help-27

[devout trench]

correct me if i'm wrong

[solar goblet]

yes that is correct

can you do this now?

[devout trench]

ohh

okay wait

i substitute the value of b in this?

or should i do a different one

like uh

oh nvm still the same

which is?

[solar goblet]

i want you to calculate it yourself

and tell me the answer

[devout trench]

i substituted the value of b^2 into the equation

idk if it's right

(a+c)^2 + c^2 = 2a^2?

[devout snow]

@devout trench Has your question been resolved?

[devout trench]

yo @solar goblet am i right?

i've been watching spiderman

lmao ghosted me

my take is 45 degrees

idk if i'm wrong

[glacial drift]

hi

[solar goblet]

oops sorry i was playing pjsk

[solar goblet]

yes you are on the right track

[devout trench]

2ac + 2c^2 = a^2

[solar goblet]

now calculate c in terms of a or b

[devout trench]

lmao i'm still watching spiderman

[solar goblet]

stop

this is not something to be proud of

do this first, THEN watch spiderman

[devout trench]

then what

[devout snow]

@devout trench Has your question been resolved?

[devout snow]

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[devout snow]

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[plain lance]

Having trouble conceptually / visually understanding how moving a negative exponent to the numerator and vice versa denominator becomes positive with integer #'s

[hollow pulsar]

u mean a^(-n) = 1/aⁿ ?

[plain lance]

Yup

[willow dew]

just think harder

[hollow pulsar]

its actually a definition

but some people prove it too

[lost laurel]

hint||: multiplicative inverses ||

So essentially if I have $a^n$ , what's its multiplicative inverse ?

[woven radish]

ƒ( wai ina teacup)= I don't know

[plain lance]

$a^-n$

[woven radish]

Lupa

[lost laurel]

$a^{-n}$

yup

[woven radish]

ƒ( wai ina teacup)= I don't know

[lost laurel]

Now, we define $a^{-1} = \frac{1}{a}$

or alternatively

multiply $a^n$ by $1/a^n$

[woven radish]

ƒ( wai ina teacup)= I don't know

[lost laurel]

what does that give you?

Isn't 1/a^n a multiplicative inverse too?

[lusty sapphire]

(you may want to talk about $a^0$ first)

[woven radish]

SWR

[lost laurel]

Fair, but I pressume OP know that that is defined to be 1

Now, multiplicative inverses are unique

what does that tell you

[plain lance]

Not sure on this one

[lost laurel]

Hmm

if I divide a number by itself

what do I get

[plain lance]

1

[lost laurel]

recall $\frac{a}{b}$ means I'm dividing a by b

[woven radish]

ƒ( wai ina teacup)= I don't know

[lost laurel]

that's better IMO

[plain lance]

Think i got it

thank you

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[naive mantle]

This is a physics problem indeed, but the work is mostly mathematical and it revolves around a few formulas from Ohm's law on a simple circuit

[naive mantle]

16. A battery powers a resistor with an electric current of intensity I = 24A. If the resistance is increased by f_1 = 25%, the current intensity decreases by f_2 = 20%. Determine:

a. The new value of the electric current intensity in the circuit if the resistance is decreased by f_1 = 25%.

b. By what percentage does the voltage across the resistor change in the case where the resistance decreases compared to the case where it increases?

c. By what percentage does the resistance of the conductor change if the temperature varies by Δt=80°C, given that the thermal coefficient of resistivity is α=6*10^(-4)°C^(-1) and the initial resistance was at a temperature of 0°C?

[merry patio]

which question are you stuck at?

[naive mantle]

Honestly, a

[merry patio]

well you know that the battery is always the same so V is constant

[naive mantle]

Yup

[merry patio]

therefore IR must be equal to IR after reduced

[naive mantle]

So I1R1 = I2R2 and we just adapt to f_1?

[merry patio]

well im not good at this so ill let someone else help

[merry patio]

that would be my guess

[naive mantle]

Sure, thanks for the efforts

For reference:

V = IR
I = E / (R + r)

[devout snow]

@naive mantle Has your question been resolved?

[devout snow]

@naive mantle Has your question been resolved?

[naive mantle]

<@&286206848099549185>

[devout snow]

@naive mantle Has your question been resolved?

[devout snow]

@naive mantle Has your question been resolved?

[devout snow]

@naive mantle Has your question been resolved?

[solid osprey]

V=IR
V_1=24R_1
V_2=24R_2
(V_2-V_1)/V_1=25/100
(R_2-R_1)/R_1=20/100
find R_1 interms of R_2 and V_1 interms of V_2 no?

[twilit goblet]

whats the problem

i think the have explained enough

[devout snow]

@naive mantle Has your question been resolved?

[naive mantle]

Yea! I think I got it. Thanks!

.close

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[karmic hill]

what kind of curve can i use to fit into this distribution?
i tried gamma and it didnt converge
gaussian area, gaussian amp, bigaussian, lorantz all look weird

[valid silo]

can you tell how you generated it?

[karmic hill]

this is from an afm image

welp id say its simular to a lorantz distribution

[devout snow]

@karmic hill Has your question been resolved?

[vivid estuary]

poisson maybe?

[sage burrow]

lognormal?

[devout snow]

@karmic hill Has your question been resolved?

[devout snow]

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[restive river]

I asked chatgpt how to find the inverse of a 3x3 but it's giving me a long method, but this question is only worth 3 marks

Is there a quicker way?

[tall stirrup]

Don't ask ChatGPT for answers, currently AI is not that advanced and can give wrong answers confidently

[karmic hill]

fastest way is just LM=1

cause you know M=L^(-1)
ML=L^(-1)L=1

[polar chasm]

I'd just use gaussian elimination

its lower triangular, so it should be quite easy

[restive river]

Ahhhh yesss

I remember this

Thank you all for your inputs

.close

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[devout snow]

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[small urchin]

$\mathbf{t}_1 = \begin{pmatrix} 1/2 \ 1 \ 1 \end{pmatrix}, \quad
\mathbf{t}_2 = \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix}, \quad
\mathbf{t}_3 = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}$

[woven radish]

Task Bot

[small urchin]

They're not linearly independent, right?

$t_1=-2t_3 , t_2=t_3, 0=0$

[woven radish]

Task Bot

[zenith jacinth]

t_1 isn't equal to -2 t_3

and t_2 and t_3 aren't equal too

[wheat cave]

btw there is an obvious linear combination between them

[zenith jacinth]

t_1 = 1/2 t_2 + 1/2 t_3

[vivid estuary]

yeah these vectors are linearly dependent

[small urchin]

Why?

T1/2+t3=0 -> T1=-2t3

[zenith jacinth]

-2 t_3 doesnt give t_1

you can calculate and see for urself

[zenith jacinth]

since we can write t_1 as a linear combination of t_2 and t_3, they are linearly dependent

[small urchin]

I don't understand where I went wrong

[zenith jacinth]

where does this come from tho

[small urchin]

This

[zenith jacinth]

t_1 definitely isn't equal to -2 t_3

its obvious

[small urchin]

????

$\frac{t_1}{2}+t_3=0$

[woven radish]

Task Bot

[dawn lintel]

Why do you say $t_1=-2t_3$

[woven radish]

Eternal Night

[zenith jacinth]

and where does this come from

[dawn lintel]

This is wrong

[small urchin]

From this

[zenith jacinth]

then it's wrong

[dawn lintel]

Bruh

[zenith jacinth]

you can verify

[small urchin]

I think I didn't make any mistakes

[zenith jacinth]

$-2 \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} -2 \times 1 \ -2 \times 0 \ -2 \times 1 \end{pmatrix} = \begin{pmatrix} -2 \ 0 \ -2 \end{pmatrix}$

[woven radish]

Herels

[zenith jacinth]

that thing is not t_1

[small urchin]

Ops

I think I asked the question wrong

Sorry

$\mathbf{t}_1 \begin{pmatrix} 1/2 \ 1 \ 1 \end{pmatrix}+ \quad
\mathbf{t}_2 \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix}+ \quad
\mathbf{t}_3 \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}=\begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}$

[woven radish]

Task Bot

[small urchin]

@zenith jacinth

[zenith jacinth]

now it makes sense

[small urchin]

So are they linearly dependent?

Because

$t_1=-2t_3 , t_2=t_3, 0=0$

[woven radish]

Task Bot

[zenith jacinth]

t_1 = -2 t_3
t-1 = - 2 t_2
t_1 + t_2 + t_3 = 0

thats what im getting

[small urchin]

Yes

[zenith jacinth]

t_1 t_2 and t_3 aren't equal to 0, so the vectors aren't linearly independant

[small urchin]

So there is no basis for the image of f?

[zenith jacinth]

it doesnt mean that

you only took 3 vectors

[small urchin]

https://math.stackexchange.com/questions/2647868/find-basis-of-image-of-a-matrix

I'm following this

For vectors to be a basis they need to be:

linear independent
span the subspace

And the first point comes off as false to me

[zenith jacinth]

well, yea since we can write this :

$$\begin{pmatrix} 1/2 \ 1 \ 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}$$

[woven radish]

Herels

[zenith jacinth]

the first vector can be write as a linear combination of the two others

would make sense that the other two vectors themselves formed a basis

since they are linearly independent

[devout snow]

@small urchin Has your question been resolved?

[devout snow]

@small urchin Has your question been resolved?

[somber heron]

A^-1=adj(A)/det(A)

2t_1 -t_2=t_3

[devout snow]

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[patent geyser]

What do y’all get

[patent geyser]

I got 68 but I feel like that’s wrong

[devout snow]

@patent geyser Has your question been resolved?

[last burrow]

You're right

You're right that you're wrong.

I got 32.

[patent geyser]

How

[valid silo]

i also got 68

you are right

i did

[devout snow]

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[valid silo]

.reopen

@patent geyser

[patent geyser]

How?

[patent geyser]

Are you 100% sure

Maybe there’s two solutions

Maybe it’s 68 or 32

[devout snow]

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[last burrow]

oh

[valid silo]

68 for certain

reopen it quickly

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[old jungle]

how do we know its a rotation of 120

[devout snow]

@old jungle Has your question been resolved?

[restive river]

just compute

A²

A ³

A ⁴

and see the difference

and how it rotates

[old jungle]

.close

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[ocean gale]

Annie and Xenas each arrive at a party at a random time between $2:00$ and $4:00$. Each stays for $45$ minutes and then leaves. What is the probability that Annie and Xenas see each other at the party?

[woven radish]

Ryxo

[ocean gale]

I’m supposed to use geometric probability somehow but idk how

[devout snow]

@ocean gale Has your question been resolved?

[ocean gale]

<@&286206848099549185>

Nvm I got it

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[restive river]

can anyone say me the answer for this question?

[restive river]

so according to me the answer should be (d) 0

because

my teacher said that we will not consider the edges with directions while calculating the degree

so can someone please say the correct thing?

please tag me if anyone answers

[restive river]

in this question they havent mentioned anything about whether we have to calculate indegree or outdegree

if it was indegree then the answer was supposed to be 1 and similarly for the case of outdegree

but how do we calculate the degree?

<@&286206848099549185>

[safe jasper]

afaik if they specified indegree or outdegree then you'd consider the directionality, but if they just say degree then you'd treat it like any other graph

as in just the number of adjacent edges

regardless of directionality

[restive river]

so what should the answer be in this case? 2?

[safe jasper]

that's what i would think

[restive river]

okay thank youu 💛

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[ripe plank]

• The expression: “ a | b^b | (-c) => a|c “ is true, sometimes true, false

[alpine python]

seems true,
you don't even need to use b to show it's true

[ripe plank]

yea i also cant find an example where its wrong

[alpine python]

a | -c means there's some k with ak = -c
so a(-k) = c
so a divides c

[ripe plank]

yea tgx

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[restive river]

Integration by parts, correct?

We would have to take 1/(cos x)^2 as the part to take the antiderivative of

bruh

as im explaining it to myself im understanding it

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[hollow pulsar]

two last questions

[olive snow]

Ça veut dire quoi les flèches

Implique ?

[hollow pulsar]

nn nn

les limites sont égales

en +inf

car ce sont des suites

[olive snow]

Ok

Bon déjà Un est constante

[hollow pulsar]

et donc?

ouf, mais non, j'ai écris les suites en faux

[olive snow]

Pour la deux

[hollow pulsar]

je me suis trompé

[devout snow]

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[hollow pulsar]

oh wait no

ill get another channel

[olive snow]

Refais en un

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[blazing crest]

i dont know what to do

[supple knot]

Do you know derivative rules

,w .diff rules

[blazing crest]

to a certain extent

i feel like they just asked me the same question thrice

[supple knot]

,tex .diff rules

[blazing crest]

but just worded differently

[woven radish]

riemann

[blazing crest]

idk Leibniz but the rest im familiar with

[supple knot]

Just use power rule then

[blazing crest]

do i have to say for x smaller than 1 the derivative is 2x

and for bigger than 1 the derivative is also 2x

[supple knot]

Yes

[blazing crest]

damn

and for b do i use the definition of derivative

f(x) - f(c) / x-c and x approaches c

for both cases, 2x when x approaches 1 is 2

is that it?

[supple knot]

Missing parentheses but sure that works

[blazing crest]

what are those

[supple knot]

(a-b)/(c-d)

() are parentheses

[blazing crest]

Oh ye mb

And how does c work?

[supple knot]

Read b) and compare to the definition of derivative in your book/notes

c should be inferred by matching the definition

[blazing crest]

im lost

i get the definition but how does this correlate with c

[supple knot]

Oh you don't need the definition of derivative for b)

just plug in 2x for f'(x) into the limit

[blazing crest]

can we do

f(1) = 1

and lim x goes to 1 of x^2 + 1 = 2

so there is a jump

and not continuos

not continuos on 1=> not differentiable on 1

[supple knot]

That works yea

[devout snow]

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[thick schooner]

is the wording on this ambiguous?

aren't they asking for maximum spanning (length) interval

i actually don't know how to phrase it

[feral agate]

they are

[thick schooner]

but would you say this question is posed well?

i'm guessing they want 1 < x < 7 but okay i'm more concerned about how they phrased the question

[feral agate]

it would be better if they said
If the possible values of x lie in (a, b) then find b-a

[thick schooner]

yeah okay smth like that

maybe all possible values instead?

[thick schooner]

does this question work as is though

ngl makes no sense to me

[feral agate]

it does

[thick schooner]

if someone didn't know triangle inequality

they'd understand it?

i guess maybe yeah

[alpine python]

i feel like they should just ask for the maximum and minimum possible values of x

[devout snow]

@thick schooner Has your question been resolved?

[topaz axle]

it doesn't make sense

failure of communication

[thick schooner]

it doesn’t?

[thick schooner]

yeah i think that’s what they’re going for

there’s also the bias that i kinda know what they’re testing

i wanted to clarify whether the way they’ve asked the question could be handled by someone unfamiliar with this topic

[topaz axle]

it's absurd

[arctic field]

it sure is poorly phrased

[devout snow]

@thick schooner Has your question been resolved?

[thick schooner]

alright thanks guys

[devout snow]

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[devout snow]

@restive river Has your question been resolved?

[orchid axle]

i’ll take a look

[tall stirrup]

I think I know the answer but not sure how to articulate this

[orchid axle]

problem is wrong

[restive river]

Nope

[orchid axle]

impossible

[restive river]

Why?

[tall stirrup]

This is possible the answer i got is 30

[restive river]

Yup, that's the answer

[orchid axle]

wait or not

i missed something

[tall stirrup]

I had solved a similar problem while ago

Can you draw a diagram

[orchid axle]

answer is 30???

[tall stirrup]

Yeah

I got the problem I solved its from Indian National Mathematics Olympiad 2016

[restive river]

1/(1/5 - 1/6)

[orchid axle]

i was messing around in geogebra

and constructed this

A is (0, 160)

decimals might be off but there is a triangle that satisfies the problem around here

[tall stirrup]

https://olympiads.hbcse.tifr.res.in/wp-content/uploads/2016/09/sol-inmo16.pdf

You can refer to any solution to Q.5

[restive river]

thank you so much

🙂

.close

[devout snow]

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[orchid axle]

can you construct the triangle and send it

please

turns out it actually looks like that

A and C do not lay on rational coordinates

[devout snow]

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[sand haven]

How’d you do this?

[sand haven]

^3sqrt(64) is 4

Sqrt(625) is 25

[orchid axle]

so its sqrt(4/25)

which is sqrt(4)/(sqrt(25)

which is 2/5

[sand haven]

Omg!!!

Thank you

This is sooo easy

[devout snow]

@sand haven Has your question been resolved?

[devout snow]

@sand haven Has your question been resolved?

[sage burrow]

why not solved then?

[devout snow]

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[harsh drum]

Where does the 2root2 over 3 come from?

[tall stirrup]

sin x =1/3

[woven radish]

Pro_Hecker

[devout snow]

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[faint gorge]

Let $G$ be an abelian (commutative)$^{}$ group and $n \in \mathbb{N}.\$
Show that the map [ \Phi_n : G \to G, \quad \Phi_n(u) := u^n ] is a group-homomorphism.
$\ \ \textbf{Proof.}\$
Let $a,b \in G.\$
Then [ \Phi_n(ab) = (ab)^n \overset{}{=} a^nb^n = \Phi_n(a)\Phi_n(b) \quad \text{for all } a,b \in G. ]
Hence $\Phi_n$ is a group - homomorphism.$\ \$
Which elements of $G$ form the kernel of the map $\Phi_n? \$
Proceed to give a non-trivial group $G$ and natural numbers $n_1, n_2 > 1 \$ such that $\Phi_{n_1}$ is $\textbf{ an isomorphism }$ and $\Phi_{n_2}$ is $\textbf{not an isomorphism.} \$

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

I need help on how to tackle the next part with the kernel and isomorphism stuff

My problem is I don't know whether G is finite or not too

And I am afraid I would need to do cases

The kernel is [ \ker(\Phi_n) := {u \in G \mid u^n = e, : n \in \mathbb{N} } ] where $e$ denotes the neutral element of $G.$

[woven radish]

𝔸dωn𝓲²s

[stone stump]

you are allowed to choose the group G for your examples

[hot steeple]

Without any more info about G it is fine to keep the kernel non specific

[faint gorge]

So before I start, I choose an abelian group and then proceed my work from there?

[hot steeple]

For this you can just choose a G

For this you gave the answer as specific as is possible without more information about G

[faint gorge]

So I literally just write down the definition of the kernel

[hot steeple]

well no

You specified that the elements have to be to the power of n and be equal to the identity

instead of saying

the kernel of f is all u for which f(u) = e

you say

[faint gorge]

ok i get it

[hot steeple]

the kernel of f is all u for which u^n = e

[faint gorge]

ok

good

you know you remember yesterday

where we specified n

like n = ord(a)k, k in N or something like that

[hot steeple]

That works if the order of a is finite

[devout snow]

@faint gorge Has your question been resolved?

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

I hope this is fine

[hot steeple]

ye

That's fine

Also a not obvious choice i like it

[faint gorge]

thank

.solved

[devout snow]

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[devout snow]

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[faint gorge]

,, 19^{290} \overset{37}{\equiv} : ?

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

I am kind of clueless how to approach this

[frozen aurora]

37 is prime

[frozen aurora]

you should try and use fermat's little theorem here

[faint gorge]

ok will

[devout snow]

@faint gorge Has your question been resolved?

[faint gorge]

28

I spent so much time cant believe it I hate numbers

It's not even about being genius

[ 19^{290} = 19^{36 \cdot 8} \cdot 19^2 = \underbrace{(19^{37-1})^8}_{\overset{37}{\equiv} 1} \cdot 19^2 = 361 = 37 \cdot 9 + 28 \overset{37}{\equiv} 28 ]

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

rip

,w 19^(290) mod 37

[woven radish]

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Requested by adoniszgks

[faint gorge]

.close

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[devout snow]

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[restive river]

anyone?

[restive river]

group terms with common factors

[restive river]

the mark scheme grouped it like this

but that doesnt make sense to me considering it grouped the ABC term twice

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[undone sequoia]

Just curious on what the optimal approach for questions regarding permutations and combinations is.
Also I'd like to understand multinomial expansions and how to solve questions such as:-

the remainder of (7^2022 + 3^2022)÷5

Last digit of 3^104

[undone sequoia]

Last two digits

My teachers done something like (10-1)^52

Binomial expanded and I don't know what after. No idea why though

[alpine python]

i would try something like

[crisp tapir]

10^x gives you 0 in the units place thats why they do that

so essentially only the last digit matters

[undone sequoia]

I could also use some kind of pattern could I not?

this method is quite interesting though

[crisp tapir]

wdym

oh right for 3 you can i get you 3 9 7 1 right?

[undone sequoia]

3^2 has last digit as 9, 3^3 has 7, 3^4 has 1, 3^5 has 3 again

[undone sequoia]

Yeaaaah

[crisp tapir]

and 104 is 4k type so it will be 1

[alpine python]

$7^{2022}+3^{2022}=\left(7^4\right)^{505} 7^2 + \left(3^4\right)^{505} 3^2 \equiv 1^{505} 4 + 1^{505} 4\ (\textrm{mod}\ 5)\equiv 3\ (\textrm{mod}\ 5)$

[undone sequoia]

can I use that for last two digits?

Also does the method we're talking about work for say 10-2 or 9-3? Or is it valid only for this situation?

[woven radish]

Axe

[undone sequoia]

What'd you do after the second part

[alpine python]

look at the powers of 7 mod 5:
7 is 2

multiply 2 by 7 to get 14 which is 4

multiply 4 by 7 to get 28 which is 3

multiply 3 by 7 to get 21 which is 1

so powers of 7 go in a cycle: {2, 4, 3, 1}

and 7^4 is 1

so u just replace 7^4 with 1 in the equation

[crisp tapir]

you can use bt for the last 2

[undone sequoia]

Oh alright. give me a second I'll write it down

Got it

[undone sequoia]

I'll read this tomorrow morning I'm so sorry

Thank you so much both of you

[devout snow]

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[devout snow]

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[devout snow]

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[upbeat stone]

Aaa

[upbeat stone]

I was trying to find derivative df/dx of f(x) = sqrt (x) by using geometry

So if u make a square with sides sqrt (x), the area of the square is x

So if u increase the sides by a little amount (df), we wanna know how much we increase the area (dx)

So we added df.(sqrt(x)) + df.(sqrt(x)) + df^2 which is equal to dx

But df is approaching 0, not dx, and idk what to do

May anyone help please?

[olive snow]

uhm

[sage burrow]

i dont even understand what you are trying to do.

[upbeat stone]

Aaa

My Wi-Fi is not working

[upbeat stone]

Hmhmhm

Idk how i can explain in a different way

[upbeat stone]

Yep

[wicked turtle]

try rearranging the above equation to get d(sqrt(x))/dx

[wicked turtle]

with the usual observation that (d something)^2 is very small and can be neglected haha

[upbeat stone]

Ye if i do this it works

But idk why

[olive snow]

indeed

[upbeat stone]

I dont think it could be ignored but idk

[wicked turtle]

well you don't know why because it's not rigorous, it's a physicist-style argument

[olive snow]

it can

:)

[wicked turtle]

to understand why you need to formulate it properly in terms of limits

[upbeat stone]

Breh

[upbeat stone]

I would understand it if you had dx / df

Then df approaches 0 and u can eliminate everything that has df

[upbeat stone]

Can you? Please

[wicked turtle]

you want me to write down the limit you need to evaluate?

[upbeat stone]

I did not learn limit so idk

[olive snow]

d.sqrt(x) -> 0 and (d.sqrt(x))² goes faster to zero

[wicked turtle]

oh well then it won't help you if i do that

[upbeat stone]

Indeed but...

[olive snow]

without limit its the idea to have in mind

[upbeat stone]

Thats not so rigorous

[wicked turtle]

but dx is the thing you're making small here, right?

[olive snow]

otherwise i don't know any sol without this

[upbeat stone]

No

[wicked turtle]

it is, if you're finding df/dx

[upbeat stone]

Yeah

[wicked turtle]

which is your goal

[upbeat stone]

But df is the small thing

[wicked turtle]

they're both small

rearrange your equation to get it in the form df/dx instead of dx/df

[upbeat stone]

Df is the one approaching 0

[wicked turtle]

they're both approaching 0

that's how derivatives work

you want to find out what limit their ratio approaches

[upbeat stone]

Hmmmm

Thats true since dx has df multiplying it

So it also approaches 0

So can i find dx / df and then do 1 / it?

[wicked turtle]

yep

that's valid in general as long as the derivative is not zero

[upbeat stone]

Yeah yeah that makes sense

So i think it makes sense like this

Thx

[wicked turtle]

yw

[upbeat stone]

I still wont close cuz i might need help still lol

[wicked turtle]

sure

[upbeat stone]

So dx = 2.sqrt(x).df + df^2, but saying that df^2 is perfectly equal to 0 because df approaches 0 does not seem rigorous

So we find dx/df = 2sqrt(x) + df. Since df approaches 0, 2sqrt(x) + df approaches 2sqrt(x). So df/dx will approach 1/(2sqrt(x))

[devout snow]

@upbeat stone Has your question been resolved?

[upbeat stone]

Is this correct?

[devout snow]

@upbeat stone Has your question been resolved?

[devout snow]

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[solid osprey]

[lusty wing]

Ok

[solid osprey]

well for a cant you take x=2^2013 -1

[lusty wing]

Lmao this was my primary school task…gosh I miss so much that book.

[solid osprey]

if you take the mod that gives -1^(some massive odd number+1 mod 2^2013
-1+1=0 mod 2^2013 so its divisible

[cyan surge]

Can't we use mathematical induction?

[solid osprey]

please enlighten me

heres for the first few terms:
the smallest x for any n i will express in (n,x)
(1,1)
(2,9)
(3,7)
(4,6)
(5,5)
(6,5)
(7,4)
(8,4)
(9,4)
(10,4)
(11,4)
(12,4)
(13,4)
(14,3)
(15,3)
(16,3)

assuming desmos isnt bugging cause the numbers are massivr

[stone stump]

time to learn better tools

[solid osprey]

dpes modulsr aritmethic break at e31

[stone stump]

x can definitely not be 4

4^(...)+1 is certainly not divisible by 2^n

[solid osprey]

ok fair

[stone stump]

just use for example python

easy to use and can handle large ints without problems

[solid osprey]

like i can code 🙏

[stone stump]

learn it

[solid osprey]

a

ok yeah doesent work

how big of numbers can python handle

[cyan surge]

let me try by proving specifically for 2013

[devout snow]

@solid osprey Has your question been resolved?

[solid osprey]

ok i coded it in python and it only showed 1,3,5 being strong i think the other numbers (if they are even strong) has too big of an x as it halted

[cyan surge]

I've done my calculations. But I have to get ready for work.
So I'll be able to explain in the afternoon or evening.
I can send you my rough work though

[solid osprey]

alright

[solid osprey]

sorry man i have no idea what im looking at

[devout snow]

@solid osprey Has your question been resolved?

[torn bane]

you want to find y so that 2^n|(y^ny + 1), hence we must also have 2^n|(y^{2ny} - 1)

now you can apply euler's theorem mod 2^n

not sure if this is enough, but it's where i would start

[solid osprey]

???

[solid osprey]

eulers totient theorem?

[torn bane]

yeah

[solid osprey]

uhh

my brain is not braining rn

totient of 2^y is 2^(y-1)

[torn bane]

sorry, should be 2^n, not 2^y

[solid osprey]

2^n and 2ny isnt coprime tho

[torn bane]

what does euler's theorem say

[solid osprey]

if a and n are coprime then a^totient(n)=1 mod n

[torn bane]

yeah, so why is it relevant if 2^n and 2ny aren't coprime

[solid osprey]

if they are coprime then a^2ny=1 mod 2^n no?

wait so is ny=2^(n-1)

[torn bane]

??

[torn bane]

where'd this come from

[solid osprey]

uhh

is that not a property

[torn bane]

are you trying to cite euler's theorem?

you need y and 2^n to be coprime for euler's theorem, not 2^n and 2ny

you always have y coprime to 2^n since otherwise we definitely have that 2^n does not divide y^{ny} + 1

[solid osprey]

oh ok

im o sure y^{2ny}=1 mod 2^y, by eulers 2ny is a multiple of 2^(y-1), ny=2^(y-2), since y is coprime to n then n=k*2^(y-2)

or am i stupid

[torn bane]

wait

why are you claiming that 2ny is a multiple of 2^y-1

2^y-1 isn't necessarily the order of y mod 2^y

[solid osprey]

cause 2^(y-1) is the totient(2^y) no?

wait

[torn bane]

yeah, but that doesn't make it the order (least number k so that y^k = 1 mod 2^y)

[solid osprey]

or is it the opposite

[torn bane]

if the order was that, then yeah i would agree

we don't know either way

all we can say is that both 2^n | y^{2^(n-1)} -1, and that 2^n | y^{2ny}-1

[solid osprey]

i saw somewhere the order divides the totient

so the order<=totient?

[torn bane]

yeah, but we don't know what the order is

[solid osprey]

hm

[torn bane]

the way to proceed is that this tells us that $2^n | y^{gcd(2ny, 2^{n-1})} -1$

[woven radish]

Bair

[torn bane]

since $o_{2^n}(y) | gcd(2ny, 2^{n-1})$

[woven radish]

Bair

[solid osprey]

how can you say y^gcd(2ny,2^(n-1))-1 is divisible by 2^n

[torn bane]

if $m|a^k-1$, then $o_m(a) | k$

[woven radish]

Bair

[torn bane]

so if $m|a^{\alpha} - 1$, and $m|a^{\beta} - 1$, then $o_{m}(a)|\alpha$ and $o_{m}(a)|\beta$, hence $o_m(a)|\gcd(\alpha, \beta)$

[woven radish]

Bair

[solid osprey]

o

alr so gcd(2ny,2^(n-1))=2*gcd(n,2^(n-2)) as y is coprime to 2^n

$o_{2^n}(y)\mid 2\gcd(n,2^{n-2})$

[woven radish]

skissue.in.a.teacup

[devout snow]

@solid osprey Has your question been resolved?

[torn bane]

yep, maybe you can let $n = 2^p \cdot q$, so then $o_{2^n}(y) \mid 2^{p+1}$

[woven radish]

Bair

[torn bane]

so o(y) = 2^k for some k <= p+1

actually it seems like we need something more here :(

the other idea was to do some sort of expansion with the binomial theorem

so could try that

[solid osprey]

hmm

how can we apply binomial theorem here?

[devout snow]

@solid osprey Has your question been resolved?

[crude wasp]

if m is even, is m strong?

hint: ||mod 4||

[solid osprey]

idts

[crude wasp]

so we only need to deal with the m = odd case for part b

[solid osprey]

[crude wasp]

the easiest way i can think of to do part b is lifting the exponent lol

but i think basically the idea of the solution is to "prove" lifting the exponent without directly quoting it

(so factorise it)

[solid osprey]

errrrrr so like
y^ym+1^ym=(y+1)(y^(ym-1)-y^(ym-2)+y^(ym-3)-.....+1)?

wait wrong var

do you want to do like v_2(something)>=m or smth?

thats the only lte usage that makes sense to me

[crude wasp]

yeah

what can you say about your other bracket?

[solid osprey]

so 2|y+1=>y=1 mod 2 (duh)
y^(ym-1)-y^(ym-2)+y^(ym-3)-...+1=y^2m=m=1 mod 2

so y^ym+1^ym=y+1 mod 2

[crude wasp]

well all the prime factors of 2s are contained in y+1

so in order to have 2^m divide y^(ym) + 1, 2^m divides y+1

so you can finish

(this is basically you prove LTE btw)

[solid osprey]

whar (note i have no idea how wiki makes the jump from the factor to proving v_p(x^n+y^n)=v_p(x+y))

v_2(y^ym+1^ym)=v_2(y+1)>=m

[crude wasp]

well as in for a smaller example

y^3 + 1 = (y+1)(y^2 - y + 1)

y^2 - y + 1 is odd

so if 2^n divides y^3 + 1, 2^n divides y+1

[solid osprey]

oh

so id the smallest y just 2^m -1?

i seeeee

.close tyy!

[devout snow]

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[crude wasp]

nw!

[solid osprey]

mr/ms/smthelse ly

[crude wasp]

(but yeah this was just to prove LTE basically in case u didn't want to just quote LTE, in a contest u can just quote LTE)

[solid osprey]

national contest?

[crude wasp]

yeah

for any contest, as long as u say LTE or lifting the exponent, it'll be fine

[solid osprey]

hype, so hype

are there any well known ""lemma""'s that i can use without proving it

[crude wasp]

i mean as long as the proofs of the theorems you use for contests aren't crazy, then as long as u quote it properly, u can use anything

(so i'd try to avoid stuff like Mihalescu's theorem/Catalan conjecture)

(i think that's the only one lol)

[solid osprey]

cataland opening?

[crude wasp]

no it's the x^y - a^b = 1

[solid osprey]

:p wb the zsigmondy

[crude wasp]

zsigmondy is fine

for bigger theorems, if you use them i'd try to make sure you quote them properly (just so whoever marks your work knows u aren't bluffing)

you may want to avoid using big theorems like these when u do practice problems but that's up to you

[solid osprey]

agh i wanted to ask something but i forgor

[devout snow]

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[flint obsidian]

I need help with this which approach should I use

[devout snow]

@flint obsidian Has your question been resolved?

[devout snow]

@flint obsidian Has your question been resolved?

[narrow spindle]

can this be done using an example?

[devout snow]

@flint obsidian Has your question been resolved?

[flint obsidian]

how?

[wheat pawn]

well, the simplest way would be to brute-force compute the determinant

if you dont wanna die computing it tho, you would replace one of the lines or columns by the sum of all 4 lines or columns as first step, then dividing the determinant in two using their properties

[flint obsidian]

I did not get how, I mean which operation would I use do so

[wheat pawn]

what, to brute force it? or with properties?

i'm gonna call a1, a2, a3 and a4, a b c d for simplicity (loading mspaint:)

i'd start like that

then on both, i'd substract the first row from the other 3

that should simplify the determinant quite a lot

[flint obsidian]

okay so you firstly added 4 to the first column right? and then using properties you separated the first column into two

right?

[wheat pawn]

no

[flint obsidian]

so how you getting 4 + a?

I mean by which operation

[wheat pawn]

.

if you add a linear combination of the other columns to a column, the determinant is not modified

on this case, to column 1, i added column 2, column 3 and column 4

[flint obsidian]

oh yeah

I got it now

[flint obsidian]

can you write the actual operation for me please? Other 3 means you would sum R2,R3,R4 first and then substract it to row 1? like R1 -> R1 - (R2+R3+R4)?

[wheat pawn]

C1'=C1+C2+C3+C4

then, on both:
Rn'=Rn-R1

[flint obsidian]

what does ' this refers though?

[wheat pawn]

the new line

to distinguish it from the ones you originally had

[flint obsidian]

Oh got it

damn thanks man it really helped me it got a lot simpler

[wheat pawn]

honestly it's probably not the approach *i* would use on a test, but it's one of the recommended ones.
I would most likely only go, from the original, Rn'=Rn-R1 and then bruteforce it

[flint obsidian]

I see I also thought of doing substracting R1 from R2,R3,R4 and then taking a1 common from column 1 but That approach did not work

or maybe I did mistake somewhere calculating it

[wheat pawn]

you dont have a1 as common factor on column 1, if you do that

because the first element is (1+a1), while the others are (-a1)

[flint obsidian]

yeah that's where I messed up lol

I am learning maths again after 3-4 years, now even I forgot the basics of it

I took a1 common from 1+a1 too

[narrow spindle]

wait which grade math is this

[wheat pawn]

where i live this is taught on the last year before uni

[flint obsidian]

its my uni last year

but its actual 12th grade maths

I was a commerce main thats why after 10th never touched maths

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[forest palm]

beggining of last line why do we have a seperate integral for just e^u?

[supple knot]

Because the integrated integral treats other variables as constants

int int f(u) g(v) dudv = int f(u) du * int g(v) dv

[forest palm]

oh so its just partial integration?

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[amber tapir]

i am pretty sure about this being 4, but why?:

a = 2²⁰¹¹ + 2⁻²⁰¹¹
b = 2²⁰¹¹ - 2⁻²⁰¹¹

what is a² - b²

[frozen stump]

can you write out the expansion for a^2

just in indice form

[amber tapir]

what do you mean by indice form

[frozen stump]

just

like to the power ofs

eg 2^(2*2011)

[amber tapir]

a=2^(2011)-2^(-2011)

like that?

[frozen stump]

ye but for a^2 pls

[frozen stump]

also this is b, not a

[amber tapir]

(2^(2011)+2^(-2011))^2

[amber tapir]

sorry

[solid osprey]

use the diffrence of squares no?

[frozen stump]

OH YEA

i was just gonna get them to expand and cancel out like terms

but thats probs better lol

[amber tapir]

and that goes like...

[solid osprey]

both works

[restive river]

a^2-b^2 = (a+b)(a-b)

[frozen stump]

not +

[faint zinc]

a^2 + b^2 = a^2 - (ib)^2 = (a + ib)(a - ib)

Oh lol yeah there you go

[frozen stump]

lmao

[frozen stump]

anyway, try this and youll get your answer

[amber tapir]

and how do I start solving a (or b)

[frozen stump]

nah, just sub your values of a and b in

[faint zinc]

a and b are given

This form just happens to be much easier to compute

(due to a lot of things canceling)

[amber tapir]

so its like (2^(2011)+2^(-2011) + 2^(2011)-2^(-2011)) * (2^(2011)+2^(-2011) - 2^(2011)-2^(-2011))

[frozen stump]

thats a pain to look at but yep

you forgot the last closing bracket tho lol

[frozen stump]

oh

this is wrong

a bit

the last term souldnt be -2^(-2011), it should be +2^(-2011)

[faint zinc]

Instead of writing out 2^2011 and 2^(-2011) let's write c and d

So we have a = c + d, b = c - d

[amber tapir]

yes

[faint zinc]

So a + b = 2c and a - b = 2d, yes?

[amber tapir]

yea

[faint zinc]

Additionally, we have d = 1/c

So a^2 - b^2 = (a+b)(a-b)

Then what?

In terms of just c and d

[amber tapir]

2c*2d

[faint zinc]

Which simplifies to

[amber tapir]

oh yea

i get ir

22^(2011) * 22^(-2011)

[faint zinc]

It's generally nicer to work with letters than numbers when the situation arises.

[amber tapir]

huge thank you

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[vague obsidian]

7 men, 6 women; 5 people are selected. At least 3 men has to be selected, how many ways can this be done?

[vague obsidian]

Why is 7C3 * 10C2 a bogus solution?

My logic was: I already chose 3 men in the 7C3. So it doesn't matter how many men/women I choose next, right? Or no? So I can now choose 2 people from the rest that are not chosen

[lament kraken]

Are the men distinct

[vague obsidian]

It wasn't stated in the problem

From a group of 7 men and 6 women, five persons are to be selected to form a
committee so that at least 3 men are there on the committee. In how many ways can
it be done?

[lament kraken]

If the men aint distinct then the sol is right I think

[vague obsidian]

Apparently the correct solution is

(7C3 * 6C2) + (7C4 * 6C1) + (7C5)

[lament kraken]

Hmmm

[arctic field]

you've assumed an ordering by doing 7C3 * 10C2

[vague obsidian]

wdym

[arctic field]

from this, you've labelled the 3 men chosen first as having been chosen first

[vague obsidian]

oh

[arctic field]

and hence you are overcounting

[vague obsidian]

sorry, i don't understand. why is this wrong?

[arctic field]

because the choice of 5 people is ultimately unordered

say you've got M1 to 7 and W1 to 6

you choose M1, M2, M3, then M4, W1

but this is the same as choosing M1, M2, M4, then M3, W1

[vague obsidian]

OH

[arctic field]

in doing 7C3 * 10C2, you count these as distinct

[vague obsidian]

yeahhhh

that's why

thanks

but if it was stated that the men and women are distinct, my solution would be correct right?

[arctic field]

no

[vague obsidian]

oh

[arctic field]

the men and women are already distinct

it is the set of people chosen that ultimately matters

not the order in which they are chosen

[vague obsidian]

OH GOT IT

there's no "first choice" and "second choice"

[arctic field]

yes

[vague obsidian]

gotcha

thanks a lot

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[sage pawn]

I need help with this question. I don't know how I would describe it but I tried for A, A={x in S : 1 ≤ x < 4} so I suppose this way it includes 1,2 and 3 but excludes 4

[fair juniper]

you dont need to have the condition that x<4 since no elements are greater than 4

greater than or equal to 4

[devout snow]

@sage pawn Has your question been resolved?

[sage pawn]

What about 1≤x≤3

[fair juniper]

you dont need an upper bound

[solar goblet]

!nosols

[devout snow]

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.

[low oriole]

Oh i didnt know it was some kind of homework

[solar goblet]

you generally don't send direct answers in help channels, homework or not

[sage pawn]

I mean, is 1≤x≤3 technically right? 1 is included, 2 is between 1 and 3 and 3 also included

[solar goblet]

it is technically right, but it's unnecessary

[low oriole]

Not technically, it is right but we need with only one condition

[solar goblet]

like LocalLunatic said, you don't need the upperbound

your universal set, S, does not contain any element above 3, so adding the restriction of smaller or equal to 3 is unnecessary

[sage pawn]

Does the condition on x involve a inequality

[fair juniper]

you can call me loon

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[fair juniper]

yes, only one

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[sage pawn]

Well the elements 1,2,3 are all positive

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[fading sorrel]

i expanded the function and made it under one fraction

and got -1/16

but ion get how Moe is wrong lmao

what he said does make sense ngl

[graceful stone]

But the other expression evaluates to undef as x->4

So it's a type of undetermined limit of the form 0*inf

[fading sorrel]

oh we need L, M
both need to be finite for the limit laws to work ye?

[graceful stone]

Exactly

[fading sorrel]

thanks g

[graceful stone]

Np

[fading sorrel]

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[fading sorrel]

wait what i hthought they had to be finite?

[graceful stone]

This can happen if f(x)+g(x) add to be finite

[fading sorrel]

so infinite + infinite = finite?

thats trippy tho ngl

cuz isnt infinite + infinite determinate that its infinite?

[graceful stone]

Calculus do be like that tho

[fading sorrel]

ye man should have done like bachelors of media or something 💀

[graceful stone]

Only sorta related

Imagine your job is to lay down an infinite number of crumbs for a trail

And you start and lay down 5 by the time i notice

My back is feeling big so i pick up the crumbs at the same rate as you lay them down

[devout snow]

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[fading sorrel]

so the case of infinity - infinity
which is indeed indeterminate

we dont know which is approcahing infinity faster

[fading sorrel]

so are you saying that even tho we got f(x) + g(x)
we can possibly have a -(gx) or -(fx) which can result in infinity - infinity

[graceful stone]

Sorta

[fading sorrel]

some sort of cancellation happening

[graceful stone]

If you think about it tho, after infinite time, i pick up infinite crumbs and you lay down infinite crumbs but they will only be 5 crumbs on the floor at any one time

This cancellation can only happen if the infinites grow at the same rate

[fading sorrel]

yoooo thats a good analogy
but still incredibly trippy tho

but from what i learned, infinity - infinity like what we are doing now is indeterminate

[graceful stone]

If i picked up the crumbs faster than you lay them down, i will catch up. If you lay them down faster than I pick them up, there will be infinite crumbs on the floor

This showcases that even if some functions grow to infinity, some infinities are bigger than others

[fading sorrel]

so remember how we had 0*inf earlier and we said we cant do anything with it cuz we are not sure what it is and Moe just said its 0. but what if Moe said the limit is indeterminate. would that be correct

same with that here no? we dont know what inf - inf is, it could be finite like you said but it could be infinite as well. we dont know

so in this question here, it says provide an example. im just providing an example where inf - inf is finite?

but i should be aware that if lim fx, and lim gx is non existant

lim (fx + gx) could be either existant OR non existant?

[graceful stone]

So when a limit is in indeterminant form, we have to manipulate/simplify the expression inside before we say what the value is

[graceful stone]

So for this problem, things should be simplied into one fraction before evaulating inside the limit statement

[fading sorrel]

ye i did that
ye solving the limit is easy
Im just getting confused by indeterminate, determinate

cuz im finding it weird how infinite limits can combine to become finite, which im sure is infinite - infinite case

there has to be some cancelling i think

i got the example

(x^2 / x-3) - (3x/x-3)

lim to 3 doesnt exist for each of the three functions separately but exists for them combined

i think i get it now kind og

of

[graceful stone]

That works

[graceful stone]

I think so

[fading sorrel]

ye thank you man
that grain analogy was goated

thanks man

[graceful stone]

Np

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@fading sorrel Has your question been resolved?

[devout snow]

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[fading sorrel]

i think this is false

?

using limit laws
we would get a finite number +- infinite

which is infinte, hence limit doesnt exist

[sage burrow]

are the [] brackets just brackests or do they have a special meaning?

[fading sorrel]

combine fx + gx
THEN take the limit

[snow raptor]

normal brackets unless stated

[fading sorrel]

wait whats the go with brackets? is there something important im missing

?

its just parentheses aint it?