help-49

[worldly pine]

then why with 3%

[lyric charm]

yknow i wonder why you didnt ask me this 5 days ago

yes that is what i did.

[worldly pine]

i forgot

[lyric charm]

12000 * 12% * 3 if you want

only i wrote 12% directly as 0.12

[slender walrus]

12800

[worldly pine]

ok,but

[lyric charm]

there is no 3%

[worldly pine]

why with 3

[lyric charm]

there never was a 3%

3 because 3 years.....

3 years!!

[worldly pine]

got it

ok

[fathom onyx]

Alright I'll ask you a counter-question: why is this "3" unfamiliar to you?

You're investing 12,000 RP on 12% simple interest per annum, over 3 years

[worldly pine]

because we already multiplied with 3 in 12800*12%*3

oh it went italic

[fathom onyx]

(don't use asterisks here, you upset the markdown)

[worldly pine]

12800 x 12% x 3

[fathom onyx]

Okay

But this is the same 3 that Ann has?

[worldly pine]

i think yes

[fathom onyx]

So you got confused with us multiplying more than two numbers together

Admittedly I think this was the spanner that was in the works

[slender walrus]

put spaces before and after the * to avoid italicisation

[worldly pine]

i got it

[worldly pine]

i meant solution here

thank you everyone

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[graceful ferry]

how to swtich the order of integration?

[graceful ferry]

I get the shape (x-2)^2=4-y^2 or (x-3)^2=1-y what if i dont know how to graph these

[mortal falcon]

the first one of those is a circle, and the second is a parabola

but, you don't always have to do graphs, you can often do some algebraic manipulation/reasoning to figure out the bounds, although that's sometimes harder

referring to a general case where you maybe aren't sure how to graph the expressions, it goes something like this

[midnight plank]

@graceful ferry Has your question been resolved?

[mortal falcon]

look at that third integral

3 + sqrt(1-y) <= x <= 2 + sqrt(4-y^2), while 0 <= y <= 1

first, we can get the range of x by plugging in the y bounds

the minimum value of x is 3 when y = 1
the maximum value of x is 4 when y = 0

overall, this means 3 <= x <= 4 are our new bounds

now to bound y in terms of x:

(x-3)^2 >= 1 - y

y >= 1 - (x-3)^2

and using the second inequality,

(x-2)^2 <= 4 - y^2
y^2 <= 4 - (x-2)^2
-sqrt(4 - (x-2)^2) <= y <= sqrt(4 - (x-2)^2)

we can use our x bounds to simplify these.

since 3 <= x <= 4,

y >= 1 - (x-3)^2 means that y >= 0, which means the -sqrt(4 - (x-2)^2) bound is redundant. so we're left with

1 - (x-3)^2 <= y <= sqrt(4 - (x-2)^2)
and 3 <= x <= 4

as a general algorithm:

using the outer constant bounds and the inner variable bounds, determine the range of values the inner variable can take. that'll specify the new constant outer bounds. From there, solve for the new inner variable using the old inner inequalities, and you may have to use information from your new outer bounds to narrow these inequalities down or combine them. that'll give you the correct resulting bounds

of course if you're able to graph it it's easier to visualize and switch that way, but I do find the algebra helpful when functions are ugly and hard to work with

[graceful ferry]

How do I determine the range of values the inner variable can take? For example you said x=3 when y=1 and x=4 when y=0, did you plug in the values of y on the edges and see which is smaller and which is bigger is the minimum and maximum?

Btw I appreciate the explanation I mostly understood it!

[mortal falcon]

yup basically

this can be pretty dependent on what you're doing though

generally, it should work to minimize the lower bound and maximize the upper bound

but most generally, you would want to minimize both bounds to get the true minimum, and same for maximizing (in case the bounds ever cross or something)

as an example, say our integration bounds were -2 to 2 for x, and -sqrt(4 - x^2) to sqrt(4 - x^2) for y

then plugging in the bounds just gives us 0 and 0 in both cases, which isn't helpful

what we really care about is the min and max of each bound. the lower bound's min is -2 when x = 0, and the upper bound's max is 2 when x = 0, so our new constant y bounds would be -2 to 2

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[graceful ferry]

okay ty!

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[chrome swift]

need to verify

[chrome swift]

let's say we got 1/(2n+1)

if we want to find n+1 you do 1/(2(n+1)+1) so it's 1/(2n+3)

right

[scenic wyvern]

you mean you want to substitute n+1 into the expression?

[chrome swift]

yeah

[scenic wyvern]

then yes

[chrome swift]

preciate it

wish me luck for my exam

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[candid gull]

Hi again! "describir por extensión" significa "lista todas las combinaciónes de (x,y) en R", verdad?

[tidal turret]

YES

[candid gull]

entonces, necesitas listar todas las combinaciónes que satisfacen el condición [por ejemplo en i:, R tiene (1,1)]

(que significa "par"? no sé ésta palabra)

[tidal turret]

even

[candid gull]

👍

[tidal turret]

i) R = {(1,1),(1,3),(1,5),(1,7),(2,3),(2,5),(2,7)}
ii) R = {(2,1), (3,1)}
iii) R = {(2,1),(2,3),(2,5),(2,7)}
iv) R = {(1,7),(2,5),(2,7),(3,5),(3,7)}

.solved

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[tidal turret]

I appreciate the help

[candid gull]

ofc!

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[twilit field]

We begin by defining a basis of $Ker(T)$ to be $e_1,e_2,\dots,e_n$. We then extend this to a basis of $V$: $e_1,e_2,\dots e_n,\dots, e_m$. We thus have the vectors $e_{n+1}$to $e_m$ mapping to non-zero vectors. Thus we have $m-n$ basis vectors mapping to non-zero vectors. Thus there will be $m-n$ non-zero vectors in the matrix of $T$.

[grand pond]

wai

[lavish venture]

#help-49 message

[twilit field]

Oh right

thanks!

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[viral dagger]

#help-45 message

[viral dagger]

find all positive integers $n$ such that there exists a positive integer triplet $(a,b,c)$ such that
$$a^3+bn$$
$$b^3+cn$$
$$c^3+an$$
are all perfect cubes

[grand pond]

ihave<skissue>

[viral dagger]

$$a^3+bn=x^3$$
$$b^3+cn=y^3$$
$$c^3+an=z^3$$

[grand pond]

ihave<skissue>

[viral dagger]

$$bn=(x-a)(a^2+ax+x^2)$$
$$cn=(y-b)(b^2+by+y^2)$$
$$an=(z-c)(c^2+cz+z^2)$$

[grand pond]

ihave<skissue>

[viral dagger]

!occupied

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[vocal raft]

ignore name, I might be wrong here but have a look

[hard sigil]

Can someone help me solve my induction problem : 1 square + 3 square + 5 square + ............ + (2n-1) square = n(2n-1)(2n+1)/3

[vocal raft]

!occupied

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[viral dagger]

how did you get this? also what about the case of a=b=/=c? (and their permutations)

[vocal raft]

I compared LHS and RHS

If x=a, y=b, z=c, it satisfies this equation.
Yea I was wrong here, there may be more solutions to this, some other x,y,z may also satisfy this.

[sullen ember]

hii

[midnight plank]

@viral dagger Has your question been resolved?

[unkempt skiff]

<@&286206848099549185>

[vocal moon]

!15m

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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.

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[viral dagger]

ok you know what i give up i have other problems to ask 😭

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[molten bay]

x+y+z=3

I want to understand all the combination

[molten bay]

x,y,z>=0

And xyz>0

so first condition gives

3+3-1 C 2

Second one gives n-1Cr-1

[fresh sparrow]

integer solutions?

[molten bay]

Am I right?

Yeah@fresh sparrow

Beggers method

Bars and stars

[pure oasis]

Given a parallelogram ABCD. Let K and I be the midpoints of sides AB and CD , respectively. Let M and N be the intersections of AI and CK with the diagonal BD , respectively. Prove that:
a) triangle ADM=triangle CBN
b) ^MAC=^NCA and IM//CN
c) BN=NM=MD
d)AC,BD, and IK are concurrent.
Help question d pls

[austere blade]

!occupied

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[molten bay]

Huh??

[frozen talon]

are you trying to show that your answer is correct?

sorry, xyz > 0 just tells us that neither x, y, or z are zero

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[molten bay]

.reopen

[midnight plank]

✅

[molten bay]

Yes

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[umbral shuttle]

Hello, I don't understand part b at all, can anyone help? Thank you.

[midnight plank]

@umbral shuttle Has your question been resolved?

[fresh sparrow]

what is it that you don't understand

the question or how to do it?

[umbral shuttle]

How to do it

[umbral shuttle]

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[sudden yacht]

<@&268886789983436800>

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[keen harbor]

In this example from a book, the highlighted -1 seems wrong to me. Did the author make a mistake?

[runic hamlet]

that seems fine

why do you think its wrong

[keen harbor]

hmm, faulty logic, I guess.

Thank you

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[quaint shoal]

If I wanna find the limit as x—>-6

would it be -9?

[lyric charm]

why minus 9

[quaint shoal]

sry I meant 9

[lyric charm]

yes it's +9

[quaint shoal]

okay, and then for a one sided limit, it’ll just be the same thing? lim x—>-6^- =9

[scenic wyvern]

is that a left-side limit as x -> -6?

[quaint shoal]

yea coming from the left

[lyric charm]

coming from the left it doesnt exist

[scenic wyvern]

there's no graph to the left of -6 though

[lyric charm]

there's nothing to the left

[quaint shoal]

meant right

mb I mess up left/rights

🥀

[scenic wyvern]

you wrote - there

so just to clarify you really mean right? because that would be +

[quaint shoal]

yea I meant right

[scenic wyvern]

then yes

[quaint shoal]

okay, then lastly just to make sure, when ur finding the lim x—>1, it’ll be -2? even though the line ends 1?

[scenic wyvern]

two-sided limit?

if two-sided, DNE

[quaint shoal]

that’s because it doesn’t approach 1 from both sides?

[scenic wyvern]

yes

the right-side limit DNE

[quaint shoal]

okay, so if u did lim x—>0 =-2 that’d be fine?

[visual tiger]

there's a little distinction to make here

true the "two-sided" limit in 1 doesn't exist because right-side limit DNE

but the "limit" in 1, if we don't specify, does exist

to approach x -> 1 with this function, you can only approach 1 from the left

and so $\lim_{x\to 1} f(x)$ technically does make sense

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

and is equal to -2

I'll just check to make sure but

this should be okay since

"x->1" can only mean one thing according to the domain of f

(it's implied x -> 1- but it makes sense)

[visual tiger]

understand that this is not a two-sided limit

[quaint shoal]

Instead its a one sided limit coming from the left?

[scenic wyvern]

i see. so if unspecified and the domain only has points to one side of the limit value, we can assume a one-sided limit from that side

[visual tiger]

yes, it depends on the domain

[scenic wyvern]

got it, noted this down

thanks for the insight

[quaint shoal]

okay I see how it works

thanks for the help guys!!!

[scenic wyvern]

!done

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.close

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[molten bay]

Hello guys i want to understand some concepts mixing with permutation combination (groups) partition and distribution

[molten bay]

okay so first doubt is

x+y+z=3 and solution is integer

x,y,z>=0

WHAT CAN BE THE ANSWER?

I know the formula but i want to realise it how it works

And formed

[lethal owl]

So you want the number of sets {x,y,z} that solves this question such that x,y,z are integers

@molten bay ?

[molten bay]

Yes

Okay let's do it with bars and stars method

1 1 1

102,201
012 021

003 030 300

[lethal owl]

Wait you said set not tuple

[molten bay]

What?

You can put comma

[lethal owl]

In a set the solution {0,1,2} is the same as {2,1,0}

For example

[molten bay]

No in the set

Then

[lethal owl]

If you want them to be different then we call it tuples (x,y,z)

[molten bay]

Okay

Tuples

If the correction and definition is complete will you see my work

I left two things

Total will be 10

210, 120

[lethal owl]

Ok so you set x = 0
Then count the number of pairs (y,z) that it have which is n-x+1 where n is just 3 in your case (the result of x+y+z)
Then do the same with x=1 then x=2 then x = 3

[molten bay]

Okay so formula is working fine

n+r-1 C r-1

[lethal owl]

So the formula is
$$\sum_{x=0}^{n}(n-x+1)$$

[grand pond]

Sherif Player

[lethal owl]

Which can be simplified into
$$(n^2+n)+(n+1)-\frac{n(n+1)}{2}$$

[grand pond]

Sherif Player

[lethal owl]

You can further simplify it if you wanted

So according to the formula there is 10 tuples (x,y,z) that solve your question

@molten bay understandable?

[molten bay]

Yeah

[lethal owl]

You understand how we got this?

If you want the most simplified formula then it is $$\frac{(n+2)(n+1)}{2}$$

[grand pond]

Sherif Player

[lethal owl]

Need anything else @molten bay ?

[molten bay]

.close

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[molten bay]

Thanks

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[floral ruin]

$\int\frac{sinx + 3cosx +1}{sinx - 3cosx - 1}dx$

[grand pond]

Prathmesh

[floral ruin]

How do I start with this?

[twilit field]

tan half angle?

[floral ruin]

sounds nice. lemme try

$\int\frac{2tan(u) - 2 tan^2u + 4}{2tanu + 2tan^2u -4}dx$

[grand pond]

Prathmesh

[floral ruin]

where u=x/2

I don't think if it got any easier

[feral sedge]

you do it like this

[floral ruin]

I don't understand this.

[feral sedge]

you replace sinx with 2t/(1 + t^2)
you replace cosx with etc…

[floral ruin]

that's what i did

tan half angle

[feral sedge]

no

your integrand should have no trigonometric expressions in it

[floral ruin]

ok so I have to substitute tan

it becomes $\frac{1}{2}\int\frac{t- t^2 +2}{(t+t^2+2)(1+t^2)}dt$

[grand pond]

Prathmesh

[floral ruin]

.close

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[hallow meteor]

How far reaching and general is the result that "the greedy algorithm leads to a set Y (subset of X) which maximise w(Y) for some weight function w if and only if pair (X, I) is a matroid"

[hallow meteor]

Is it true for all problems where greedy approach works(since that's what the "if and only if" seems to imply)

[mortal falcon]

https://cstheory.stackexchange.com/questions/21367/does-every-greedy-algorithm-have-matroid-structure

[hallow meteor]

[mortal falcon]

based on what this says, no, but matroid embeddings sound pretty close

[hallow meteor]

Yeah I'm reading up on matroid embeddings now

Does that mean this theorem is kinda wrong?

[mortal falcon]

i think there's an assumption of structure there

a generic greedy problem might not be able to fit in that maximum weight setting?

[hallow meteor]

[mortal falcon]

but for the ones that you can formulate like that, the matroid equivalence applies

is my guess

[hallow meteor]

Dijkstra has a min weight setting

By dijkstra I mean sssp problem

But is there matroid structure underlying it

[mortal falcon]

https://www.reddit.com/r/computerscience/comments/xrpka1/is_dijkstras_algorithm_really_a_matroid/

i suppose not 😭

maybe there is still a matroid embedding though....

[hallow meteor]

hmmm

[mortal falcon]

to be fair i think greedy algorithms are weird to characterize sometimes

[hallow meteor]

The usual cut paste argument for greedy choice property feels very neatly similar to the augmentation property of matroids

The wikipedia page on matroid embeddings is woefully short lmao

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[woeful turret]

could someone check my work? im pretty sure what i got is wrong

nvm i seem to have missed an x in the denominator

i need help solving this

given f(1) = 2

<@&286206848099549185>

anyone ? 😔

no?

okay

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[woeful turret]

how to do this using derivative

[blissful pier]

you can preform some algebra but the general term of the sum is [a_n=3\cdot\frac{3n^2-n}{6^n}]

[grand pond]

PajamaMamaLlama

[woeful turret]

oh so u have to find the general term

[blissful pier]

with this in mind, rewrite the sum:

[S=9\sum_n\frac{n^2}{6^n}-3\sum_n\frac{n}{6^n}]

[grand pond]

PajamaMamaLlama

[blissful pier]

recall that [\sum_n x^n=\frac1{1-x},\quad|x|<1]

[grand pond]

PajamaMamaLlama

[woeful turret]

yeah

[blissful pier]

now here's where the derivtive comes into play

[\sum_n nx^{n-1}=\frac{1}{(1-x)^2}]

[grand pond]

PajamaMamaLlama

[blissful pier]

multiply both sides by x

[\sum_n nx^n=\frac{x}{(1-x)^2}]

[grand pond]

PajamaMamaLlama

[woeful turret]

should there be a - sign or am i going crazy

[blissful pier]

let (x=\frac16) thus:

[S_1=\sum_n\frac{n}{6^n}=\frac{\frac16}{\qty(1-\frac16)^2}=\frac{1}{6}\cdot\frac{36}{25}=\frac{6}{25}]

[woeful turret]

oh ok im with u till here

[grand pond]

PajamaMamaLlama

[blissful pier]

it's 1/1-x

so when we use cahin rule

the -x deriv will cancel out the - from the reciprocal power

[woeful turret]

oh mb yeah

[woeful turret]

oh that is pretty cool

how did u see that

[blissful pier]

I studied this class of series (\sum_n\frac{n^p}{\alpha^n}) a while ago in like Sophomore year of HS, so I already had an advantage with all the tricks

[grand pond]

PajamaMamaLlama

[blissful pier]

[woeful turret]

ohhh damn

thanks a lot bro

[blissful pier]

oh and for the n^2/6^n, just take derivative of both sides again :)

[woeful turret]

although i dont know how much it will help me in the future because i wouldnt be able to notice this 💀

[blissful pier]

me and my buddy used the method I was showing the other guy for like many months

before I saw this

[woeful turret]

are u in college now?

[blissful pier]

so the intuition definitely takes time, but it's amazing when it hits

[woeful turret]

yeah agreed

well thanks again

.close

[midnight plank]

Channel closed

Closed by @woeful turret

Use .reopen if this was a mistake.

[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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[robust ibex]

alr i just need someone to double chekc my anwsers plz

[robust ibex]

im working with inverse functions (english is my second language so please correct me if something is confusing)

[pearl hull]

yeee

[robust ibex]

let me send a pic rq

oop the pic is upside down

uhrm

[pearl hull]

,rccw

[livid python]

,rrcw

[mortal mirage]

,rotate

[robust ibex]

thats like the first part of it right

i would like to now if im right?

and especially with the g`(x)

[keen widget]

its not in english i cant understand whats written

[robust ibex]

like do i just remove the 0/3 since it gives 0?

Uhrm lemme one sec i have athingy

[novel flax]

idts that the way u find inverse functions

[keen widget]

what ur doubt?

[robust ibex]

yeah its just finding the inverse functin

im unsure if i did it right

[keen widget]

ok

[robust ibex]

and i would like help understand the second part

[keen widget]

first of all u write the given expression and equate it to 'y'

like y=f(x)

[robust ibex]

the one im struggling with is f(x)=-x+1

[keen widget]

then u find value of x in terms of y

then u interchange x and y again

simple as that

[robust ibex]

oh

[keen widget]

yessir

[robust ibex]

im so confused T_T my teacher told me to do it like this

[keen widget]

like what

[livid python]

just one note write the inverse as f^-1(x) and not f'(x)

[keen widget]

yeah

[robust ibex]

he like made me find the inverse to the exact funktion

[keen widget]

f'(x) is derivative

[robust ibex]

like i show in the second pic

oooh wooops mb xd

i suppose those are meant to to be the inverse

[keen widget]

how did u write this?

[robust ibex]

but then he told me to like "input" the functions into those inversed functions

lemme send a pic

[livid python]

you mean as in (f^-1 o f) (x) to find if it gives x ?

[keen widget]

@robust ibex try doing the way i told

[robust ibex]

This is one of the ways

[robust ibex]

sorry i dont understand T_T

this is one of the earlier tasks he told me to solve and said i did right

[keen widget]

thats wrong

[stark tartan]

Do you mean f-1(x) instead of f'(x)?

[keen widget]

u should be getting f^-1(x) = log(x/3) with base 2

[robust ibex]

yes

[robust ibex]

yep

i misunderstood the f´´(x) to be the same as f^-1(x)

i mean f^-1

[robust ibex]

how?

[keen widget]

try the method i told

[keen widget]

starts from here^^

y=3 x 2^x

[livid python]

technically it's not, just bad use of notation/variables

[keen widget]

his final answer is wrong

it should f^-1(x) = log(x/3) with base 2

[livid python]

not really use log base change and you'll see it is exactly log_2(x/3)

[keen widget]

oh

yeah right

[livid python]

$\log_y{x}=\frac{\log{x}}{\log{y}}$

[grand pond]

qimmah

[keen widget]

yeah got that

[robust ibex]

ooop

[keen widget]

ur answer is correct but a lil bit clumsy in my opinion

[livid python]

understandable

[keen widget]

what u can do is
y= 3x2^x
y/3=2^x
x=log(y/3) base 2
interchange x and y
y=log(x/3) base 2, where y=f^-1(x)

[keen widget]

@robust ibex do u understand this method?

[robust ibex]

Wait so this is correct?

[robust ibex]

when u say base what do u mean? sorry im just trynna understand 😅

[keen widget]

uh the number written below log

[robust ibex]

oh ok okok ye i understand then

ty

[keen widget]

yup

[robust ibex]

for the correction

[robust ibex]

was this right then?

[keen widget]

how did u write the 2nd step

[robust ibex]

OH wait i gotta change -x to x

i forgor to change the - and +

right?

[keen widget]

ur given f(x) = -x +1
equate it to y
y=-x+1
x=y+1
interchange x and y
y=x+1, where y = f^-1(x)

[keen widget]

this the right way

[robust ibex]

oh

[keen widget]

its very easy and once u get it u will be fast with it

[robust ibex]

why do u not change the negatives/positives of the x and y when u interchange them?

[keen widget]

where?

[robust ibex]

here

like when u first change the x over u make it positive

[keen widget]

u just have to switch x and y with eachother dont worry about the negative or positive signs

[robust ibex]

but when u change it back u dont change it

[keen widget]

r u talking about the 3rd and 4th line?

[robust ibex]

im talkin bout both

like in the 3rd when u interchange them u change -x to x

and then in the 4th u dont change the values

it just stays the same

[keen widget]

i didnt interchange x to -x

i just transposed

as i told we need to find value of x in terms of y

so thats why i transposed to get x on one side and y on the other

then i just replace x and y with eachother

replacing meaning write x wherever y is written and write y wherever x is written

replacing and transposing is different

[robust ibex]

ah

i think i get it now

ty

[keen widget]

yeah

maybe try one more question and send a pic

and then u can talk to ur teacher about this method

this method is pretty easy

[robust ibex]

oki doki

[keen widget]

ping me whenever ur done

[robust ibex]

alrighty

[keen widget]

@robust ibex u did any question?

[robust ibex]

sorry got a bit caught up in it

this is right ye?

[keen widget]

yessss

[robust ibex]

WOOP

[keen widget]

amazing

[robust ibex]

ty for explaining xd

[keen widget]

u got it

[robust ibex]

and for the help

[keen widget]

no problem!

[robust ibex]

i was so confused by my teacher😅

[keen widget]

haha it happens

[midnight plank]

@robust ibex Has your question been resolved?

Channel closed

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

As a first year undergrad, how do I write this?

[livid python]

first off what is the definition of continuity ?

[languid veldt]

Yeah, the epsilon delta thingy. Can't I just directly write lhl=this thing and rhl=this thing

And both must be equal?

Or do I need to do some epsilon-delta stuff to obtain b?

[livid python]

no no just the lhl and rhl are enough

[keen widget]

just equate f(-2^-) = f(-2)

[languid veldt]

Ah okie.

Wait, so just to confirm,
$\lim_{x \to -2^{+}}f(x)=\lim_{x \to -2^{-}}f(x) \implies\lim_{x \to-2^{+}}bx^{2}=\lim_{x \to -2^{-}}x \implies 4b=-2 \implies \box{b=\frac{-1}{2}}$

Wait wut ._.

[scenic wyvern]

how did the limit get up there

[languid veldt]

Ahh I confused between the sum and limit thingies

[livid python]

try {x \rightarrow -2^+} without the ^

[keen widget]

b=-1/2

[livid python]

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[grand pond]

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

[languid veldt]

Ye..

Can I just write this?

[livid python]

yep looks good

[keen widget]

yesss

[languid veldt]

Thanks.

Thenk you all, mates. 🫡

.close

[midnight plank]

Channel closed

Closed by @languid veldt

Use .reopen if this was a mistake.

[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.

[floral ruin]

$\int\frac{sinx}{sin2x + 2}dx$

[grand pond]

Prathmesh

[floral ruin]

how to start with this?

[peak walrus]

Denominator can be simplified to ( cosx+sinx)^2 + 1 if that helps

[floral ruin]

i did that but what next?

[gaunt nimbus]

[ \int \frac{\sin x}{(\sin x + \cos x)^2 + 1} \dd x]

[grand pond]

k

[peak walrus]

hmm

Maybe t=tanx/2

[floral ruin]

I don't think so that it will simplify this ahead

[peak walrus]

Sorry?

[floral ruin]

I meant that it will become lengthy

[peak walrus]

,w integrate \frac{\sin x}{(\sin x + \cos x)^2 + 1}

[grand pond]

Results provided by WolframAlpha

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

Requested by yooyooyoyooyooyooyoo

[austere blade]

[peak walrus]

Just try a T=tanx/2 sub

[austere blade]

yeah t = tan(x/2) looks good

[floral ruin]

$\int\frac{4t(1+t^2)^2}{(2t+1-t^2)^2+ (1+t^2)^2}$

[grand pond]

Prathmesh

[floral ruin]

this is what I got

<@&286206848099549185>

.close

[midnight plank]

Channel closed

Closed by @floral ruin

Use .reopen if this was a mistake.

[digital relic]

hi guys

[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.

[unkempt skiff]

the top and bottom are thicker than the side of the can. How much more exact can we get?
is there anyway i can get the ratio of the r to h of the LID? then relate it even more to R to H of the can itself?

[past jay]

Hmm

Are there details to the question?

[unkempt skiff]

nope.

[scenic wyvern]

huh

is this a continuation of your other channel?

[austere blade]

it is probably

[scenic wyvern]

you probably should include more context, OP.

[unkempt skiff]

yeah

[unkempt skiff]

how?

i dont even have much context w

im just thinking of ways

of making it work

[scenic wyvern]

like, what can? what's its shape?

perhaps a diagram?

[past jay]

Is it a hollow cylinder with an open top that has an insertable lid, whose radius is R and height is H?

If so, should it have a uniform thickness? Perhaps a composite shape?

[unkempt skiff]

cylinder

[unkempt skiff]

yeah i guess

bro honestly

idk anything

cuz like

i ws given basically nothing

[past jay]

Hmm

Can you post the original question?

Not the one you typed

[unkempt skiff]

uh yea

[past jay]

The one from the paper you're trying to solve

[scenic wyvern]

a screenshot would help!

[unkempt skiff]

onesec

[past jay]

So is it like so?

Oh this is an optimization question

[unkempt skiff]

whats the top rectangle?

[unkempt skiff]

yep

[unkempt skiff]

is this a birdseye view lmao

[past jay]

That's the lid

[past jay]

This is a sectional elevation

[unkempt skiff]

wait whys it a rectangle

[past jay]

When you put the can in front of you, you see its silhouette as a rectangle

[unkempt skiff]

uhhh

oh

ohhhhhhhhhh

i see what you mean

yea okay

[past jay]

So this question seems like a research project of sorts

[unkempt skiff]

yea it sure is

💀

im so lost lol

like i did some work but so far basically nothing calc related, i just did some geometry stuff

[past jay]

Could you please post it?

[unkempt skiff]

my work?

[past jay]

The geometry stuff will help

[unkempt skiff]

sure

yikes

i odnt have nitro

one sec

[past jay]

Mood

[unkempt skiff]

the yap is for the TAs

im trying to be as specific as i can T-T

this literally comes down to basic algebra T-T

i feel like it's way too easy

and now im stuck at the thickness part

[past jay]

Hmm

You made a small mistake when talking about the percent wasted part

The top and bottom parts are cut in which arrangements, not in which sheet of metal

[unkempt skiff]

oh really

[unkempt skiff]

huh wdym

[past jay]

This is the rectangular/square grid

This is the hexagonal grid

The question aims to see which configuration produces the least waste

Yep I double checked the question and it means what I showed you

[unkempt skiff]

yea

so

1 circle vs 6 circle isnt ittt the sameeeeeeeee

[past jay]

Hmm

Let's assume square a metal sheet of side length 10R

In the square configuration, you can fit around 25 circles (10 radii / 2 radii per circle = 5 circles per side).

[unkempt skiff]

yea

[past jay]

The hexagonal configuration is a bit more complex. Lemme try to illustrate it

[unkempt skiff]

no worries

[past jay]

Alright this is the square grid

[unkempt skiff]

woww

ty

[midnight plank]

@unkempt skiff Has your question been resolved?

[past jay]

This took a long time. Here's the hexagonal grid @unkempt skiff

You'd need to calculate the loss in both cases. In the square config case, it will be the area of 1 circle multiplied by 20 deducted from the area of the square.

[unkempt skiff]

do i have to do it this way.. 😭

i appreciate it

[past jay]

Well, if you can find the area of 1 loss between 3 circles, you can multiply it by the amount you can count and the rest will be some small losses you can calculate with a bunch of trigs.

Now on to the cylinder body itself

If we wanted to find the surface area of the cylinder, what would it be?

[unkempt skiff]

2pir^2 +2pirh

[past jay]

$2\pi r^2 + 2\pi rh$

[grand pond]

VulcanOne

[past jay]

Ok these are 3 components

The top and bottom, which are each πr^2, and the cylinder body, which is 2πrh

Let's focus on the body

2πrh

From the looks of it, it's a rectangle with height h and length 2πr right?

@unkempt skiff lemme know if you're stuck anywhere in the problem

[midnight plank]

@unkempt skiff Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[scenic wyvern]

<@&268886789983436800>

[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>.
• Type the command .close to free the channel when you're done.
• 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.

[molten bay]

n^2+2n+4 is divisible by 7....find number of n less than 120

[zinc prism]

,close

.close

LOL

I thought I opened a help channel by mistake

Sorry for the interruption

[molten bay]

Lol

[lyric charm]

progress?

also is n natural or integer? (ie do we want to count only positive n)

[molten bay]

n is positive integer

I'm little bit scared of Ann, last time I got mutdr for 24 hours. And I couldn't ask questions

[zinc prism]

I can help as well

[mortal falcon]

as a hint, start with a modular equation

[zinc prism]

I just don't want to interrupt another fellow helper

[zinc prism]

% vibes

[zinc prism]

Yeah well, I would just start of writing a set that contains all possible values that are within (0, 120)

[lyric charm]

ngl the mute is easy to avoid. it was made clear to you why you got muted back then.

anyway n^2+2n+4=(n+1)^2+3

consider this

thinking in terms of modulo will be good

[molten bay]

Yeah but that guy's behaviour was not appropriate. It was like egoistic

[molten bay]

Yes. I was thinking like that

(n+1)^2 mod 7 is 4

n=1,4,8...

[zinc prism]

I would use a set here.

Do you know how to use sets?

[molten bay]

Sure

[zinc prism]

Cool

I don't see where you need help; you are doing everything right

[lyric charm]

he is a moderator he has authority

[molten bay]

I don't care if they are egoistic and not have proper behaviour. Same words to you also

I don't accept such behaviour. I feel bad

And I know how do you talk in helpers chat Ann....many people Dm me about your behaviour

[jaunty ivy]

The gratitude shown towards volunteering helpers

[molten bay]

Definitely no

[carmine sigil]

@molten bay if you feel like a moderator abused their authority, you may contact modmail. If you have a problem with another user, consider doing the same. But it's extremely counterproductive to make your help thread about your personal beefs.

and frankly it makes helpers not want to help you.

[zinc prism]

I do want to help him

[molten bay]

Thanks. For all the information. Yes I'm new here so people are roasting my behaviour for fun

[zinc prism]

That's Discord for you, mate

[hazy oracle]

Wait can anybody type in here?

[carmine sigil]

I assure you, that "roasting people for fun" is not something that normally happens on this server.

[zinc prism]

Hello. Yeah, anyone can

Just be respectful because you might interfere with someone else's help

[hushed mauve]

considering you have been warned about not making those types of statements at least three times afaik....

[hazy oracle]

Oh ok lemme not type in here

[molten bay]

I don't know why people stop,mute... I don't think it will change something

Everyone should have freedom to learn!!

Thanks everyone!! I will joke, fun with other user

Who will be comfortable in it!!

Maths is tough, hard sometimes it is depressing

So i do some little jokes so that it makes me feel good

[zinc prism]

Agreed

[pearl hull]

Can’t agree more.

[zinc prism]

but then you show off at your college and you feel better all of a sudden

[lyric charm]

let the record state that the reason for your mute was:

• you made a joke that i found rather disrespectful (you joked about me beating you with a hammer)
• i told you not to do it, as clearly as possible, 4 times
• you kept doing it anyway

[molten bay]

Should I make a list of possible outcomes

(n+1)^2 mod 7

[molten bay]

The joke was over me, over my mind

[carmine sigil]

it might be more helpful do a proof by induction

prove that equality holds for n=1

[lyric charm]

what induction

[molten bay]

4+3=7

[lyric charm]

he's being asked to find WHICH n make it work

[carmine sigil]

then assuming it works for some n, show it works for n+1

[molten bay]

n=0 fails

[lyric charm]

omni you are off the mark

[carmine sigil]

oh, I thought he was saying that it worked for all n (presumably 1 or larger)

[molten bay]

We are searching for (n+1)^2=4 mod 7

[carmine sigil]

and I didn't verify

[lyric charm]

it doesn't work for all n

[last slate]

[zinc prism]

No, he already found the solution

[molten bay]

next one is 4...8

[zinc prism]

(I think)

[zinc prism]

try to add them

[molten bay]

it will take 3 hours if i check all the numbers

[zinc prism]

add the possible values for n

[zinc prism]

oh

[molten bay]

n=1,4,8...pattern is 3,5??

[carmine sigil]

well, the numbers mod 7 are a field, so you just need to solve for n within the field, and then you can extrapolate to all values n + 7k

[zinc prism]

no

[molten bay]

1,4,8 satisfied no?

[zinc prism]

yes

[carmine sigil]

the next working value ought to be 11

can you see why?

[zinc prism]

try with n=1 and then n=4

those two should give you a pattern

[molten bay]

I tried

[carmine sigil]

(assuming that 1, 4 are the only working values between 0 and 6 inclusive.)

[molten bay]

z_7?

Which is cyclic group

7 is prime

120/7=17 value so we have 34 totals

[zinc prism]

YES

[molten bay]

Thanks andy

Andy with andy☺️

[zinc prism]

lol

and @carmine sigil

[carmine sigil]

ann also helped.

[molten bay]

Enjoyed!! Thanks Ann and omnipotent

☺️ 😊

Actually i have read the field, rings

But i don't know much like how we use in other problems

[zinc prism]

Well

[molten bay]

.close

[midnight plank]

Channel closed

Closed by @molten bay

Use .reopen if this was a mistake.

[zinc prism]

oops

[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>.
• Type the command .close to free the channel when you're done.
• 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.

[flat spire]

crazy name

[lyric charm]

casual chatter goes in #discussion or #chill, also definitely dont want a nickname like that on here

[olive matrix]

i am

and it's inappropriate

[lyric charm]

i am not "lil" nor your "bro".

do not call me "bro".

[olive matrix]

.close

[midnight plank]

Channel closed

Closed by @olive matrix

Use .reopen if this was a mistake.

[lyric charm]

did someone ping me here

oh looks like the trash took itself out

[midnight plank]

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[twilit jetty]

the multiple choice answers here heavily suggest doing u = sec x + tan x

[woeful turret]

yeah agreed

but i was not able to proceed after that

because its differentiation is not present right

[austere blade]

wdym by that?

[lyric charm]

what is d/dx (sec(x) + tan(x)) anyway

[woeful turret]

sec x tan x + sec^2x

[lyric charm]

u*sec(x) then

[woeful turret]

im sorry?

[lyric charm]

$\dv{u}{x} = u \sec(x)$

[grand pond]

Ann

[woeful turret]

oh ok yeah but we need sec^2 x right

[lyric charm]

yes but we can get another sec(x)

$\sec(x) = \frac12 \sqb{(\sec(x)+\tan(x)) + (\sec(x)-\tan(x))} = \frac12 \paren{u + \frac1u}$

[grand pond]

Ann

[twilit jetty]

oh wow

this is using the fact that $\sec^2-\tan^2=1$

[grand pond]

mtt

[lyric charm]

which is another rather unpleasant and dirty trick

but it cooks

[woeful turret]

oh damn 💀

thats kinda tough to notice

i think i can solve it now tho

thanks a lot

[woeful turret]

any inputs for how i could have got to this myself

[lyric charm]

god idk

[woeful turret]

alr alr 👍

ty

.close

[midnight plank]

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[midnight plank]

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[austere blade]

!status

[midnight plank]

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

[knotty lichen]

1

[past jay]

Start by calling a side to the small upright square

[midnight plank]

@knotty lichen Has your question been resolved?

[lethal path]

(or you could let x = 1, since it's asking about the fraction of area)

[lethal path]

1. Pythagoras to find the area of the largest square
2. area of triangle = 1/2 * base * perpendicular height

[midnight plank]

@knotty lichen Has your question been resolved?

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[woeful turret]

i took x sin x + 5 cos x = t

but i was not able to simplify numerator

[lethal path]

oh it's IBP again

[woeful turret]

oh man 😭

[flat spire]

ggwp

[woeful turret]

how to use ibp on this

nvm wait let me try

um

im not getting it using ibp

:(

[midnight plank]

@woeful turret Has your question been resolved?

[woeful turret]

ykw imma just close this

thanks anyway

.close

[midnight plank]

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[midnight plank]

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[viral dagger]

a board has $1,2,\dots,22$ written on it, you can do a "move", where you erase 2 numbers $a,b$ where $b\geq a+2$ and replace it with $a+1$ and $b-1$, find the maximum amount of moves possible

[grand pond]

ihave<skissue>

[viral dagger]

lemme write the solution

We will prove it generally, where you find the maximum amount of moves on a board with $1,2,\dots,n$, which would be $\frac{1}{6}\floor{\frac{n-1}{2}}\ceil{\frac{n+1}{2}}\cdot n$, where in our case the answer will be $440$

[grand pond]

ihave<skissue>

[viral dagger]

Define $s=(s_1,s_2,\dots,s_n)$ where $s_k$ is the amount of the number $k$ written on the board, here the starting case is $S=(1,1,\dots,1)$. We will define a value of $s$ as $f(s)=\sum_{k=1}^{n}s_kk^2$. An $s$ can change into $t$ with a move with choosing $a$ and $b$ where $a<b$, then you will get
$$f(s)-f(t)=a^2+b^2-(a+1)^2-(b-1)^2=2(b-a-1)\geq 2$$
with equality at $b=a+2$

[grand pond]

ihave<skissue>

[viral dagger]

with the inequality above and $f(t)>0$, it will end at some case $T$ (in all honestly, i dont really get this either, but im p sure its basically saying it ends at $T$)
at the end case $T=(t_1,t_2,\dots,t_n)$, then either all of the numbers are equal or theres a pair neighbour (like they are side by side) that has all the numbers. If $n=2m$ is even then we can let $t_m=t_{m+1}=m$ and $t_k=0$ for $k\notin{m,m+1}$, and if $n=2m+1$ is odd then $t_{m+1}=n$ and $t_k=0$ for $k\neq m+1$. From the inequality, we get that the upper bound is
$$\frac{f(S)-f(T)}{2}=\frac{(m-1)(m+1)\cdot 2m}{6}$$
from the claim at the start, we can show that wether $n$ is odd or even, this upper bound is reachable

[grand pond]

ihave<skissue>

[viral dagger]

im not going to lie this is terribly written, heres op just in case

what im asking is the last line, how do you prove the upper boud is reachable

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

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[viral dagger]

gtg sleep sorry

[chilly cobalt]

buh

[midnight plank]

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[past jay]

<@&268886789983436800>

[midnight plank]

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[hallow meteor]

For a relation to be a function

[hallow meteor]

Does every element of domain have to be mapped to exactly one element in co domain?

If yes, are partial functions not true functions?

[lavish venture]

yes

[lavish venture]

yes

as in partial functions are functions

[hallow meteor]

?

They are?

Or are not

[lavish venture]

they are

[hallow meteor]

Then it doesn't match this definition

[lavish venture]

sure it does

the domain is just a subset of what you had beforehand

[hallow meteor]

"what you had beforehand" as in?

[lavish venture]

if you have some set X such that f does not define a function with X as the domain then you can restrict the domain of f to some subset A of X so that f is a function on A

[hallow meteor]

We can never know what A is though beforehand

[lavish venture]

i mean for example f : R to R, f(x) = 1/x is nonsense because f isn’t defined at x = 0 so we can simply restrict our domain to R \ {0}

i don’t know what you mean know what it is beforehand

[hallow meteor]

Yeah i was working with the same example

[lavish venture]

we remove problematic points

[hallow meteor]

I mean we may have a function defined over a domain,

But not know it's behavior over the entire domain

So we can never truly restrict

[lavish venture]

this seems like you’re asking a computational question

[hallow meteor]

[lavish venture]

yea wikipedia says there is no algorithm for this

[hallow meteor]

Yea that's the halting problem yes

Hmmm

So basically the partial function idea in math only exists to have some sort of uniformity

Like defining square root function uniformly with the way we define square

Both over R-> R

Instead of squeezing the domain of sqrt function acc range of square func

I suppose this is correct line of reasoning?

[lavish venture]

i’m not sure what you mean by uniformity here

[hallow meteor]

I mean to have less hassle

With funcs and their inverses

But in reality there's no reason to have partial functions if we aren't talking computability

We can just define the domain exactly so that all elements map to something

If we want

[lavish venture]

i just view it as making it well defined but of course there are reasons for making such restrictions as you mentioned

[glacial slate]

most of the set theory courses I have seen define functions as functional (elements of the domain are related to at most 1 element of the codomain) and total (same, but at least 1 element) relations

so partial functions would not qualify as functions under that definition

I think as long as you're clear and consistent about your definitions you can use whichever convention

[hallow meteor]

Hmm this was my fear

[hallow meteor]

Makes sense

[glacial slate]

what's the context in which this distinction matters?

because if you're doing computability theory, that's perfectly fine

and you can write "total function" to disambiguate

[hallow meteor]

Like my doubt was regarding of "partial function" has any merit in math aside from computability

If*

[glacial slate]

set theory, I guess

[hallow meteor]

In set theory that won't be a function no?

[glacial slate]

sure, but we're just changing the name of something?

doesn't make it any different or less useful

[hallow meteor]

Makes sense

But "partial functions are not true functions" is a true statement yes?

[glacial slate]

if you take "function = functional + total relation", then yes

if you're in computability theory, you might disagree (e.g. general recursive functions are partial functions N->N in general)

it depends on what you're doing

[hallow meteor]

Hmm

Fair enough

Thanks

.close

[midnight plank]

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[midnight plank]

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[hallow meteor]

Is there any proof/intuition as to how we can express all information within any sort of data structure like a graph or something as a sequence of natural numbers?

[lyric charm]

"all information"?

"express"?

this has got to take the award for vaguest question ive seen this month

[hallow meteor]

I mean how do we encode any data structure as a sequence of natural numbers

[lyric charm]

i might regret asking this but: what do you understand by "data structure"

[hallow meteor]

A way of storing data

And why would you regret asking

[lyric charm]

cause that was completely meaningless

like you apparently want formal proof but theres no definitions for any of the words you're using

and if you want intuition, then i guess uhh

https://www.youtube.com/watch?v=PMpNhbMjDj0

[delicate sage]

We can't actually, the cardinality of the set of real numbers doesn't equal the cardinality of the set of naturals.

If there is no bijection, it's impossible to define such a mapping. Well, this follows from cantors theory of cardinality.

[lyric charm]

a single real number admits a representation as an infinite sequence of naturals

but i strongly doubt that this is what op was looking for

[delicate sage]

Yeah, i should've asked for clarity, nvm

[hallow meteor]

For an example as to what I was wondering: how would we encode a weighted graph as a sequence of natural numbers

[lyric charm]

are its weights also natural

[hallow meteor]

Hmm I would assume they are not real, so from any set with bijection to naturals (countable)

Rationals would count I suppose

[lyric charm]

here is a very crude encoding then

- 1st number: vertex count [vertices are implicitly numbered from 1 to vertex_count]
- 2nd number: edge count
- a series of 3*[edge_count] numbers, arranged in trios representing each edge: starting vertex, beginning vertex, weight

[delicate sage]

What do you mean, by 'not real'

[hallow meteor]

I mean not from the set of Reals

Or any set which is uncountably infinite

[hallow meteor]

[delicate sage]

Uhh, real numbers are the superset of rationals,integers and naturals

[delicate sage]

All the mentioned sets are uncountable and infinite, though.

[hallow meteor]

No?

[lyric charm]

does this answer your question @hallow meteor

[hallow meteor]

Naturals, rationals, etc are countably infinite

[hallow meteor]

Yes I'm thinking over it

Sorry

[lyric charm]

theres kinda nothing to think about is there

it is laid quite bare

[delicate sage]

Oh well, I overlooked

[hallow meteor]

I mean I have a follow up

[lyric charm]

ok then ask your follow-up

[hallow meteor]

Which is that how would we encode something like dijkstra as a composition/primitive recursion/unbounded minimisation over succ zero and proj functions over that sequence of naturals (in a crude way/ even crude explanation works, I just want a sort of understanding)

[lyric charm]

why would you want to do that explicitly

[hallow meteor]

Not explicitly, but even a vague idea

Just so that I can kinda relate mu recursion to usual computation

[lyric charm]

it is probably going to take an unpleasantly long time to do that

[hallow meteor]

Hmm

[lyric charm]

recommend against it

or i mean if you wanna waste your time go right ahead

[hallow meteor]

Is there any example(which shows up in usual computation) which would require a pleasantly short time

Hmm well thanks for your help, I guess I'm going off in a wrong direction here

.close

[midnight plank]

Channel closed

Closed by @hallow meteor

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[glacial slate]

I would recommend reading through the arithmetisation chapters of the open logic textbook if you want to see general recursive functions written out explicitly

[hallow meteor]

.reopen

[midnight plank]

✅

[hallow meteor]

Which textbook?

Open logic?

[glacial slate]

just Google open logic project

I'm not at my computer but it should be the first result

[hallow meteor]

Also I had another question

Will restricting the axioms to just the 3 elementary functions(proj, zero, succ) and just function composition, yield presburger arithmetic?

[glacial slate]

as in, no primitive recursion?

[hallow meteor]

Yeah

And ofc no unbounded minimisation

[glacial slate]

you can't define addition either then

so you can't get presburger arithmetic