help-38

[lean thunder]

Then maybe some pattern will emegre

There is the sequence and there is the number m itself changing

[ripe valley]

I think if you bound your function (that's actually what I was suggesting) then you can eventually see that atleast 1 term will lie between two multiples of m, which are guaranteed to have a square between them

[lean thunder]

I have tried till m=5 but couldn’t find any significant pattern

[ripe valley]

Actually nvm scrap this idea it is stupid

[ripe valley]

U wont

[lean thunder]

The number 5 is the first number that doesn't appear in the series as a starting value.

[ripe valley]

U have to experiment with floor(sqrt(n))

[lean thunder]

2,3,4 are parts of series when m is 1,2,3,4

[lean thunder]

Before that we need to find a relation

What are we gonna prove with floor(√n)?

[ripe valley]

what I mean is see the behavior of those terms as u apply f(n) repeatedly

[lean thunder]

Yep there is clear pattern though

[ripe valley]

I can't really try anything right now, maybe when I am free I will try it myself

[ripe valley]

Sorry for that 😐

[lean thunder]

Umm wait

Maybe I got a pattern

For m=n^2+1 the term f2(m) is a perfect square where 2 indicates the second term

You can check it

[ripe valley]

Well we need to prove something like that for all m. Maybe try assuming a weaker form of m like m=4k +1

[lean thunder]

But I think I can extent that to formula. Where if m=n^2+b then fb(m) is perfect square

[ripe valley]

U will have to try that out (induction if it turns out be true)

[lean thunder]

Except there is one problem. When b=2 its true but when b=3 its not

[ripe valley]

Well then that's an invalid strategy

[violet gust]

I think I have a rough idea. Lets start with p = floor sqrt(m). Then,p^2 < m < p^2 + p + p+1
Then, If m < p^2+p + 1, we have p^2 + p + 1 < f(m) < (p+1)^2. So we reduce the problem
from p^2 < m < p^2 + p + p+1
to p^2 + p + 1 < m < p^2 + p + p+1

So lets assume p^2 + p + 1 < m < p^2 + p + p+1 for rest of the calculations.
Then f(m) = (m + p) and f(f(m)) = m+p + p+1

As such, we can notice that for each successive f(m,r) denoting f(f(...(m)..) r times, we have f(m,r) = f(m) + p * floor(r/2) + r/2*(r/2+1)

Or in other words the increments go as p, p+1, p+1, p+2. p+2, p+3 ... until you hit a perfect square

[lean thunder]

How did you get f(m)<(p+1)^2?

[violet gust]

If m < p^2+p

[lean thunder]

Ok I see

[violet gust]

floor sqrt m = p

I just reduced the problem so I could show one case can transform into the other

[lean thunder]

Ok

[violet gust]

So, if a perfect square of p+k, k>0, exists in the sequence, then f(m,r) = (p+k)^2 = p^2 + k^2 + 2pk

And I am somewhat lost at this point

[lean thunder]

Can you explain how you got f(f(m))=m+2p+1?

[violet gust]

f(m) was reduced to m + p

and by my previous assumption of m > p^2+p + 1, I know f(m) > p^2 + 2p + 1 making it more than m + p. And that pushes floor sqrt f(m) to p+1

so f(f(m)) = m+p + p+1

Ohh, I see. I need it m > p^2 + p + 1. My bad

The rest of the argument still stands btw

[lean thunder]

You just assumed here m<p2+p+1 then you stated that m>p2+p+1 after some line

How is that

[violet gust]

I took case 1: m < p^2 + p + 1
So, f(m) becomes m + p
That still leaves f(m) < p^2 + 2p + 1.
But thats equivalent to the second case where m is: p^2 + p + 1 < m < (p+1)^2

so my case 1 after one step reduces to case 2

so, from there on after I treat it all the same

coz case 1 to case 2 trasformation is trivial

Proving for case 2 now means we proved for both cases

[lean thunder]

I see

[lean thunder]

Me also 🙂

But you made serious a progress

There are some hints came with this problem. It tells.
if m=n^2 + 1 or m=n^2 + 2, then
finitely many iterations of m will result in a perfect square.
And
If m has remainder b, then f2(m) has remainder b − 1

@violet gust I can prove first two one by assuming n^2<=m<(n+1)^2 and doing f2 iterations

But I can't prove the last conjecture

[violet gust]

m has remainder b means, m = n^2 + b?

[lean thunder]

Yes

if m is 2 more than a perfect square, and
m> 1^2 +2=3, then two iterations of f(m) yields a number that is 1 more than a perfect square

[violet gust]

if b>n+1, we just derived f2m becomes m+2n+1, so it still has remainder b tho

[lean thunder]

For m=1 f2(m) is perfect square

[violet gust]

so i think b has to be smaller than n+1

[lean thunder]

No 2n+1

That's my calculations tells

[violet gust]

if b > n+1, then f(m) = m + n, and f2m = m + n + n+1 = m + 2n + 1 = n^2 + 2n + 1 + b

In that case, the remainder is still b coz f2m becomes (n+1)^2 + b

[lean thunder]

n^2 ≤ m< (n + 1)^2
Then
g(n^2 + b) = n if and only if 0 ≤ b < 2n + 1 where g(m)= floor √m

[violet gust]

But, if if you reduce the problem to b < n+1 which is opposite of how I reduced the problem, it works, coz then f2m becomes m + 2n which leaves remainder b-1

[violet gust]

yes, but f2(m) means f(f(m)) right so doing f once more means, you can have the floor sqrt (f(m)) as either n OR n+1

but yea, if you restrict b to b < n+1, then every 2nd step, the remainder reduces by 1, so in the end it would reduce to 0, giving you a perfect square

[lean thunder]

f(m) = m+ g(m) = n^2 +1+ n.
Iterating the function f once more,
f2(m) = f(n^2 + n + 1) = n^2 + n +1+ g(n^2 + n + 1)

n2 + n +1+ n = n2 + 2n +1=(n + 1)2

[violet gust]

and b > n+1 is just one step ahead of b < n+1

[lean thunder]

So if m=n^2+1 then f2(m) is perfevt square is proved. Similarly it can be proved for m=n^2+2 then the perfect square term will be 1 less then f2(m)

But I find difficulty proving the last conjecture and go along with it

[violet gust]

for m = n^2 + 2, the square would be f4(m) right? coz if you take m = 11, the sequence becomes 11, 14, 17, 21, 25

remainder reduces by 1 every second step

[lean thunder]

Yeah

[lean thunder]

Thats how it should supposed to be end

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[scenic basin]

A catapult is set up on an open plain facing a castle – the castle and catapult are at the same height. If the catapult launches a stone with a velocity of 100 m/s at an angle of 30 degrees, how far away is the catapult from the
castle?

[shrewd ridge]

you want the vertical speed

it lets you know the time until the stone vaporizes a castle defender

[scenic basin]

how would i get that

[sour tree]

could be 1m away or 10000m away the question is shitty

does he hit it or not?

what is the area of the castle?

[scenic basin]

it hits

[sour tree]

where does eh hit the castle?

do we assume the castle is a singular point in a plane?

[scenic basin]

bro this is projectile motion in 2 dimensions

[sour tree]

does the castle have two or three dimensions?

so the castle has a height?

how tall is it?

[scenic basin]

the castle and the catapult are at the same height

[scenic basin]

yeah i drew something like that

[sour tree]

ok cool

[shrewd ridge]

100 times sin 30 i think

[scenic basin]

isnt 100 the velocity and not the distance travelled?

[sour tree]

thats the component vector for horizontal axis

[shrewd ridge]

yes, 100 is the velocity

it comes from adding horizontal and vertical speed of the stone

[sour tree]

then multiply by time taken

[late laurel]

did it hit

[sour tree]

that's what I asked, so much missing shit

[scenic basin]

it did

[shrewd ridge]

sin 30 is 0.5, so it's the same as launching vertically up at 50 m/s and horizontally at some other speed at the same time

[sour tree]

why r u using sin

why not cos

[shrewd ridge]

the time it takes to hit the ground is the same as if it was only launched vetically

[scenic basin]

is it because the horizontal velocity is constant?

[shrewd ridge]

no

[scenic basin]

why theh

i did not get this lesson well at all

[shrewd ridge]

it just works like that, the horizontal speed doesn't do anything to the distance between the stone and the ground

[scenic basin]

ok then

[sour tree]

I swear this is a function

[scenic basin]

i would assume i caluclate the time of flight next?

[shrewd ridge]

yes, you solve the problem of launching vertically

[sour tree]

I think all you need is to keep in mind that the acceleration acting upon the body is 9.8m/s/s

[shrewd ridge]

at 50m/s you find the time

and that's most of the work

[scenic basin]

yeah then multiply velocity and time

right

[sour tree]

nah

It's not that

[scenic basin]

then what would it be

[sour tree]

First off the time isnt provided

[scenic basin]

bro did u read everything

[sour tree]

i thought it was component vectors initially too but you need a time to calculate displacement

... yes

no time

[scenic basin]

no i already got the time myself

[sour tree]

how

it could be wrong

then you find the horizontal component vector X time

[scenic basin]

by doing sin(30)*100 and plugging that into equation

wait i need to type it out

[sour tree]

why is x the opp? shouldnt it be the adj?

then you need cos(3)

30

[shrewd ridge]

it's vertical

it's y if you don;t like x

[sour tree]

wtf?

[scenic basin]

yeah so i plug sin(30)*100 into the 2nd kinematics equation

[dry copper]

@scenic basin did you guys derive the formula for the range a projectile?

[scenic basin]

huh

no

[sour tree]

? isnt it this?

[scenic basin]

yeah thats what i drew

[shrewd ridge]

so it's 5 seconds roughly

[sour tree]

tht's what i'm saying i dont think you can do this with component vectors

[sour tree]

brev then you're finding the horizontal, but even then i dont think this is the right approach

[dry copper]

the range has a fixed formula

maybe you didnt yet show this in lecture

[scenic basin]

yeah we havent learned the formula for the range of a projectile yet

[dry copper]

you simply substitute v and theta

[scenic basin]

ive seen it before but he hasnt taught us yet so i assume we shouldnt be able to use it

[shrewd ridge]

then you use cos 30, multiply by 100, by 5 secodnds, 433 m

[dry copper]

false answer, false method, no

[scenic basin]

what

[dry copper]

it is the way to go

but you must derive it

since you are not allowed to use it

derive it from Newton's 2nd law of motion

project on y axis, and on x axis

obtain y(t) and x(t)

[shrewd ridge]

well yeah, maybe they will hate the method, i care about the answer

how tf do you know it's the wrong answer you can only tell it's the wrong methiod

[sour tree]

so it's 884m away?

[dry copper]

cuz i did this exercise

i did the physics stuff and law

[shrewd ridge]

oh yeah it is

[dry copper]

884 yes

[scenic basin]

yeah thats what i got as well

[shrewd ridge]

10 seonds not 5

[sour tree]

yeah bruv i swear the fucking horizontal vectors looked sus

[scenic basin]

ok cool

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[sour tree]

:)

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[cloud mountain]

the distance formula is d=√((x2 – x1)² + (y2 – y1)²) but here it looks different

[cloud mountain]

can someone explain?

[night patio]

it looks the same to me?

[cloud mountain]

what is x2 cuz like it should be -1

[night patio]

(x2-x1)^2 = (x1-x2)^2

x2 x1 are just arbtirary

the order that they are picked doesnt matter

[cloud mountain]

;-;

so here pa = pb is that why the d formula is same

[night patio]

yes they tell u that they are equidistant

[cloud mountain]

ohh i get it

first pa then pb so for pa x2 is 1-x and for pb its -1-x

ty

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[digital hull]

I need to do this

[digital hull]

Using this formula

Which is

As (2)' is 0 = h'(x), g(x)*f(h(x))h'(x) = 0

So, let's just work with second part

Is it ok?

[limpid dawn]

what is this formula doing?

[digital hull]

The exercise gives you those two formulas above, and tells you to solve the 3 integrals below without trying to calculate integrals, that is, using the two formulas above. And I understand that I must use that formula that I sent to calculate the third integral.

6. a) are the formulas, b) are the integrals, I've already solved i) and ii)

<@&286206848099549185>

[summer parcel]

I wish I knew integrals 😔

[wraith hinge]

lol

[digital hull]

Don't worry, I'm sure someone will know :)

[limpid dawn]

Yea I understand that but what you send isn't the formula

[summer parcel]

But I am grinding for that helpful role 🦾

[digital hull]

Really? I may have made a mistake in some calculation, in that case, please show me how would be the correct way to do it

[limpid dawn]

This like just some term

doesn't tell me what it is equal to

but if you did a) you might show a)

So I can see how you derived them

[digital hull]

This is what the formula is equals to

[wraith hinge]

awesome

[limpid dawn]

Oh my bad I overlooked it

[digital hull]

Perfect, wait me a minute and I send that too

[dry copper]

the formula works if the lower bound is a number and the upper bound is a function of x

you have the inverse here, lower is function and upper is number

you cant apply it right away

[limpid dawn]

yeah thats no problem

you can switch the bound with a minus

[dry copper]

$\int_{x^4}^2 (x-2)^2 , \sqrt{t} , dt = - \int_{2}^{x^4} (x-2)^2 , \sqrt{t} , dt$

[solid kiln]

Emily

[dry copper]

now you can apply it

[digital hull]

Perfect, lets try with this

[dry copper]

first term will not be 0, cant be

[digital hull]

Is is ok now?

[limpid dawn]

left side should have a d/dx ?

[dry copper]

looks good

[digital hull]

I think that as it is, it should be fine.

[limpid dawn]

well you applied this you said

[digital hull]

I understand that the exercise does not end until the integral on the left is done, right?

[limpid dawn]

that is equal to this, and this has derivative notation ()'

[digital hull]

Oh yes, because the exercise consists of finding the derivative of the integral that they give you.

[dry copper]

lol i only checked right side, didnt even look at the left

ofc missing the ( ) '

[digital hull]

Like this?

[dry copper]

no, this is the derivative 😛

[limpid dawn]

I think I see how you derived this formula lol

[dry copper]

the ()' must be on the left side

[digital hull]

This?

[limpid dawn]

,, \frac{\dd}{\dd x} \int_a^{h(x)}g(x)f(t) : \dd t = g(x)f'(h(x))h'(x) + \int_a^{h(x)}g'(x)f(t) : \dd t

[solid kiln]

bacc

[digital hull]

Perfect, wait me a minute pls so I write this

Nice!

Let's continue

[limpid dawn]

the very left side equal sign is not correct

[digital hull]

Perfect, corrected

[limpid dawn]

ok now you may continue

[limpid dawn]

question, is this already it or what is really the point of this exercise?

Do you have to do another iteration for the second integral?

[digital hull]

I understand that it ends when there are no more integrals and this is the derivative of the function, although perhaps it has already finished with the first derivative and it doesn't matter that there is an integral there in the middle, now that you mention it, it would make sense for it to be like that.

The exercise simply says: Without attempting to calculate the integrals, find the derivative of the following functions

So, I was doing more calculations, but I realized that it gets very long, and I think it actually ends after differentiating it only once.

[cinder badge]

u just want derivative of the functions in b part right?

[digital hull]

Yes

[digital hull]

Yes, the problem is that we haven't given Leibnitz in my course yet, so I can't use it.

[cinder badge]

oh

[digital hull]

Also, that's for formula a), and I already did that in class, however, thank you so much for showing it to me, if I could use it, it would have been very helpful <3

[limpid dawn]

that is basically the same as your "first formula" on the left

[digital hull]

Yep

In that case, it would be done, thank you very much bacc, Emily and Blaze, I hope you have a nice day :)

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[eager hinge]

Prove that there doesn't exist a pair of numbers a,b so that $\sqrt{3-\sqrt{2}} = a-b * \sqrt{2}$

[solid kiln]

24tudor24

[eager hinge]

a,b in Q (sorry)

[rough goblet]

do you mean (a-b)*sqrt 2

[night patio]

could do proof by contradiction

[rough goblet]

or a - (b*sqrt 2)

the latter, right?

[eager hinge]

no

[eager hinge]

yes

[night patio]

suppose there exists a,b such that the equaltiy is true

it probably crunches down to a point where a negative number appears in the square root when computing a or b

[solid kiln]

ren

[nova spire]

hint, (a-bsqrt2)^2 = ...

[eager hinge]

reply: mepq

Not rlly?

[eager hinge]

a^2 - 2abSQRT(2) + 2* b^2

idk how this helps me

[nova spire]

the rational and irrational parts must match

so 2ab = ?

and a^2 + 2b^2 = ?

[eager hinge]

3-sqrt2 = a^2 - 2abSQRT2 + 2 * b^2

3-sqrt2 + 2abSQRT2 = a^2 + 2b^2

[night patio]

u should set up a simultaneous system

[eager hinge]

never heard of that in my life

[night patio]

1. 3 = a^2 + 2b^2
2. 2ab = 1

[nova spire]

you have ... + ...sqrt2 = ... + ...sqrt2

the sqrt2 parts match

the rational parts match

[eager hinge]

OH RLLY?

damn

[night patio]

and then ur goal is to just find something wrong with this set up to prove that a and b cannot exist

[eager hinge]

so 2ab = 3?

[nova spire]

?

[eager hinge]

wait what

[nova spire]

.

this is equal to 3 - sqrt2

3-sqrt2 = a^2 + 2 * b^2 - 2absqrt2

[eager hinge]

.

[nova spire]

yes

[nova spire]

make the "sqrt2" coefficients equal

and make the rational rest match

[eager hinge]

so erm... ... either ab and 2 is 1

so like a*b is 1/2

[nova spire]

yes ab = 1/2

and the other equation is...

[eager hinge]

?

left or rigth

[nova spire]

right now you have left + left*sqrt2 = right + right*sqrt2

we already got ab = 1/2 from the sqrt2 things

you still need to make the other left = right

[eager hinge]

one sec

so like...
3 = a^2 + 2b^2

[nova spire]

yes

[eager hinge]

but a = -b/2

wait no

[nova spire]

no

still not

still not

[eager hinge]

wtf

[nova spire]

ab = 1/2

how do you get b for example

[eager hinge]

a = 1/2b

[nova spire]

yes

a = 1/(2b)

so now

[nova spire]

replace a

by its value

and find b^2

(hint : quadratic)

[eager hinge]

3 = 1/(4b^2) + 2b^2

[eager hinge]

erm.... y-yeahhhh about that

[nova spire]

multiply by 4b^2

you can let c = b^2 for example

uhhhhh no

[eager hinge]

oh shit

[nova spire]

that's bad multiplication

[eager hinge]

yeah

12b^2 = 1+ 8 b^4

[nova spire]

yes

solve for b^2

[nova spire]

.

[eager hinge]

litelarly no ideea

12c-8c^2 = 1

yeah i didnt learn quadratic in school

[nova spire]

really

you gotta use it here

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[eager hinge]

Thx

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[west crescent]

Where is the origin of this I know this is probably easy to you lots but I have no idea what these weird symbols mean

[west crescent]

i am new to english math

[tender sluice]

I suppose bar above p and q are for negation

[red loom]

[
\begin{array}{|c|c|c|c|c|}
\hline
p & q & \overline{q} & p \land \overline{q} & \overline{p \land \overline{q}} \
\hline
T & T & & & \
T & F & & & \
F & T & & & \
F & F & & & \
\hline
\end{array}
]

[solid kiln]

PajamaMamaLlama

[red loom]

suppose you have this table

the bar above the q is the negative of q

that is, whatever the input for q, and output will be the opposite

so if I input T then the output would be F

[west crescent]

can you simplify

im having trouble understanding quit a bit

[red loom]

lol sure

so we have two inputs, p and q, yes?

[
\begin{array}{|c|c|}
\hline
p & q \
\hline
T & T \
T & F \
F & T \
F & F \
\hline
\end{array}
]

[solid kiln]

PajamaMamaLlama

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[west crescent]

yea

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[west crescent]

i understand well

https://cdn.discordapp.com/emojis/1150137120028110929.gif?size=48&quality=lossless&name=SCHIZO~1

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[iron valley]

how can i find the intervals at which f(x)=sec(pix/4) is continuous

[thorn heart]

Find its points of discontinuity

Those points will be the extremes of the intervals on which the function is continuous

[iron valley]

so that is when cos is 0

[thorn heart]

Yup

[iron valley]

but then where do i go from there

[thorn heart]

In this case, though, there are infinite intervals, since the function is periodic

So you probably wanna find a way to express them in terms of a single interval

Do recall the solutions of a trigonometric equation are always a real(or complex) number+2πn, whereas n is an integer

[iron valley]

kind of

is it because cos is 0 at pi/2 and -pi/2

[thorn heart]

That's because it's symmetric

[iron valley]

so do i just say its undefined at number+ pi/2n

[thorn heart]

But it will also be discontinuous at those points +2π, +4π, +6π and so on

[iron valley]

wait is this correct then

[thorn heart]

Actually, let me see, Haven't solved it myself yet

Though I don't think that One's right

[iron valley]

wait then how would i solve

im confused

[thorn heart]

Alright first of all you should solve the trigonometric equation cos(πx/4)=0, have you done so already?

[iron valley]

yeah

ohhh wait i think i get it

so i know the cos function is 0 when its pin+ pi/2

so then i just set pix/4 equal to that and solve?

[thorn heart]

Yup, for now do that

[knotty oar]

yall know how to do matrices?

[thorn heart]

Yeah but don't see the relation between them and her question

[knotty oar]

nah am asking

if uk how to do it i dunno what yall was on

[thorn heart]

Yeah, I know some things about matrices but if you wanna ask an unrelated question go to an unoccupied channel

#help-21｜아리스킨충1 for example

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[wraith dragon]

Guys I don't know how to start solving this question , I tried substituting n as 1,2,3and 4 it shows that it's clearly divisible by 64 , I actually want to know how to solve this problem

[granite cove]

If you want to show this is divisible by 64 you should calculate mod 64 ofc. Here are some hints:

9^8 = 1 mod 64
8*8 = 0 mod 64

So you only need to look at the first 8 cases

[wraith dragon]

Wdym by mod 64 ?

%64?

[granite cove]

Modular arithmetic

[granite cove]

Yes, % is used by programmers. In maths you write mod

[wraith dragon]

1%64 is one right ?

[granite cove]

Yes

[wraith dragon]

But 9^8 is not 1

I don't understand

[granite cove]

,w 9^8 mod 64

[solid kiln]

Results provided by WolframAlpha

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Requested by error_5506

[granite cove]

Where is this from?

[wraith dragon]

??

[granite cove]

This question

[wraith dragon]

The question on my textbook

I have seen the textbook solution they have taken 9^n+1 as (1+8) ^n+1

And expanded

[granite cove]

That's savage

[wraith dragon]

I'm cooked

[wraith dragon]

And they have cancelled the terms out and they had taken 8^2 as common and said it is divisible by 64

But i don't understand anything

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[wraith dragon]

<@&286206848099549185>

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Please don't occupy multiple help channels.

[torpid lantern]

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[solar shoal]

During a meeting of a committee of ten people, four friends want to sit together. How many ways can they sit at the same 10-seat table? Don't forget, the chair occupied by each person is important.

[solar shoal]

i thought it's 4! * 6!

but it's 4! * 6! * 10 and i don't understand why

[cosmic meadow]

Is it * 10 or * 7?

[solar shoal]

*10

perhaps the correction has a mistake

why would you assume *7?

[cosmic meadow]

Without looking too much I would say it is * 7 but I have no time now, so better wait for someone else to help

[cosmic meadow]

From 4 to 10

[vernal briar]

Should be times 7

[cosmic meadow]

Sits 1 2 3 4; 2, 3, 4, 5 etc

[lusty delta]

it’s * 10

the table is circular, meaning something like 9 10 1 2 would be allowed

[vernal briar]

It's circular?

Where was that mentioned

[cosmic meadow]

Where does it say that?

[lusty delta]

nowhere, but it’s also not mentioned that the table is not circular

[solar shoal]

it's probably circular yeah

[lusty delta]

so i’m providing an explanation that is consistent with the answer given

[cosmic meadow]

That makes sense

[solar shoal]

ohhh ok it makes sense now

thanks

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[molten comet]

Hello

[molten comet]

How did they get the overlapping intervals from the table?

[bright quarry]

so you want the function $\frac{x+2}{x+1}$ to be $\leq$ 0

[solid kiln]

knief

[bright quarry]

it can only be zero when x+2 = 0

and to be negative either the numerator or denominator is negative but not both

so if you look at the table they determined the sign of each the numerator and denominator

then multiplied/divided the signs

to get the sign of the quotient

[molten comet]

I understand that part

[bright quarry]

well it’s clear that only one interval satisfies the condition of being negative

and only one value of x makes it 0

[molten comet]

Yes

[bright quarry]

are you asking why they used -2,-1 etc

if you understand it then what’s your question

[molten comet]

How did they get -2 ≤ x < -1 from looking at the table?

Did they take the positive value or

[bright quarry]

because -2<x<-1 is the only interval where it’s negative or <0

and x=-2 is included because at x=-2 the quotient is 0

[molten comet]

So basically every signs except undefined?

[bright quarry]

so the resulting interval is $-2\leq x <-1$

[solid kiln]

knief

[bright quarry]

what?

no consider x<-2

that has a sign

[bright quarry]

yet it’s not in the interval

because it’s not negative

or zero

only the zero and negative intervals

/points

because in the question they asked for it to be less than or equal to zero

[molten comet]

Ohhhhh

I now see that

Thank you so much for the help!

[bright quarry]

you’re welcome

[molten comet]

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[shut forum]

Hi, honestly not really to sure where to begin. I was thinking about maybe a proof by contradiction, where suppose lambda and v were both not equal to 0

[zinc ginkgo]

just prove directly using axioms of vector space

[shut forum]

Here's an example of my proof for the previous question, if an idea for a format helps

[shut forum]

Alright, lemme take a crack at it

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[shut forum]

Hi sorry, so I understand the idea of using axioms to prove a theorem, but in terms of proofs I am not sure how I would approach a problem like this, in terms of "Show x OR y is true"

I am going to maybe look for a proof like that where "show x or y" is demonstrated

[zinc ginkgo]

It's just two cases

Assume one is true at a time

[shut forum]

but how would that show that one of them MUST be true?

I feel like that would only show that case 1 would work, or that case 2 would work

Not sure if that would prove that the case of $\lambda \neq 0$ and $v \neq 0$ isn't possible

[solid kiln]

bucky

[frail drift]

sorry but i don't know how

[shut forum]

lol ur good

[zinc ginkgo]

You're told to prove them

If lambda isn't zero, then show v=0 and vice versa

If you're still stuck just do it your way with contradiction

[shut forum]

Okay I see, I think this approach is good. I'll try it out

[shut forum]

Hi sorry, so I wrote and rewrote some stuff but I'm not sure if I'm missing an axiom or something or some relevant theorem that can bridge the gap

This is what I have so far, it feels pretty redundant though and like I'm going in circles

There's just some missing link ah

If I could just do $\lambda v = 0_v = 0_\mathbb{F}$ and then $\lambda = 0_\mathbb{F}$ that would be so nice

[solid kiln]

bucky

[shut forum]

maybe ill try proving that first and that'll help

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[molten lava]

I need help with turning standard form into vertex form(quadratic equations)

[molten lava]

Our teacher gives us the answer and my vertex is wrong

[spare orchid]

Unfortunately you made a mistake right at the start with how you factored -2/3 out

the x term should be -3x instead of +3x

since when you factor -2/3 out, one way to think of it is that you divide each term by that and leave that inside the parentheses. Dividing the positive 2 by the negative -2/3 leaves you with a negative number instead.

[molten lava]

Ok thanks

I will solve this again

I got the same thing

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[frail knoll]

doing partial fraction decomposition I got A/x^2 + B/1+x^2 and it's wrong but I don't know how else to break it down

[tulip violet]

If you factored the denominator, what do you have?

[frail knoll]

x^2(1+x^2)

[tulip violet]

The different with this PFD is x^2 is x * x, aka repeated root

[frail knoll]

so it would be more like x(x(1+x^2))?

[tulip violet]

Not exactly

Do you know how repeated roots affect PFDs?

[frail knoll]

no

[tulip violet]

You would have fractions with denominators, start from x^1 to x^n

[frail knoll]

like A/x + B/x^2 + C/1+x^2?

[tulip violet]

Yes

Wait

I'm pretty sure that last fraction is wrong

[frail knoll]

okay

[tulip violet]

Because the numerator goes up to n - 1 power based on the the demoninator

So if the denominator's highest power is 2, then you would have terms up to the power of n - 1 or 1, so Cx + D

[frail knoll]

okay I see

thank you

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[stray drum]

Make a system

[vernal halo]

i did

idk how to solve teh system

im braindead rn

xz=1/42, yw=1/72, z+y=7/24, z+w=16/63?

[lusty delta]

it is probably better to make the system using the variables the problem defines

[stray drum]

If that’s the correct system you can solve

[lusty delta]

it is not a correct system as the letter z appears 3 times

[vernal halo]

oh yea my bad

[stray drum]

Ok yeah probably use a b c and d

[vernal halo]

i didnt read the last bit

[lean hamlet]

Now get two equations for two variables

[vernal halo]

so y=7/24-x as an example?

using my variables

[lean hamlet]

Yes

[vernal halo]

the plug it back in?

but i dont see how that works...

[lean hamlet]

In yw=1/72

Substitute y

Do it again for z+w=16/63

[vernal halo]

so (7/24)w-xw=1/72?

[lean hamlet]

And just like this you'll get two equations for two variables

[vernal halo]

wait

[lean hamlet]

One original and other which u get by substituting

[vernal halo]

and you isolate w from this?

then sub w back in for z+2=16/63?

w*

[lean hamlet]

I think u wrote wrong equations

There are 3z

[vernal halo]

hun

i dont see how theres 3

[lean hamlet]

.

[vernal halo]

since z appears in one column and row?

[vernal halo]

oh i typed it wrong

[lean hamlet]

One should have been x

[vernal halo]

its correct on my paper lol

js typed it wrong ig

[lean hamlet]

Just use the given variables in paper

[vernal halo]

yea its x+y

alright

but like same idea right?

[lean hamlet]

Yes

[vernal halo]

alright thx

[lean hamlet]

Idea is correct

[vernal halo]

i think i got it

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[fresh pendant]

$$f(x,y)=\begin{cases}\left(x^2+y^2\right)\sin\frac{1}{x^2+y^2},&(x,y)\neq(0,0)\0,&(x,y)=(0,0)\end{cases}.$$
Prove:\

1. $ f(x, y) $ is continuous at $ (0,0) $.\
2. The partial derivatives of $ f(x, y) $ exist at $ (0,0) $\
3. The partial derivatives of $ f(x, y) $ are not continuous at $ (0,0) $.\
4. $ f(x, y) $ is differentiable at $ (0,0) $\

[solid kiln]

Laya

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[sweet grotto]

$$\left(-1\right)^{S}\cdot2^{\left(E-\left(\frac{2^{x{bits}}}{2}-1\right)\right)}\cdot\left(1+\frac{M}{2^{\left(m{bits}\right)}}\right)$$
I wanna know How this formula works

[solid kiln]

Salmon Sushi

[sweet grotto]

ok let me explain

what i know

so it's for IEEE 754 floating point numbers conversion

x_bits is the number of bits allocated for the exponent 8 in this case
m_bits is the number of bits allocated for the mantissa 23 in this case
S is the sign bit if it's zero the number is positive if it's one number is negative

Now the main input values

E is decimal form of binary stored in the memory for exponent
M is decimal form of binary stored in the memory for mantissa

ok lets say we have
37.92

after converting it into binary we get 100101.1110101110000101001

now converting it into scientific notation we get 1.001011110101110000101001 x 10^101 2^5

[ripe valley]

whats mantissa? and exponent is not the usual thing ig?

[sweet grotto]

we drop the first one in mantissa because all binary numbers start with one so why store it

[sweet grotto]

1.001011110101110000101001

[ripe valley]

as if i can read binary like that

i assume thats the part after the decimal point?

[sweet grotto]

the part of a floating-point number which represents the significant digits of that number. ~google

anyways

[sweet grotto]

from here we take the 23 bits after 1.

[sweet grotto]

store them in red boxes

[sweet grotto]

and the exponent 5 we add 127 to it since those values are used to store negative exponents and we convert it back into binary its gonna naturally become 8 bits

[sweet grotto]

agian we store them in the green boxes

[sweet grotto]

and blue bit box is 1 for negative numbers and 0 for the positives

now to convert back decode or call it what ever we use the formula

$$\left(-1\right)^{S}\cdot2^{\left(E-\left(\frac{2^{x{bits}}}{2}-1\right)\right)}\cdot\left(1+\frac{M}{2^{\left(m{bits}\right)}}\right)$$

[solid kiln]

Salmon Sushi

[sweet grotto]

(-1)^S makes sense its just the sign

rest of it is kinda confusing

[sweet grotto]

we can generalize it for 32 bit floating point numbers

$$\left(-1\right)^{s_{bit}}\cdot2^{\left(E-127\right)}\cdot\left(1+\frac{M}{2^{23}}\right)$$

[solid kiln]

Salmon Sushi

[sweet grotto]

I cant figure out why it works

(1+M/(2^23)) is like normalizing M it ranges from 1 to 2 but that explains nothing

[sweet grotto]

E-127 is just an undo

but again it makes no sense

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[hollow schooner]

15^2+d^2=39^2
b+c=d
c=a+9
15^2+b^2=a^2
how do i approach this sys of eq. to avoid degrees higher than 2? i am just gonna start substituting from top down

[lusty delta]

you can solve for d first

[hollow schooner]

yes its alot easier since it was a geometry probelm but i got it

[lusty delta]

🎉

[hollow schooner]

https://tenor.com/view/rage-emoji-rage-gif-2922587064166244739

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✅

[lusty delta]

better way is this:
from equation 1, d = 36

from equations 2 and 3, a + 9 = 36 - b, which is a + b = 27

from equation 4, 225 = (a - b)(a + b), so a - b = 25/3

from here you quickly get a = 53/3, b = 28/3, c = 80/3

[hollow schooner]

i see

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[frail knoll]

I don't understand this

[night patio]

which part of it

[frail knoll]

like how am I supposed to turn x^2/a^2 + y^2/b^2 = 1 into y=b/a sqrt(a^2 - x^2)

[night patio]

are you able to make a start

like are u stuck because u dont know how start or on a particular step

[frail knoll]

i don't know how to start

[night patio]

are you familiar with transformations?

[frail knoll]

i don't think so

[night patio]

you can do it via transformation but im not sure if you care for how rigorous it is

have u done integration>

[frail knoll]

yes

[night patio]

you can pick your poison here and decide which way you want to integrate

whether along x-axis or along y-axis

because it doesnt matter which direction you choose to sum the values, u get the same thing at the end

like adding vertical strips or horizontal strips

[frail knoll]

okay I see

[night patio]

it doesnt require anything outrageous but would be a tedious exercise

[frail knoll]

hmm ok

to make the equations equal eachother should I make the equation equal itself but like / 4

[night patio]

wdym?

[frail knoll]

like idk how to show the equations equal eachother

[night patio]

oh wait i read the wrong question

i thought u had to find area

oh my bad

sorry

[frail knoll]

oh lol no it's ok

[night patio]

make y^2 the subject first

see what you get on the right hand side

u should get
||y^2 = b^2 - b^2x^2/a^2||

[frail knoll]

i got y= sqrt(b^2 - x^2/a^2

yes that

[night patio]

you are missing a b^2 with your x^2

you can factorise this now

[frail knoll]

ok yeah oops

[night patio]

you can take out b

and then multiply right hand side by a/a

and see what happens

[frail knoll]

b sqrt(x^2/a^2) a/a

sorry was that the right thing to do

[night patio]

what happened to your other term inside the square root

should be b sqrt(1-x^2/a^2) * a/a

move the a in the numerator into the square root and the denominator a will come to the front of the sqrt root

and u get the desired result

[frail knoll]

why did it turn to 1-x^2

[night patio]

because ur square root originally contained b^2 - x^2b^2/a^2

[night patio]

^

[frail knoll]

okay i see

if I had like b^2 - b^2x^2/a^2 and if I took out the first b^2 wouldn't there still be one in the square root

[night patio]

you are factorising the entire expression

not removing a b^2 out

for example

sqrt[ b^2(1+2+3+4+5+6) ] = b sqrt(1+2+3+4+5+6)

it is not equals to bsqrt(b^2(2+3+4+5+6))

[frail knoll]

I see thank you

[night patio]

nw

[frail knoll]

so going back to this if you multiply everything by a/a it would have extra things like ba/a sqrt(1/a^2 - x^2)

[night patio]

it should turn into the expression they want u to find

because when u multiply by a/a

[night patio]

this is otherwise correct if u get rid of the extra a

should be b/a not ba/a

since u already multiplied the terms inside the square root with a

[frail knoll]

but b * a/a is ba/a where does that extra a go

[night patio]

wdym

your current expression is
bsqrt(1-x^2/a^2)

multiply by a/a

you have b/a * sqrt(a^2 - x^2)

[frail knoll]

like if everything is multiplied by a/a in the sqrt it's a^2/a^2 - a^2x^2/a^2 but then the outside of the sqrt the b has nothing to cancel out with it's just ba/a

sorry lol

I messed up on the inside sqrt but i fixed it

but the outside b I don't really get why it isn't ba/a

oh nvm

i forgot the the y side

thank you :)

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[olive egret]

Stuck here

[tall kernel]

f(x) = 3x - 1
g(x) = x^2

f(g(x)) = 3x^2 - 1

how are you stuck? all your work is right

[olive egret]

What do I do after

Or is that the answer

I thought I have to solve

[tall kernel]

do they give you a particular x value

[olive egret]

Nvm

[tall kernel]

or just find f(g(x))?

[olive egret]

Thought u I’d to distribute

[meager bloom]

3x²-1
That's the answer

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[olive egret]

Good job boy

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[vague walrus]

how do i prove this i have a general idea

[lusty delta]

consider when f(x) is increasing and when it is decreasing

for a proof, you could think of this with the intermediate value theorem

[vague walrus]

not sure what to assume

any hints?

[lusty delta]

are you supposed to prove your answer to this?

[vague walrus]

not really just solve

i am wondering if it is possible to prove

[lusty delta]

alright that is better

[vague walrus]

i can draw the graph and form some logic based off of it

[lusty delta]

You are familiar with what the sin(x) function looks like on the interval from (0, pi) yes?

[vague walrus]

yes

its gonna be shifted 2 units above and stretched k times

on the y axis

after that transformation

[lusty delta]

more or less yes

[vague walrus]

and discontinous at points of integers

since it is inside floor function

[lusty delta]

that’s also good yes

tbh it looks like you have the main ideas of this problem down

[vague walrus]

i do

we have 3 sets of disc at the start and end

at the peak

since k is an integer

but im not sure how to formalise

it

[lusty delta]

can you describe the behavior of sin(x) on the interval (0, pi/2) ?

[vague walrus]

and we need 6 more so there shouldbe 3 integers in total between 2 and k

i think

[vague walrus]

increasing

[lusty delta]

yes good it just increases

[vague walrus]

concave down

[lusty delta]

and on the interval (pi/2, pi) it just decreases

[vague walrus]

yes

[lusty delta]

so let us say we have something like floor(2 + sin(x))

[lusty delta]

ok roughly that is fine yes

[vague walrus]

hmm not sure how someone will do it with math tho

this is just intuition

[lusty delta]

well yes that is kind of the point

[vague walrus]

tru

[lusty delta]

in general pre-university problems involving floor function you probably expect to use a fair bit of intuition

[vague walrus]

i guess i should leave it at that

[lusty delta]

so you know what answer is?

[vague walrus]

yea i just thought how a math person would prove it

with definitions and such

[lusty delta]

a proof can be written using the concepts we mentioned earlier

i.e. the function is increasing on interval (0, pi/2) and decreasing on interval (pi/2, pi)

[vague walrus]

hmm so it takes on the integer value twice and is discont twice for a given integer

but it still borrows some from the knowledge of sin graph

[lusty delta]

sin being an increasing function on (0, pi/2) should be a fact that can be stated without justification

[vague walrus]

well yea we can just prove that also using derivatives not necessary true

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[keen sable]

Hello, I don’t understand where negation of Q(x) ^ negation of P(x)came from? Which line and why? So in the problem you have to negate the expression

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[tidal gale]

can anyone help me with this i have been trying to figure this out all week (im not good at math) and I have a test tommorrow

[keen sable]

.close

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[urban copper]

Let the series $A = \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^3+1}}$ and $B = \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+1}}$ verify convergence.

[solid kiln]

ඞඞඞ

[somber ginkgo]

for B, intuitively we're roughly summing up 1/sqrt(n)

so it shouldn't converge

to turn that into a proof u just basically compare it to a multiple of 1/sqrt(n)

A is a little harder intuitively cus we're roughly summing up 1/(n+ a little bit more)

[potent patio]

can anybody help me

[somber ginkgo]

!occupied @potent patio

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[somber ginkgo]

you can create ur own help channel with one on the side here

[potent patio]

i cant understand how

[somber ginkgo]

but we're summing up sqrt(1/(n^2+1/n))

and remember that if we can compare it to something like 1/linear in n, we'd be done too

[somber ginkgo]

i think you just type a message in one of the channels i listed

[somber ginkgo]

but to answer ur question here

the x just gives you a factor of 2 so it's only the y and z integrals you have to worry about

it is pretty easy geometrically so you could just do it all geometrically and ignore the integrals

you could try using polar co-ordinates, that might simplify things

or you can just do the y integral first and have the z bounds depend on y

[potent patio]

i tried with polar one but didnt get the answer

[potent patio]

btw thanks for directing me

[somber ginkgo]

nw!

[wise dagger]

what have you tried?

[urban copper]

I am thinking maybe direct comparison

[somber ginkgo]

did my hints here make sense @urban copper ?

[urban copper]

I haven't tried much

[wise dagger]

they do give me some great insights for me

[pulsar dust]

i dont wanna verify.../j

[urban copper]

I did not understand

[wise dagger]

and I am thinking using 🥪 theorem because of the hints LY gave us

[urban copper]

Like i need more handholding

Squeeze theorem?

[wise dagger]

yep

[wise dagger]

as LY has stated B is probably divergent and A is funny, so we can use squeeze theorem for A

[somber ginkgo]

||A actually diverges too||

[urban copper]

How do i

[left oriole]

"verify convergence" - most misleading question ever haha

[somber ginkgo]

ahaha true

[urban copper]

Sorry

I translated

[wise dagger]

no worries lol

[left oriole]

ohh ok

[pulsar dust]

indeed what blasphemy is this to ask me to verify

[somber ginkgo]

i think B is easier cus it's more obvious what to do

so like i said, intuitively it should diverge since we're roughly summing up 1/sqrt(n)

so if we want to show a series diverges, it suffices to show that it's 'bigger' than a series we know diverges

[urban copper]

Roughly 1/n

[somber ginkgo]

i'm talking about B here

[urban copper]

Yes

[somber ginkgo]

A is roughly 1/n which makes it harder cus it's less intuitively obvious what we should do

[urban copper]

I was referring about B

[somber ginkgo]

are you sure?

[urban copper]

Idk maybe I'm misunderstanding

[somber ginkgo]

n/(n^2+1) is roughly 1/n

so if we square root it it'll be roughly 1/sqrt(n)

[wraith hinge]

what does verify convergence mean?

[urban copper]

,, \frac{n}{n^2+1} \implies \frac{1}{n + \frac{1}{n}}

[solid kiln]

ඞඞඞ

[urban copper]

Ok

[somber ginkgo]

we know that $\sum_{n=1}^{\infty} \frac{1}{k\sqrt{n}}$ diverges for all constants $k$

[solid kiln]

LY

[urban copper]

I mean cant we can direct comparison with harmonic series or something

Don't know if I'm tripping

[wraith hinge]

?

[somber ginkgo]

so if we can show that $\sqrt{\frac{n}{n^2+1}} \geq \frac{1}{k\sqrt{n}}$ where $k$ is some nice constant we've chosen we'd be done

[solid kiln]

LY

[somber ginkgo]

you could

[somber ginkgo]

in fact, you can show this diverges by comparing it to the harmonic series

but in general, you shouldn't have a fixed 'this is what we're gonna compare it to' before you do a question

sometimes it's more natural to compare our series with a different series

[wise dagger]

which will roughly be:

Suppose we have two series $\sum a_n$ and $\sum b_n$ with $a_n \leq b_n$ for all n, and $$\lim_{n\to\infty}\sum a_n = \infty$$
then $\lim_{n\to\infty}\sum b_n=\infty$ .

[solid kiln]

Biscuity

[somber ginkgo]

roughly means to show that these infinite sums actually equal some number

[wise dagger]

haha, you guys have already gone through a lot while i was typing lol

me slow typer

[urban copper]

To know what

[wise dagger]

$\frac{n}{n^2+n}\leq\frac{n}{n^2+1}$

[solid kiln]

Biscuity

[wise dagger]

this

for all n

all good till here?@urban copper

[urban copper]

Yes since n starts from 1

[wise dagger]

nice, so with this, we can safely say that B diverges to infinity

[limpid dawn]

check if the series converges or not

[urban copper]

How so

Which is the infinity

[wise dagger]

did you try to simplify these? (with the summation of course)

of course, I was assuming you know this

[urban copper]

Is it similar to p series?

[wise dagger]

yep

[wise dagger]

try simplify first, you'll get a better picture

[urban copper]

,, \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+n}} \leq \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+1}} \ \sum_{n=1}^{\infty} \sqrt{\frac{n(1)}{n(n+1)}} \leq \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+1}}

[wise dagger]

yea, that's a great start!

[urban copper]

One sec I'm on mobile

[wise dagger]

no problem, me too!

ohhh, you don't have to simplify the RHS

it can stay what it is

[urban copper]

Sorry

[solid kiln]

ඞඞඞ

[wise dagger]

afterall we are trying to show it's in it's original form and diverge after comparing

oh dear.... you modified the LHS

[urban copper]

Im stupid

And on mobile

Sorry

[wise dagger]

i understand

[wise dagger]

no worries

[urban copper]

,, \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+n}} \leq \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+1}} \ \sum_{n=1}^{\infty} \sqrt{\frac{n(1)}{n(n+1)}} \leq \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+1}} \ \sum_{n=1}^{\infty} \sqrt{\frac{1}{n+1}} \leq \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+1}} \ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}} \leq \sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+1}}

[wise dagger]

yay

[urban copper]

N is positive so is valid

[wise dagger]

so, we have
sum from n=1 to infinity √(1/(n+1))

[urban copper]

Oops

[solid kiln]

ඞඞඞ

[wise dagger]

which is actually equal to
sum from n=0 to infinity 1/√n
which is
1 + sum from n=1 to infinty 1/√n

,align
&\sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+1}}\
\geq&\sum_{n=1}^{\infty} \sqrt{\frac{n}{n^2+n}}\
=&\sum_{n=1}^{\infty} \sqrt{\frac{1}{n+1}}\
=&\sum_{n=0}^{\infty} \sqrt{\frac{1}{n}}\
=&1+\sum_{n=1}^{\infty} \sqrt{\frac{1}{n}}

[solid kiln]

Biscuity

[wise dagger]

or you can even add one more line:$$\geq\sum_{n=1}^{\infty} \sqrt{\frac{1}{n}}\$$

[solid kiln]

Biscuity

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[wise dagger]

lol

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[crude pelican]

what does this even mean?

[crude pelican]

what could the sinx inside the L shapes possibly mean?

is that not just a nonexistent notation?