help-49

[dreamy lichen]

this should be changed i think

Also you should use full stops and proper grammar

E.g. without a full stop, I wouldnt know whether this is some assertion, or it's the i+1 th case, or if its something completely different

[twilit field]

got it, I'll work on that

[dreamy lichen]

otherwise the proof structure is quite good

just work on the wording and proving inequalities

[twilit field]

thanks

thanks a lot

.close

[midnight plank]

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[acoustic owl]

Question 45 I’m completely lost

[acoustic owl]

I understand how they arrive at the integrals they present but I have no clue how to evaluate those

And idk where f(t) is coming from

@helper

<@&286206848099549185>

[midnight plank]

@acoustic owl Has your question been resolved?

[acoustic owl]

<@&286206848099549185>

[acoustic owl]

<@&286206848099549185>

[acoustic owl]

<@&286206848099549185>

[midnight plank]

@acoustic owl Has your question been resolved?

[acoustic owl]

<@&286206848099549185>

[acoustic owl]

Yes how

[acoustic owl]

How do I evaluate the integral of its composed if undetermined f(x) or f^-1(y)

[small jasper]

You don't need to evaluate the integral

[acoustic owl]

How am I supposed to find a point they agree on then

[small jasper]

inspection

put t = a

W(a) = S(a) = 0

[acoustic owl]

Okay they are both 0

Yes

Okay then I take the derivative?

[small jasper]

yes

[acoustic owl]

Okay the top one would be

$$\pi((t)^2-(a)^2$$

[grand pond]

Nathan

[acoustic owl]

Right?

And the bottom would be

0?

The S(t)

[small jasper]

Hint: Use the ||fundamental theorem of calculus||

[acoustic owl]

Yes you can do that but you can also just sub in t

[small jasper]

what

the functions are already in terms of t

[acoustic owl]

The interior function is in terms of x

and t?

Uh

Okay what

[acoustic owl]

How am I supposed to get the antiderivative of that

[small jasper]

You don't.

Like I've said, you're not evaluating the integral outright.

[small jasper]

As I was referring to here, differentiate both functions with respect to using the fundamental theorem of calculus.

[acoustic owl]

Is the top one right

W(t)

[small jasper]

$W'(t)=f'(t)(\pi)((f^{-1} (f(t)))^2 -a^2)$

[grand pond]

Civil Service Pigeon

[acoustic owl]

Huh

Okay

So it’s

f’(t)(pi)(t^2-a^2)

okay but how do you do that with the bottom

[small jasper]

you're doing the same thing

[acoustic owl]

Then how is it not 0??

Yes I know

Neither was f(t)

[small jasper]

$S(t)=\int^{t}{a} 2\pi x f(t) \dd{x}-\int^{t}{a} 2\pi x f(x) \dd{x}$

[grand pond]

Civil Service Pigeon

[small jasper]

$S(t)=2\pi f(t) \int^{t}{a} x \dd{x}-2\pi \int^{t}{a} xf(x) \dd{x}$

[grand pond]

Civil Service Pigeon

[small jasper]

k this works

[acoustic owl]

but f(t) is not a constant

It’s a variable

[small jasper]

the integral is with respect to x, not t

so f(t) is a constant w.r.t x

[acoustic owl]

Yeah it’s still 0

[small jasper]

Did you use the product rule on the first term?

[acoustic owl]

Wait okay

It’s

$$2\pi(f’(t)t)-2\pi(f(t)t)$$

[grand pond]

Nathan

[acoustic owl]

Is this right

[small jasper]

????

$\frac{d}{dt} f(t) g(t)=f'(t) g(t)+f(t) g'(t)$

[grand pond]

Civil Service Pigeon

[small jasper]

where $g(t)=\int^{t}_{a} x \dd{x}$ in this case

[acoustic owl]

Mm yep

[grand pond]

Civil Service Pigeon

[acoustic owl]

One second

$$\pi(f’(t))(t^2-a^2)$$

[grand pond]

Nathan

[acoustic owl]

I simplified it a bit

[small jasper]

and now you're done

[acoustic owl]

But they aren’t equal

[small jasper]

um

they kinda are tbh

[acoustic owl]

No W(t)=pi(t^2-a^2)

[small jasper]

uh

are you ok

you need to use the chain rule on the upper limit of the f'(t)

which I wrote right here

[acoustic owl]

f(t) is the variable

[small jasper]

with respect to t.

and you're differentiating the function with respect to t

hence you need to use chain rule on it

[acoustic owl]

It’s with respect to y

??

dy

f(t) is a y value

[small jasper]

You're differentiating with respect to t tho

#help-49 message

the only reason why I said this was to justify why I was factoring f(t) out of the integral

but that's separate from the differentiation itself

@acoustic owl are you there

b/c I gtg soon

Uh

http://www.geofhagopian.net/M1B/diskAndShellEquilalence.pdf

take this ig

I gotta dip

👋

[acoustic owl]

Sorry I had to do something

I’ll look at it thanks

[midnight plank]

@acoustic owl Has your question been resolved?

[midnight plank]

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[inland patio]

From Durret's book on probability:

[inland patio]

I'd be grateful if someone could explain why the function Durret defines as l(x) satisfies l(x)<=phi(x) for all x?
I.e. why does a(x-c)+phi(c)<=phi(x) hold?

[humble torrent]

I think a simple drawing may help here?

[midnight plank]

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[inland patio]

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[void pivot]

@inland patio see https://math.stackexchange.com/questions/3807417/first-order-condition-for-a-convex-function

[midnight plank]

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[last slate]

For this question, will the answer be: v= -180 + v1 + n⋅360 OR v= 180 + v1 + n⋅360

[sage olive]

the first one

the negative gets distributed to (180 - v1) so you get -180 - (-v1) or -180 + v1

another way to think about it is that a negative flips the order of subtraction, so -(a-b) = b-a

and -(180 - v1) = v1 - 180

[last slate]

Ohhh okay, ty :D

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[slim hazel]

i was just wondering if someone could look over these and see if my integration looks right

i believe id have to use the washer method for all three revolutions because of the excess solid generated, and id have to split up the integrations because of the different graphs

[old storm]

It passes my eye test

[slim hazel]

thank you!

i just lack confidence doing these problems lol

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[young spire]

I’m not sure how to find the exact decimal

I’m able to graph the lines.

Q5

[blissful trench]

do you know elimination and substitution?

[young spire]

Ik substitution

I think

[blissful trench]

well yea the question is asking you to estimate

so you gotta use elimination

think of a number that makes the y-axis same so you can cancel them and solve for x

[young spire]

0

[blissful trench]

not quite

how about looking at x-axis

[young spire]

1

[blissful trench]

not quite

there are two equations and you need to multiply one of them by a number

[young spire]

3

[blissful trench]

its wrong but you're close

[young spire]

What am I trying to eliminate

[blissful trench]

the second equation

multiply by something to eliminate the x-axis

[young spire]

4

[blissful trench]

yes!

now multiply everything by 4 on second equation and then solve for y

https://cdn.discordapp.com/emojis/1112022027847606383.gif?size=48&quality=lossless&name=catcombing

[young spire]

Y=-2

I still don’t get it

I still need help

[wary trail]

You probably need a bit more gold for your archer towers

[young spire]

?

But it didn’t say use elimination method

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[wary trail]

@young spire @young spire @young spire

[young spire]

Mate

U said u got first in math

[wary trail]

indeed

well

first year

[young spire]

First year meaning?

High school?

[wary trail]

alcoholism

[young spire]

?

[wary trail]

don't worry about it

[young spire]

<@&286206848099549185>

.close

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[distant cape]

Hi please don't spam the tag it is rather annoying

[midnight plank]

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[twilit field]

While deriviing newton's formula, namely that $c(n,r) \cdot (k,r) =c(n,r) \cdot c(n-r,k-r)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

can I make any assumptions about n+k, k+r etc?

[violet storm]

what assumptions specifically?

[twilit field]

like are they equal

$c\left(n,k\right)\cdot c\left(k,r\right)=c\left(n,r\right)c\left(n-r,k-r\right)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I simplified this to yield

$\frac{n!}{k!\left(n-k\right)!}\cdot\frac{k!}{\left(k-r\right)!r!}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$\frac{n!}{\left(n-k\right)!}\cdot\frac{\left(n-r\right)!}{\left(k-r\right)!r!\cdot\left(n-r\right)!}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$c\left(n,r\right)\frac{\left(n-r\right)!}{\left(k-r\right)!\cdot\left(n-k)\r\right)!}$

now what

[grand pond]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

nvm, got it

thanks chat!

.close

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[drifting delta]

I need help with fractions

[glass kindle]

What is your question?

[drifting delta]

So its 3/7 x 8/9

[glass kindle]

Are you asking how to multiply fractions in general? Or are you having trouble with this in particular?

[drifting delta]

Multiply

[glass kindle]

Do you have some kind of textbook/course where you're learning about fractions? Did you read about the topic there? Do you have some questions about it?

[drifting delta]

So its khan academy im bad at fractions

[young spire]

I needed help and no one answered

[weary dirge]

make a new help and i can help

@young spire

[glass kindle]

What have you learned about fractions so far? Did you learn about reducing fractions? (i.e. going from 3/6 to 1/2)

[midnight plank]

@drifting delta Has your question been resolved?

[midnight plank]

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[stable mirage]

i erased the 8x since x^2 is bigger

and then it's 7x+7x

so i tried infinity and it was wrong

negative infinity*

and 0 was wrong also

[slender walrus]

multiply by conjugate/conjugate

can't just dismiss terms like that

[stable mirage]

wdym conjugate

[slender walrus]

is this your first time hearing the word conjugate

[stable mirage]

in math, outside imaginary numbers, yes

[slender walrus]

wdym by outside

[stable mirage]

i know what the conjugate of an imag. number is

[slender walrus]

same idea applies for stuff with square roots

in general a+b and a-b are conjugates of each other

[stable mirage]

multiply what by its conjugate then? the whole expression?

[slender walrus]

multiply by conjugate/conjugate

[stable mirage]

ok

8x/(sqrt(49x^2+8x)-7x)

so like

isnt it 0

[slender walrus]

no

how are you getting 0 from here

[stable mirage]

denom looks way bigger than numerator

or

[slender walrus]

"looks like" is not sufficient,
do the algebra

from here divide numerator by |x| = sqrt(x^2) = -x

[stable mirage]

-8/((sqrt(x)*sqrt(49x+8)-7x)/-x)

im not getting it

[stable mirage]

well i cant divide the numerator alone so i did both

otherwise it's not the same expression

[slender walrus]

wait

$\lim_{x\to -\infty} \frac{-8}{\frac{\sqrt{49x^2+8x}}{\sqrt{x^2}} - \frac{7x}{-x}}$

[grand pond]

ℝαμΩℕωⅤ

[stable mirage]

then $\frac{\sqrt{49+\frac{8/x}}{\sqrt(x^2)}$

[grand pond]

MWB117
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[stable mirage]

oof

bottom left is

sqrt(49+8/x)

correct?

and bottom right is 7

now can i cancel 8/x?

[slender walrus]

that component will approach 0, yes

[stable mirage]

-8/17?

[slender walrus]

why 17

[stable mirage]

14

mb

[slender walrus]

yes, -8/14
though that can be simplified

[stable mirage]

i see

thanks

for this one i tried to follow the rules i know from rational functions

but it seems like these dont work with exponential numerators/denominators

[slender walrus]

you've mixed up vertical with horizontal

[stable mirage]

bruhhhhh

ok thanks

looks like 3 isnt right still

[slender walrus]

there's more than one horizontal asymptote,

you'd need consider both x→inf and x→-inf

[stable mirage]

oh im supposed to evaluate the limits

used a grapher to check my answer and the right one is 0, how would i come up with that?

[slender walrus]

both y= 0 and 3 will be asymptotes

evaluate the limits as x→inf and x→-inf

for the limit as x→inf, divide numerator and denom by e^x

and limit for x→-inf can be evaluated directly

[stable mirage]

how can x->-inf be evaluated directly?

[slender walrus]

what does e^x →
as x→-inf

[stable mirage]

OH

for this one am i just not familiar enough with the graphing tool

or is it wrong

[midnight plank]

@stable mirage Has your question been resolved?

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[twilit field]

In the sequence of triangular numbers find two triangular numbers who's sum and diffrerence are also triangular numbers

[twilit field]

so here's what I've tried

$\frac{\left(m\right)\left(m+1\right)}{2}+\frac{n\left(n+1\right)}{2}=\frac{p\left(p+1\right)}{2}$
\
$\frac{\left(m\right)\left(m+1\right)}{2}-\frac{n\left(n+1\right)}{2}=\frac{q\left(q+1\right)}{2}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

now what

I can find m and n in terms of p and q

but does that help?

[hollow oyster]

Try that

[twilit field]

,w solve m(m+1)= (p(p+1)+q(q+1))/2 for m

[grand pond]

Results provided by WolframAlpha

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

Requested by physicsrocks

[twilit field]

eek

just a hint please

oh, hi esca

[hollow oyster]

Esca rescuing you fr

[fleet moss]

hi :3

hold on lemme think about it

[hollow oyster]

,w solve n(n+1)= (p(p+1)-q(q+1))/2 for n

[grand pond]

Results provided by WolframAlpha

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Requested by convergence0091

[hollow oyster]

Both are worse

[twilit field]

,w solve for m and n $\frac{\left(m\right)\left(m+1\right)}{2}+\frac{n\left(n+1\right)}{2}=\frac{p\left(p+1\right)}{2}$

$\frac{\left(m\right)\left(m+1\right)}{2}-\frac{n\left(n+1\right)}{2}=\frac{q\left(q+1\right)}{2}$

I guess this is a quadratic Diophantine equation?

I think I just have to use trial and error

[fleet moss]

i mean why do you have to solve for m here? you can use this as is

[twilit field]

I substitute in random values of m and n?

[fleet moss]

nono

you have written m(m+1) in terms of p and q

[twilit field]

*yes

[fleet moss]

this has "information" from both equations "encoded" in it, so what happens if you plug it back into one of the equations?

[twilit field]

so I substitute one equation into the other?

[fleet moss]

yeah, frame it like a normal system of linear equations

$\begin{center}
a+b=y\
a-b=z
\end{center}$

[grand pond]

esca (@ with reply)
Compile Error! Click the reaction for more information.
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[twilit field]

yeah os a= y+z/2

and b=y-z/2

[fleet moss]

right now rewrite it in terms of n, m, p, and q

well i guess that doesnt really help you does it

[twilit field]

yeah

I think I saw something on MSE, let me check

https://math.stackexchange.com/questions/127514/finding-pairs-of-triangular-numbers-whose-sum-and-difference-is-triangular

[fleet moss]

says pells equation

which is something from diophantine stuff

which i know nothing of

[twilit field]

yeah, never heard of it until now

[fleet moss]

i know slayla has worked with it

[junior flower]

ooo what's the question

[fleet moss]

wow insta summon

[twilit field]

this

[junior flower]

i spawn in whenever pell's equation is mentioned

do you just need any two?

[twilit field]

yeah

I know 15 and 21 are one such pair, but how would i compute them ?

[junior flower]

hmmm

maybe you can just get them with some factoring and a little brute force

[twilit field]

, is this what number theory actually is like

[junior flower]

yep

[twilit field]

I guess I'll skip this exercise for now then, and continue on with proofs

thanks!

[junior flower]

this one is kinda hard

[twilit field]

yeah, I guess knowing more number theory may help

thanks again!

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[last slate]

Help getting solution

<@&286206848099549185>

[slim hazel]

m!15m

[pearl hull]

!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.

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[humble flint]

why'd u delete the msg btw

btw the length of the sides should be 1/2 each prolly

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[severe glade]

I used quadratic formula to solve this problem but not sure where I went wrong?

[midnight plank]

@severe glade Has your question been resolved?

[viral dagger]

can you even just use quadratic formula on this

wont it be tlike 75(x^2)^2+37(x)^2+4=0

eh ifk

like let a=75x^2, b=37x, c=4

would that work

[sly horizon]

your answer isnt wrong

oh wait

+-

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[severe glade]

Oh

[severe glade]

.close

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[twilit field]

prove that the nth pentagonal number is given bu $p_n = \frac{n(3n-1)}{2}$ where $p_n=p_{n-1}+(3n-2)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so I was thinking of using induction here

or is there a way around induction

Using induction I'm getting

$\frac{n(3n-1)}{2} +3(n+1)-2$

now what

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

This gives me

$\frac{n(3n-1)+6n-2}{2}$

not sure what to do after this

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

That gives me

$\frac{(n+2)(3n-1)}{2}$ which is wrong me thinks

[hollow oyster]

This step what did you do?

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

oops

yeah

made a mistake there

[hollow oyster]

Yeah

[twilit field]

yup, now I expand and factorise, right?

[chilly condor]

shouldn't you have $\frac{n(3n-1)+6n+2}{2}$ ?

[grand pond]

Valsorim

[twilit field]

Yeah, my bad

sorry

so that gives me $\frac{(3n+2)(n+1)}{2}$ which is right

[grand pond]

ƒ(Why am. I here)=I don't Know

[hollow oyster]

shouldn't you have $\frac{n(3n+2)+6n+2}{2}$ ? For n+1 case

[grand pond]

convergence

[hollow oyster]

Or am I missing something

[twilit field]

$p_{n+1}-p_{n}=3(n+1)-2$

right?

[grand pond]

ƒ(Why am. I here)=I don't Know

[chilly condor]

yes

[twilit field]

so I add this to $p_n$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

in my inductive step

[strange jewel]

What's a paradox ?

[twilit field]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[strange jewel]

#1021175428326633542

[chilly condor]

you can add this to p_n you will just get p_n+1 back

[twilit field]

yes

[chilly condor]

the hard part is porbably about the pentagonal numbers

[twilit field]

which is what I'm trying to obtain using induction

I'm trying to prove adding this back gives $\frac{(n+1)(3n+2)}{2}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which I did

so I think I'm done

thanks!

[chilly condor]

ok nice well done

[twilit field]

.close

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[chrome epoch]

I’m not sure I understand why the coordinates of Q are (cos(A+B), sin(A+B))

That is my main hurdle here.

[chrome epoch]

It doesn’t seem obvious to me why those are the coordinates, but he writes it very confidently.

[fallen galleon]

shouldnt it be -cos(A+B) tho

[viral dagger]

juh the x coordinate for q is negative, and cos(something) when something>90 is negative no?

[fallen galleon]

yeah thats what im saying

the book writes the x coordinate to be cos(A+B)

[viral dagger]

yea isnt that correct tho?

clearely a+b>90 from the graph

[fallen galleon]

uh huh

oh i see

oh ok thats right then

[main current]

Basically the red angle is the angle A, combined with the angle B

So it's A+B

The coordinates of a point with angle θ is:
(cosθ, sinθ)

[midnight plank]

@chrome epoch Has your question been resolved?

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[pearl hull]

Find the domain of x

[bleak pier]

Domain will be x<2

[pearl hull]

Yeah

Then why don’t you pick it ==

[bleak pier]

I see

Thanks

[midnight plank]

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[shadow sparrow]

if √11-√7 / √11+√7 =a-b√77 ,find a and b

[shadow sparrow]

pls help me regarding this question

someone help me

noone's here help meeeeeeeeeeeeeee

if √11-√7 / √11+√7 =a-b√77 ,find a and b help me with this question

.close

[midnight plank]

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[twilit field]

show that for every $\alpha \in \mathbb{C}$ there exists a unique $\beta \in \C$ such that $\alpha+ \beta=0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[glass kindle]

What axioms (or definitions) were you given?

[twilit field]

These

so what I was thinking was I can assume $\alpha =a+bi$

[grand pond]

ƒ(Why am. I here)=I don't Know

[glass kindle]

LADR?

[twilit field]

yup

$\beta =c + di$

[grand pond]

ƒ(Why am. I here)=I don't Know

[sweet wing]

To show uniqueness, it usually useful to assume solutions b and b', then show that they are the same

[glass kindle]

This should also be also relevant

[twilit field]

yeah, I think I know how to do this, but as it's been a while since I've done LA i need a sanity check

[twilit field]

yeah, thanks!

[glass kindle]

(that's from the previous page I think)

[twilit field]

we have to prove the additive inverse property, I think

we can't use it

$a+c + (b+d)i=0$ (say this is true)

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

then it implies $a=-c$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

and $b=-d$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so if an additive inverse exists , $a=-c$ and $b=-d$

[grand pond]

ƒ(Why am. I here)=I don't Know

[glass kindle]

Yes, you can use the additive inverse properties of reals

[twilit field]

OK, thanks!

now to prove uniqueness

[glass kindle]

You can use the additive inverse uniqueness of reals 😛

[twilit field]

I can?

Cool, let me TeX this and send it as proof rq

[glass kindle]

Uniqueness proofs are usually constructed as:

1. Suppose I have two things, x and y, such that both satisfy <some property>
2. (something)
3. Therefore ,x = y, so the thing that satisfies <some property> is unique

so in this case, suppose that you have c + di and x + yi that are both inverses of a + bi.

(a + c) + (b + d)i = 0 + 0i
(a + x) + (b + y)i = 0 + 0i

(a + c) = 0 and (a + x) = 0, therefore c = x from the uniqueness of inverses on reals
(b + d) = 0 and (b + y) = 0, therefore b = y from the uniqueness of inverses on reals

[twilit field]

ah

thanks!

.close

[midnight plank]

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[midnight plank]

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[bleak pier]

Hints please

Yes

Both

Hints

@red tangle

[fossil horizon]

21-C

[midnight plank]

@bleak pier Has your question been resolved?

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[twilit field]

use the division algorithm, show that the square of any integer is of the form 3k or 3k+1

[twilit field]

so the division algorithm states that $a=bq+r$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so I have $a^2= bq+r$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

for now let's assume the reminder is 0

then we have $a^2=bq$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

could I have a hint

[hard umbra]

divide a by 3

[twilit field]

a or a^2?

[hard umbra]

a

[twilit field]

ok

so that gives me $\frac{a^2}{3}= \frac{bq}{3}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[hard umbra]

thats definitely not what i meant

[twilit field]

ah, here?

[hard umbra]

yes

a = 3q + r

[agile lichen]

consider any natural number a. For the remainder when you divide by 3 you have 3 cases: 0, 1 and 2. Work through each case.

[twilit field]

I think I need to think about this a bit more, thanks for all the help!

.close

[midnight plank]

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[hollow oyster]

First condition for a number to be a square number is that if :x$\in \mathbb{N}$ then $x=(0,1)mod(4)$

[grand pond]

convergence

[hollow oyster]

Try using this

It may help @twilit field

[twilit field]

I don't know modular arithmatic

got it

thanks!

[hard umbra]

why do you need that

the question is about mod 3 not 4

[hollow oyster]

Fair enough

I'm just stating

[hard umbra]

if you know what mod 3 is the result is immediate

and is just as good a necessary condition for a number to be square as your mod 4 condition

[hollow oyster]

Fair enough

I told what i knew

[hard umbra]

ok

[hollow oyster]

:)

[midnight plank]

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[shut canyon]

f is set F ⊆ A x ø
ø has no elements, so its direct product with A has no relations, or elements either

nevertheless F exists as a subset of A x ø

i don't know why A is empty

can A be not empty and A x ø be ø?

for example, F ⊆ {1} x ø

F ⊆ ø

[outer pilot]

a property of functions is that for every a in A there is f(a) in the empty set...

[shut canyon]

ø is the relations of A and ø

[carmine crystal]

A has to be empty because if there exists e in A then f(e) cannot be defined

Because theres no element in the empty set to assign it

[shut canyon]

thanks! so {1} → ø is not a function?

but ø → ø is

[outer pilot]

no cause f(1) needs to be defined

[wary trail]

a meaningless function but yes

[shut canyon]

thank you

[sweet wing]

Could it be considered the trivial function?

Mapping nothing to nothing?

[shut canyon]

the relationship of nothing with nothing

[glass kindle]

If you have A = {1} → ø, then A does describe a set of functions. But the set is empty

[glass kindle]

Yes, the intersection of ø with any set is always ø.

(because there are no elements that belong to both sets)

e.g. {1, 2, 3} & {1} = {1}, {1, 2, 3} & {} = {}

[shut canyon]

(∀a ∈ ø)(∃b ∈ ø)[(a,b) ∈ F]

[glass kindle]

that is a true statement

[shut canyon]

F is subset of A x ø, the cartesian product

[glass kindle]

ah yeah that's a brain fart

But the cartesian product of any set with ø is always ø as well

[shut canyon]

and a function doesn't map a non-empty set to an empty set

i don't yet see why

[glass kindle]

What is the definition of a function?

[shut canyon]

a function is a subset F ⊆ A x B, such that for all a in A exists a unique b in B with (a,b) in F

[glass kindle]

Right. Do you agree that the cartesian product of A and ø is ø?

[shut canyon]

there isn't a unique b in the ø for every a in A

unless A is ø, in which case there exists a unique b for every a in A, even though there are no existing elements in A, nor in ø

[glass kindle]

Yes. For example, if you had a function f : {1} -> ø, it would need to produce something from ø if you "called" f(1)

Oh, is your question about whether the A = ø case is different? nvm I mixed something up

[shut canyon]

why can't any function map everything to nothing?
and why can another map nothing to nothing?

[glass kindle]

A function from a non-empty set (e.g. {1}) to an empty set does not exist, because it can't produce any value when you give it an argument

[shut canyon]

but what are the values of f: ø → ø?

vacuously everything

[glass kindle]

Let's consider f = ø

A function is a subset F ⊆ A x B, such that for all a in A exists a unique b in B with (a,b) in F

• A x B = ø, so f ⊆ A x B does hold
• "for all a in A exists a unique b in B with (a,b) in f" does also hold, because A is empty.
  therefore, f is a function mapping ø to ø

(∀a ∈ ø) can look confusing, but a property that looks like (∀a ∈ ø. <something>) is always true, because there's no a that would violate the condition. In other words, (∀a ∈ A. Property(a)) must be equivalent to not (∃a ∈ A. not Property(a))
https://en.wikipedia.org/wiki/Universal_quantification#The_empty_set

There is only one subset of ø x ø, therefore f = ø is also the only function mapping ø to ø

[shut canyon]

thank you! i am thinking about this, i see your reasoning, but i still struggle to understand it
not (∃a ∈ A. not Property(a)) rephrasing helps to look at it from a different perspective

not (∃a ∈ ø. not Property(a))

[glass kindle]

I know, it does seem a bit insane

[glass kindle]

for every d in doctors: d has a medical license is the same as there's no d in doctors such that: d doesn't have a medical license

[shut canyon]

does this mean everything has every property in ø?

[glass kindle]

Yes

(∀a ∈ ø. (a is married) & (a is a bachelor))

[shut canyon]

is the same as (∀a ∈ ø implies (a is married) & (a is a bachelor)) ?

[glass kindle]

yes

[shut canyon]

but a in ø is true?

[glass kindle]

it's never true

[shut canyon]

it is an implication, the antecedent is false, the implication is true, nevertheless the consequence can be true or false

[glass kindle]

Yes, the antecedent is always false (nothing can be in ø). Therefore the whole implication is always true

[glass kindle]

Having this nonsense identity function can help in certain cases where you're manipulating more abstract functions where the types are not known (so one of the sets could be ø). For example, you could have:

TryBoth: (A -> C, B -> C, A | B) -> C
TryBoth(f, g, x) = f(x) if x ∈ A
                   g(x) otherwise

And then in some context you can apply TryBoth to f: R -> R, g: ø -> ø, x: R and it will work out

[shut canyon]

okay, this is very interesting! thank you for your help @glass kindle 🙌
i'll rest now
do you see it the empty set works intuitively or it still baffles you?

or how it works logically

[glass kindle]

It is a bit unusual that every property is true for an empty set, yes

Sometimes you have to give up parts of your intuition and just use the formal definitions 🙂

If you like thinking of functions as formulas, you can define f: ø -> ø as f(x) = x

[glass kindle]

I like to think of it as: the behaviour of this special element is defined such that it won't break the rules for other elements.

Like x^0 = 1, so that x^(a + b) = x^a * x^b keeps working when a = 0. Even though intuitively, "multiplying x by itself 0 times" makes no sense

[shut canyon]

lately i find that if i learn how the definitions are built, i get to understand concepts, as if there was a structure and one could see how the structure is built

[glass kindle]

yes, just learning definitions by heart is not useful

but definitions can work in weird ways when applied to edge cases

[shut canyon]

perhaps if 0 is ø, 1 is {{}}, and x is some set A, then A^ø = {{}} for some reason

[glass kindle]

fun fact: if A and B are finite sets, the set of functions $A \rightarrow B$ has $|B| ^ {|A|}$ elements (and 1 element when $|A| = |B| = 0$)

e.g. there are 2^3 = 8 functions of type {1, 2, 3} -> {foo, bar}, and 3^0 = 1 functions of type ø -> {1, 2, 3}

[grand pond]

fix error

[midnight plank]

@shut canyon Has your question been resolved?

[midnight plank]

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[twilit field]

prove that no integer in the sequence $11,111,1111,\dots$ is a perfect square

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I was thinking of using the reminder theorem somehow here

[misty gorge]

this is like when i set an alarm so that i don’t forget

[twilit field]

so it's wrong?

I know no modular arithmatic if that's what you're getting at

[misty gorge]

it’s not wrong it’s just spelling

you said reminder theorem

[twilit field]

oh

[misty gorge]

sorry for detailing a bit

i just found it funny

[twilit field]

so each number can be expressed in the form $x=5q+1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

hmm

give me a minute

hmm, so a^2= 5q+1

so $\frac{a-1)(a+1}{5} =q$

[misty gorge]

you can’t prove that this is impossible

btw

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

right

actually, I may be able to

[misty gorge]

you won’t

[twilit field]

for a number to end in 1

oh right

[misty gorge]

notice that 11^2=121

[twilit field]

yes

just realised

hmm

maybe I try with another divisor

say 4

,w 108/4

[grand pond]

Results provided by WolframAlpha

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Requested by physicsrocks

[twilit field]

ok

so I'm guessing the reminder when any of these numbers are divided by 4 is 3

so $a^2=4q+3$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so $a^2-1=2(2q+1)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

hmm, don't think this helps either

[misty gorge]

you can just check the squares mod 4

[twilit field]

this is supposed to be solved using the division algorithm, haven't learnt modular arithmetic yet

[misty gorge]

here is a theorem that you might like

if you multiply two numbers and take the result mod 4, it’s the same as taking both numbers mod 4, and then multiplying them, and then taking the result mod 4

you could prove that this works without the division algorithm

oh

i misread

[twilit field]

I mean this is easy enough to disprove

$\frac{a^2-1}{2} =2q+1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

the left hand side will always be even

my bad

nah, this doesn;t work

wait

yeah

the left hand side will always be even

(it will end in 0)

and the right hand side will be odd

we have thus arrived at a contradiction

[misty gorge]

no

[twilit field]

oh right

110/2=55

my bad

[misty gorge]

the LHS will indeed be even though

if it is an integer at all

[twilit field]

yeah

and the right hand side will be odd

for all q

[misty gorge]

you need to prove this

[astral garnet]

where are u coming to this conclusion from?

is a any integer?

[misty gorge]

i’m saying that if the LHS is an integer

then it’s also even

[astral garnet]

ah right

[misty gorge]

do you see why? (don’t spoil it)

[twilit field]

so it will be of the form $11111\dots 0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[humble torrent]

whats this theorem called again

[misty gorge]

uh idt there’s really a name for it, i would just say multiplication descends to the quotient

[humble torrent]

oh i see

[misty gorge]

what will

[humble torrent]

for some reason i thought of the sum of two squares lemma thing with mod 4

[twilit field]

wait

when divided by 2

the number is always of the form $55 \dots 5$

[grand pond]

ƒ(Why am. I here)=I don't Know

[misty gorge]

we already said that this was a result of 11…1 being 3 mod 4

which is fine. we’re already almost done

[misty gorge]

you just need to show this fact

[twilit field]

so I have to show that $\frac{a^2-1}{2}$ is always even?

[grand pond]

ƒ(Why am. I here)=I don't Know

[misty gorge]

if such an expression is an integer yeah

[twilit field]

let's see

if it ends in 3,5,7,9, that follows trivially

if it ends in an even number or 0, it won't be an integer

not sure of how to deal with the case where it ends in 1 though

[misty gorge]

no need for cases, just stare at the expression for a sec

[twilit field]

hmm

nothings stand out to me

[misty gorge]

and ||use difference of squares|| spoiler

[twilit field]

I thought of that, but does that actually help

oh

it just prove a^2-1 is divisible by 2

doesn't it

[misty gorge]

what do you know about a

[twilit field]

I know that a^2 is odd

[misty gorge]

yeah and therefore a must be odd

[twilit field]

oh yeah

right

makes sense

[misty gorge]

okay now the factors

you can consider the factors and get a stronger result (about the whole thing)

[twilit field]

one minute

I don't quite follow

[twilit field]

this bit

[misty gorge]

i mean consider a+1 and a-1

[twilit field]

yeah, they are both even

[misty gorge]

great

[twilit field]

but even/2 can be odd

[misty gorge]

that’s right

but lucky for us we have two such even factors

[humble torrent]

i think maybe start backwards? suppose this was to be odd

if that helps sorry for intervening

[twilit field]

oh right

[misty gorge]

so now we’re finally done, though it’ll become way easier when you learn about modular arithmetic because it’s (surprisingly) powerful machinery

[twilit field]

yeah, this was easier than I expected though . Thanks a lot for the help!

.close

[midnight plank]

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[tawdry epoch]

can someone tell me if im even close to doing this right lol, im trying to prove area of an arc segment and im using an integral but im kinda doubting i set this up right. sqrt r^2 -x^2 is just formula for a circle rewritten

[sweet wing]

arc segment of a circle or length of the arc?

like 1/2 pi radians of a circle

[tawdry epoch]

arc segment like the area

[sweet wing]

the area

[tawdry epoch]

idk if thats the right word

[sweet wing]

alright

well, if oyu really want to use an integral

i'd recommend using a double integral in polar coordinates

[tawdry epoch]

Ah

[sweet wing]

where r goes from 0 to R

[acoustic owl]

Called a sector

[sweet wing]

and theta goes from 0 to arc length

[tawdry epoch]

idk how to do that lol im trying to prove this because im studying polar coordinates rn

[tawdry epoch]

oh thanks

[sweet wing]

btw, this channel is still occupied

[tawdry epoch]

oh

[sweet wing]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tawdry epoch]

i thought it was closed

[acoustic owl]

Should help me in #help-13 😎

[sweet wing]

yeah, youve gotta wait for it to reset

[tawdry epoch]

ah

okay that makes sense

[sweet wing]

under the available category, you can see which channels are open

[tawdry epoch]

thank you

[midnight plank]

Available help channel!

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[twilit field]

If n is an odd integer, show that $n^4+4n^2+11$ is of the form 16k

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

ok, so if it's odd it's of the form $2q+1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so I just substitute that in and solve?

[sweet wing]

yeah, i imagine that wold be one of oyur first steps

[twilit field]

this is going to be a rather nasty problem

,w (2q+1)^4+4(2q+1)^2+11

[sweet wing]

XD

lmao

[grand pond]

Results provided by WolframAlpha

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Requested by physicsrocks

[sweet wing]

welp

there you go

[twilit field]

the expanded form isn't divisible by 16 though

[sweet wing]

are you sure?

[twilit field]

let me write it down

[hollow oyster]

It is

[sweet wing]

obviously the q^4,q^3 and q^0 terms are divible by 16

[twilit field]

oh, right

[sweet wing]

so you only need to show 40q^2+24q is divisible by 16

[hollow oyster]

All of them are

[twilit field]

yeah

[sweet wing]

yeah but its not immediately clear

[twilit field]

this can be done using induction IMO

[hollow oyster]

Fair enough

[sweet wing]

yeah or that

a lot of NT can be done with induction, no?

[twilit field]

true

thanks!

without induction though

I'll need to prove 5q^2+3|2

[sweet wing]

is this a different question?

[hollow oyster]

No

[sweet wing]

oh right

same thing

you divided by 8

[hollow oyster]

Taking a factor of 8

[sweet wing]

well, its quite easy to show

[hollow oyster]

Yes

[sweet wing]

mby i shouldnt say easy

but, its possible to show without too much effort

[twilit field]

using induction not too bad, yeah

but without induction

hmm

another sub

[sweet wing]

you are trying to show that odd* even* even + odd is even or odd* odd* odd + odd is even

[twilit field]

$q=2p+r$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

,w 5(2p+1)^2+3

[grand pond]

Results provided by WolframAlpha

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Requested by physicsrocks

[sweet wing]

stop using wolfram

you can do this quite easily

[hollow oyster]

Use mods

It also works

[zealous schooner]

This isn’t true for all q, only for odd q

[twilit field]

oh

right

yeah, then it's done

thanks everyone!

[hollow oyster]

:)

[zealous schooner]

No, it is not done

[zealous schooner]

I said it’s true for all odd q, not all odd n

Remember that n=2q+1

What you actually need to prove is that 5q^2+3q | 2

Not sure why you omitted the second q

[zealous schooner]

You need to prove this for both odd and even q

[hollow oyster]

Oh yeah you missed it mate

[cobalt sigil]

n⁴+4n²+11 = (n²+ 2)² +7 = (n²+2)²-9 + 16
= (n²-1)(n²+5) + 16
= (n+1)(n-1)(n²+5) +16

[twilit field]

hmm, that works too

[twilit field]

right

[hollow oyster]

5q²+3q= 3q(q+1) + 2q²

[zealous schooner]

Yeah this was what I thought of

[hollow oyster]

This must be easy to prove

[twilit field]

yeah, with q=2p , that's fairly easy to prove

thanks

[zealous schooner]

why is q=2p?

[twilit field]

when it's even it's 2p

no?

[zealous schooner]

Ah, I suppose you’re taking q=2p and q=2p+1?

[twilit field]

yeah

[zealous schooner]

Alright that works

[twilit field]

thanks!

[zealous schooner]

This is a much easier solution though

Imo

[twilit field]

yeah, thanks !

[zealous schooner]

Typically adding/subtracting multiples of the ‘modulus number’ to get a better expression is a good idea

[hard shard]

honestly i wouldve gone with plug 2k+1 and go from there

[hollow oyster]

Learn congruency it'll help

[twilit field]

mods and congruence is chapter 4 in my book, still in chapter 2, going to be a whil until then

[hollow oyster]

Oh I see

Then use these methods

[twilit field]

yeah

thanks

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[twilit field]

.reopen

[midnight plank]

✅

[twilit field]

@hollow oyster you were saying?

[hollow oyster]

Nah like practice factorisation it'll help

Many elementary proof are based on that

[twilit field]

I know lol, I didn't care about my studies until 8t grade

so my fundamentals are weak

I'll work on it

thanks

[hollow oyster]

Same

:)

[twilit field]

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[zealous schooner]

That’s because studies till 8th grade don’t matter

In the long run

[cobalt sigil]

I think the 12 th grade rigorous calculations arent necessary too

[hollow oyster]

Well he's still not in uni

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[drowsy thunder]

can someone pls tell me where i made a mistake

[raven abyss]

Ok

[drowsy thunder]

the answer is supposed to be ln(14)-1

[raven abyss]

Oh nvm I see integration by parts I can’t check that 😭

Oh

Why did you say that this = 1

[drowsy thunder]

that is -1

[raven abyss]

uhhhh wait hold on

How did you get that

I see that a approaches 0

[drowsy thunder]

[x(ln14x-1] x=1 x=a

[raven abyss]

So ln(14a) would be approaching ln(0)

which is negative infinity right

So negative infinity * 0

[drowsy thunder]

ye but it a(ln14a)

and what would be 0(ln(14*0))

which would be 0

[raven abyss]

No because 0 * infinity is undefined

Don’t you have lhopital rule or something

[drowsy thunder]

oh

[raven abyss]

I’m gonna graph it to check if that’s the mistake

[drowsy thunder]

i think ur right

[raven abyss]

Yes that’s the limit

It’s 0

but you have to figure that out

Without graphing

unless it’s a graphing calc problem

[drowsy thunder]

ok thank you very much

.close

[midnight plank]

Channel closed

Closed by @drowsy thunder

Use .reopen if this was a mistake.

[raven abyss]

Np, using lhopital rule works here btw

[midnight plank]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[sage helm]

@last slate Sorry if it's a little earlier than you expected but

[sage helm]

I don't think I'm smart enough for this

So what I did think was just to prove $(b_n) \to b \implies (b_n^{-1}) \to b^{-1}$

[grand pond]

neon

[sage helm]

and use the ab thing

[ \left | \f{1}{b_n} - \f 1b \right | = \f{|b_n - b|}{|b||b_n|} ]

[grand pond]

neon

[sage helm]

okay

[surreal moon]

[sage helm]

hello SWR

proving the a_n/b_n limit thing

or atleast trying to