help-49

[twilit field]

$g(x+h) - g(x) = \int_{a}^{x+h} f(x) dx - \int_{a}^{x} f(x)dx$
\
$\frac{g(x+h) - g(x)}{h} = \frac{\int_{a}^{x+h} f(x) dx - \int_{a}^{x} f(x)dx}{h}$
\
$g'(x) =\frac{ \int_{x}^{x+h} f(x) dx}{h}$

[grand pond]

A dense set(Ping when reply)

[twilit field]

Is what I've done so far

now to procced I suspect I'll need MVT

not sure how to apply it though

[zealous schooner]

g' is the limit of that expression as h tends to 0 (if it exists)

[carmine void]

You should get that the integral is h•f(a_n) for some a_n in (x,x+h)

[hard umbra]

that's assuming you know the derivative

which you don't

you must use the continuity of f to conclude

your notation is very poor for this purpose

[twilit field]

As f is continuous, the integral is between min(f) (h) and max(f) h

but that doesn't really help much

$min(f)h \leq \int_{x}^{x+h} f(t) dt \leq max(f)h \implies min(f) \leq\frac{ \int_{x}^{x+h} f(t) dt}{h} \leq max(f)$

[visual tiger]

also I wouldn't write f(x)dx

[hard umbra]

that is what i mean by poor notation

it will confuse you

[visual tiger]

when x is the point where you're trying to differentiate on

[twilit field]

okay, so ftdt

[grand pond]

A dense set(Ping when reply)

[visual tiger]

min and max of f taken on which domain?

[twilit field]

I'll now have to prove that $min(f(x),f(x+h)) =f(x) = max(f(x), f(x+h))$

[grand pond]

A dense set(Ping when reply)

[twilit field]

as h tends to zero

[hard umbra]

uh no

[visual tiger]

wut

first of all

how did you reduce min(f) to min(f at two points)?

and even so, why should the min be equal to f(x)?

[twilit field]

right

[visual tiger]

(same questions for max)

[twilit field]

well, it's the min and max on the interval [x,x+h]

[visual tiger]

ok

do we agree that for any h > 0, it won't be equal to f(x)

so what do we expect to use now?

[hard umbra]

(maybe it might)

[visual tiger]

generally xd

[twilit field]

I feel I need to squeeze it somehow

[hard umbra]

squeezing this way will not be fun

you are missing some steps that make your life much easier

[visual tiger]

well you need to prove min(f) on [x,x+h] converges to f(x) when h-> 0

same for max

but you could have done it alternatively

for example finding |(g(x+h)-g(x))/h - f(x)|

[twilit field]

well, for that i have to prove $\lim_{h \to 0} f(x+h) = f(x)$ no?

[grand pond]

A dense set(Ping when reply)

[visual tiger]

again, is f(x+h) the only point that matters in [x,x+h]?

[twilit field]

no

$f(x+(h/2))$ matters too for instance

[grand pond]

A dense set(Ping when reply)

[visual tiger]

yes

[twilit field]

so $f(x+a) = f(x) \forall a \in [0,h]$ is what what i need

[grand pond]

A dense set(Ping when reply)

[visual tiger]

no

that would require f to be constant on an interval to the right of x

not all continuous functions are like that

[hard umbra]

i think constant functions are like that

maybe (g(x+h) - g(x))/h - f(x) might be clearer...

continuity is gonna sneak into this somehow

[visual tiger]

yup

[twilit field]

hmm

[carmine void]

Using the bounds you gave and the continouity of f you're showing the lemma I suggested to use.

[twilit field]

I don't understand this

[visual tiger]

you already computed that $\frac{g(x+h)-g(x)}{h} = \frac 1h\int_x^{x+h}f(t)dt$

[grand pond]

rafilou is not not born in 2003

[twilit field]

yes

[visual tiger]

subtract f(x) to that

[twilit field]

$g(x+h) - g(x) = \int_{a}^{x+h} f(x) dx - \int_{a}^{x} f(x)dx$
\
$\frac{g(x+h) - g(x)}{h} = \frac{\int_{a}^{x+h} f(x) dx - \int_{a}^{x} f(x)dx}{h}$
\
$g'(x) =\frac{ \int_{x}^{x+h} f(x) dx}{h}$
\
$g'(x)-f(x) = \frac{1}{h} \int_{x}^{x+h} f(t) dt - f(x)$

[grand pond]

A dense set(Ping when reply)

[visual tiger]

"g'(x)" is again dubious

there's some h in there that is unwarranted

and we don't even know if g'(x) exists since we haven't proved differentiability yet

[twilit field]

the statement says differentiable on (a,b)

[zealous schooner]

Also g' is the limit of that ratio as h tends to 0

[hard umbra]

g might not be

you must prove it is

[visual tiger]

that's literally what YOU have to prove

using something that remains to be proven in its own proof is... circular

[hard umbra]

okay so we'll have to prove the min and max go to f(x) seems good 👍

i didn't give the bounds tho

[twilit field]

okay, so

$\frac{g(x+h) - g(x)}{h} - f(x). = \frac{1}{h} \int_{x}^{x+h} f(t) dt - f(x)$

oops

[visual tiger]

_

[carmine void]

Yeah, just wanted to mention that I didn't intend to use differentiability to show differentiability

[grand pond]

A dense set(Ping when reply)

[hard umbra]

on the same page now

[visual tiger]

ok, any way to put the f(x)... inside the integral?

[twilit field]

hmm

[hard umbra]

(just do it)

[twilit field]

instead subtract $\int_{0}^{x} f(t)dt$ from both sides?

[grand pond]

A dense set(Ping when reply)

[hard umbra]

hint: $\ds f(x) = \f1h\int_x^{x + h} f(x) \dd t$

[twilit field]

what

how

[hard umbra]

just evaluate it and see

[visual tiger]

how does a constant get integrated

[twilit field]

oops

yeah

makes sense

[visual tiger]

is \f an existing latex abbreviation or is it a personal-made command? I still don't know how to do those

[hard umbra]

it's mine

[twilit field]

hmm, okay

ykw, I'm kind of lost, I think I'll do some examples first, and then prove it?

[visual tiger]

not sure if examples will show you the proof

[hard umbra]

this proof is probably one of those you get more used to as you do a bunch of different proof examples in real analysis

doing calculations won't really elucidate it

because the idea itself is quite abstract

[twilit field]

This is for an introductory calc course

😭

[hard umbra]

where proofs are dispensed with

[twilit field]

This is the book's proof

[hard umbra]

blackboxed for future study

[twilit field]

It felt very handwavey so I asked here

[hard umbra]

there's definitely some detail to be filled

[hard umbra]

this is the meat

one proves this much more carefully in a real analysis class

[twilit field]

So for now I just go with this, right

[hard umbra]

sure why not

[twilit field]

okie

thanks

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[rapid tapir]

is the integral over (a, b) and [a, b] the same? if so why?

[midnight plank]

@rapid tapir Has your question been resolved?

[patent glen]

I dont think so, i believe the first interval might be a variation of ⟨a,b⟩, where it includes all the numbers between a and b, without including a, b

While the other one [a,b] includes both a and b

Oh, you were talking about integrals, not intervals, srry

But regarding integrals, i would say it depends on the function, and the values of a,b

eg if the indefinite integral is 1/x

the ranges (0, b), and [0,b] could be different

But for like fully continuous functions, i would say their similarity doesnt matter since the result doesnt change alot

[visual tiger]

if f is defined over [a,b], then if f is integrable on one of them, f is integrable on the other and the integral is the same

the point is that the "mean" over an interval, an uncountably infinite amount of points, won't change due to counting the value of 2 more points

[rapid tapir]

why not? wouldn't (a, b) means that you are excluding a, b from the area?

[fresh sparrow]

{a} and {b} on their own contribute 0 area

[visual tiger]

does removing two thin lines (of thickness = 0) change the area?

[rapid tapir]

no but is it really 0

[visual tiger]

yes

[rapid tapir]

isn't it like infinitely small?

[visual tiger]

the thickness is exactly 0 isn't it

so the area is exactly 0

another way to view it: if you view the integral as the limit of the riemann sum

then no matter if you account for f(a) and f(b)

since the coefficient in front is similar to 1/n

so I could very much add or remove those two

when n goes to infinity, the impact adding/removing f(a) or f(b) goes to 0

so taking the limit, there is NO difference at all

[rapid tapir]

which coefficient

[midnight plank]

@rapid tapir Has your question been resolved?

[visual tiger]

have you seen riemann sums?

[patent glen]

,ask 0.99...

[visual tiger]

(b-a)/n * sum(f(a+k(b-a)))

[grand pond]

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[patent glen]

I think same logic is used here

[visual tiger]

I don't think it's directly related to 0.(repeating9), they just share "sum that increases number of terms at each iteration"

it's the (b-a)/n that makes everything

sure, your f(a) might make a difference at the first iterations, when n = 1, n=2 or other small values

but when n gets big

like infinitely big

no one is gonna notice if you get rid of a single term in the sum

in the 0.(repeating 9) = 0.9 + 0.09 + 0.009 + ...

if I get rid of a single term in this sum

for example 0.009

the result is immediately changed

we would obtain in this example 0.991

[patent glen]

The difference between 0.9... and 1 is basically zero due to the infinite repeating nature of 0.9..., now if you have two intervals r=(a,b), p=[a,b]. You can think of p_1 as (r_1-1).999, so a = (r_1 - 1).99999..., a = r_1 - 1 + 0.99..., a = r_1 technically

So the first element of the interval r is basically very near to the first element of p, to the point where the difference between the two is zero

Now when you have a fully continuous function thats integrable on both the intervals r and p, the idea that a = r_1 basically suggests that integration over any of the two intervals is basically equal

,ask r_1 - 1 + 0.99...

[grand pond]

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[midnight plank]

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[rapid tapir]

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[stable halo]

$\frac{dx}{dt} = -\frac{k}{2\pi} [\frac{1}{x(x-1)}]$

[grand pond]

Mortta

[stable halo]

erm how to solve for t?

[steep hinge]

Separation of variables

[stable halo]

so

x(x-1)dx = -k/2pi dt?

integrate and solve?

[steep hinge]

yes

[midnight plank]

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[safe onyx]

I know that ln(e^x) = x, but it didn't like that either 🤔

[grim vector]

When you have h = int from a to b of f(x) dx then h' is just f(b) - f(a)

[vast swan]

??

That woulb be a number

[safe onyx]

it wants the first part not the second

[vast swan]

Oh

[grim vector]

Oh

Indeed

[safe onyx]

previous for example

you just swap the variables and remove int

[grim vector]

Then i think it just ask you to simplify it

ln(e^x) = x

[vast swan]

3x

No way other answer

Is right

[grim vector]

Soz i read it diagonnaly

[safe onyx]

:(

let me see their video

,, h'(x) = \frac{d}{dx} \int_{1}^{e^x} 3ln(t) dt

[grand pond]

smeagol

[grim vector]

Yes

[safe onyx]

he says do u substitution
u = e^x , du/dx = e^x

,, u = e^x , \frac{du}{dx} = e^x , \frac{dh}{dx} = \frac {dh}{du} \frac {du}{dx}

[grand pond]

smeagol

[safe onyx]

,, h'(x) = \frac{d}{du} \int_{1}^{u} 3ln(t) dt \cdot \frac{du}{dx}

[grand pond]

smeagol

[safe onyx]

,, = 3ln(u) \frac{du}{dx}

[grand pond]

smeagol

[safe onyx]

,, = 3ln(e^x) \cdot e^x

[grand pond]

smeagol

[safe onyx]

so
= 3x e^x (the hw marks this as correct)

I don't fully get it

[safe onyx]

why is du/dx added on the end

ohh I think I see

,, \frac{d}{dx} = \frac{d}{du} \cdot \frac{du}{dx}

[grand pond]

smeagol

[safe onyx]

u sub is needed when you have a function of x as b

thank you all

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[safe onyx]

How can you go from
a - b to (a+b)^-1 ?

[tacit rose]

Multiply by the conjugate

And you'll see that the math for that problem will do that

[grim vector]

Hint : indentity

[safe onyx]

@grim vector i tried (a+b)(a-b) but i'm still confused

[tacit rose]

It's not always going to be 1 in the numerator

It was just a coincidence for that problem to have the math work out to be 1

[tacit rose]

The concept that was applied here was just multiplying by the conjugate

[safe onyx]

so you have 1 / a + b
then you multiply by conjugate
so you have (a-b) / (a+b) (a-b)
but how do you simplify it

sorry i am confused

[tacit rose]

What you have here is what it simplifies to

You just need to do the math

[safe onyx]

oooh in this problem it just happens that

,calc (8721sqrt(3))^2-(10681sqrt(2))^2

[grand pond]

Result:

0.99999994039536

[safe onyx]

it's nearly 1

[tacit rose]

Yes, exactly

[safe onyx]

also with
sqrt(2) - 1 = 1 / sqrt(2) + 1

,calc (sqrt(2))^2 - (1)^2

[grand pond]

Result:

1

[safe onyx]

is also one

that was another one that confused me

thank you both

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[carmine tide]

how do i go about factoring x^3 +3x^2 -54 = 0?

[steep hinge]

x=3 is a root -

then just divide it by x-3

or perhaps you could use cardano's formula https://en.wikipedia.org/wiki/Cubic_equation

[dawn dagger]

If you turn it into a depressed cubic equation then yes

[steep hinge]

can't you do it all the times ?

[grand pond]

𝔸dωn𝓲²s

[dawn dagger]

Kinda insane

[steep hinge]

right

[midnight plank]

@carmine tide Has your question been resolved?

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[tepid parcel]

hello can someone help me with this

[tepid parcel]

Here is my working but im stuck

[midnight plank]

@tepid parcel Has your question been resolved?

[tepid parcel]

<@&286206848099549185>

[ripe gulch]

You didn’t square the denominator

[tepid parcel]

for to=he quotient rule i feel like the numerator is too complicated as well is there something wrong with my calculations?

[ripe gulch]

For the numerator you forgot to distribute the -6 to the x

-6(6-x) = -36 + 6x

[tepid parcel]

Like this?

[ripe gulch]

I’m confused about the black and red writing

Why are they different

[tepid parcel]

sorry just forget about the red writng i was just testing like different workings

[ripe gulch]

Actually your calculations look fine the first one

[tepid parcel]

what shouls i do next

im so confuse

[ripe gulch]

When you tested f”(5) you said 6-5=-1

[tepid parcel]

OH WAIT YEAH

SO CONCAVE DOWN IS (-Infinity, 6) right?

[ripe gulch]

Yeah that seems right

I guess it would be (-Infty, 6]

[tepid parcel]

Here is my final workings

Here is my friends

[ripe gulch]

Your original f”(x) was right

[tepid parcel]

The red writing?

[ripe gulch]

This one

The black

I was looking at the red which is why I was confused

[tepid parcel]

Oohh okk

[ripe gulch]

It should be
$-6\sqrt{6-x} - (-\frac{12-3x}{\sqrt{6-x}})$

[grand pond]

parm

[ripe gulch]

It’s not -12 it’s $-(\frac{12-3x}{\sqrt{6-x}})$

[grand pond]

parm

[ripe gulch]

And then you subtract that whole thing

When you got x=8 it was right

[tepid parcel]

oohh okok

man my head is about to burst i havent slept, sorry if i kept on making mistakes, ill close now, THANKS

.close

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[midnight plank]

@grim monolith Has your question been resolved?

[astral garnet]

i dont think you should be finding their mass

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[vagrant mica]

How do I solve this?

[tawdry laurel]

@vagrant mica you can close this

[vagrant mica]

.close

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[elder ivy]

when you graph a derivative and when its positive is it possible for the second derivative to be negative?

[lavish venture]

yep

if f is concave down and increasing then f’>0 and f’’<0

[elder ivy]

ohhh positive is not same as increasing ._.

why do i keep messing that up

[elder ivy]

oh yeah

thx

but i get it now idk what i was doing

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[outer beacon]

need help and on some other questions

[gaunt nimbus]

y + 4 > 2x

y > 2x - 4

the line should be y = 2x-\4

and the area is above that line

[tawdry laurel]

you can graph y = 2x - 4

and use logic

[gaunt nimbus]

is y > 2x - 4

[outer beacon]

a b c or d?

[gaunt nimbus]

a

[outer beacon]

can u help me on this too?

[outer beacon]

yo

[gaunt nimbus]

?

[last heron]

Hmm.

[outer beacon]

can u help me wtih the above image

[last heron]

Bro I'm in the unit right below yours 😭

I would help/

[last heron]

Systems of Equations

Is where I'm at.

[outer beacon]

ohhh

[last heron]

Your in HS I presume?

[outer beacon]

nah ms

8th

[last heron]

Me to.

[outer beacon]

can anybody help me with this

[midnight plank]

@outer beacon Has your question been resolved?

[midnight plank]

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[sharp prairie]

Little bit of problems on these

[sharp prairie]

I had to skip the lesson today for math because i was sick

So i dont exactly understand some of these topics

for the simulation question i got 562.34 for the initial value and 1.78 for the common difference

the other 2 i just5 tried using desmos to try and approximate an answer

(calculator active)

[midnight plank]

@sharp prairie Has your question been resolved?

[lethal path]

,w (-2,11), (0,6), (4,2), (8,1) exponential model

[grand pond]

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Requested by sooouth.

[midnight plank]

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[lethal path]

.reopen

[midnight plank]

✅

[lethal path]

,calc 6.19205 * exp(-0.283359 * 4) - 2

[grand pond]

Result:

-0.0066229486083098

[lethal path]

I don't get the question

if you are supposed to find a model that passes through exactly 2 points

which 2 points are you even supposed to choose

also your question 3 is obviously wrong

the population is 50,000 or less is literally in the question

start by finding the equation

it's y = 1000 * a^(x - 2) for some real number a

(what happens when you sub in x = 2 into this?)

use x = 6, y = 10000 to find a
then sub in x = 4 to find the population on day 4

,w exponential model (2,5.63), (3,8.44), (4,12.66), (5,18.98)

[grand pond]

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Requested by sooouth.

[lethal path]

,calc 2.50384*e^4.05118

[grand pond]

Result:

143.8837315194

[lethal path]

yeah your question 6 answer should be correct though

[midnight plank]

@sharp prairie Has your question been resolved?

[midnight plank]

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[modest fox]

where this came from?

[icy adder]

ig its unrelated to the question

[modest fox]

wdym?

[icy adder]

that second part isn't related to the question i think.. you already got the answer right.. that x = 0 y 0 and z = 0 are the only solution to the equation

[modest fox]

so for homogenous determinent x,y,z is always 0?

[icy adder]

no no

it depends upon delta

[modest fox]

delta here is 40 so how come x,y,z is 0?

[icy adder]

do you know Cramers rule? smth like that

[modest fox]

yeah

[icy adder]

read it again probably that will help... it mentions the conditions for homogenous equations as well

[modest fox]

ok man lemme try again

thx for ur help

.close

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[twilit field]

$\frac{d^2}{dx^2} \int_{0}^{x} (\int_{1}^{sin(t)} \sqrt{1+u^4} du) dt$
\
differentiating once , we get
\
$\int_{1}^{sin(x)} \sqrt{1+u^4} du)$
\
Differentiating again we get
\
$cos(x) \sqrt{1+sin^4(x)}$

[grand pond]

A dense set(Ping when reply)

[twilit field]

Does this seem right?

[hard umbra]

it does

but your latex is atrocious

[twilit field]

wdym

[hard umbra]

i'm not sure if you're also looking at the same picture as i am

because i'm seeing some pretty

the mathematics is fine though

probably

[twilit field]

hmm, okay

[hard umbra]

Calculate [ \dnv 2 x \int_0^x \parens [\bigg] {\int_1^{\sin(t)} \s {1 + u^4} \dd u} \dd t. ]
Differentiating once, we get [ \int_1^{\sin(x)} \s {1 + u^4} \dd u. ]
Differentiating again, we get [ \cos(x) \s {1 + \sin^4(x)}. ]

[hard umbra]

compare the quality of the typesetting ^

[twilit field]

Yeah

I probably need to get a few TeX packages I suppose

thanks

[hard umbra]

this has nothing to do with packages

you can do it all with the default preamble

[twilit field]

.close

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[prime hornet]

Calculate
$$\frac{d^2}{dx^2} \int_{0}^{x} \left(\int_{1}^{\sin(t)} \sqrt{1+u^4} du\right) dt$$

Differentiating once, we get
$$\int_{0}^{1}{\sin(x)} \sqrt{1+u^4} du)$$

Differentiating again, we obtain
$$\cos(x) \sqrt{1+\sin^4(x)}$$

[grand pond]

higher!

[hard umbra]

$$ L

i can't believe you're a $$ user

[prime hornet]

wdym

[hard umbra]

using $$ in latex is a crime

[prime hornet]

I pretty much never use $$

I am an align user

[hard umbra]

the correct thing to say is "i never use $$"

[prime hornet]

:kekhands:

Calculate
\begin{align*}
\frac{d^2}{dx^2} \int_{0}^{x} \left(\int_{1}^{\sin(t)} \sqrt{1+u^4} du\right) dt

Differentiating once, we get
$$\int_{0}^{1}{\sin(x)} \sqrt{1+u^4} du)$$

Differentiating again, we obtain
$$\cos(x) \sqrt{1+\sin^4(x)}$$

[grand pond]

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

[prime hornet]

ack

oops

[hard umbra]

use \[ \]

[prime hornet]

fat fingered

[xyz]

[grand pond]

higher!

[prime hornet]

[\frac{d^2}{dx^2} \int_{0}^{x} \left(\int_{1}^{\sin(t)} \sqrt{1+u^4} du\right) dt]

[grand pond]

higher!

[prime hornet]

Calculate [\frac{d^2}{dx^2} \int_{0}^{x} \left(\int_{1}^{\sin(t)} \sqrt{1+u^4} du\right) dt]

[grand pond]

higher!

[prime hornet]

[hard umbra]

[midnight plank]

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[gusty falcon]

since no other context is given

A, B, C, and D can be anywhere right?

as long as they’re along the circumference of the circle?

the question was to determine whether the area of the quadrilateral ABCD > 40 or not but well we can’t really tell?

[rain wasp]

A, B, C and D can be anywhere, yes

given arbitrary A, B, C, and D on a circle of radius 5, can you determine whether the area of ABCD will always be bigger than 40?

if no, give a counterexample. if yes, then prove it

[viral dagger]

cant we like.. make A,B,C,D arbitrarily close

[rain wasp]

yes, an extreme case is when A, B, C and D overlap

then the area is 0

[rain wasp]

!xy

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[gusty falcon]

ABCD can’t overlap

it’d not be a quadrilateral then

it’s be a degenerate one

[gusty falcon]

that is the actual question? i just wanted clarification

[rain wasp]

how about they are really really close

[gusty falcon]

yeah okay so it can be close to the area of a circle

or close to zero area

40 is not the circle area so it’s not the upper bound for the area of a circle

so the assertion is false

right?

[gusty falcon]

yes

okay then

thanks

.close

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[twilit field]

Can I have a hint as to how I'd go about this

I'd start by consider the set of integral lower and upper bounds

then what

I'm wondering if WOP can help us in anyway here

[elder zephyr]

wop ?

[twilit field]

well ordering principle

[astral bough]

m is not unique

conside x = 1

then 0 <= x <= 1 and 1 <= x <= 2

one of those inequalities would need to be sharp

[twilit field]

oops

We now prove if $x \in \R$ there is a unique integer $m$, such that $m \leq x < m+1$

[grand pond]

A dense set(Ping when reply)

[runic hamlet]

yes WOP works here

[twilit field]

We first assume $x>0; x \notin \Z$

in that case the set of all integral upper bounds are subsets of the naturals, and thus have a least element, let this element be $m$

[grand pond]

A dense set(Ping when reply)

A dense set(Ping when reply)

[twilit field]

so $x<m$.

[grand pond]

A dense set(Ping when reply)

[twilit field]

We now prove there exists $m$ such that $m-1<x<m$

[grand pond]

A dense set(Ping when reply)

[twilit field]

This is equivalent to proving there exists $m$ , such that $\frac{m-1}{m} <x/m<1$

[grand pond]

A dense set(Ping when reply)

[twilit field]

ykw, I'm lost

[astral bough]

Does it not follow directly from the principle?

Define $S = {a \in Z; a > x}$

[twilit field]

Hmm, and by WOP, this set has a least element

[grand pond]

LooseEthics

[astral bough]

Then by the wop there exists $a_{min}$

[grand pond]

LooseEthics

[twilit field]

yup

[astral bough]

Let m = a_min

[twilit field]

so $x<m$

[grand pond]

A dense set(Ping when reply)

[twilit field]

we now assume $m-1>x$

[grand pond]

A dense set(Ping when reply)

[twilit field]

and try to arrive at a contradiction

[astral bough]

m is the least element of S, done

[twilit field]

yes

I'm now trying to prove that $m-1<x$

[grand pond]

A dense set(Ping when reply)

[twilit field]

let to the contrary $m-1>x \implies m>x+1$

[grand pond]

A dense set(Ping when reply)

[astral bough]

you don't need to, that side allow equality

[twilit field]

I'm assuming $x \notin \Z$

[grand pond]

A dense set(Ping when reply)

[astral bough]

why?

[twilit field]

I feel it makes the problem simpler

[runic hamlet]

it doesnt

it changes nothing relevant

[astral bough]

yeah, not really

all that's left now is to prove all this holds when x < 0

[runic hamlet]

x<m and m is the smallest such integer. so x<m-1 cannot hold, so x>=m-1 has to

and thats it

[twilit field]

Why can't $x<m-1$ hold

[grand pond]

A dense set(Ping when reply)

[runic hamlet]

because then m-1 is in S

but m is the smallest element of S

[twilit field]

ah yes

[runic hamlet]

and clearly m-1 < m

[twilit field]

got it

Thanks!

.close

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[carmine latch]

This is the question

Some values are given

And i need to solve for x

I have no idea how to get it

Mayb give some hints on how to solve?

The things in tube are 2 different liquids with different densitys*

[lethal path]

,rotate

[lethal path]

idk how to do the problem; just helping others answer

[carmine latch]

me neither

i have no idea how sin37 could be helpful in any way

nor the density of said 2 liquids

[lethal path]

but looks like some kind of tricky mamometer problem

[carmine latch]

it is*

[midnight plank]

@carmine latch Has your question been resolved?

[patent glen]

Is q_2 the volume of the circle liquid

Or volume of the parallelogram prism thing

[carmine latch]

no

circle liquid

both of them mention the density of liquids

in the tube

i js want

hints

on what to do

not the answer itself

[midnight plank]

@carmine latch Has your question been resolved?

[carmine latch]

someone answer

maybe!?!?!?

[midnight plank]

@carmine latch Has your question been resolved?

[carmine latch]

??????

[weak gate]

u can ping the helpers

[carmine latch]

Oh

Bruh

[weak gate]

idk much

but i think you have to equate the pressures on both sides

and w the help of the angles, u can find the height of the liquid

[ancient plank]

let me try solving

x is the height of liquid 2, correct?

[carmine latch]

Yes

[keen saddle]

The pressure of yellow and blue parts will be equal

Also sin is useful because we only look at the vertical height of the liquids

Not diagnol

[ancient plank]

My answer comes out to be 10cm. I'm not sure if it's correct though.

[keen saddle]

@carmine latch

Still need help?

[midnight plank]

@carmine latch Has your question been resolved?

[long dagger]

equalize pressures

[long dagger]

they are at different levels

you cant say that those pressures are equal

[keen saddle]

Ooo

I was thinking like 1 • g • 9 ⅘ = 1 • g • x ⅗ + 8/10 • g • (13-x)⅗

Is this wrong?

Guess I'll ping you @long dagger

[long dagger]

what the hell is this minuscle fraction

,w minuscle

[grand pond]

Results provided by WolframAlpha

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Requested by _nofluxgiven

[long dagger]

minuscule

forgive me

[keen saddle]

Somehow I can type fractions on my keyboard lol

[steep brook]

lol

[carmine latch]

I am

Understanding

Nothing

Meh ill leave it to a proffessor or something

Sorry for taking so much time to answer

[midnight plank]

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[fossil ember]

Well, I still can't find average of all distances from a point to a polygon 3+D using Green's Theorem, square root makes it too complex and finding distance squared is not an accurate thing, what can I do? This is for draw order of a 3+D drawing engine, anything else that would work? Polygons are planar, closed, not self intersecting, and all straight lines. To mitigate inaccuracy, staying within minimum and maximum distances of all vertices and centroid might be good. What else? X E.

[fossil ember]

In case it helps, I can consider polygons like 2d. X E.

I plan cutters so all shapes will be not intersecting except edges and not intersecting except edges on 2d view plane, and not intersecting planes of each other. I don't have that yet. X E.

I have really efficient and accurate to computer precision cutters and point in polygon algorithms. X E.

Maybe I should try min distance and max distance and keep it within those? How do I do that?

Some things to know, average of all distances should be greater than or equal to distance to centroid, average of vertices distances should be greater than or equal to average of all distances. I think I am correct, am I? X E.

With that I can generally mitigate inaccuracy but is this last one correct?

Should I close and make that last one my question?

.close

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[neat silo]

Given the polynomial ( P(x) = x^2 - 4x + 6 ). Define the polynomial ( Q ) by
[
Q(x) = P(P(x)) - x.
]
The sum of all real roots of ( Q ) is . . .
[
\text{a. } 4 \quad \text{b. } 5 \quad \text{c. } 6 \quad \text{d. } 7 \quad \text{e. } 8
]

[grand pond]

m13

[neat silo]

Expanding gives x^4 - 8x^3 + 24x^2 - 33x + 18

Factoring gives (x - 2)(x - 3)(x^2 - 3x + 3)

Observe that, x^2 - 3x + 3 has a discriminant of -3, which means it has no real roots. Therefore, the only real roots are 2 and 3 and the sum is 5.

Is there another approach to solving this problem, other than just expanding and factoring

[surreal moon]

For a polynomial $x^n+ax^{n-1}+bx^{n-2}+...$, the sum of the roots is $-a$

[grand pond]

SWR

[neat silo]

Well

Gotta be careful with that

The problem asks for real roots, thats why i don't use Veita

Q(x) = (x - 2)(x - 3)(x^2 - 3x + 3),

(x^2 - 3x + 3) does not have real roots. By using, Vieta, it's going to be a mistake.

[surreal moon]

oh all real roots, I missed that

Yeah that's a bit more tricky

[neat silo]

Yea

Expanding works

I'm just looking for another method to solve it

[midnight plank]

@neat silo Has your question been resolved?

[surreal moon]

@neat silo yeah I see no better way. The real roots caveat really screws you

I couldn't dig up any research that relates to that.

There are ways to tell how many real roots a polynomial will have, but you can't get their sum

[neat silo]

How so

I'll just compute the sum them

Knowing how many real roots is crucial

[surreal moon]

How can you compute the sum if you only know how many real roots there are?

[surreal moon]

https://en.wikipedia.org/wiki/Sturm's_theorem

It's definitely not an easier way though

[midnight plank]

@neat silo Has your question been resolved?

[midnight plank]

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[royal venture]

i need help with question 6, genuinely don’t know how to start nor continue. the answers r in blue but idk how to get them

[nova drum]

if f(x) = x^2 then with f(x-b) you want to replace x with x-b wherever it comes up so f(x-b) = (x-b)^2

once you do that the general equation should look more familiar

[royal venture]

but where do i go from there ? i just genuinely have no clue how to do this

[nova drum]

well what you have is a general equation for completing the square, so at x= b you have the minimum and y=c is the minimum if i remember correctly

[royal venture]

am i supposed to multiply (x-b)^2 out

[nova drum]

well if you use the table you can find b and c. Once you've done that you can take another value from the table and use that to find a so say i took x=1 the table says y= 39 so a(1-b)^2+c = 39. You know what b and c are so you can put those numbers in and when you do you get an equation to get a

[nova drum]

look for which row y is the lowest number

[royal venture]

at (-3,7) ?

[nova drum]

yeah

so you have b = -3 and c = 7

its like for (x-5)^2+10 the minimum point will be at (5,10)

[royal venture]

so i can choose any point to find what a is as long as b = -3 and c = 7

[nova drum]

yep

[royal venture]

besides (-3,7)

[nova drum]

yeah exactly

if you choose -3,7 i think youll get a 0 somewhere or smthn like that

[royal venture]

yea

ohhh i see

okay

thank youuu

[nova drum]

np

[royal venture]

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[midnight plank]

@timber arrow Has your question been resolved?

[midnight plank]

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[quaint field]

Is there an intuitive explanation (without using calculus) for why the area under the sine curve from 0 to π/2 is exactly one?

[midnight plank]

@quaint field Has your question been resolved?

[midnight plank]

@quaint field Has your question been resolved?

[west iron]

Probably not :(

[quaint field]

:(

how can it be so hard to justify why something is an integer

[hard umbra]

there's that animation on youtube

https://youtu.be/gO8AwBmQK5Q

skip to mathologer's proof

[midnight plank]

@quaint field Has your question been resolved?

[quaint field]

this is awesome

[midnight plank]

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[upbeat plinth]

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[upbeat plinth]

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[grand pond]

Kenzo

[upbeat plinth]

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[grand pond]

Kenzo

[upbeat plinth]

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[lavish venture]

do \sqrt{}

[upbeat plinth]

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[lavish venture]

and for fractions do \frac{}{}

[lavish venture]

{} not ()

it’s next to the p

where the brackets are

[upbeat plinth]

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[lavish venture]

$\frac{a}{b}$

[grand pond]

knief

[lavish venture]

are you finding y’

[grand pond]

Kenzo

[upbeat plinth]

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[lavish venture]

nice the parentheses next to dy/dx are unnecessary though

[upbeat plinth]

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[lavish venture]

mhm

[upbeat plinth]

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[lavish venture]

so $y = \pm \sqrt{16-4x^2} = \pm 2\sqrt{4-x^2}$

[grand pond]

knief

[upbeat plinth]

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[grand pond]

Kenzo

[upbeat plinth]

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[upbeat plinth]

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[lavish venture]

yep

[upbeat plinth]

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[lavish venture]

how’d you get this

[upbeat plinth]

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[lavish venture]

if it’s explicit you won’t have two dy/dx terms nor will the coefficients of dy/dx be anything other than 1

it’ll just be dy/dx = …

[upbeat plinth]

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[lavish venture]

right so how did this happen

[upbeat plinth]

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[lavish venture]

how is y a coefficient or how is there a 2 and all of that

[upbeat plinth]

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[lavish venture]

right but when you differentiate the side with just y the derivative will be dy/dx

that’s it

nothing else

[upbeat plinth]

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[lavish venture]

the derivative y isnt y dy/dx

[upbeat plinth]

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[lavish venture]

i think you’re missing the point

[upbeat plinth]

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[lavish venture]

suppose we have an explicit function y = g(x) then we will always get dy/dx = g’(x) where g’(x) and g(x) are simply functions of x and x alone

[upbeat plinth]

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[lavish venture]

huh

[upbeat plinth]

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[lavish venture]

i’m not sure what you didn’t understand

[upbeat plinth]

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[lavish venture]

^^^^

[upbeat plinth]

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[lavish venture]

🤔

why is there a dy/dx on the right sir

do you mean d/dx

[upbeat plinth]

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[lavish venture]

ok then correct that

[grand pond]

Kenzo

[lavish venture]

yea sure

[upbeat plinth]

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[lavish venture]

yep

[upbeat plinth]

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[lavish venture]

see this is strange though because we will get two different derivatives since the relation given doesn’t represent a function

are you sure they asked for explicit

[upbeat plinth]

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[lavish venture]

ok then we will ignore the \pm

and only consider the positive root

[upbeat plinth]

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[lavish venture]

,rotate

[lavish venture]

ok

take home test?!??

oh no kenzo

[upbeat plinth]

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[lavish venture]

🤔🤔

[upbeat plinth]

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[lavish venture]

ok

[upbeat plinth]

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[lavish venture]

looks good just simplify a little

[upbeat plinth]

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[lavish venture]

2s cancel

and it should be y’

[upbeat plinth]

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[lavish venture]

but

again

[upbeat plinth]

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[lavish venture]

uhh

[grand pond]

Kenzo

[upbeat plinth]

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[lavish venture]

sure but i don’t like how there’s a pm tbh usually depending on the context we choose which branch we are dealing with

[upbeat plinth]

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[lavish venture]

can i see the graph

[upbeat plinth]

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[upbeat plinth]

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[lavish venture]

yea i mean maybe he wants the pm but tbh id split it to be more clear

[upbeat plinth]

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[lavish venture]

like write it with the + and indicate which branch it’s for and the same for the - branch

[lavish venture]

yea

[upbeat plinth]

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[lavish venture]

the the +sqrt branch vs the -sqrt branch

[upbeat plinth]

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[lavish venture]

no below zero

usnsndn - is below zero

[upbeat plinth]

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[lavish venture]

brother

i’m not referring to y’

i’m referring to y

[upbeat plinth]

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[lavish venture]

i know y’ is positive for the - branch

[upbeat plinth]

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[lavish venture]

ok

[upbeat plinth]

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[lavish venture]

factor out sqrt(4)

[upbeat plinth]

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[upbeat plinth]

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[grand pond]

Kenzo

[lavish venture]

yes then slope intercept form

[lavish venture]

and for y’

[upbeat plinth]

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[lavish venture]

f(x)-f(a) = f’(a)(x-a)

[upbeat plinth]

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[lavish venture]

you use both

[upbeat plinth]

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[lavish venture]

he said find the two tangent lines

[upbeat plinth]

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[lavish venture]

hmm

[upbeat plinth]

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[lavish venture]

do you ever do these yourself first

[upbeat plinth]

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[lavish venture]

because you resort to looking at solutions or getting help without thinking about it

hence why you don’t learn

[upbeat plinth]

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[lavish venture]

then think about it for a while

or look at notes

but don’t look at notes for awhile

problem solving doesn’t consist of immediately just writing the solution every time

[upbeat plinth]

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[lavish venture]

hence why you don’t get much better

[upbeat plinth]

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[lavish venture]

alright

[upbeat plinth]

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[midnight plank]

Channel closed

Closed by @upbeat plinth

Use .reopen if this was a mistake.

[midnight plank]

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[viral dagger]

prove that a set with n positive integers always have atleast one subset that the sum is divisible by n

[viral dagger]

taking mod n seems good here but im not sure how i can continue it

[modern sapphire]

is there some additional constraint on the n integers?

Coz you can always find n primes

[viral dagger]

wait

the sum inside the subsets

[modern sapphire]

Ohh, okay. I see

[midnight plank]

@viral dagger Has your question been resolved?

[lavish gazelle]

mod n can only have n-1 remainders if not divisible and there must be two subsets with the same remainder

[viral dagger]

makes sense

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[patent glen]

Is just the subset {n} allowed

[viral dagger]

n isnt always in the set

[patent glen]

So not consecutive integers, just random

[viral dagger]

yes

[midnight plank]

@viral dagger Has your question been resolved?

[visual tiger]

order the elements x1,...,xn

and let Sk be the sum from x1 to x_k

there are n such sums in total

if one of them is 0 mod n, you win

if none are 0 mod n, pigeonhole principle...

[viral dagger]

php would be this, but i dont understand what that can do

or rather, there must be atleast 2 with the same remainder

[exotic pelican]

say you have 1, 2, 3, 4, 5, 5, 6 mod 7, then the partial sums are 1, 3, 6, 10, 15, 20, 26 -> 1, 3, 6, 3, 1, 6, 5 mod 7. None are 0 here, but you are guaranteed that two partial sums are equal to each other, in this case 6 = 20 mod 7 for example, so 1+2+3 = 1+2+3+4+5+5 mod 7, then just take the difference between these two, which is 4+5+5 = 0 mod 7

[viral dagger]

sorry what

what is a partial sum

og