help-49

[mint ravine]

how do you get the central angle in the first place

[past horizon]

vibes

do u have a question

[mint ravine]

what

no my question

[ebon mica]

google is usually good for short and sweet questions like this

[mint ravine]

ok satvik

[past horizon]

hahahaha

[mint ravine]

.close

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[midnight plank]

Please don't occupy multiple help channels.

[lyric charm]

<@&268886789983436800> scam

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[coral belfry]

Hello. Suppose we have proven that for a,b in N, there exists a d in N such that ax+by=d, and then d is a common divisor between a and b. If gcd(a,b) = 1 , then can we automatically assume that the d in question is 1? without proving beforehand that d is the gcd anyway

[coral belfry]

nvm i think its pretty obvious

oops

.solved

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[restive briar]

How do i approach this problem? Do i use formulae or try differentiating it directly

[mellow sand]

Start by simplifying your trigonometric terms and then go from there, it will be a lot simplier to do it after simplification

[restive briar]

Its already simplified isnt it? I dont think i can simplify if further

Oh nvm i can simplify tan as sin/cos

[compact copper]

try breaking tan^5x

[restive briar]

Yea

Alr ty for yall help

[midnight plank]

@restive briar Has your question been resolved?

[slim dagger]

guys if the lim is approaching infinity, will we say it exists?

[gloomy lark]

no

[slim dagger]

what if it is approaching lets say pos infinity from both sides, does this mean it exist?

[slim dagger]

but they write infinity as if it was a number or smt i mean smt like lim(x->3)=infinity

[sudden cove]

infinity isn't usually a number

[gloomy lark]

for limit to exist the tendency should be a finite number

[sudden cove]

unless you're working in the extended reals or something a limit which tends to infinity definitionally does not exist

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[radiant roost]

how do you solve recurrence relations with a characteristic polynomial?

[dawn ruin]

you find the roots of the characteristic equation

[mystic condor]

you mean like $T_{n+2}-T_{n+1}-T_{n}=n?$

[grand pond]

ImOakley

[radiant roost]

yeah, and the T's can have coefficients

[mystic condor]

yeah sorry

ok

[radiant roost]

would the characteristic polynomial be x^3 - x^2 - x = 1 ?

[mystic condor]

like galaxy said you find the roots where T n+2 is x^2, T n+1 is x and T n is 1

the number added to n determines the power

[radiant roost]

oh so x^2 - x - 1 = 0 ?

[mystic condor]

yeah

[sage olive]

maybe this would help as a general algorithm

the roots will be your guesses for the solution as a sum of distinct exponential terms, and if a root is repeated, it involves building up the exponential terms by multiplying them by polynomials

if you've taken a differential equations class, this is an almost identical approach to how one might solve a homogenous linear diff eq with constant coefficients

[radiant roost]

the recurrence has to be linear right? is that what homogeneous means?

like i can't have T_n + T_(n-1)^2 ...

[sage olive]

correct

and this can be modified for nonhomogenous linear relations by just adding a particular solution

[radiant roost]

i have an example i want to try

$f(x) = 2 f(x - 1) - f (x - 2) + 2$

[grand pond]

Axe

[radiant roost]

f(0) = 1 and f(1) = 2

so x^2 = 2x-1
(x-1)^2 = 0
double root x = 1

[sage olive]

yes

keep in mind that this is nonhomogenous btw

[radiant roost]

what makes it non-homogeneous?

[visual tiger]

+2

[sage olive]

so you're first finding the general solution and then taking a guess at a particular solution

[sage olive]

homogenous equations have all non-recurrent terms equal to 0

[visual tiger]

Homogeneous is "Linear combination of f(x-k) = 0"

The homogeneous version of your equation would be $f(x) = 2 f(x - 1) - f (x - 2)$

[grand pond]

Rafilouyear2026

[sage olive]

but yeah focus on the general solution first by treating it as a homogenous equation, what does the double root of 1 tell you about the form of your general solution according to this?

[visual tiger]

And so you should start by finding the general solution of that related homogeneous equation

[radiant roost]

so i'm still working with x^2 = 2x - 1 right?

[sage olive]

yes

and you got that it had that repeated root at x = 1

[sage olive]

look at part c of this algorithm to see what kind of term that contributes to your general solution

[radiant roost]

f(x) = A + Bx

(1^x = 1)

[sage olive]

yes

if you had gotten that x = 2 was a triple root, the term would be a quadratic polynomial times 2^n

etc

[sage olive]

so substitute this guess into your homogenous equation and use your initial conditions to determine A and B

[visual tiger]

huh?

You need to find the general solution to the non-homogeneous equation first

[sage olive]

oh whoops i forgot the original was nonhomogenous

[visual tiger]

so, Linear functions are the solutions to the homogeneous equation

to get a constant = 2 to show up on the RHS

maybe as a test for a particular solution, ||increasing the degree could work||

[sage olive]

here's a table of useful guesses at particular solutions depending on what's remaining on the RHS btw

[radiant roost]

ok so this is not strictly algorithmic, the next step says "take an educated guess at the form"

oh i see, i have a constant so this table tells me what to do

[visual tiger]

ehhhhhhhh

Fact is, you already know linear functions are the solutions to your homogeneous equation

so don't expect a constant to work for the non-homogeneous one here

That's the problem of the double roots case, or when x = 1 is a root of the characteristic here

[sage olive]

oh right there are sometimes problems when the form of the stuff on the RHS matches up with a characteristic root

[visual tiger]

so um, nothing to say more than intuition here

[radiant roost]

you don't happen to know how this algorithm is justified do you?

maybe it's with linear algebra

[visual tiger]

Have you already seen solution method for differential equations?

[radiant roost]

no, not really

[visual tiger]

linear ODEs of some rank

well there's a method that usually works called variation of constant

it does have to do slightly with linear alg

The space of solutions of a linear homogeneous equation is a linear subspace

so any solution can be written as some linear combination of "basis" solutions

Variation of constant says: for the non-homogeneous case
Replace the coefficients in the linear combinations by functions

and look at the equations those functions verify

After some simplifying, you usually get for those functions a linear equation with one less degree

so if you're solving for y'' + 2y' + 3y = ...

the coefficients solve something like y' + ...y = ...

hard to rigorously explain without losing you atp xdd

[radiant roost]

🤷

yeah

thanks though

[visual tiger]

Bottom line, there is some linear alg thingy behind because of the homogeneous equation

and there is a method that usually works no matter the RHS

but it's very hard to pull through

so it's sometimes easier to guess

[radiant roost]

oh

[visual tiger]

for the particular solution

[radiant roost]

there's a method where you compute the jordan normal form of a matrix

i got the eigenvalue 1, this might be the same as the root of x^2 = 2x - 1

anyway thanks i'll close this

.close

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[scarlet snow]

<@&268886789983436800> idk if the person deleted the message or already got removed

[tender trench]

i think automod worked

[sharp coral]

no that was me

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[inland patio]

I have a question which is a bit stripped away from context. Don't know if it makes a 100% sense. Suppose there are some $n\times n$ matrices $A_1,\ldots,A_k$ such that $A_1+\cdots+A_k=I_n$ and also $$n=\operatorname{rank}(A_1+\cdots+A_k)=\operatorname{rank}(A_1)+\cdots+\operatorname{rank}(A_k).$$ Is it also true that $$\operatorname{rank}(A_2+\cdots+A_k)=\operatorname{rank}(A_2)+\cdots+\operatorname{rank}(A_k)?$$

[grand pond]

psie

[grim vector]

I think it is true

[keen sinew]

peak maths

[grim vector]

A1 = I_n - (A2 + A3 + ... )

Let B the second thing

A1 = I_n - B

[inland patio]

For the interested reader, I picked up on this in the comments to this answer. They claim that rank(A2 + ... + Ak) = rank(A2) + ... + rank(Ak).

The answer tries to prove the general case of this lemma:

The author of the screenshot proves the lemma for k = 2 and says that the k > 2 follows from induction.

[floral apex]

what if

\begin{align*}
rA_1
&= r\qty( I_n - \qty[A_2 + \dots+A_k ] ) \
&\leq rI_n - r\sum _2 A_i \
&= n - r \sum _2 A_i \
&= r\sum _1 A_i - r\sum _2 A_i \
&\leq rA_1
\end{align*}

then uhh

i think this is enough

between the first and second lines

im wrong arent i

[grand pond]

jan Niku

[inland patio]

Thank you for your attempt. 🙂 If I'm not mistaken, rank(cB)=rank(B) for any nonzero scalar. Therefore your second line seems incorrect. It should be rank(I_n - B) <= rank(I_n) + rank(B) I think.

[floral apex]

arent we just using subadditivity

just call a_2 + .. + a_k some C

then r(I_n + C) <= n + rC

I'm now not sure that we need any of this

[inland patio]

Here's how I see the calculation going. Let A2 + ... + Ak = B. Then rank(I_n - B) <= rank(I_n) + rank(-B) = rank(I_n) + rank(B). We did use subadditivity, but rank is not linear such that we can move the negative sign out front. Unfortunately. 😔

[floral apex]

what if

$r\sum _1 A_1 \leq rA_1 + r \sum _2 A_i$ by subadditivity

[grand pond]

jan Niku

[floral apex]

then $r\sum _1 A_i \leq rA_1 + r \sum _2 A_i \leq r A_1 + \sum _2 rA_i$ again

[grand pond]

jan Niku

then $r\sum _1 A_i \leq rA_1 + r \sum _2 A_i \leq r A_1 + \sum _2 rA_i$ again

[floral apex]

$r\sum _1 A_i \leq rA_1 + r \sum _2 A_i \leq r A_1 + \sum _2 rA_i = \sum _1 rA_i$ lastly

[grand pond]

jan Niku

[floral apex]

but then all <= are =

this is enough, yea?

so choose to start from $r \sum _1 A_i - rA_1$ instead and youre there

[grand pond]

jan Niku

[floral apex]

@inland patio

[inland patio]

Ah cool! 🙂 Looks good. I guess that does it. Thank you. 👍

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[gaunt mantle]

Just need a bit of intuition

[gaunt mantle]

Intuition explaining this please

Because the proof kinda doesnt feel that clear for me

<@&286206848099549185>

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[gaunt mantle]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[gaunt mantle]

ask again maybe

.close

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[blissful pier]

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

D is the determinant of the 2x2 Hessian matrix

[raw hamlet]

https://math.stackexchange.com/questions/1985889/why-how-does-the-determinant-of-the-hessian-matrix-combined-with-the-2nd-deriva might be more useful than the wikipedia article

[scenic wyvern]

.reopen to give OP a chance at reviewing the material

[midnight plank]

✅ Original question: #help-49 message

[blissful pier]

Thanks Hanako! And good morning!

[gaunt mantle]

thx i guess

.close

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[chilly cobalt]

Let $P(x) = a_{2016}x^{2016} + a_{2015}x^{2015} + \cdots + a_0$ be a polynomial of degree 2016 with real coefficients such that [ |a_1+a_3+\cdots +a_{2015} | > | a_0 + a_2 + \cdots + a_{2016} | ] Show that $P(x)$ has an odd amount of real roots that have an absolute value less than 1. (counting roots with multiplicity)

[grand pond]

Copter

[chilly cobalt]

i don't really have an idea of beginning this other than splitting P into odd degree + even degree polynomials?

even then i don't know what i would do

[midnight plank]

@chilly cobalt Has your question been resolved?

[chilly cobalt]

<@&286206848099549185>

[limber dragon]

hopefully someone else can give you a more definitive hint, but my intuition says to use induction to prove it

[chilly cobalt]

induction on the degree of P?

[limber dragon]

yes

and I would only consider even degrees

but without knowing what course this is for, how your prof makes problems, what sorts of proofs you've done so far, or any other context, it's hard to say what your prof intends for you to do here

[chilly cobalt]

i found this problem on facebook😭

its supposed to be an olympiad problem so

[limber dragon]

ahhh

ok now I'm going to be thinking about this stupid problem all night instead of sleeping

[crimson glade]

Who pinged

[west iron]

What do you know about the values of f(1) and f(-1)?

(What would you need to know to solve it? Can you prove that?)

[chilly cobalt]

from P(1) and P(-1) and the problem statement we get that there is atleast one root in (-1,1)

other than that idk 😭

[raw hamlet]

can you do someting by adding together p(1) and p(-1)

i guess one of the sides is p(1) + p(-1) and the other is p(1) - p(-1)

[chilly cobalt]

how would i arrive to an odd amount of roots as a conclusion in the first place

[raw hamlet]

hmm i was thinking it would have to do with the values of the polynomial at -1 and 1

like, if it changes signs

then clearly it has to cross the x axis an odd number of times

-> odd # of roots

[chilly cobalt]

wait isnt there some way of counting roots that goes by an even amount

something something rule of signs

[raw hamlet]

descartes?

[unkempt sluice]

P(1) and P(-1) are of different sign. You travel from -1 to 1, P has to cross x-axis odd number of times, otherwise you end up in the same half plane

[chilly cobalt]

oh

🫩🫩

is there a more formal way of writing that

[unkempt sluice]

There is a way:

p(x)=a (Π(x-r)^m(r)) q(x) for a constantly non-negative q(x) and some roots r with multiplicity m(r). Right? I don’t assume these r is in (-1,1)
Then P(-1)P(1) has the same sign with Π((1-r)(-1-r))^(m(r))
r<-1 or r>1, (1-r)(-1-r)>0
only when -1<r<1, (1-r)(-1-r)<0
Thus -1=sign of P(1)P(-1)=(-1)^Σm(r) implying Σm(r) odd

[chilly cobalt]

ah i see

thanks! :3

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[unkempt sluice]

Np

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[late rover]

hiii integral nub here, What must I know in order to solve this?

[azure bolt]

maybe some partial fraction decomp

[modest heart]

Along with a substitution

[azure bolt]

try factoring out the denominator

[topaz epoch]

Sub sqrt(x - 1) then use Glasser's Master Theorem

[late rover]

Okay

[topaz epoch]

I'll show you

[topaz epoch]

So after the initial sub, did you obtain $$\int_{0}^{\infty}\frac{2\left(x^{2}+1\right)}{x^{4}+2x^{2}+5}dx$$?

[grand pond]

Roy

[late rover]

Uh... I think so? Hang on I haven't done it yet I'm a bit slow

[topaz epoch]

The way you notice that substitution is because you can factor the denominator as $\int_{1}^{\infty}\frac{x\sqrt{x-1}}{x^{2}\left(x-1\right)+4\left(x-1\right)}dx$

[grand pond]

Roy

[topaz epoch]

That naturally makes one inclined to substitute the entire sqrt(x - 1) so we get an entirely rational function

[topaz epoch]

The weaker case we will be using is called the Cauchy-Schlömilch transformation

[late rover]

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

Wait

[topaz epoch]

It basically says for any integrable function $F$ on the real line, $$\int_{-\infty}^{\infty}f\left(ax-\frac{1}{x}\right)dx=\frac{1}{a}\int_{-\infty}^{\infty}f\left(x\right)dx$$

[grand pond]

Roy

Fionna The Unemployed

[topaz epoch]

I'll give you a few examples

[late rover]

Ohhh

[topaz epoch]

$$\int_{-\infty}^{\infty}\frac{1}{1+\left(x-\frac{1}{x}\right)^{2}}dx=\int_{-\infty}^{\infty}\frac{1}{1+x^{2}}dx$$

[grand pond]

Roy

[topaz epoch]

{\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }{\frac {x^{2},dx}{x^{4}+1}}&=\int _{-\infty }^{\infty }{\frac {dx}{\left(x-{\frac {1}{x}}\right)^{2}+2}}\&=\int _{-\infty }^{\infty }{\frac {dx}{x^{2}+2}}\&={\frac {\pi }{\sqrt {2}}}\end{aligned}}}

[grand pond]

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

[topaz epoch]

This example I took from the Wikipedia page

Now it's known in the integration bee meta that you can also generalize any $\int_{0}^{\infty}\frac{px^{2}+q}{ax^{4}+bx^{2}+c},\dd x$ with Glasser's master theorem

[grand pond]

Roy

[late rover]

Gimme some time to look into this

[topaz epoch]

Yep no worries

[small jasper]

orz u

[topaz epoch]

This is the stronger version but we won't be using that here

So with this you could literally conclude that $$\int_{-\infty}^{\infty}\frac{1}{1+\left(x-\frac{1}{x}-\frac{67}{x+420}-\frac{250}{x+\phi}\right)^{2}}dx=\int_{-\infty}^{\infty}\frac{1}{1+x^{2}}dx$$

[grand pond]

Roy

[topaz epoch]

Or $$\int_{-\infty}^{\infty}\frac{1}{1+\left(x+\tan(x) - \sec(x) \right)^{2}}dx=\int_{-\infty}^{\infty}\frac{1}{1+x^{2}}dx$$

[grand pond]

Roy

[topaz epoch]

Because the poles (mittag leffler) expansion of tan and sec also satisfy that condition

You can also concoct some evil integrals this way by recursing x - 1/x

@late rover are we ready to move on with your integral

[late rover]

😭 Nooo, I'm watching some more example about Glasser's Master Theorem

I haven't got the hang of it

[topaz epoch]

Are you watching dr3213's video or Silver's?

[late rover]

dr3213

[topaz epoch]

Yeah I recall he describes how to generalize the quartic rational integral I sent

I have to go out in a bit, so I'll write up my solution and send it here

,texsp \begin{align*}
I &= \int_{1}^{\infty}\frac{x\sqrt{x-1}}{x^{3}-x^{2}+4x-4}dx \
&= \int_{1}^{\infty}\frac{x\sqrt{x-1}}{x^{2}\left(x-1\right)+4\left(x-1\right)}dx \
&= \int_{0}^{\infty}\frac{2\left(x^{2}+1\right)}{x^{4}+2x^{2}+5}dx && \text{Substituting $\sqrt{x - 1}$}\
&= 2\int_{0}^{\infty}\frac{x^{2}}{x^{4}+2x^{2}+5}dx+2\int_{0}^{\infty}\frac{1}{x^{4}+2x^{2}+5}dx && \text{Splitting the integral to force both integrals into GMT form}\
&= 2\int_{0}^{\infty}\frac{1}{x^{2}+\frac{5}{x^{2}}+2}dx+2\int_{0}^{\infty}\frac{1}{5x^{2}+\frac{1}{x^{2}}+2}dx && \text{Sub $1/x$ on the second integral, then divide both sides by $x^2$}\
&= 2\int_{0}^{\infty}\frac{1}{\left(x-\frac{\sqrt{5}}{x}\right)^{2}+2+2\sqrt{5}}dx+2\int_{0}^{\infty}\frac{1}{\left(\sqrt{5}x-\frac{1}{x}\right)^{2}+2+2\sqrt{5}}dx && \text{Complete the square} \
&= 2\int_{0}^{\infty}\frac{1}{x^{2}+2+2\sqrt{5}}dx+\frac{2}{\sqrt{5}}\int_{0}^{\infty}\frac{1}{x^{2}+2+2\sqrt{5}}dx && \text{Apply GMT!!}\
&= 2\left(1+\frac{1}{\sqrt{5}}\right)\int_{0}^{\infty}\frac{1}{x^{2}+2+2\sqrt{5}}dx \
&= \sqrt{\frac{1+\sqrt{5}}{2}}\frac{\pi}{\sqrt{5}} \
&= \sqrt{\frac{\phi}{5}}\pi \
\end{align*}

[late rover]

spoiler, pls

[grand pond]

Roy

[late rover]

okay I'll try to do it myself first

Ty,Roy

[topaz epoch]

Also if you're not familiar with the completing the square method here, I recommend this video
https://www.youtube.com/watch?v=V3EB0e5K0SI

[topaz epoch]

:)

no worries, good luck!

[midnight plank]

@late rover Has your question been resolved?

[late rover]

Okay he explain everything in the video

.close

[midnight plank]

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[late rover]

Kinda cool

[rich yew]

Maybe Timo H. can be a little less generous next time…

[late rover]

you integral pro

me nub

[topaz epoch]

Definitely

It’s much better than Sophie Germaine factorization and partial fractions

[midnight plank]

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[balmy cypress]

But if I have an integral f(g(x))g'(x) , e^u u du It Is ?

[balmy cypress]

It would be e^{g(x)} g(x) with g'(x)=1

f(g(x)) = e^u u

And g'(x) = 1, Right?

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[balmy cypress]

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[midnight plank]

✅ Original question: #help-49 message

[balmy cypress]

but if I have the integral which is of that form

Then I can do the integral

Without changing the extremes even if the substitution is not monotonous?

[hidden kiln]

im confused of your questions can you rephrase it

[balmy cypress]

if I have an integral f(g(x))g'(x)

Then I can do the integral

Without changing the extremes even if the substitution is not monotonous?

[hidden kiln]

We could let u = g(x), du = g'(x) dx
So INT f(g(x))g'(x) dx = INT f(u) du
If f(x)'s integral can be found then yes

[balmy cypress]

You know

You know that when you make a substitution

You can change the extremes

[hidden kiln]

Second, unless g(x) = ae^x, you most likely need to replace the limits

[balmy cypress]

Only if the function is monotonic

[hidden kiln]

hm

[balmy cypress]

u = cos x + sin x

It decreases to π/4 you have to split the integral

[hidden kiln]

mhm

yea that we need split

um...actually, if you are trying to find absolute value of the integral's final value (aka area under the graph), then yes you need to split
Or else you dont need

[balmy cypress]

Can you explain better pls

[lyric charm]

the substitution isn't required to be monotone, no

but it does need to be continuous

the bounds of integration have to be recalculated regardless though

[balmy cypress]

f(g(x)) g'(x) g derivable continuos is not enough

[lyric charm]

you need to get some punctuation

also, i said that it was necessary, not that it was sufficient

[balmy cypress]

for the substitution integral_a^b f(g(x)) g'(x) dx = integral_{g(a)}^{g(b)} f(u) du, g does NOT need to be monotone/invertible. Monotonicity is only needed for the 'inverse-function' change-of-variables form. What we DO need is that g is differentiable (e.g. C^1/absolutely continuous) so g'(x) exists

Right?

[lyric charm]

yes

[midnight plank]

@balmy cypress Has your question been resolved?

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[leaden seal]

can someone help me answer this question its a statics question

[leaden seal]

from my knowledge the max moment = areas under the point where the shar crosses the xaxis

[midnight plank]

@leaden seal Has your question been resolved?

[midnight plank]

@leaden seal Has your question been resolved?

[lofty knoll]

Calculate the area under each segment of the shear diagram and compare the values

[midnight plank]

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[mint ravine]

please help

[lusty python]

oh

[lusty python]

let's try this one

list all of the factors of 4

121

69

and 35

@mint ravine

[mint ravine]

yes im still here

[lusty python]

try to list the factors of the numbers above

positive factors ofc

hello?

[mint ravine]

? yes please wait im working on it

[lusty python]

ping me when you're done

[mint ravine]

4: 1, 2, 4; 121: 1, 11, 121; 69: 1, 3, 23, 69; 35: 1, 5, 7, 35

@lusty python

[lusty python]

so look at the square numbers

they have odd amount of factors or even

[mint ravine]

they have 3 factors

odd

[lusty python]

cool

how about the non-squares

[mint ravine]

4 factors

[lusty python]

they have an odd amount or even amount?

[mint ravine]

even

[lusty python]

cool

so basically

the ones who has odd factors are basically square numbers

and the ones who has even factors are non-squares

[lusty python]

so how would you do this exercise now?

[mint ravine]

biggest square up to 2000

and take square root and that's our answer

[lusty python]

cool

yep

the question seems complicated

but when translated correctly, it turns into an easy problem

[mint ravine]

yeah i thought i had to do all of these different cases and stuff

and i thought the trick thing only applied to squares of prime numbers

[lusty python]

lol

[mint ravine]

thanks

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[slow thorn]

is the trace of the adjoint of a matrix the same as that of the original matrix?

[runic hamlet]

do you mean adjoint in the transpose kind of way or the adjugate kind of way

https://math.stackexchange.com/questions/2065822/trace-of-adjugate

[slow thorn]

umm the simpler definition most likely

transpose of the cofactor matrix

[runic hamlet]

tranpose kind of way would be way simpler

but then the question would be trivial

[sharp coral]

have you tried any examples?

[slow thorn]

following from this question

[runic hamlet]

there is some kind of identity for adj(adj(2A))

should be something like c*A for some number A

taking the adjoint is nearly the same as inverting the matrix

so if you do it twice, you nearly end up at the same thing

[slow thorn]

yeah the solution just says tr(B) = tr(8A) though

[runic hamlet]

probably B=8A then

you know A*adj(A)=det(A)I, yes?

[slow thorn]

yeah

[runic hamlet]

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

so apply that twice

questions like these just assume that you remember all these kinds of random formulas

or at least thats what it feels like

from seeing them on here

so then the solution manual also is short and doesnt show the derivation of the formula again

[slow thorn]

alr yeahh makes sense

im getting 8A = B yu

although its just given that adj(adj(2A)) = det(2A) x (2A)

thank

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[sterile wing]

Can someone run me through this? I dont understand what D:P(U)×P(U) is supposed to mean, I get what powerset means...but how does this definition of the relation tie into ∀M,N∈P(U):MDN⇔M∩N=∅

[sterile wing]

translated*

and try to explain in simple terms - no technical terms unless necessary please

[jaunty ivy]

D is a relation that acts on the subsets of U

they define MDN to be such that MDN if and only if M and N are disjoint (have no common elements so their intersection is the empty set)

[main current]

When you say "this definition of the relation", what are you refering to?

The ∀M,N∈P(U):MDN⇔M∩N=∅ is the definition of the relation

[sterile wing]

this part

how does it tie into the function below

[main current]

That's just the sets that the relation works between. It doesn't tell you what the relation actually does

[sterile wing]

(is it called function? prolly not)

[visual tiger]

a first subset M (in P(U)) can potentially relate to another subset N (in P(U))

[sterile wing]

im confused - can you elaborate

[visual tiger]

Relations should probably more be seen as subsets of the product set than functions, though it is possible

[visual tiger]

the way this written thing should be interpreted is "D puts in relation elements from P(U) with elements from P(U)"

so not to be interpreted as a function from P(U) to P(U)

[main current]

I'm going to back it up a little, let's talk about functions.

Let's talk about a function f: A -> B.

A is the domain of the function, the set that f takes elements from.

B is the codomain of the function, the set of elements that f returns.

Note that f: A -> B does not tell us what the function actually does to elements of A. We need more info to complete the definition.

In the case of your question:
D is a relation between p(U) and p(U).

The rule of the relation is given by ∀M,N∈P(U):MDN⇔M∩N=∅.

In plain english: MDN is true exactly when M and N have an empty intersection. No elements in common.

[midnight plank]

@sterile wing Has your question been resolved?

[sterile wing]

alright so with an example because otherwise it wont click fully, lets say we have
U = {0,1,2} (powersets {∅,{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}}
and M = {1,2} and N = {0}
p(U) -> p(U) just says take a random subset within (U) and compare it to another random subset within (U)
which fufills the requirement: => ∀M,N ∈ P(U): MDN ⇔ M∩N = ∅

and the output would just be true or false? it wouldnt be stored anywhere would it?
did I word anything wrong

[main current]

A relation between A and B is basically this:
Give me an element of A and an element of B, and I'll tell you if they're "related" or not.

Where "related" is some rule that we can make up. In this case, the rule is ∀M,N ∈ P(U): MDN ⇔ M∩N = ∅
M and N are "related" if they have no elements in common.

In your example, M and N are indeed related! Since we are calling the relation "D", we typically say MDN is true.

Let's say we have M = {0,1} and N = {1,2}. Since M∩N = {1}, M and N are not related by D, and MDN is false.

[runic hamlet]

we are just putting into symbols that we can say "M and N are disjoint" or "M and N are not disjoint"

relations are a way to formalize what we do very often when we compare two objects and say things like "A and B are (parallel/congruent/same size/disjoint/...)"

[sterile wing]

I get this - my only confusion only was the top part the p(U) -> p(U) everything else youve explained makes perfect sense

[main current]

Just saying that the relation is between elements of p(U) and p(U). That's all that says.

[runic hamlet]

fwiw, its not very standard notation

[main current]

We need more info than just that to fully set up the relation.

[sterile wing]

this makes sense, Im just (or was) confused on the purpose of p(U) -> p(U), like what wouldve changed if that wasnt there, what does it tell me? but if my example is right maybe I get it now

[runic hamlet]

often you would just say that D is a relation on P(U)

[sterile wing]

mhh alright

[runic hamlet]

or D is a subset of P(U) x P(U)

[sterile wing]

I see

I think I get it now - now I gotta actually do the checks which should be easier

thanks @runic hamlet @main current

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[chilly cobalt]

Prove that if the positive real numbers $\alpha , \beta$ have the property that among the numbers [ \lfloor \alpha \rfloor, \lfloor 2\alpha \rfloor, \lfloor 3\alpha \rfloor, ... , , \qquad \lfloor \beta \rfloor, \lfloor 2\beta \rfloor , \lfloor 3\beta \rfloor, ... ] every natural number appears exactly once if and only if $\alpha$ and $\beta$ are irrationals such that $1/\alpha +1/\beta = 1$

[grand pond]

Copter

[chilly cobalt]

for the if direction i only have that α, β are irrational otherwise it would contradict uniqueness, but i dont understand how to proceed

also i assume the problem means each of the sequences {α} and {β} satisfy the condition separately right?

[olive matrix]

i feel like you've got some missing words in that statement

[chilly cobalt]

oops yea

[grand pond]

Copter

[chilly cobalt]

i apologize

i cant think of why 1/a + 1/β = 1

[eternal pawn]

This is called Beatty's theorem i think

[chilly cobalt]

wuh

,w beatty's theorem

[eternal pawn]

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

You don't need to spoil it for yourself

[chilly cobalt]

okay, thanks

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[chilly cobalt]

ill look into it

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[onyx tide]

Prove that for every odd natural number n, the number 3n^2 + 2n + 7 is divisible by 4.

[onyx tide]

Quadric formula?

Can’t I factor it in two brackets?

[lusty python]

not really quad formula is needed

modulo properties

[olive matrix]

induction is going to be much more useful here or yeah modulo stuff

[lusty python]

modulo

i can just consider $n \equiv 1 \mod{4}$ and $n \equiv 3 \mod{4}$ and im done

[grand pond]

1 divided by 0 equals Infinity

[onyx tide]

Idk what is module

Module

Mudlo

[olive matrix]

oooookay then do you know what induction is

[onyx tide]

no

Nothing

[lusty python]

yeah ik what to do here

[olive matrix]

what is your mathematical background?

[lusty python]

so since $n$ is odd, how about we let $n = 2k + 1$ for some $k \in \mathbb Z$

[grand pond]

1 divided by 0 equals Infinity

[sudden cove]

there's an even simpler way

yeah this

[onyx tide]

I’m just trying to solve tasks on a HC exam

[lusty python]

if you can't use modulo then use this

[sudden cove]

let n := (2k + 1) and do the alge

[onyx tide]

Why not k+1?

[lusty python]

n is odd

[sudden cove]

2k + 1 is the form for an odd number

[lusty python]

so $2k + 1$'s the way to go

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

or $2k - 1$ but i prefer $2k + 1$

[grand pond]

1 divided by 0 equals Infinity

[onyx tide]

But why 2?

K is also good

[sudden cove]

all odd numbers are of the form (2k + 1) for some whole number k, can you see why?

[lusty python]

an odd number is a number which divides 2 gives a remainder of 1

[lusty python]

works for integers too

[sudden cove]

what i said

[onyx tide]

3(2k+1)^2 + 2(2k+1) + 7

[sudden cove]

now, just simplify :3

[lusty python]

:3

[onyx tide]

But it’s an expression

Not equation

[sudden cove]

trust me on this one

[lusty python]

that's why it's called simplify the expression

[sudden cove]

there was no equation to begin with anyways

[lusty python]

best shot to simplify this is to expand allat out

also you need to note that $k \in \mathbb Z$ too (this will be useful later)

[grand pond]

1 divided by 0 equals Infinity

[onyx tide]

3(2k^2 + 2k + 1) + 7

[slender walrus]

Whole number doesn't include negative integers

[onyx tide]

Oh I thought you said equation

[sudden cove]

really? never knew

[onyx tide]

Is this correct

@sudden cove @slender walrus @lusty python

[lusty python]

i don't think so

[sudden cove]

those terms don't combine, expand the square first instead

[onyx tide]

3(4k^2 +4k + 1) +7

[lusty python]

you expanded the square

now you're only missing $2(2k + 1)$

[grand pond]

1 divided by 0 equals Infinity

[onyx tide]

no…

I’m not

[sudden cove]

can you show your working? this doesn't look right

[lusty python]

$3(2k + 1)^2 + 2(2k + 1) + 7 = \dots$

[sudden cove]

sorry to bug you about this lmao

[onyx tide]

(2k+1)^2 is (4k^2 +4k + 1)

[sudden cove]

yes that's the problem

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

yep

[onyx tide]

okay it’s 3(4k^2 +4k + 1) + 4k + 2 + 7

[sudden cove]

indeed

[onyx tide]

12k^2 + 12k + 3 + 4k + 2 + 7

12k^2 + 16k + 12

And I still have a fucking square

[sudden cove]

but importantly

you have a common factor

[onyx tide]

I have 16k…

And I need 12…

[lusty python]

hold up

they are asking you to prove it's divisible by 4, so what do you do?

[sudden cove]

what can you factor out of this expression?

you're really really close

[onyx tide]

The twelve?

But there’s a 16…

[lusty python]

hint: this

[sudden cove]

what is the gcd of 12 and 16

[onyx tide]

12 is divisible by 4 and 16 and 12

[sudden cove]

?

[lusty python]

4 divides 12 and 16

i think that's what OP means

[sudden cove]

nosols!!!! but yes

[onyx tide]

4…

Ok kids

[sudden cove]

now, factor this

[lusty python]

PLEASE DON'T NOSOLS ME

[onyx tide]

4(3^2+4k+3)

[lusty python]

$4(3k^2 + 4k + 3)$

[onyx tide]

YESS AND I HAVE 4 AS MULTIPLICATIOB

[grand pond]

1 divided by 0 equals Infinity

[sudden cove]

YESSSSSSSS

[lusty python]

now look at this

[onyx tide]

Wohohohoho

WOHOOHOHOHOHOH

so easy

That’s so easy

[sudden cove]

and because k is an integer this will always be divisible by 4

boom

youre goated

[lusty python]

if $k \not \in \mathbb Z$ then ts impossible

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

yessir

[onyx tide]

Because there’s a four outside the bracket

[lusty python]

@onyx tide

another exercise?

[onyx tide]

no but like

Why won’t even number fit?

[sudden cove]

that means the whole thing will always be a multiple of 4

4 times any integer is divisible by 4

[lusty python]

prove that $\forall n \in \mathbb Z$, if $3 \mid n$ then $3 \mid n^2$ and if $3 \nmid n$ then $n^2$ dividing $3$ always gives a remainder of $1$

[grand pond]

1 divided by 0 equals Infinity

[lyric charm]

try putting n=2 into the original expression

you will see that an odd number will result

[onyx tide]

That is hard

[onyx tide]

But I have to like prove that an even number doesn’t fit?

[lyric charm]

no you don't

[sudden cove]

they only asked the odd case

[lusty python]

if n is even then the expression is always odd, thus it is not divisible by 2, then it is not divisible by 4 if you're wondering

[onyx tide]

12, 4* 2, 3* 2

[lusty python]

@onyx tide new exercise for you to practice

[lyric charm]

they asked to prove what happens when n is odd

even n's are just not relevant

and here you don't have to worry about them

[sudden cove]

don't have to give them any more homework lol

[lusty python]

shush it ain't homework

[onyx tide]

But that is 16…

And if it’s even number you can’t divide with an odd..

[lusty python]

?

6/3 = 2

[onyx tide]

@lyric charm

[sudden cove]

(regardless the problem is solved now)

[lusty python]

let OP try this one so OP can get a gasp on how to do these kinds of problem

[slender walrus]

What's 16

[onyx tide]

Can you like present in a normal way

[onyx tide]

If we put n = 2

[hallow zinc]

hey guys

[slender walrus]

what are you putting n=2 into

[hallow zinc]

why 1/0= infinity? i just don't agree with this statement

[lusty python]

if $n$ is divisible by 3, then $n^2$ is divisible by 3. if $n$ is not divisible by 3, then $n^2$ divides 3 always resulting in a remainder of 1

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

!redir

• 29th bait ty

[midnight plank]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[hallow zinc]

?

[lusty python]

please move to another channel if you want to rant about my display name as the bot says

[hallow zinc]

hmm, tuff statement. i'll try to counter example it

[lusty python]

chop chop or mods mods

choose 1

if you can't help in this channel then please don't interupt us

[hallow zinc]

ok sorry. im just new no need to be hostile

[lusty python]

ik you're new, that's why im giving you a chance here

[hallow zinc]

dang.

[lusty python]

@onyx tide the problem rephrased

[hallow zinc]

did u prove it?

[lusty python]

goodness gracious

[lusty python]

yes i did but im not allowed to spit the solution out yet

[hallow zinc]

ah, what should i do if i proved it? should i submit it? and where?

[lusty python]

this is called a help channel

we help people in these kinds of channels

you don't submit anywhere or smth

[hallow zinc]

oooh, thank you for ur clarification

[slender walrus]

Review guidelines in #❓how-to-get-help and #rules

[sudden cove]

is OP still here?

[lusty python]

idk

[sudden cove]

@onyx tide

[lusty python]

-# OP is scrolling on tiktok or youtube or reddit

[onyx tide]

I was scrolling on TikTok actually

[lusty python]

lol

im not wrong

[lusty python]

anyways this works for any integer $n$

[grand pond]

1 divided by 0 equals Infinity

[onyx tide]

Is it 2?

[lusty python]

you're supposed to prove the statement above 😭

[onyx tide]

Because n^2 will always be an even number and divide with an odd number and you get odd number

3(n^2-1)

Hahah got ya!

I’m so smart

Hehehee

Smoke time

[lusty python]

nah

im not talking about dividing by 2 here

im talking about dividng by 3

😭

[lusty python]

also at n = 3 then n^2 = 9 which is not even

😭

[onyx tide]

But it’s well proven

[lusty python]

no

not even close

[onyx tide]

This

How even old r u

[lusty python]

how old are you first

[onyx tide]

17

[lusty python]

15

[onyx tide]

…

And you know this before me?

[lusty python]

at least not the vertical line test which you knew before me

[wanton spade]

-# happy belated birthday

[onyx tide]

Ha ha ha ha did you really not knew it

[lusty python]

well, i know it now because you opened a channel about how functions work lol

[onyx tide]

So you didn’t know anything about functions

[lusty python]

i did know

just didn't know the vertical line test

[lusty python]

anyways for this question, consider each remainder case for $n$

[onyx tide]

But like knowing what functions are and what isn’t is the basics of functions

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

$n = 3k$, $n = 3k + 1$ and $n = 3k + 2$

[grand pond]

1 divided by 0 equals Infinity

[onyx tide]

Ok and?

[lusty python]

then square each cases and simplify

[midnight plank]

@onyx tide Has your question been resolved?

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[grand wharf]

can anyone give me the solution of the foll qn

determine the number of positive integer solutions of the equation

x1+x2+x3=17

should me take the xn>=0 or x>=1?

[fallow scarab]

x positive means x > 0, so not x = 0

[lilac finch]

=1

[sudden cove]

x1 x2 x3 means separate variables x_1, x_2, x_3?

[grand wharf]

yes

[lilac finch]

Or what reimann said

[grand wharf]

its a combinatiosn qn btw

combinations*

[sudden cove]

what have you tried so far

[lyric charm]

it says positive integers, so 0 is not allowed unless you're French

btw we won't just give you the solution

[grand wharf]

i have solved it

i got 16C2 which is 120

i assumed all the x are greater than or equal to 1

so it becomes

y1 + y2 +y3 =17-3

since for all the x , (x-1)>=0

so y1+y2+y3 =14

n=3
r=14

we use combination with repetions to solve

i got 120

but like i recently had a test

i did the same qn
got the same solution

but my prof said its wrong and i shouldnt assume x>=1 unless they specify it in the qn

so now im here

[lilac finch]

If this is not true, then x>=1 should work

Also did it say positive or non negative solutions?

[grand wharf]

determine the number of positive integer solutions of the equation

x1+x2+x3=17

this si the exact qn

[lilac finch]

Then idk, why your prof expects to include 0

[grand wharf]

he said only to like take x>= something

if it is mentioned in the qn

but to my understanding

non-negative integer solutions means x>=0

whereas positive integer solutions does not include 0

[lilac finch]

Thats what i think

[grand wharf]

thats what everyhwere i searched on google said too

but i thought i might be missing something so i thought id come here and confirm

anyways thank you guys for u help

.close

[midnight plank]

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[midnight plank]

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[coral belfry]

hi. suppose we have the polynomials a(x)= x²-1, and b(x) an arbitrary one, both over R. their gcd can only be one of the three: x-1, x+1, or 1 right? and what if we worked on C, and a(x) = x²+1, b(x) arbitrary again?

[grim vector]

Or x² - 1

[coral belfry]

oh, that counts as a divisor too

shouldnt they have to be like prime factorization tho? bc x²-1 can be factorized

[shell wigeon]

Divisors include prime divisors

[coral belfry]

hmm alright got it

[grim vector]

but you have gcd(2x² - 2, x² - 1) = x² - 1

[coral belfry]

yeah, it works just like with numbers now i see it

[grim vector]

I think it would be the same on C but x-i,x+i and x² + 1

[shell wigeon]

If you work in C, x^2-1 is still just (x+1)(x-1)

[coral belfry]

i gave a different a(x) over C

for that reason

[shell wigeon]

Sure, x^2+1 = (x+i)(x-i)

[coral belfry]

ok so its the same principle

[shell wigeon]

Yes

[coral belfry]

thank you both:)

.solved

[midnight plank]

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[midnight plank]

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[strange dome]

how do i do this

[frozen rain]

do you know what inequalities are

[strange dome]

um maybe

[frozen rain]

with the greater than and less than

[strange dome]

oh yes

[frozen rain]

greater than or equal to

alr

[strange dome]

i understand that for 300 it will be “>300” and “less than or equal to 20”(idk how to type that)

[frozen rain]

oh

copy paste lol 😂

[strange dome]

i alr was typing it but

i answered ur question first

but just

idk how to do the rest

[frozen rain]

okay so for the first one

it says to write an inequality based on the number of minutes

what 2 numbers are minutes that is given?

or just any numbers

[strange dome]

75 and 15

[frozen rain]

alright

but he also uses these minutes to do work

[strange dome]

whos he?

ph wait

lol

nvm

[frozen rain]

alright so what does 75 represent

[strange dome]

right

[strange dome]

minutes

[lime apex]

helpies

[strange dome]

erm excuse you

[frozen rain]

!occupied

[midnight plank]

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

[strange dome]

helpjk

[frozen rain]

minutes to what

[strange dome]

ueah this is occupied

[lime apex]

sory

[strange dome]

ph yeah 1 word assignment

[lime apex]

i didnt know

[frozen rain]

k

and it says to let w be word

assigment

[strange dome]

yes

[frozen rain]

so if he does more assignments and he takes 75 minutes to do one

what would you do to like add to the minutes

sorry i suck at explaining 😭 without giving the answer

[strange dome]

what

idk

[frozen rain]

so let’s say i do 3 word assignments

[strange dome]

how much paper,

[frozen rain]

which takes me 75 minutes, each assigment

[strange dome]

?*

[frozen rain]

we need minutes right now

[strange dome]

then i add nothing

[frozen rain]

what would you do to find the total number of minutes

[strange dome]

times?

[frozen rain]

yes

[strange dome]

times what

[frozen rain]

so u would have

[frozen rain]

W

for word

[strange dome]

they didnt say how much

[frozen rain]

75W

[strange dome]

wait what

[frozen rain]

yeah u need the inequality

[strange dome]

mkay

[frozen rain]

so basically it represents the total time spent on each assignment

[strange dome]

wait huh

how?

lh wait

i get it

yes

[frozen rain]

k then do it with math

how many minutes

[strange dome]

what

[frozen rain]

• varible

[strange dome]

wdym

what variable??

[frozen rain]

no offense but did you read the problem

[strange dome]

we got 75W how do i know what to do next

[strange dome]

YES AND THATS EHY IM ASKING FOR HELP

[frozen rain]

we need to show an inequality

for the number of minutes

[strange dome]

meaning we need to know what to + it by?

wait

[frozen rain]

uhh no

[strange dome]

inequality as in

< >

[frozen rain]

yes

so

[strange dome]

we know its >

[frozen rain]

yes

and we also have 75W

[strange dome]

yes

[frozen rain]

but we also need to include the math part

because he may do math assignments

[strange dome]

do i plus it with 15M

[frozen rain]

yes

[strange dome]

yk u can say that

i will still understand

[frozen rain]

i can’t give the answer i think

[strange dome]

thats not the answer silly

[frozen rain]

!noans

[midnight plank]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[strange dome]

75W + 15M > 300

[frozen rain]

wait oops

[frozen rain]

k that’s the first one

[strange dome]

do i need to put M after 300

[frozen rain]

!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.

[frozen rain]

nope

[frozen rain]

yeah see

[strange dome]

thats not doing that tho

cus im still learning

and understanding where ur getting it

[frozen rain]

alr

[strange dome]

ok back to eqatioj

[frozen rain]

i’ll do that for the next one then

[frozen rain]

you’re finished with the first one

[strange dome]

wait fr?

TJATS ALL I GOTTA DO

[frozen rain]

so we need the inequality based on the number of sheets

[frozen rain]

yeah…?

how about you try this on your own

[strange dome]

yeah i just did and

ok give me a sec

[frozen rain]

here’s a less than and equal to sign

≤

[strange dome]

okay thabks

howd you do that

3W + 1W ≤ 20

[frozen rain]

copy paste

[strange dome]

awe okay

[frozen rain]

okay ur close

[strange dome]

…

[frozen rain]

but you need both math and writing

[strange dome]

????

oh help

sorry im dyslexic

[frozen rain]

you’re fine

[strange dome]

3W + 1M ≤ 20

[frozen rain]

[strange dome]

WOOO

wait

no

the sign is wrong

it has to be 20 max

[frozen rain]

yes

[strange dome]

it cant be greater than 20

[frozen rain]

it saying at max 20

[frozen rain]

nope

[strange dome]

OHHH

[strange dome]

wait what

myb i suck at sighs

[frozen rain]

i was saying yeah it can’t be greater than 20 sorry

[strange dome]

signs

ohhh

oh air

wait

i see

[frozen rain]

maybe i should have said nope it can’t

[strange dome]

okay

[frozen rain]

yeah

the sign says less than or equal to

[strange dome]

yes

sorry i forget that

the alligator eats the 20

[frozen rain]

you’re fine

[strange dome]

i was thinking it was the other way around

okay tyyy :)

.close

[frozen rain]

i remember when i learned about alligator in school, long time

[midnight plank]

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[strange dome]

frrr

💔

[midnight plank]

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[azure bolt]

\textbf{Principle of Complementation (CP)} If $A$ is a subset of a finite universal set $\mathcal U$, then [|\mathcal U \backslash A| = | \mathcal U| - |A|]

[grand pond]

calvin

[azure bolt]

can someone help me understand this please

[junior flower]

i would draw a picture

[azure bolt]

idk how to

[west iron]

draw U

make all the elements bananas

but maybe make them different colors or sizes or something so we know they're distinct

[azure bolt]

why not just label it {1, 2, 3, 4, 5, 6}

[junior flower]

that is also a legal declaration

[azure bolt]

icbf drawing

[west iron]

you could label the bananas with numbers instead, yeah

[azure bolt]

wait what does the | mean

[junior flower]

U\A?

[west iron]

|S| is the number of elements in S

[junior flower]

or the | | bars

[azure bolt]

||

[azure bolt]

ah okay

[azure bolt]

wait this just occured to me

why do i need help with this

thanks guys anyways

i figured it out

.close

[midnight plank]

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[west iron]

np

[midnight plank]

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[near geyser]

.close

[midnight plank]

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[lilac finch]

Do you still need help?

[midnight plank]

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[sand flume]

how can I use the epsilon delta definition to confirm that f(x+h) -> f(x) as h-> 0 means continuous. If f(x+h) approaches f(x) as h approaches 0 I see that as a limit and then we know that if we evaulate f(x+h) with h = 0 then f(x+h) = f(x). but that just shows that f(x+h) is continuous. How can I use epsilon delta to show f(x) is continuous

[runic hamlet]

f is continuous at x if f(x+h)->f(x) as h->0

[sand flume]

but why? I see the mechanical steps of his proof but I don't understand how we draw the conclusion

[runic hamlet]

and $\lim_{h\to 0} f(x+h) = \lim_{h\to 0} f(x)+h\left(\frac{f(x+h)-f(x)}{h}\right)=f(x)+\lim_{h\to0}h\left(\frac{f(x+h)-f(x)}{h}\right)=f(x)$

[grand pond]

Denascite

[runic hamlet]

continuity means $\lim_{y\to x} f(y)=f(x)$

[grand pond]

Denascite

[runic hamlet]

this is just the same thing with a substitution

writing y=x+h, aka setting h=y-x

and then instead of y->x you have h->0

[sand flume]

isn't the function that we find the limit continuous though? like if I said lim x-> 2 x^2 is 4 and x^2 at 2 is also 4 we say that x^2 is continuous at 4. we don't say anything about 4

[sand flume]

where did you get this? Arent the two ways to show that a function is continuous is showing that the limit and the value at the point match or using epsilon delta?

[runic hamlet]

the limit and the value at the point match
exactly the equality I wrote down

feel free to use eps delta but thats more painful

$\lim_{x\to 2} x^2=4$ says that the function $x^2$ is continuous at $2$

[grand pond]

Denascite

[sand flume]

we only conclude things about the left side and not the right

but if the book we make it about the right side

[runic hamlet]

what

[sand flume]

I mean that for the x^2 example x^2 is on the right side and that is what we found to be continuous. For what he is doing on the book he show the limit of f(x+h) = f(x) but then says f(x) is continuous at h=0. f(x) is on the left side