help-49

[fading ore]

Anyways, this just takes practice

There are problems that even the smartest people in the world couldn't figure it out had they not been well-practiced

[distant pagoda]

It sounds rlly hard

[fading ore]

It is

Usually when u do math in college you spend days, weeks, months, even years on problems

[distant pagoda]

Oh

[fading ore]

The years thing isn't til a doctoral program

Anyone doing a doctorate, even outside of math, spends years working on a problem

[distant pagoda]

That must be super rewarding after u solve it lol

[fading ore]

Yeah it is for sure

Honestly even the problems that only take days are super rewarding

Or when you do the Putnam exam and solve some problems there, that's also rewarding

[distant pagoda]

Whats that

[fading ore]

Generally considered to be the hardest math exam out there

There's 120 total points, 12 questions over 6 hours

The median score is a 0

[distant pagoda]

That’s crazy

[fading ore]

I did it this year for the first time

[distant pagoda]

How was it?

[fading ore]

My goal was a 3%, I think i got a 6-8%

[distant pagoda]

That’s real,y good well done

So ur like pretty smart then lol

[fading ore]

Thank you, it was very tough this year

[fading ore]

I mean i guess

There's a lot of talent in math tho

[distant pagoda]

Like genetics?

[fading ore]

More just like general skill

This years putnam exam was more challenging than usual in my opinion

There was no calculus which was disappointing bc its my strong suit

[distant pagoda]

N u still did rlly well so that’s impressive

[distant pagoda]

That’s a shame

Is there any like point to do ping the exam?

[fading ore]

It was a bunch of algebra and number theory

[distant pagoda]

Sound scary

[fading ore]

I mean its mostly just a flex for grad schools

But its not the most important thing for grad school

[distant pagoda]

Ahh

Where r u from?

I assume not uk as u call uni college

Is it an exam in every country?

[fading ore]

I'm from the US

[fading ore]

Pretty sure just the US but it may be global

[distant pagoda]

Ahh

[midnight plank]

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[deft ravine]

Is this right? If yes, what relation is this, i mean where do we use it

[buoyant yoke]

!original

[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.

[buoyant yoke]

it doesnt seem quite right

$\sum_{i=1}^n \alpha_ia_ix_i=\sum_{i=1}^n \frac{L}{i}\alpha_i=L\sum_{i=1}^n\frac{a_i}{i}=L\left(\frac{a_1}1+\frac{a_2}2+...+\frac{a_n}{n}\right)$

[grand pond]

Bonk

[deft ravine]

Ok thx

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[barren onyx]

If a chord, which is normal to the parabola $y^2 = 4ax$ at one end, subtends a right angle at the vertex, prove that:

The chord passes through the point $(4a, 0)$.

[grand pond]

riddle

[barren onyx]

i got 2/t1× 2/t2=-1

so t1t2=-4

[manic bison]

I think that should be correct

[barren onyx]

yes after that what do i do

[manic bison]

I think there was some theorem or something

[barren onyx]

o

[manic bison]

t1t2= -4 should be enough

I don't really remember it clearly though

[barren onyx]

i need the formula

[manic bison]

Lemme try to remember

[barren onyx]

ok

[manic bison]

This is all I've written before

[barren onyx]

i cant see anything

oh ok

understood

[manic bison]

Alr

[barren onyx]

are u doing jee

[manic bison]

It's a backup

[barren onyx]

oh

Imo?

[barren onyx]

why are u doing coordinate tho

it won't be asked in Olympiads

[manic bison]

Required in my institute

[barren onyx]

oh

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[pearl elbow]

can someone tell me what to do

[midnight plank]

@pearl elbow Has your question been resolved?

[midnight plank]

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[crude galleon]

What is the discriminant of ax^3 + bx^2 + cx + d?

[grim vector]

such a beautiful discriminant

[crude galleon]

Good job

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[fair plume]

On planet Porto, a day does not last 24 hours like on planet Earth. On the face of a clock on this strange planet, all the hours of a day are arranged in a circle at equal intervals. The hour hand travels the same distance between 1 o'clock and 9 o'clock as it does between 10 o'clock and 2 o'clock.

How many hours are there in a day on planet Porto?

I got 16 as an answer

[buoyant yoke]

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[fair plume]

10-2 is 8 hours, 1-9 is 8 hours, so 8+8 = 16 hours, and the 1-9-10-2 align properly on a click

clock*

i am not stuck I just want to verify

if my process is correct

[midnight plank]

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[midnight plank]

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[carmine schooner]

Hello, can anyone help me with this?

[wind oxide]

translate?

[carmine schooner]

yes, one moment

[buoyant yoke]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[carmine schooner]

1101 + 1202 means --> lamda * v1 + lamda * v2

[buoyant yoke]

Let $V$ be a vector space of all functions continuous functions on the interval [0; 1]\
On $V$ the scalar product $\left<f,g\right>=\int_0^1 f(x)\cdot g(x)dx$ is considered.\
In $V$ we now consider the plane $E$ which is spanned by the two vectors $v_1(x)=1$ and $v_2(x)=x$.\
We are looking for the best approximation $v=\lambda_1v_1+\lambda_2v_2$ within the plane $E$ to the vector $f$ with $f(x)=x^2+x-3$.\
To do this, determine $\lambda_1$ and $\lambda_2$ such that the difference vector $f-v$ is perpendicular to $v_1$ and $v_2$ (w.r.t. the given scalar product).

[grand pond]

Bonk

[carmine schooner]

Nice. How can i solve that?

[buoyant yoke]

first of all, what does it mean for two vectors to be perpendicular with a given scalar product?

[carmine schooner]

It means, that their scalar product is equal to zero.

[buoyant yoke]

so then, we need to find $\left<f-v,v_i\right>=0\quad,\quad i=1,2$ right?

[grand pond]

Bonk

[carmine schooner]

Yes, but what is $\left<f-v>$ ?

[grand pond]

John4320

Yes, but what is $\left<f-v>$ ?
```Compilation error:```! Missing \right. inserted.
<inserted text> 
                \right .
l.49 Yes, but what is $\left<f-v>$
                                   ?
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[buoyant yoke]

we are given the definition

$\left<f-v,v_i\right>=\int_0^1 (f-v)(x)\cdot v_i(x)dx$

[grand pond]

Bonk

[carmine schooner]

Yes, but i don't know what "v" is.

Is it v1 * v2?

[carmine schooner]

Ohh, my fault...

[buoyant yoke]

$v=\lambda_1v+\lambda_2v_2$

[grand pond]

Bonk

[carmine schooner]

So we have for v1: $(x^2+x-3-lambda_1 + lambda_2x)*1$, right?

[grand pond]

John4320

[buoyant yoke]

no, $v_1(x)=1$ and $v_2(x)=x$

[grand pond]

Bonk

[buoyant yoke]

oh i see what you mean

[carmine schooner]

$(x^2+x-3-lambda_1 + lambda_2x)*x$, right?

[grand pond]

John4320

[buoyant yoke]

$f-v=x^2+x-3-(\lambda_1+\lambda_2x)=x^2+(1-\lambda_2)x-3-\lambda_1$

[grand pond]

Bonk

[buoyant yoke]

there we go

and now plug that into the scalar product

[carmine schooner]

So, we need the antiderivative function.

[buoyant yoke]

what?

[buoyant yoke]

do this

calculate that

for both v_1 and v_2

[carmine schooner]

Don't we need the integral?

[buoyant yoke]

this is the integral....

[carmine schooner]

Yes, and we need a antiderivative function, or we use it directly and putting in the 1.

Okay, so we have: $-1-\lambda_1+\lambda_2$

[grand pond]

John4320

[buoyant yoke]

$\int_0^1 (x^2+(1-\lambda_2)x-3-\lambda_1)\cdot 1dx$ and $\int_0^1 (x^2+(1-\lambda_2)x-3-\lambda_1)\cdot xdx$

[grand pond]

Bonk

[buoyant yoke]

this is what you need to solve

set them both equal to 0, and find lambda 1 and 2

[carmine schooner]

First: $(-3-\lambda_2-\lambda_1)-3$ = 0

[buoyant yoke]

=0?

[grand pond]

John4320

[carmine schooner]

Second: $-1-\lambda_2-\lambda_1$=0

[grand pond]

John4320

[buoyant yoke]

now you have a system of equations that you can solve

and then you have your lambda 1 and lambda 2 and are done

[carmine schooner]

Yes, nice. But i think, the first equation is invalid.

[buoyant yoke]

because?

,w -3-x-y-3=0,-1-x-y=0

[grand pond]

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[buoyant yoke]

ah, i see

[carmine schooner]

I can't find the fault.

[buoyant yoke]

,w int (x^2+(1-a)x-3-b)dx from 0 to 1

[grand pond]

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[buoyant yoke]

,w int (x^2+(1-a)x-3-b)xdx from 0 to 1

[grand pond]

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[buoyant yoke]

thus $3\lambda_2+6\lambda=-13$ and $4\lambda_2+6\lambda_1=-11$

[grand pond]

Bonk

[buoyant yoke]

,w 3a+6b=-13,4a+6b=-11

[grand pond]

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[carmine schooner]

How did I come up with my solutions?

[buoyant yoke]

seems liek you made a mistake in your integrals

[carmine schooner]

Thanks a lot for your help. Nice job!

[midnight plank]

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[midnight plank]

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[obtuse totem]

i don’t get how they got these two values

so i was thinking, to check ratio

so for 24 marbles, since 16 are black, the ratio is 2/3

and i just do 2/3(48 outcomes)

but like

i dont think thats what I was taught (so not a good way to do it)

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

its ok

.close

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[main stone]

What's the proper name for 40?

[main stone]

i forgot the name for that number

meaning

[fallow scarab]

forty?

[main stone]

no

[fallow scarab]

,w forty

[main stone]

like

[grand pond]

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[main stone]

OMFG

like

lets say 40 people were given the following grades

[main stone]

is this sample?

overall sample

[lavish venture]

word

[main stone]

<@&268886789983436800>

[fallow scarab]

Totally necessary ping

[misty gorge]

hello

[fallow scarab]

Smay day

[heady plume]

Total sample? Cumulative frequency?

[misty gorge]

please don’t ping the mods for math help

[main stone]

I´m trying to simply explain my problem, they give me completely wrong answers.

I´m not.

[misty gorge]

oh

[main stone]

I´m asking you to tell them to stop this behavior...

[lavish venture]

you asked what the proper name of 40 was?

[fallow scarab]

Lol our answers are correct

[formal blade]

People are asking you to give more info as well

[main stone]

I´m doing statistics right now.

[formal blade]

Better to just move on and give more info on your q

[misty gorge]

the 40 is the sample size

[main stone]

I need PROPER name for that.

[fallow scarab]

If you can't express your question correctly, then that's not our problem

[main stone]

OVERALL sample sizeÞ

?

isnt it

Σf?

@misty gorge

[heady plume]

That is the cumulative frequency of the sample ._.

[main stone]

yes, marked "f"

number of what responders/population gave to an option. @heady plume

[heady plume]

No, marked "Σf"

40 = Σf

[main stone]

f = frequency

f for FREQUENCY

[main stone]

i said Σf

Σf = 40

sigma is SUM.

[heady plume]

Now, what next?

[misty gorge]

the reason it’s labeled sigma f is because you’re summing the frequency of each bin.

[main stone]

ikr

variables is like 40-48?

@misty gorge FYI, i didn´t abuse the ping... i just wanted to call someone to warn them cause they were giving me the definition of a regular number '40'.

[misty gorge]

i don’t understand what you’re asking here.

[misty gorge]

that’s fine.

[misty gorge]

a variable is a letter that stands for some number

[main stone]

ill show u rq

[fallow scarab]

🙂‍↔️

[main stone]

categorical variables, ordinal categorical variable, categorical variable, numerical variables

@misty gorge

[lavish venture]

holy pings

[misty gorge]

the number sigma f is a numerical variable.

it stands for the total frequency.

[main stone]

ordinal categorical variable is like

Strongly agree
Agree
Disagree
Strongly Disagree

categorical variable

is like

Strongly Disagree
Agree
Strongly agree
Disagree

continuous variables

is like

How many times do you eat a day?

1, 2, 3

discrete variables

is like
How much do you weight?

"55,81 kg"

@misty gorge

[misty gorge]

i don’t know why you’re pinging me with this. i don’t know what you want me to answer with.

[spring isle]

hello! i need some help in numerical analysis using python. Is anyone good with it?

[misty gorge]

see #❓how-to-get-help , this help channel is occupied by someone else.

[main stone]

idk

[main stone]

X is called?

or stands for **

.close

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[faint granite]

given f(x) = 2x^4 + 6x^3 - 21x - 5 what is the slope of this function in P( 1 , f(1) )?

[grim vector]

What would you do ?

[faint granite]

put x = 1 and got -19 so f(1)= -19

[grim vector]

Ok

What would it gives u about the slope ?

[faint granite]

I assume taking the derivative of f(1) would give the slope

[grim vector]

You assumed well

[faint granite]

thank you, so derivative of f(x) is 8x^3 + 18x^2 - 21 so derivative of f(1) = 8 + 18- 21 = 5

[grim vector]

Parfait

[heady plume]

Derivative of f(1) is 0. But f'(1), the derivative of f at x = 1 is 5, yes

[faint granite]

I see

[grim vector]

Yeah arya is right, be aware of wording

All good ?

[faint granite]

yup

one more question can you get from a derivative back the coordinates without integrals?

[grim vector]

You want to go from derivative to the function without integrals ?

[faint granite]

yup

[grim vector]

Unfortunately not

[faint granite]

ok, still thank you very much

[grim vector]

Nw

[faint granite]

.close

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[faint granite]

was there a point to calculating f(1) for f(x) is 8x^3 + 18x^2 - 21 if I want to know the slope at f(1)?

[grim vector]

.reopen

[midnight plank]

✅

[grim vector]

No

Would be useful for the whole tangent equation

But for the slope of it only its not neccessary

I let you reclose it if you have no questions left

[faint granite]

then can I calculate if there are more points with the same slope in the equation?

[grim vector]

You want to find more point (a,b) where f'(a) = 5 ?

[faint granite]

yes

[grim vector]

Probably

Kinda hard tho

Im not sure you end up with cool values of a and b

But with intermediate value theorem you show that there is others point like this yeah

(Having a result is not include in my argument)

,w graph 8x^3 + 18x^2 - 21 = 5

[grand pond]

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[grim vector]

,w solve 8x^3 + 18x^2 - 21 = 5

[grand pond]

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[grim vector]

Ok, you will not have more points (a,b) such that f'(a) = 5

[faint granite]

yea, at least no points in the domain of the real numbers, it would seem

still quite interesting 🙂

[grim vector]

Intermediate value theorem have some issue in the complex also

And differentiation in complex is an issue in itself

Anyway

[midnight plank]

@faint granite Has your question been resolved?

[faint granite]

that is a problem indeed

I kind of still have questions but my original question has been answered, so still thank you very much

[midnight plank]

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[hybrid grail]

Can someone help me with this question/

[hybrid grail]

first i used kx = mg which after plugging in i got 3k = 1960which then left me plug the value into the formula which gave me s''(t) + 1960/6 * s(t) = 0 and then i solved the general solution by using the quadratic formula and then oibtaitning the general solution of Acos(wt) + Bsin(wt) and then applied the inital condition at the end which gave me 3cos(18.07t)cm.

[midnight plank]

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[midnight plank]

@hybrid grail Has your question been resolved?

[midnight plank]

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[grave crater]

How would I do b) ?

[hearty rune]

can you find either the x or y coordinate with what you found in a?

[grave crater]

Yup

I can

So would I use -b/2a ?

[hearty rune]

no, that would be the vertex of the parabola

it wants the vertex of the triangle

think about what it says at the bottom of the diagram

and the fact you know the points A and B

[grave crater]

Alright I got it

Thanks

.close

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[rigid turret]

$\lim_{h\to0} \frac{2(x+h)^5-5(x+h)^3-2x^5+5x^3}{h}$

[grand pond]

SLURPZZZ

[rigid turret]

whats the first step in evaluating this

since direct substitution not work

[dusty portal]

do you know the limit definition of the derivative

if not then (painfully) expand or if you know the algebraic method, factor (x+h)^5-x^5

[radiant roost]

you can selectively expand

[rain wasp]

separate it into two limits, one with exponent 5 and one with exponent 3

for each one, apply binomial theorem

[fading ore]

We dont need to do binomial theorem

[radiant roost]

for example from the first term you get 2x^5 + 10x^4 h, all other terms will have more h's that will go to 0

[fading ore]

We genuinely should just use the limit definition

[dusty portal]

well, we dont know if OP knows it yet

[fading ore]

Well I suppose if you want to directly derive what the derivative of this would be

[dusty portal]

thats why im waiting

[fading ore]

Then might as well

[rain wasp]

that's a bit silly though, i assume this limit came to be because of the derivative itself

[fading ore]

Yeah I can understand how that's circular

[rain wasp]

so we are assuming power rule is not proven yet

[fading ore]

If we want to be rigorous about it

[dusty portal]

not 100% the case? this could be a practice exercise for prep

besides, just factoring (x+h)^5-x^5 suffices here

i think binomial will do the job but is just painful

[rain wasp]

not at all?

[dusty portal]

same with the cubed terms

[rain wasp]

2x^5 will cancel, and as h approaches 0, you only get 1 term

it's not at all painful

[radiant roost]

you just need to know the first and second terms of the binomial expansion

the rest will have h^2 or higher powers of h

[midnight plank]

@rigid turret Has your question been resolved?

[rigid turret]

f'(x) = f(x+h)-f(x) / h

[dusty portal]

perfect here then

identify what f is here

[dusty portal]

also should be lim h->0 of that expression

[rigid turret]

yea yea

[dusty portal]

what is f?

[rigid turret]

um

ohh

i think ive done something like this before

[dusty portal]

lel

[rigid turret]

its 2x^5 - 5x^3

[dusty portal]

nice

now take its derivative and you're done

[rigid turret]

oh

ohh

its like reverse engineering

[dusty portal]

yes

[midnight plank]

@rigid turret Has your question been resolved?

[midnight plank]

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[hybrid grail]

can someone help me with this question

[midnight plank]

@hybrid grail Has your question been resolved?

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[viral dagger]

ABC is a triangle where AB>AC, P is on the extension of BA such that AP+PC=AB, M is the midpoint of BC, and Q is on AM such that AM and CQ is perpendicular, prove that BQ=2AP

[viral dagger]

ok so BCP is icoseles

wait hold on

[wind oxide]

More geometry 😭

[viral dagger]

true

is this just not true at all

[wind oxide]

I have no idea what language is that...

But yeah you shud check

Start with an isosceles triangle, extend it and then make the relevant constructions

[viral dagger]

like this does not look right

oh wait hold on whoops

huh how am i supposed to make P

its at the extension of BA

[wind oxide]

What?

That doesn't make sense...

[viral dagger]

like smth like this

[wind oxide]

Okay I thought pc <0 with that for sm reason

[modern sapphire]

Make angle A to be obtuse, I think that would make the figure a bit better

[wind oxide]

What if we make P on the side of BC?

[wind oxide]

It would make the figure better but you can't use that because it's for a general triangle

[modern sapphire]

Make the figure better is my entire point of making A obtuse 😭

[wind oxide]

Kk

[modern sapphire]

Also, makes it to understand that PBC is not isosceles

since P is outside AB

[modern sapphire]

i think you meant PBC is isosceles? I dont think it should be

[wind oxide]

Ye back then i thought p on ab

[viral dagger]

yeah mb on that

[midnight plank]

@viral dagger Has your question been resolved?

[heady plume]

Should probably fix P. Reflect B about A to A'. Draw perp bisector of CA'. Intersection of this perp bisector with BA extended is P

[wind oxide]

Okay I have made some pointless progress, i found bq in terms of AB and ac

[heady plume]

I found ∆B'BC ~ ∆ABM

[viral dagger]

ik P here is eyeballed but the intersection doesent look like half BQ

[heady plume]

If that's so, then the Q is wrong

[viral dagger]

why would it be at the intersection

oh wait its PC?

[heady plume]

Because as you can see, AB = AP + PC, in the above figure. And we constructed BQ just as instructed

[viral dagger]

huh

hmmm

[heady plume]

Yes. AP + PC = AP + PA' = AB

[viral dagger]

yeah the quuestion is maybe wrong

[heady plume]

It's wrong ✓ or maybe some info is missing

[viral dagger]

alr ill close this ty

.close

[midnight plank]

Channel closed

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[midnight plank]

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[neon hatch]

why is number 3 true

[neon hatch]

whats the relation between rows and the right/left inverse

<@&286206848099549185>
im mentioning the helpers cuz i asked for help in another channel and the bot just shut down the channel bruh

[sharp coral]

next time make sure you react to the bot message

[neon hatch]

ye i was afk

can you please help me

im acc so lost

how they just put up a relation between rows and columns based on left and right inverse

[sharp coral]

if A has a left inverse then it has linearly independent columns. this is only possible if c <= r

[neon hatch]

you mean that A must have a free variable right?

[sharp coral]

no

[neon hatch]

do you know what this property is called

ill find a yt vid

[sharp coral]

another interpretation is that in rref that each column must contain a pivot

[muted panther]

Hi

[neon hatch]

lmao the username

but anyway

[neon hatch]

ye and if more r than c then A has non leading columns ye?

no wait

non leading row

if more rows than columns, then non leading row (possibly)

[sharp coral]

note that the number of pivots is at most max(r,c)

• if it has a left inverse then there is a pivot in every column, so c = number of pivots
• if it has a right inverse then there is a pivot in every row, so r = number of pivots

[neon hatch]

is this sorta just rote learning or is there a proof somewhere I can find?

[sharp coral]

you can find a proof of pretty much every important theorem in linear algebra online without too much trouble

[neon hatch]

i have been trying to find for this

but I cant 😭

[sharp coral]

the statement to look for is "a matrix with full column rank has a left inverse"

[neon hatch]

bro thank you so much

i found a relevant video

[midnight plank]

@neon hatch Has your question been resolved?

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[modern holly]

how to do c) part? i'm confused

[modern holly]

no idea where to start

[neon hatch]

lmao wait is this hsc

for the limit

plug in t = infinity

basically, as time moves to infinity, what does the height of the plant appraoch

[modern holly]

huh but how would you solve that

nvm

i got it

.close

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[narrow edge]

i decided to assume that bd = df and ec = ef in the beginning but now after finishing the proof how should i prove that bd = df and ec = ef? also my proof kinda feels wrong

[narrow edge]

<@&286206848099549185>

[midnight plank]

@narrow edge Has your question been resolved?

[narrow edge]

.cloose

.close

[midnight plank]

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[narrow edge]

i was trying to prove the midpoint theorem, in the beginning i assumed that be = df and ec = ae but now after finishing the proof how should i prove that be = df and ec = ae? also is my proof correct?

[narrow edge]

guys pls respond its been an hour

[hollow pilot]

just tag the helpers

[narrow edge]

<@&286206848099549185>

bro does everyone here hate geometry or smth

[midnight plank]

@narrow edge Has your question been resolved?

[heady plume]

That is circular reasoning. To prove midpoint theorem, you're only given a triangle and two midpoints. Then onwards, you can only make constructions and deductions, not assumptions

[narrow edge]

what is circular reasoning

[narrow edge]

how to proove this

[heady plume]

are you done proving Midpoint theorem?

Then DE = 1/2 BC = BF, BE ≠ DF idk who told you that

[narrow edge]

oh im sorry

bd = df

[narrow edge]

im using bd = df and ec = ae to prove the midpoint theorem

is midpoint theorem the only method to prove that bd = df and ec = ae?

[heady plume]

That's also not true unless ABC is isosceles

[midnight plank]

@narrow edge Has your question been resolved?

[midnight plank]

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[gaunt jetty]

My professor skimmed over this a bit, but why is this true? If Gamma is a function of omega, then you can express it as a rational of the polynomials M and N

[blissful talon]

Is that meant to be the gamma function?

[gaunt jetty]

It is not. But here is a more concrete explanation

Gamma is the so-called reflection coefficient, but I don't think that context really matters here as this seems to be a mathematical observation rather?

[blissful talon]

No, gamma is important

Most functions can't be written that way

[gaunt jetty]

I see. Thank you then. I will close this then as it would not be relevent to mathematics anymore

.close

[midnight plank]

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[midnight plank]

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[barren onyx]

TP and TQ are tangents to the parabola $y^2 = 4ax$ and normals at $P$ and $Q$ meet at a point $R$ on $y^2 = 4ax$. Prove that the center of the circle circumscribing $\triangle TPQ$ lies on the parabola $2y^2 = a(x - a)$.

[grand pond]

riddle

[barren onyx]

I am completely stuck

no idea

<@&286206848099549185>

[buoyant yoke]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[barren onyx]

1

[buoyant yoke]

did you try making a drawing?

[barren onyx]

yes but i think that's wrong

[buoyant yoke]

just show it regardless

[barren onyx]

ok

1s

i drew this

but the triangle had more than 180degree in it

[buoyant yoke]

where is the parabola?

nvm

[barren onyx]

cool

also i made them as t1 t2 t3

[delicate sage]

What is the use of R?

[barren onyx]

intersection points of the normal

i wrote alpha= a(2+t1^+t2^2+t1t2) and since it lies on the parabola

it must also be equal to at3^2,2at3

[delicate sage]

We have to proof for triangle TPQ right ?

[barren onyx]

yes

[delicate sage]

Where is R ?

[barren onyx]

lies on the parabola

also mentioned in the question

So i figured id draw

[delicate sage]

Oh I see !

The question is literally so tiresome and boring !

[barren onyx]

oh can u give me steps

Ill solve it myself

maybe give hints

[delicate sage]

(y – y1) = (-y1/2a)(x – x1) is the eqn of normal you know

So just consider two points, ( x1,y1) and (x2,y2) in the form (at²,2at) solve for intersection, satisfy it with the equation of parabola and generate a relation between t1 and t2

[barren onyx]

y+tx= 2at+at^3

ik all the results

t1t2=2 and t1+t2+t3=0

[delicate sage]

What is t3 ?

[barren onyx]

R coordinates

[delicate sage]

Why would you consider the coordinates of R, R is to be represented in terms of t1 and t2

[barren onyx]

t3= -(t1+t2)

[delicate sage]

Ok you got t1×t2 = 2 right ?

[barren onyx]

i got the relation

yes

because normal equation has 3 roots there has to be 3 points

t1 t2 t3

[delicate sage]

Yeah

[barren onyx]

yes

how do i solve further

[delicate sage]

Wait

Have you find T in terms of t1 and t2 ?

?

[barren onyx]

No

[delicate sage]

Can you find it, idk but the result may be necessary

Tangent eqns given by, yy1 = 2a(x+x1)

[barren onyx]

yes

(x+x1)/2?

[delicate sage]

I've written 2a

[barren onyx]

oh ok

mb

so u get y-2at2= 2a(x-at2^2)

and same for t1

[delicate sage]

We have t1 = 2/t2

[barren onyx]

yes

[delicate sage]

So solve them for the coordinates

[barren onyx]

ok after that?

[delicate sage]

I'm thinking, if you get some sort of right angled or equilateral triangle

[heady plume]

??

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[heady plume]

Where are we +_+

[delicate sage]

I think the rest of work is tiresome too and we need to proceed by definition

Find the Eqn of line perpendicular to TP and passing through center of TP, and similarly for TQ and find the point of intersection that is the circumcenter!

I was wondering if there are any other simple ways ! I've forgotten a lot of 2d Coordinate geometry !

[barren onyx]

hi am back

[barren onyx]

2

there must be a easier way we are missing probably

[delicate sage]

I've written how to do it, and it will give you the desired result, but quite lengthy!

[barren onyx]

yea but my exam requires for quick solutions

[heady plume]

Did we reach T[(a(t+t'), att')], P(t), Q(t'), R[(a(2 + t² + tt' + t'²), -att'(t + t'))]?

[delicate sage]

Yeah also tt' = 2

[heady plume]

Welp we don't need R so screw that

[delicate sage]

We need it ah?

[barren onyx]

oh

[delicate sage]

R gives the reln between t and t', so it is necessary ig

[barren onyx]

also we have t3=-(t1+t2)

[heady plume]

Oh, coz R lies on y² = 4ax?

[barren onyx]

yea

[heady plume]

Hmm, so you got tt' = 2?

[barren onyx]

yea

[delicate sage]

Now he has two ways as i see 🥲

[barren onyx]

i want short method

[delicate sage]

The one i mentioned, or finding the eqn of circle through three points and note the center !

I'm missing on some properties surely !

[heady plume]

No no, you find midpoint between R and T

[barren onyx]

conic sections are making me insane

[heady plume]

That's circumcenter

Of TPQ

[barren onyx]

till binomial it was good

[delicate sage]

JEE right !

[barren onyx]

ye wby?

[barren onyx]

oh right

[delicate sage]

Haha i appeared previous yr !

[barren onyx]

oh cool

rank?

[heady plume]

Mhmm, you just go (T + R)/2

[delicate sage]

Mention not ! Mathematics percentile was ≈ 99.5

[barren onyx]

ok

[delicate sage]

Chem fcked me up 😭

[barren onyx]

wait can we get a circle

[barren onyx]

i hate chemistry too

[barren onyx]

thats great!

[barren onyx]

what will this give us

the circumcentre?

[heady plume]

Yes

[barren onyx]

what will that accomplish getting?

[heady plume]

Huh?

Didn't the question ask for locus of circumcenter?

[barren onyx]

ohh

the questions wording threw me off guard

so TR is the diameter

[heady plume]

Yes

[barren onyx]

ok thnx

[barren onyx]

how do u prove circle e tpq passes through r?

[heady plume]

Center C = intersection of perpendicular bisectors of TP and TQ, Now RP perp to PT, RQ perp to QT => perp bisectors are parallel to PT and QT respectively => midpoint theorem gives you intersection point = midpoint of TR

[barren onyx]

oh ok thanks

.close

[midnight plank]

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[midnight plank]

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[alpine oracle]

plse help me <@&286206848099549185>

[alpine oracle]

bro plse go to other

[chrome drum]

sorry

[buoyant yoke]

!15m

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[alpine oracle]

$dim(\sum _ { i=1}^{n-m} V_i ) =n-1$

[grand pond]

math geek

[alpine oracle]

all V_i are subspaces of V , i could only able to think of this

<@&286206848099549185>

[visual tiger]

have you seen basis completeness theorem?

[midnight plank]

@alpine oracle Has your question been resolved?

[midnight plank]

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[alpine oracle]

can u state it once?

[coral fiber]

I assume for this, you start with some basis $e_1, \ldots, e_m$ for $U$, extend it to a basis $e_1, \ldots e_m, e_{m+1}, \ldots, e_n$ for $V$, and define $V_i$ to be spanned by ${e_1, \ldots, e_n} - { e_{m+i} }$ for $i \in {1,\ldots, n-m }$

[grand pond]

Nayan

[midnight plank]

@alpine oracle Has your question been resolved?

[midnight plank]

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[granite cipher]

Why is this wrong

[dusty portal]

you let u=3+4tan(5t) right

[granite cipher]

I did reverse chain rule

[dusty portal]

that isnt' wrong

????

[lavish venture]

C

[wind oxide]

Yes yes that is what he means

[granite cipher]

The website says I’m wrong

[wind oxide]

.

[lavish venture]

,w \int \frac{20sec^2(5t)}{3+4tan(5t)}

same shit

[dusty portal]

its 20

[granite cipher]

It’s an indefinite integral where is C coming from

[lavish venture]

ahh right

[dusty portal]

+C

[grand pond]

Results provided by WolframAlpha

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

Requested by kniefy

[dusty portal]

you're supposed to write the constant of integration

[granite cipher]

How do u calculate +c for indefinite integral

[dusty portal]

you dont

oop

[granite cipher]

What’s

[dusty portal]

https://en.wikipedia.org/wiki/Constant_of_integration

[granite cipher]

Is the way to find it in there

[dusty portal]

thats there so you understand the concept

[granite cipher]

Okay how do I find it

[dusty portal]

you dont

[granite cipher]

So this is integrated wrong

[dusty portal]

its literally an arbitrary constant

it can be anything

[granite cipher]

I see

[dusty portal]

you integrated correctly

you just didn't add +C

[granite cipher]

Let me try that

Okay question

[dusty portal]

yes?

[granite cipher]

Do I always mod Ln when I integrate ?

Like the inside

Ln |x|

Always right?

[dusty portal]

depends

if the inside is always positive you dont have to

[granite cipher]

Okay

[dusty portal]

like how sqrt(x)>0 for the reals

ln(sqrt(x)) suffices

[granite cipher]

Okay

But

Can’t sqrt x = 0 too

[dusty portal]

yeah

didnt have notation for that sr

y

[granite cipher]

Okay no problem thanks a lot

You made me understand

See you around bye

.close

[midnight plank]

Channel closed

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[midnight plank]

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[static jay]

My sister had an inquiry about the following question

[static jay]

please help asap 🙂

.close

[midnight plank]

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Closed by @static jay

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[dusty portal]

ok, do you still need help?

[static jay]

No I figured it out

Thanks

[dusty portal]

also i recommend you...don’t close a post after 5 minutes

k

[static jay]

The only reason I closed it is bcz I figure it out

[dusty portal]

ok

[midnight plank]

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[last slate]

what do they mean by corresponding power

like for the first one being 2^-1 ?

[buoyant yoke]

yeah i think thats what theyre asking

[last slate]

so 4 will be 4^1?

damn that easy

.close

[midnight plank]

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[buoyant yoke]

orrrr 2^2

[midnight plank]

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[royal scaffold]

Hello i dont understand how we can reach x + 2 in the numerator, could someone explain it to me how i do the polynomial division to lower the rank of p(x)

[buoyant yoke]

do you know how to do regular division?

[royal scaffold]

i think so

but my brain isn't braining right now, how can we still have x^3 in the denominator while we have divided the numerator

[runic hamlet]

they split the fraction

[dusty portal]

$\frac{x^3-x+x+2}{x^3-x}=\frac{x^3-x}{x^3-x}+\frac{x+2}{x^3-x}=1+\frac{x+2}{x^3-x}$

[runic hamlet]

$\frac{x^3-x+x+2}{x^3-x} = \frac{x^3-x}{x^3-x} + \frac{x+2}{x^3-x} = 1 -\frac{x+2}{x^3-x}$

[grand pond]

;(

[dusty portal]

lol

[runic hamlet]

damn

[dusty portal]

we all need a laugh amirite

great minds think alike

[royal scaffold]

ahh lol i see it now

[midnight plank]

@royal scaffold Has your question been resolved?

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[slim portal]

If f(x + 1) = 2x² + 4x - 1, then f(x) is...

[slim portal]

Step by step solution?

[thorn sorrel]

try getting $2x^2 + 4x - 1$ only in terms of $x + 1$

[grand pond]

ashy!

[slim portal]

I dont understand

[buoyant yoke]

do you know polynomial division

[heavy falcon]

if f(x)=2x^2+4x-1 what would f(x+1) be

[slim portal]

What is that 😅

[buoyant yoke]

yeah nevermind

[thorn sorrel]

factor the quadratic

[uneven sandal]

you can also solve for linear factors

[slim portal]

2(x+1)²+4(x+1)-1

[heavy falcon]

?

[heavy falcon]

ok so you just added one to each x

[heady plume]

Note that $f(x) = f((x-1) + 1)$ Plug $x = y - 1$ in given function to get your result

[grand pond]

Arya

[heavy falcon]

now you are going from f(x+1) to f(x)

so what should you do

[slim portal]

Is x - 1 the reverse of x + 1? And then I plug x - 1 into the function f(x+1) right?

[slim portal]

I need to minus x by 1?

[heady plume]

Since given function is true for f(x + 1), plugging x = y - 1 reduces it to f((y-1) + 1) = f(y) which is just wat you desire

[heavy falcon]

yes

[heavy falcon]

which is the same thing that this is saying

[slim portal]

My solution is the original function

So hard

[heavy falcon]

thats incorrect

[slim portal]

What part of it?

[heavy falcon]

you are misunderstanding something

but you can do something similar to what you did in order to have a clear idea

f(x+1)=2x^2+4x-1

now you can replace each x by (x+1-1) right ?

what will f(x+1) become after doing this

[slim portal]

Yes exactly like that

I think it is doing nothing

Since the result is just the same like f(x+1)

Yeah

[heavy falcon]

yes thats your mistake

but thats exactly what i am asking you to do

[heavy falcon]

to rewrite f(x+1) using this

[slim portal]

Ahhh I see

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

Bruh Im so dumb

I still can't link the dot.....

[dusty portal]

remember subsitution i told you about yesterday?

or do u not understand

[slim portal]

I understand but I want further explanation

[heavy falcon]

ok so now

this is f(x+1)

you need f(x)

[slim portal]

Yeah

[dusty portal]

k after hes done

[heavy falcon]

so what would you do now

[dusty portal]

ping me when ready cause im working rn

[heavy falcon]

you see a x+1 a bunch of times

thats what f(x+1) gives

so if you want f(x) what should you do

[slim portal]

It's alright, iscalculusfun is here. Thanks tho

[heavy falcon]

?

[slim portal]

I need to minus it by 1?

Smiliar to like inverse

[heavy falcon]

what should you do that to

[slim portal]

To the x right

If x is over there, I need to bring it here by using - 1

Hmmmm

[heavy falcon]

can you rewrite it

[slim portal]

OHHH

[heavy falcon]

so that i get exactly what you mean

[slim portal]

2(x - 1)² + 4(x - 1) - 1

Right

Is this right

2(x² - 2x + 1) + 4x - 4 - 1
2x² - 4x + 2 + 4x - 5
2x² - 3

So the conclusion, if x is over there by + 1 wich is f(x +1). When I asked f(x), I need to subtract x each by 1 to get f(x)

I see

.close

[midnight plank]

Channel closed

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[heavy falcon]

exactly thats it

[midnight plank]

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[uneven sandal]

\[
\text{let } f \in C^\infty[0, \infty) \text{ with } f(0) = 0 \text{ and for } x > 0: \quad 
f(x) = \frac{1}{x} \int_{\frac{1}{2} x}^{\frac{3}{2} x} f(y) \, dy.
\]
\[
\text{Question does this imply: } \exists c \in \mathbb{R} : \, f(x) = cx
\]

[grand pond]

tobi

[uneven sandal]

I have been asking myself this question for a while now (6 month+)
I have only proven this for analytic functions, which is not to difficult.
If there is a theorem that could help me please let me know.

[gaunt imp]

I don't know if it could help ya, but it does seem like the integral mean value theorem could be useful

Since it seems that subtracting the bounds returns exactly x

[uneven sandal]

but this just gives me f(x) = f(alpha), where alpha is in [1/2x,3/2x]

but thanks this for sure gives me one more equation to work with thanks

well thats trivially the case for alpha = x

so maybe one can show that this is the only alpha in [1/2x,3/2x] that satsifies this

[midnight plank]

@uneven sandal Has your question been resolved?

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[midnight plank]

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[viral dagger]

find the amount of 5 digit numbers abcde such that a<b>c<d>e and a=/=0

[viral dagger]

whar

[round parcel]

Well, let's say it was just a<b>c with a not 0.

What can you say about the relationships between the digits?

[viral dagger]

a<b and b>c

[round parcel]

OK, what else can you get from that?

[viral dagger]

cant think of anything

a and c shouldnt have a relationship except if its smaller than b

[round parcel]

Right, so b > max(a, c).

What are the allowed values for a, b, and c?

[viral dagger]

1-8, 2-9, 0-8

[round parcel]

Can b be 0?

OK, can a be 9?

OK, looks good.

So, one thing I notice is that if you set b, it decides what values are allowed for a and c.

Like if b = 5, then a can be in [1, 4] and c can be in [0, 4].

How many combinations are there for a and c there?

[viral dagger]

is it not like 8×9=72

or are you refering to b=5

[round parcel]

I mean, if b is forced to be 5, how many total combinations are there for a and c?

[viral dagger]

4×5=20

[round parcel]

Right.

OK, what is it if b is left as a variable?

[viral dagger]

ir would be (b-1)b

[round parcel]

OK, and what's the sum of b(b - 1) for b from 2 to 9?

[viral dagger]

240?

[round parcel]

Right.

So, that's the answer for 3 digits.

It gives us hints about how to solve for 5 digits.

[viral dagger]

hm

for a it would be b-1 and e would be d-1
for c it would be min(b,d)-1

[round parcel]

Almost.

min(b, d) - 1.

[viral dagger]

the min trips me up

[round parcel]

Well, before, we fixed b.

Here, it's more complicated.

What if instead of setting the higher value, b or d, you instead set c?

[viral dagger]

wait so d would be [c+1,9]

[round parcel]

Yes, that would be the correct range.

[viral dagger]

oh for b it would also be [c+1,9] right

[round parcel]

Well, with a caveat.

[viral dagger]

c>0

[round parcel]

Remember that b can't be 1 even though c can be 0.

No, c can be 0.

It's just that it's [max(2, c + 1), 9] for b.

But we can ignore that if we handle c = 0 as a separate case.

After that, c = 1 and above, it works perfectly with [c + 1, 9].

Sorry, min(2, c + 1), not max.

No, it was right the first time.

[viral dagger]

for c>0 then it would always be 2 no if its min?

[round parcel]

Yeah, you're right.

It was max.

So, let's handle c = 0 by itself.

What are the allowable values for b and d?

[viral dagger]

is it not this

[viral dagger]

and this

[round parcel]

Right, but with c = 0, what are they explicitly?

[viral dagger]

holdup

is it 364536

[viral dagger]

[2,9]

[round parcel]

OK, that's b, and d is [1, 9].

[viral dagger]

$\sum_{c=0}^{8}\left(\sum_{b=\max(2,c+1)}^{9}(b(b-1))\cdot\sum_{d=c+1}^{9}(d(d-1))\right)$

am i crazy

[round parcel]

That's not bad, but it's wrong with the b(b - 1) for example.

[grand pond]

skissue.in.a.teacup

[round parcel]

Let me think.

[viral dagger]

why is it wrong?

[round parcel]

I'm not sure it's wrong. Let me think about it for a minute.

Because you have the summation going from b = 2 to 9 or whatever.

So, that gives you 8 terms added together.

One for each of the bs.

Then, you should just add together the number of as for each b.

Like 1 + 2 + 3 + 4 + 5 + 6.

If you do 1(2) + 2(3) + ..., it's way over the right answer.

[viral dagger]

9 cause its from c=0 to 8 which is 9

b(b-1) is cause b is for the b, and b-1 is for the a

[round parcel]

You're already handling the b in the summation.

You don't have like 72 two-digit combinations that start with 8.

2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 is a huge overcount of two-digit numbers.

[viral dagger]

wgar

i mean its more like the count of 3 digit numbers

[round parcel]

Well with the summation of c from 0 to 8, the summation itself takes care of c.

The summation of b takes care of b. Same with d.

The only ones you need to handle specially are a and e.

Like in the three-digit variant, we had a summation of b, which handled b, and then b(b - 1) were for a and c.

[viral dagger]

$\sum_{b=\max(2,1)}^{9}(b(b-1))\cdot\sum_{d=1}^{9}(d(d-1))$

[grand pond]

skissue.in.a.teacup

[viral dagger]

is this better?

[round parcel]

No, that has two summations and four more numbers.

That might work for a six digit number.

In our three-digit variant, we had (\sum\limits_{b = 2}^9 (b(b - 1))).

[grand pond]

Chai T. Rex

[round parcel]

Adding up the count of summations and factors, you get three, which is fine because it's a three-digit problem.

You have two summations plus four factors, which isn't fine, because it's a five-digit problem.

You're stuck in trying to reuse the three-digit problem solution.

You shouldn't do that.

[viral dagger]

what should i do onstead then

[round parcel]

You had the right idea here.

It's just that the value being summed inside the b and d summations should have a single factor.

[viral dagger]

single?

wait is it just b-1 and d-1 in em?

[round parcel]

Something like that.

[viral dagger]

7212?

[round parcel]

I need to check.

[round parcel]

No, that's close though.

Think about why you had b - 1.

[round parcel]

Note here, where you have b and b - 1.

Why is one b and the other b - 1?

[midnight plank]

@viral dagger Has your question been resolved?