

lgkoo

[next meteor]

what do you mean

can i do that ?

[winter thistle]

then taking exponentials on both sides

[winter thistle]

as long as c is real yh

[next meteor]

then he just called it lncx

[winter thistle]

yh, he wrote e^c as c

which is not good notation imo

[next meteor]

and that canceled the ln

[winter thistle]

yh

[next meteor]

and it became cx

[winter thistle]

the one thing I'm not sure though

[next meteor]

where c is actually c

[winter thistle]

is that you can just drop the modulus on both sides without more info/context

[next meteor]

now it's just ln y-1/y=cx

what now

where do i got from here

[winter thistle]

?

how did you get that

[next meteor]

does the e here cancel the ln

[winter thistle]

no

you have to take exponential on both sides

[next meteor]

wym

[winter thistle]

do e^(...) to both sides

to get rid of ln

[next meteor]

but

[winter thistle]

because $e^{\ln(t)} = t$

[twin meteor]

lgkoo

[next meteor]

how'd he do ln e^c in the first place then

[winter thistle]

you can re-write any constant c as $c = ln(e^c)$

[twin meteor]

lgkoo

[winter thistle]

or $c = e^{\ln c}$

[twin meteor]

lgkoo

[winter thistle]

exponential and logarithm are inverse of each other

[next meteor]

ok thanks

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[steep vigil]

I am a bit confuse about this type of exercise

Should It be like the first or the second

Do you mean to solve it?

OK give me 1 min

[stone gazelle]

I suggest you sketch the region R, then it will be obvious which of these is correct

[steep vigil]

Is this R?

@stone gazelle

[stone gazelle]

well, the top line is y=2x not x=2y

[stone gazelle]

There are two ways of integrating over this region. Either scanning a horizontal line vertically, or scanning a vertical line horizontally

Notice that scanning a horizontal line means you would need to split the region, since the bottom side is made up of 2 equations. So it is better to scan a vertical line across

This means dy first then dx

[steep vigil]

ok, think I get it

[stone gazelle]

Can you tell which of these is right now then?

[steep vigil]

second?

ok thx

i will close channel now

!close

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[split cradle]

Guys help me with it

[hallow plover]

,rotate

[split cradle]

Well the area with ink is x

Any ideas

?

[hallow plover]

Idk, looks sadistic

probably start by loging it or something

[split cradle]

Well

What is( log5x)³ is

Can we break it or something

<@&286206848099549185>

[scenic ravine]

well, maybe take log base 5 on both sides

and then sub $log_5(x)=y$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[split cradle]

What sub means

[tough valve]

Substitute

[split cradle]

?

Oooo

Then it will be y³

Is it

[scenic ravine]

you'll then get 1$6y^3-68y+16=0$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[split cradle]

Wait

[scenic ravine]

now find the product of the roots

[split cradle]

Were this 16 came from

Logging both side?

[scenic ravine]

yes

[split cradle]

Let me do ot

It*

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@split cradle Has your question been resolved?

[split cradle]

@heplers

<@&286206848099549185>

Still not getting

Help me out

Here is the question again

[astral pilot]

,rotate

[astral pilot]

take log_5 of both sides

which will give
$$(\log_5{x})(16(\log_5{x})^3-68\log_5{x}) = -16$$

[twin meteor]

Pro_Hecker

[astral pilot]

substitute y = log_5{x} for y

[vast shale]

@astral pilot i will be lucky if u will check ur dm btw sorry for bothering u

[split cradle]

Guys I did logging but I don't know where to proceed

<@&286206848099549185>

<@&286206848099549185>

Guyzz?

Help me

[hollow lark]

dont ping multiple times

[subtle wadi]

Yo Kishan

I got the answer

Kishan

It's 5

I hope that works

#help-17

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[quartz beacon]

Inside a circle of radius 1, n circles of radius rn are drawn so that these circles are tangent to each other and are also all tangent to the large circle. In the figure below, this situation is drawn for some values of n as an illustration. Which of the following relationships holds between radius rn and n? You may assume here that n ≥ 3.

[vocal sleet]

Please don't occupy multiple help channels.

[quartz beacon]

:/

It’s a diff question

Why would this be wrong?

Answer is D but is there a way to get there w my thing

[scenic ravine]

is this an olympiad question ?

[quartz beacon]

kinda

[scenic ravine]

might want to ask in the olympiad server

[quartz beacon]

Where do I find the link?

[scenic ravine]

there was a link here somewhere

just a min

#competition-math message

[quartz beacon]

Ok tyty

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[fresh cave]

Follow up question to #help-29 message, it is not needed as context though.

Does "(...) T> (...)" mean the same thing as " '(...) -> (...)' is a tautology"?

Let T> denote tautological implication and (...) just be some arbitrary formula

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@fresh cave Has your question been resolved?

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@fresh cave Has your question been resolved?

[fresh cave]

Do ping me if you reply btw

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@fresh cave Has your question been resolved?

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@fresh cave Has your question been resolved?

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@fresh cave Has your question been resolved?

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[crisp ermine]

how would we go about finding interval of convergence for $\sum_{n=2}^\infty\frac x{1-n}$

[twin meteor]

Vѳrtєx-

[river minnow]

Is the x supposed to be to the power of n?

[crisp ermine]

no

just x

its harmonic series-esque but the 1-n is throwing me off

[river minnow]

You might as well put it in front of the sum for the case when it's nonzero

[crisp ermine]

oh ok

$x\sum_{n=2}^\infty\frac 1{1-n}$

[twin meteor]

Vѳrtєx-

[river minnow]

You can flip the sign and then compare to the harmonic series

[crisp ermine]

oh i guess $\sum_{n=2}^\infty\frac 1{1-n}=1+x+x^2+\cdots$

oh it starts at n=2 but yeah

[river minnow]

No, that's not what I meant

[desert hornet]

This is just wrong

[crisp ermine]

oh no it definitely doesnt

sorry lmao

so you compare $-\sum_{n=2}^\infty\frac 1{1-n}$?

[twin meteor]

Vѳrtєx-

[river minnow]

1/(1 - n) = -1/(n-1)

So you should have 1/(n-1) in the sum

And then compare that series to the harmonic series

[desert hornet]

Reindexing the sum to start from n=1 would directly give you the harmonic series

[crisp ermine]

oh i see

i suppose you can also write $x\sum_{n=2}^\infty\frac 1{1-n}$ as $x\sum_{n=1}^\infty\frac {-1}{n}$ now that i think about it

[twin meteor]

Vѳrtєx-

[crisp ermine]

cool thank you!

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[vast shale]

HELP

[vast shale]

ok so uh

[torn timber]

yea

[vast shale]

not rlly a problem specifically

[torn timber]

k

[vast shale]

but is there any trick to

factoring any quadratic equation

that is more efficient

than doing it manually?

no, i don't think

than the A/C method or other traditional methods

[vast shale]

google!!!!!!!!!!!!!!

[heavy yoke]

well the only method that works 100% of the time is the quadratic formula, but that can be a fair bit more calculation than just guessing (at least, if the factors are integers)

[torn timber]

well u can practice to recognise patterns

[vast shale]

tell me more

[torn timber]

do more problems

that helps in it

[vast shale]

ik but what type of patterns lol

i can solve most that are a=1

[torn timber]

lol

[vast shale]

in my head in like

4 secs

im talking abt where a= like 4,5,6,7

[torn timber]

it will be better with time

[vast shale]

ok

thx

[torn timber]

slove more factorising problems it will help alot

[torn timber]

welcome

[vast shale]

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[fresh tide]

why is it not correct?

I've tried this too and its no good

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[spring tapir]

Question two

Is that absolute value of t

Or is it outer measure

T is a scaler? So it's absolute value

[sharp lynx]

t is scalar so it would be absolute value yes

[spring tapir]

Oh okay so in the no trival case

Where both t and a are non zero and finite

I'd take the infinitim of the sum of open sets that cover a

Because I can move abs of t out of the summation it become

|tA| <= |t||A|

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[golden pecan]

Hello. I would like to get the expression on top ( $5\sqrt{4x}$ ) to be written without the squareroot. However I'm not sure how to go about with this, I know what I've done is wrong

[twin meteor]

Reuben

[heavy yoke]

you can split the square root: $\sqrt{4x} = \sqrt{4} \cdot \sqrt{x}$

[twin meteor]

cloud

[golden pecan]

Oh right, I forgot 🤦‍♂️. Thanks, I have no idea why I've been getting exponents messed up these days

[wet tulip]

5x2x x^1/2

[golden pecan]

It's always the easy simplifications asw, I just correctly simplified this and now I did this

[wet tulip]

u forgot to sqre the 4

[golden pecan]

mhm

thanks

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[violet compass]

Second one

[violet compass]

Is this correct

Cause im not getting answer

.clos3

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[elder scaffold]

u solving for x?

[shy olive]

yes

[elder scaffold]

first multiply both sides by x-5, since we'd like to remove the dominator

[formal yew]

Before you start

set parameters

What can x not be equal to

[shy olive]

I don't understand how to multiply it like that

as in change the equation to x²+x+x-150 ÷ x+x-25 =1 ?

[formal yew]

no

x^2+x-30=1(x-5)

[shy olive]

ohh i get it now

so if you multiply the left side by itself it cancels out??

[formal yew]

If it’s on the denom

ye

[shy olive]

ohhh thank you

[formal yew]

like 10 x 1/10

is just 1

same as (x-5) x 1/(x-5)

Is just 1

[shy olive]

omg thank you

what next?

[formal yew]

x^2+x-30=1(x-5)

equal to 0

x^2-25=0

[torn timber]

u get x=-5 as x=\5

[formal yew]

no just -5

5 is extraneous

[shy olive]

wait so when multiplying the numerator by x-5 do you just subtract it??

[torn timber]

yes thats what i meant

[vast shale]

wdym subtract it

are u talking about after you get

x^2 + x - 30 = x - 5

[shy olive]

ya

[vast shale]

and the n subtract x

ah

[shy olive]

and then it becomes x^2 -25

[vast shale]

yeah just cancel out the x's and solve for x^2

yessir

so that's x^2 - 25 = 0

you'll get two values of x, but you need to plugboth of them into the original equation to see if they work

[shy olive]

ohh ok give me a second to try and work it out

[vast shale]

go ahead

[shy olive]

wait so what do I do with the right side

1(x-5)

[vast shale]

yeah

you'll get

x^2 + x - 30 = x - 5

you already simplified

the right side becomes 0

[vast shale]

subtract both sides with x

x^2 - 30 = -5

add 5 to both sides

x^2 - 25 = 0

[shy olive]

let me write that down

[vast shale]

yeah go ahead

[shy olive]

wait so will it be 20-25=-5?

and sqr of 20?

[vast shale]

no

how di d u get this

[shy olive]

because 20 - 25 = -5 so I assumed then x must be the square root of 20

[vast shale]

wait show me all of your steps from the beginning

[shy olive]

to be honest i didn’t really understand what was going on so my working out might not make sense

then -5 + 25 + 20

[vast shale]

it's not x^2 - 25 yet

it's x^2 - 30

[shy olive]

oh

let me redo that

[vast shale]

on the top you wrote
x^2 + x - 30 = x - 5

and ten u subtracted x

so it should be
x^2 - 30 = -5

[jagged sky]

Who know trigonometry

[vast shale]

open a help channel #❓how-to-get-help

[shy olive]

ohh i got confused because before the person was saying that when u multiply by x-5 it becomes x^2 - 25 =1(x-5)

but i got messed up

sorry im not that good at this

[vast shale]

issok

[shy olive]

okay so how do i multiply both sides by x-5?

i genuinely don’t understand

ok wait i got it nvm

is it x^2 + x - 30 = 1 (x - 5) ?

ohh and then subtract x-5 from both sides?

so x^2 - 25 = 1?

wait nvm

subtract x?

so x^2 - 30 = -5

<@&286206848099549185>

[torn timber]

yes?

[shy olive]

i think the person hhelping me is offline

[torn timber]

this step is right

[shy olive]

sorry it is taking so long 🥶🥶

for me to get it

[torn timber]

np

[shy olive]

ok so do i add -5?

[torn timber]

we all start at diff points

[shy olive]

so x^2 -25 ?

[torn timber]

yes

now see from this eq we can kknow x^2 is less than 30 as in rhs is negative

now we add +30 in both sides

we get x^2=25

[shy olive]

oh!

[torn timber]

now sqrt on both sides

[shy olive]

so it’s either 5 or -5?

[torn timber]

we get x=5,-5

but

we must know x is not 5

[shy olive]

hhow do we know that

[torn timber]

do you remember that u multiplied x-5 on both sides?

[shy olive]

ya

[torn timber]

we do that assuming that x is not equal to 5

or else it will break math

in this case

[shy olive]

ohhhhh

because then it will equal to 0

and can’t divide by 0

[torn timber]

yea

see u get it

[shy olive]

thank you soooooooooo much

[torn timber]

welcome 🙂

[shy olive]

now i can be the smartest in my class 😉😉😉😉😉😉😉😉

well bye for now

[torn timber]

lol if u practice then u got a shot

[shy olive]

i will practice 👌👌👌👌

👋👋👋👋

[torn timber]

good practice improves

[shy olive]

.close

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[blissful panther]

how did the "yellow highlighted" part get cancelled out, I don't get it

[blissful panther]

The objective here is to simplify

[sullen shoal]

What is x^2 - x^2?

[blissful panther]

0

[sullen shoal]

And what is left

In

[blissful panther]

Nothing but where did you get x^2 - x^2

[sullen shoal]

1 + x^2 - x^2

[sullen shoal]

The yellow oart

Part*

[blissful panther]

oh wait so the yellow part, 1 remains

[sullen shoal]

Yes

U got it ig?

[blissful panther]

I get it, didn't notice it

thank you very much!

.close

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[inner cove]

anyone can help me w this function if the results is gonna be higher than 21 000 at some point ?

[inner cove]

they do an equation like this but i don t understand why ( affirmation 2)

[true grail]

well, they want to find if f(t) exceeds 21, right?

and since e^3 is smaller than 21, then in order to exceed 21 f(t) has to exceed e^3

the equations are them showing that f(t) doesn't do that

[inner cove]

yeah but i don t get how they do it w e^3

like why even

the sign changes as well > why

[inner cove]

and like e^3 is originally in the function

@true grail

[true grail]

they're examining different parts of the function

so they've taken the fact that $e^{-0,5t^2 + t + 2} > 0$ bc $e^x > 0$ for all x

[twin meteor]

Out Of Nosh

[true grail]

and then using that to say that if that expression is subtracted from any other, it will necessarily be smaller

it might help to think of $f(t)$ as $e^{3} + (-e^{-0,5t^2 + t + 2})$

[twin meteor]

Out Of Nosh

[true grail]

find your own channel. #help-28 is free

[inner cove]

how do ı know that e^3 is smaller / bigger than the other part

ı m so confused

[true grail]

I'm not sure i understand

[inner cove]

sorry my brain ain t working

[true grail]

you're not trying to find whether e^3 is bigger/smaller than the other part

you only need to find (for affirmation 2) whether the whole expression is bigger than 21

[inner cove]

21 000

so how do ı do that

[true grail]

well, f(t) works in units of 000s anyway

[inner cove]

ohhhhhh

[true grail]

because of this bit

so there's more than 21000 bacteria if f(t) > 21 for some t

[inner cove]

okay that s already smth

and so how do ı prove that

ık the language and i still can t figure it out LMAO

[true grail]

well, first step is to find e^3

just use a calculator for that

[inner cove]

yeah

ohhhh

and so 20 - whatever can t be more

is it that logic

[true grail]

i think so? I don't understand perefectly what you're asking

sorry for not speaking french

[inner cove]

no worries lmao

[inner cove]

and so just why do we only start by one part of the function here

and put smaller than 0

[true grail]

well, it's the only part that changes

e^3 is fixed, after all

think of it like 8 - sin(x)

if you wanted to know if that could go over 8,8, you'd start by saying that -1< -sin(x)<1 and then using that to show that 7 < 8-sin(x) < 9

we're doing similar, but we don't care about the lower bound

and it's exponential rather than sinusoidal

do you see @inner cove

[inner cove]

and i can do that w whatever thing that is fixed?

[true grail]

i don't understand

[inner cove]

like whatever thing that has a fixed value

[true grail]

hold on, let me start over

[inner cove]

no no ı got it from here

[true grail]

so you know how e^x can't go lower than 0 for any real x, right?

[inner cove]

yes

so e^x > 0

then

[true grail]

which means that if you subtract e^x from something it can only get smaller

here the "something" is e^3

so e^3 - e^x is always less than e^3

which, in turn, is less than 21

[inner cove]

yeah but what if the whole other expression is smth that is added

is it impossible

[true grail]

what?

here's the thing: you can just exchange x for -0,5t^2 + t + 2

[inner cove]

yeah

[true grail]

-0,5t^2 + t + 2 is always real if t is, and t is always real

so all this stuff holds, and f(t) = e^3 - e^{-0,5t^2 + t + 2} is less than 21 for all t

[inner cove]

ohhhhhhhhhhhh

that s logic

okay thanks and last question

[true grail]

mm?

[inner cove]

the sign > changed to < because ?

knowing that e^x can t be <0

ohhhh

multiplied by -1 ?

[true grail]

yes

[inner cove]

it s very slay but way too compliclated

[true grail]

or subtraction of the whole expression, if you prefer

[inner cove]

well thannks for putting up w me lmao

u helped

[true grail]

💜

[vocal sleet]

@inner cove Has your question been resolved?

[vocal sleet]

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[scarlet bough]

how would i do this question? i rearanged x=6sin(t) so its t = arcSin(x/6) but if i sub that into y i dont get it in the form they ask for

[dull bear]

See the parametric integration comment I made the other day, $\int y \dd x = \int y \dv{x}{t} \dd t$

[twin meteor]

@dull bear

[scarlet bough]

oh right, i forgot about that formula

[dull bear]

Of course, limits and stuff and all

[dull bear]

(It’s basically an application of the integration by substitution thing, so there’s that )

[scarlet bough]

so i got 30sin(2t)cos(t), but its still not in that form

[dull bear]

Much better, can you change anything in there?

[scarlet bough]

oh ye double angle

didnt realise they dont have 2t

[dull bear]

Yea there you go

[scarlet bough]

are the bounds in terms of t already?

[scarlet bough]

wait they are right

[dull bear]

You’d need to think about the bounds a little bit in general though

[dull bear]

The bounds of the original, in terms of x, are 0 and 6; you want the lower t bound to correspond to x = 0 and the upper t bound to correspond to x = 6 (but of course, as 6sin(t) is increasing between 0 and pi/2, it’s easy here)

In other examples it may be more subtle, but, cross that bridge when we get to it, tl;dr “treat it like any other integration by substitution problem”

[scarlet bough]

i just thought about y=0 so arcSin(0)=2t=0 and then (pi-0)/2 = pi/2

[dull bear]

Yeaaa that works in this case at least

[river kettle]

charttttttttttttttttttttttttttttttttttttt

[dull bear]

Water beeeeeaaaaammmmmmmm

[river kettle]

[scarlet bough]

sorry to interupt guys but now im trying to intergrate it and im doing it by parts. i got $$u=sin(t) , \frac{du}{dt} = cos(t)$$ $$\frac{dv}{dt}=cos^2(t) , v=(0.25sin(2t)+0.5t)$$ and now i have $$60[(0.25sin(2t)+0.5t)(sin(t)-0.25 \int (sin(2t) +0.5t)(cos(t)) dt]$$ how would i intergrate the 2nd time?

[twin meteor]

morphine_addiction

[dull bear]

Why did you choose u = sin(t) though

[scarlet bough]

whats the difference?

[dull bear]

(see the reverse chain rule section)

[scarlet bough]

i try my best to avoid that at all costs

[dull bear]

Oh wait are you doing it by parts-

Ouch

With that second integral it’s not very “nice” without e.g. cancelling out the work you’ve done or doing more than you need to

[scarlet bough]

what would be the best way to intergrate it?

[dull bear]

Though the reason they give it to you in the form they do is because it should be immediately apparent that if you e.g. do it by substitution, setting u = cos(t) or by recognition

(Incidentally, there are product-to-sum identities, noting yesterday’s conversation, but unfortunately they’re not given in the formula book )

[scarlet bough]

so the best way to do it would be $60 \int sin(t)u^2 dt$?

[twin meteor]

morphine_addiction

[dull bear]

Yep, though remember what happens when you change variable too!

[scarlet bough]

i think im overcomplicating it again but i got -cosec(t) du = dt, arcCos(u) = t, so should it be \int sin(arcCos(u)) * u^2 * -cosec(t) du? this def dosnt seem correct

[dull bear]

Yeeeeaaa overcomplicating it a tiny bit remember when you do integration by substitution you want to find du/dx (or dx/du) and rearrange so you can replace dx with [something] du

In this case, u = cos(t), and t plays the role of x, so you can find du/dt and then rearrange that to get a nicer integral

(You should ideally also change the integral limits, though not super essential tbf, provided you change everything back if you don’t)

[scarlet bough]

ohh du is - sin(t) dt ao i can just sub in -du into sin(t)

so i got 20 as my final answer

[dull bear]

Nice

[scarlet bough]

ty again for the help

[vocal sleet]

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[toxic birch]

How would you solve this

I can't figure it out

[scenic ravine]

x=cos(u) probably

[toxic birch]

What's the thought process?

[errant briar]

x= cos2t should be better i think

[toxic birch]

I've only seen trig sub be used when there is a square under a sqrt

But ill try doing it this way

[scenic ravine]

(1-cos(2x)=2sin^2(x)

[toxic birch]

I c

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[hollow radish]

help

[hollow radish]

with implicit diffrentiation

[vocal sleet]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[hollow radish]

find dy/dx
lny^2 + cos^2x=1-y

i derived it and got an answer of

(1/x)y^2*2ydy/dx -2sinx=-dy/dx

[astral pilot]

$\ln y^2 + \cos^2 x= 1 - y$

[twin meteor]

Pro_Hecker

[astral pilot]

Is this correct?

[hollow radish]

yes

this is the equation

i need to find dy/dx

[astral pilot]

How did you get 1/x?

[hollow radish]

d/dx = ln = 1/x

.close

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[toxic holly]

I need some help with permutations, powerindex and a few other things

[toxic holly]

help

with banzhafs and shapley-shubiks power index

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[scenic ravine]

consider a 3 by 3 matrix A satisfying $A^2-5A+7I=0$

[scenic ravine]

if $A^5=aA+bI$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

find a and b

so $A^5=A^4A$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[merry python]

hi

f(why am I here)=I don't know

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

hi

or

$(5A-7I)^2A=A^5$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[signal pendant]

Yes

[scenic ravine]

now this can further be simplified to give

[merry python]

you must escape the matrix🐺 🔥

[scenic ravine]

$A^5=(25A^2+49I-70A)A$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

which is again

[steep finch]

matrix algebra suksssss

[sudden compass]

Crazy

[scenic ravine]

It does, this is simple but I made some calculation mistakes twice

so I'm posting it here

[sudden compass]

Nvm got it

[scenic ravine]

as I work better blah blah

[merry python]

gather all your 17 uncles and form a tower to escape the matrix🐺

[scenic ravine]

or $A^5=(25(5A-7I)+49I-70A)A$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

ok, this is getting out of hand

I'm having to sub in A^2 multiple times

anyway around this?

[sudden compass]

Why are u doing that

[scenic ravine]

doing what?

[sudden compass]

Oh

U need a and b

[scenic ravine]

yes

is there no way around this?

if not, I'll just abandon this problem IMO

[sudden compass]

Unlikely

[sudden compass]

Also this is easy to simplify

[cobalt crypt]

,w divide x^5 by x^2 - 5x + 7

[scenic ravine]

yeah

[cobalt crypt]

smh

,,<polynom>, \polylongdiv{x^5}{x^2 - 5x + 7}

[sudden compass]

Bot dying

[scenic ravine]

the bot can do this?

wow

[sudden compass]

U can do this too

Divide x^5 by the quadratic

That will give u x^5 in terms of the quadratic (hopefully)

Then u just sub in

[steep finch]

WHAT THE HECK BOTS CAN DO THIS

[scenic ravine]

there's a reminder here

that;s going to make it messy

[cobalt crypt]

the remainder is the answer

[hallow plover]

bots big w right here

[scenic ravine]

wait, what

[sudden compass]

Huh?

[cobalt crypt]

you're trying to find A^5 mod A^2 - 5A + 7

[sudden compass]

What no

[cobalt crypt]

the answer is 149A - 385

[astral apex]

how do i do this cool stuff with the bot i didnt know that u couldndo that

[scenic ravine]

how so?

[cobalt crypt]

A^2 - 5A + 7 = 0 no?

thats what it means to mod A^5 by A^2 - 5A + 7

[sudden compass]

It is?

[sudden compass]

That makes sense

[sudden compass]

Ohhhh

I get it now

[scenic ravine]

ah

thanks

.close

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[scenic ravine]

this is just an application of the factor therom ?

[cobalt crypt]

no i would rather say its an application of division

just as how if 5 = 0, then 37 = 2 mod 5

[scenic ravine]

ah

ok

[scenic ravine]

5=0

what?

[cobalt crypt]

if you set 5 = 0, then 37 = 7*5 + 2 = 2

this happens when you mod 5

[scenic ravine]

oh

okay

so this is modular arithmatic basically

[cobalt crypt]

well yes

[scenic ravine]

thanks

.reopen

[vocal sleet]

✅

[scenic ravine]

I still don't follow

how can you set 5=0

[sudden compass]

Its congruence

She's using = to denote equivalent

[scenic ravine]

hmm

[sudden compass]

And also I think they forgot to add mod 5

[signal pendant]

Polynomial long division will allow us to write: A^5=(A^2-5A+7)(Q(A))+P(A) (P must be linear, or constant). Since A^2-5A+7=0,
A^5=P(A)=Linear in A, which us what you want

[scenic ravine]

wait, how does this conclude A^5 is linear in A

[signal pendant]

I edited it. I hope its more clear now

[scenic ravine]

OO

yeah

stupid of me to have missed that

sorry

thanks a lot

[signal pendant]

👍 this is just equiv. To looking at A^5 mod A^2-5A+7, which is what snoseeph had said

[scenic ravine]

got it

thanks

.clopse

.clsoe

.close

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[feral light]

Can someone explain to me coneptually what a triple integral even mean

like I know that a double integral calculates the volume of the surface

under the surface

but what is a triple integral? does it calculate a 4th dimension object under a volume??

[twin horizon]

Maybe this might help https://byjus.com/question-answer/what-does-triple-integral-represent/

[stiff dawn]

integrate one thing then integrate again

[feral light]

This doesnt rly help me understand 😭

[feral light]

what

[stiff dawn]

ye

it's just simple

[feral light]

what are u saying

[stiff dawn]

..

[feral light]

.....................................................

.clsoe

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[tardy wind]

Can algebra and indices co-incide with each other?

[tardy wind]

for example, perfect squares and stuff

[grim lotus]

yes

[tardy wind]

Ahh all good then thanks

Don't need to add unnecessary stuff to my study guide

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[steep vigil]

I discover the region

But I am not sure how should I write the integral

[potent gorge]

first off do your boundaries

[steep vigil]

Whats the correct variant.

[potent gorge]

neither, try to imagine the regions they trace out

[steep vigil]

I am not sure I understand

I am quite new to the subject

And I have a lack of knowledge

[potent gorge]

aight

so the x ranges from -2 to 2

that’s correct

but if you sketch it out

the y goes from x^2 to 4

[steep vigil]

So I mistake the region?

[potent gorge]

yeah

also are you allowed to switch dxdy to dydx in the problem?

[steep vigil]

I think yes

[potent gorge]

aight

[steep vigil]

Can you draw the region?

[potent gorge]

sorry, your drawing is right

just your numerical boundaries are wrong

[steep vigil]

What should the boundaries be ?

[potent gorge]

x is between -2 and 2
y is betwen x^2 and 4

[steep vigil]

That s make sense

Yes is because the y is on to of the x^2

Thanks a lot

Can I show you another one?

Just to be safe

@potent gorge is this good?

The region is between y=0, x=pi/2,y=rad x

[potent gorge]

yup!

[steep vigil]

Thanks

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[south ember]

Hello, good afternoon, I need help with this combinatorics exercise please. I don't know how to approach the exercises.

Ex.20) The elevator of a building carries 10 passengers and can stop at any of the 12 floors of the building.
a) Distinguishing between people:

1. in how many ways can the 10 passengers descend?
2. In how many ways if on the 10th floor exactly 3 people descend?
3. In how many ways if at most one passenger descends on each floor?
   b) If no distinction were made between people, what would be the answers to the questions posed in a?

[dusty timber]

a)i think 12^10

Each of the 10 passengers can choose any of the 12 floors to descend. Since the choices are independent, we can use the multiplication principle

b)binomil coef thing

(10)
( 3) ,The remaining 7 passengers can descend on any of the remaining 11 floors, which can be done in 11^7

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[south ember]

yes, is correct

[south ember]

I have another answer from this

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[south ember]

<@&286206848099549185>

[onyx flax]

I think we have the solutions to the question, we may have added some restrictions/assumptions on your question though

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[warm horizon]

im kinda confused with using implicit diff

i got the equation but im not sure what that really means tbh

i got x(dx/dt) + y(dy/dt) = 0

[sullen shoal]

dx/dt is rate of slipping of bottom

dy/dt is rate of slipping of the top

u basically got the whole formula

u gotta substitute values

@warm horizon

[warm horizon]

alr

so for part a do i make dx/dt the subject?

sorry no dy/dt is the subject right

and dx/dt = 2cm/s? and x = 200?

[sullen shoal]

yep

wait

yea dy/dt

if he gave one of them is 2 u can find the other one

cuz u know the equation

dx/dt is given

so ull get dy/dt

@warm horizon

[warm horizon]

ok

i got dy/dt = 2/15 * (sqrt15)

and for the second one i think its 14/15 * (sqrt15)?

also why dont u change the units ?

cos for part a the rate of change is in cm/s but the distance is 2m

i would assume you would have to convert rate of change to m/s or distance to 200cm

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[sand grove]

How to find p?

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[sick tide]

hey guys how do you show that any 3 non-zero vectors which are orthogonal one to each other are linearly independant? I pretty sure it includes constructing a matrix and proving it for one case then using the 'without loss of generality' but im not too sure from there, i tried and can send the actual question if anyone needs

[sick tide]

sorry i am unsure how to rotate it

[mild flower]

,rccw

[sick tide]

thanks!