i would but idk how if that makes sense
#help-13
livid dirge
ik the answer is 4 now
and how did you get 4ft.?
proper marsh
but i did it kinda just mental
proper marsh
and how did you get 4ft.?
2 feet tall is 1 foot wide so i thought what divided by 2 is 2
livid dirge
most exams expect you to show your work, though, or at least some justification of what you did.
proper marsh
most exams expect you to show your work, though, or at least some justification ...
im practicing for the PSAT/SAT
livid dirge
what bearing does that have on my advice?
livid dirge
does the PSAT/SAT allow you to just put down the answer with no working?
livid dirge
... can't say I like the idea of not showing working, but if that's the case, sure.
and yes, 4ft. is correct.
proper marsh
lets do the comparing ratios so its easier for me for other things
livid dirge
consider that the two triangles are similar.
that means that all three sides of one triangle must be proportional in length to the corresponding side of the other triangle.
proper marsh
yes
livid dirge
one of those sides is the hypotenuse, which isn't given, so we don't care about it.
but we have the length of the two shadows, and the height of one tree, and are tasked to find the height of the other.
the two shadows are obviously related to each other, as are the two trees, and since they are similar, their ratios must be the same.
this is the key observation. once you realize that the ratios of the lengths of the two shadows and the heights of the two trees must be the same (due to similarity), you can set up an equation to solve for x.
proper marsh
idk how to set up an equation for this ive never done it before
x/2 = 10/5 or something
livid dirge
x/2 = 10/5 or something
and this is exactly correct!
proper marsh
or uh maybe x/10 = 2/5
livid dirge
eh wait, no.
proper marsh
darn
livid dirge
or uh maybe x/10 = 2/5
this is correct.
not the other one.
in short, what you want is something like $\frac{height_a}{height_b} = \frac{shadow_a}{shadow_b}$.
wraith daggerBOT
Hyacine
livid dirge
sure.
proper marsh
**Hyacine**
ima save this
livid dirge
glad to help!
proper marsh
.close
cedar kilnBOT
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hexed vortex
im confused by the proof of 6
cedar kilnBOT
im confused by the proof of 6
wraith daggerBOT
Branshi
hexed vortex
why is that a set?
smoky cliff
Preimage
cedar kilnBOT
hexed vortex
.reopen
cedar kilnBOT
✅ Original question: #help-13 message
hexed vortex
Im confused on $\phi(g) = \phi(x)$
wraith daggerBOT
Branshi
hexed vortex
im confused by the proof of 6
maybe its this g
mmm
idk tho
oh I think g' is being used as phi(g)
and so it just comes from that
ok makes sense
.solved
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urban marsh
I have a question regarding nonlinear systems of odes. If, as a parameter in the equation is varied, the determinant of the Jacobian (at an equilibrium) goes from positive to negative, what type of bifurcation is that classified as? In class we only learned of pitchforks, saddle nodes, and transcritical bifurcations but none of those seem to apply here. All of those seem to require either two solutions colliding or the number of equilibria changing, which is not what has happened in my case. Not to mention we only did examples of them applying to single ODEs of one variable, its quite confusing when we increase the dimensions...
cedar kilnBOT
I have a question regarding nonlinear systems of odes. If, as a parameter in the...
@urban marsh Has your question been resolved?
cedar kilnBOT
@urban marsh Has your question been resolved?
cedar kilnBOT
@urban marsh Has your question been resolved?
cedar kilnBOT
@urban marsh Has your question been resolved?
topaz pewter
I have a question regarding nonlinear systems of odes. If, as a parameter in the...
have you missed a fixed point? Because it seems like your fixed point is changing stability but if you've only got one fixed point then it can't be any of those types of bifurcations
can you post the question
cedar kilnBOT
@urban marsh Has your question been resolved?
urban marsh
have you missed a fixed point? Because it seems like your fixed point is changin...
apologies, I was extremely busy so I didn't see this msg until now
tulip cave
I have a question regarding nonlinear systems of odes. If, as a parameter in the...
The kind of bifurcation you're describing—where the determinant of the Jacobian flips sign (goes positive ---) negative) without the number of equilibria obviously changing—is the hallmark of a Saddle-Node Bifurcation in higher dimensions (like 2D or 3D).
Here's the quick breakdown of why that determinant matters: The Core Rule: For any of the simple bifurcations (saddle-node, pitchfork, transcritical) to happen, the Jacobian matrix must have an eigenvalue that passes through the imaginary axis.Saddle-Node's Condition: For a Saddle-Node, the crucial condition is that a real eigenvalue passes right through zero.The Determinant Link: The determinant of a matrix is the product of all its eigenvalues. If one real eigenvalue hits zero, the whole determinant must hit zero. When it flips sign (like from $+$ to $-$), it confirms that an odd number of real eigenvalues (usually just one) has crossed that zero point.
wraith daggerBOT
papaop
tulip cave
was this the kinda answer you were looking for ?
i found it in the math help book i used to study on
urban marsh
i see, that explanation seems reasonable enough
tulip cave
is there anything else i can help u with?
urban marsh
The kind of bifurcation you're describing—where the determinant of the Jacobian ...
when I look at diagrams for the 1 variable case though, a saddle node always appears as just a place where two equilibria kinda "meet" and disappear. How does this relate to the higher dimension "saddle node" that is being described here?
tulip cave
The key to a Saddle-Node Bifurcation in any dimension is that a real eigenvalue of the Jacobian passes through zero ($\lambda = 0$).
In 1D, the Jacobian is the eigenvalue, so $\frac{df}{dx} = 0$.In $n$D, the determinant is the product of all eigenvalues, so $\det(J) = 0$. That's why the sign flip in $\det(J)$ is the universal signal!
The Center Manifold Makes it 1DWhen $\lambda=0$, the system has one critical direction where the dynamics become slow. The Center Manifold Theorem lets us mathematically ignore all the other fast-moving variables.
It guarantees that the complex $n$-dimensional system locally simplifies to the famous 1D saddle-node equation:$$\dot{u} = \mu \pm u^2$$where $u$ is the variable in that critical direction.
The Collision Still HappensBecause the system locally behaves like $\dot{u} = \mu \pm u^2$, it must locally involve the creation or annihilation of two equilibria.In 2D, this is usually a Saddle point and a Node/Focus colliding (which is how the name "saddle-node" was originally coined!).The only difference is that the collision is happening along that one critical "track" (the center manifold), while the rest of the phase space is just flowing in or out along the stable and unstable directions.So, the $\det(J)$ sign flip is just a mathematical alarm.
wraith daggerBOT
papaop
tulip cave
i love you texit
@urban marsh
tell me if its not comprehensible i come from an engineering course
urban marsh
Do you mind explaining what the Center Manifold Theorem is? Also, when you say it must locally involve the creation and annihilation of two equilibria, that seems to not be what I am seeing in my case, I might send over the question
tulip cave
while i explain it can you send the questionù
urban marsh
yep
I have the following equations, and we are fixing p=q.
the question is as follows, I clarified with my teacher, that he wants the curves where the determinant/trace of the jacobian equal to zero
tulip cave
Think of the Center Manifold Theorem as a fancy way to simplify a complex, multi-dimensional problem.Imagine you have a big $n$-dimensional system of ODEs, and you hit that bifurcation point where one eigenvalue is zero ($\lambda=0$).Stable/Unstable Directions: Most variables are doing something simple (either quickly flying away or quickly collapsing in) because their eigenvalues are not zero. These dynamics are straightforward.The Center Manifold: The theorem says we can mathematically ignore those simple directions. It isolates a small, lower-dimensional "surface" (the center manifold) that only contains the complex, slow, interesting dynamics associated with the $\lambda=0$ eigenvalue.In short: It lets us reduce a complex system ($\dot{\mathbf{x}} = \mathbf{f}(\mathbf{x})$) near a bifurcation down to a much simpler 1D equation ($\dot{u} = g(u)$). This simpler equation is what actually governs the bifurcation, and for the Saddle-Node case, that 1D equation is always the one that creates/destroys two equilibria ($\dot{u} = \mu \pm u^2$).2. Why You Might Not Be Seeing Two EquilibriaYou're right to question it! If the Saddle-Node Bifurcation (SNB) is all about collision and annihilation, why might you only be seeing one equilibrium change stability?There are a few possibilities:You're on the Edge: The SNB mechanism means two equilibria (say $E_1$ and $E_2$) exist on one side of the parameter value ($\mu < \mu_c$), collide exactly at the critical parameter ($\mu = \mu_c$), and vanish on the other side ($\mu > \mu_c$). If you are observing a single equilibrium transition from stable to unstable (or vice versa) as the parameter changes, you are most likely watching that one equilibrium approach the collision point. The second partner equilibrium ($E_2$) must exist somewhere but might be far away or hard to find.It's the Ghost: When a saddle-node pair disappears, the dynamics near where they used to be get incredibly slow. This region is sometimes called the "ghost" of the fixed points. Your single equilibrium might be transitioning through this slow region just before it meets its partner.Global Bifurcation Masking: Less commonly, the SNB might be complicated by a global event. For instance, an unstable limit cycle (a periodic orbit) could be involved in a separate event, making the local picture messy.The bottom line is that the zero determinant ($\det(J)=0$) is a local condition that forces the system to locally behave like a 1D Saddle-Node collision
wraith daggerBOT
papaop
tulip cave
the question is as follows, I clarified with my teacher, that he wants the curve...
lemme a sec so i can look at it goof
urban marsh
sure, ill also share what I have so far
tulip cave
wait ima do it real fast
urban marsh
okok
urban marsh
**papaop**
your explaination of the theorem makes good sense
urban marsh
pdf works
urban marsh
kk
tulip cave
oh i almost forgor
Transcritical/Pitchfork (TC/PF)$\lambda_1 = 0$, $\lambda_2 \ne 0$$$\mathbf{s = \frac{1-3p}{1-2p}}$$
wraith daggerBOT
papaop
tulip cave
Saddle-Node (SN)$\lambda_1 = 0$, $\lambda_2 \ne 0$$$\mathbf{s = \frac{1-5p}{1-4p}}$$
wraith daggerBOT
papaop
tulip cave
Hopf (H)$\text{Re}(\lambda_{1,2}) = 0$$$\mathbf{s = \frac{1-4p}{1-3p}}$$
wraith daggerBOT
papaop
tulip cave
urban marsh
ok, there are other equilibrium points that mathematica has found, but reading through I somewhat get the method you're doing, a couple questions
tulip cave
tell me , im all yours
urban marsh
Can you explain where these cases come from? What exactly do you do that differentiates the saddle node from the Transcritical/Pitchfork?
also how u write all that up so fast lmao
tulip cave
The Algebraic Source: Factoring the DeterminantWe found the determinant of the Jacobian at the equilibrium $E_0(0, 0)$ (assuming $p=q$ and letting $A=1-s$) was:$$\det(J) = [A(1-3p) - p]^2 - (Ap)^2 = 0$$This is the form $X^2 - Y^2 = 0$, where:$X = A(1-3p) - p$$Y = Ap$The only way for $X^2 - Y^2$ to equal zero is if the factored form $(X - Y)(X + Y) = 0$ is true. This forces the determinant equation to split into two independent linear equations, which we called Case 1 and Case 2.Case 1: $X - Y = 0$Case 2: $X + Y = 0$The Dynamical Distinction: Why They Are Different BifurcationsAlthough both cases satisfy $\det(J)=0$ (meaning one eigenvalue $\lambda$ is zero), they involve different eigenvalues and therefore correspond to different dynamical events in the phase space.Case 1: The Saddle-Node (SN) Condition$$\mathbf{X - Y = 0}$$$$[A(1-3p) - p] - Ap = 0 \quad \implies \quad A(1-4p) = p$$What it Means:Eigenvalue Behavior: This condition corresponds to the $\mathbf{J}$ matrix having a single zero eigenvalue, but the other eigenvalue is non-zero.Equilibria Created/Destroyed: This is the characteristic condition for a true Saddle-Node bifurcation, where two non-trivial equilibria ($E^$) are created (or destroyed) away from the trivial equilibrium $E_0(0, 0)$. When these two non-trivial solutions meet, the determinant goes to zero along that curve.Stability: This is where the number of solutions changes.Case 2: The Transcritical/Pitchfork (TC/PF) Condition$$\mathbf{X + Y = 0}$$$$[A(1-3p) - p] + Ap = 0 \quad \implies \quad A(1-2p) = p$$What it Means:Eigenvalue Behavior: This condition corresponds to the $\mathbf{J}$ matrix also having a single zero eigenvalue, $\lambda_1 = 0$, but it's occurring exactly at the trivial equilibrium $E_0(0, 0)$.Equilibria Collision/Exchange: This is the characteristic condition for a Transcritical or Pitchfork bifurcation. Here, the trivial equilibrium $E_0$ is colliding with the symmetric non-trivial equilibrium $E^$ at the origin and exchanging stability. The number of equilibria is preserved through the bifurcation point (though their identities change).Stability: This is where $E_0$ loses or gains stability to the non-trivial solutions.
wraith daggerBOT
papaop
tulip cave
Summary of the Differentiation
The algebra forces the determinant to split into two curves, and the physical meaning of those curves differentiates them:
Case 1 (SN)$A(1-4p) = p$Two non-trivial equilibria collide away from $E_0$.YesChanges (e.g., 2 $\leftrightarrow$ 0)
Case 2 (TC/PF)$A(1-2p) = p$The trivial equilibrium $E_0$ collides with a non-trivial equilibrium $E^*$.YesStays the same (they exchange roles)
wraith daggerBOT
papaop
urban marsh
why is this true?
tulip cave
The Full Solution StructureRemember, the system is symmetric, which means that the non-trivial equilibria $\mathbf{E^*}$ must be $\mathbf{(\bar{n}, \bar{n})}$. When you solve for the fixed points for this model, you find two types of non-trivial solutions:Symmetric Solution Branch: $\bar{n}A = \bar{n}B = n{sym}$. This branch collides with the origin $E_0(0,0)$ at the Transcritical/Pitchfork curve (Case 2: $X+Y=0$).Asymmetric Solution Branches: $\bar{n}A \neq \bar{n}B$. These two branches are created/destroyed away from the origin in a Saddle-Node bifurcation.2. The Eigenvalue BehaviorBoth Case 1 ($X-Y=0$) and Case 2 ($X+Y=0$) satisfy the general bifurcation condition $\det(J)=0$, meaning one eigenvalue is zero ($\lambda_1 = 0$).For the symmetric case ($p=q$), the eigenvalues of the Jacobian $J$ at any symmetric equilibrium $\mathbf{E} = (\bar{n}, \bar{n})$ are:$$\lambda{1, 2} = J{11} + J{12}, \quad J_{11} - J_{12}$$
wraith daggerBOT
papaop
tulip cave
Case 2 (TC/PF)$X + Y = 0$$\lambda_1 = (X + Y) - p - Ap$$\lambda_2 = J_{11} - J_{12}$Case 1 (SN)$X - Y = 0$$\lambda_1 = (X - Y) - p - Ap$$\lambda_2 = J_{11} - J_{12}$
wraith daggerBOT
papaop
tulip cave
$$\det(J) = (X-Y)(X+Y)$$This means that:If $X+Y=0$ (Case 2): $\det(J)=0$. One eigenvalue is zero. This happens when the symmetric branch collides with the origin $E_0(0,0)$.If $X-Y=0$ (Case 1): $\det(J)=0$. One eigenvalue is zero.The key to the SN distinction is that the $X-Y=0$ curve also describes where the two asymmetric branches $\mathbf{E^}$ collide with each other (like the apex of the parabola in your 1D diagram ).SN Mechanism: When two equilibria (a stable one and a saddle one) collide and annihilate, the zero eigenvalue occurs at the point of collision. This collision point is defined by the SN curve: $s = \frac{1-5p}{1-4p}$. Since this curve exists away from the origin in the bifurcation diagram, it must be where the number of solutions changes (2 solutions $\leftrightarrow$ 0 solutions), which is the definition of the Saddle-Node.In Simple TermsTC/PF (Case 2: $X+Y=0$): The pre-existing solution at the origin collides with a new solution coming toward it. They exchange stability. $\rightarrow$ Number of solutions stays the same (locally).SN (Case 1: $X-Y=0$): Two separate solutions collide and annihilate, disappearing from the phase space altogether. $\rightarrow$ Number of solutions changes (e.g., $2 \to 0$).The condition $X-Y=0$ is derived from the generic SN normal form when a center manifold reduction is performed on the asymmetric branches $\mathbf{E^}$.
wraith daggerBOT
papaop
tulip cave
nice , do u have any doubts or can i close? (if u need any help js dm me)
urban marsh
I guess, what should I do in a situation where the determinant can't factorise neatly like that?
tulip cave
Solve the Non-Factorable $\det(J) = 0$ EquationYou treat the entire, complicated expression for $\det(J) = 0$ as the single equation that defines all possible $\lambda=0$ bifurcations (Saddle-Node, Transcritical, and Pitchfork).For your previous problem (where A=1−s and p,q are parameters):$$[A(1-2p-q) - q][A(1-p-2q) - p] - A^2 pq = 0$$If this didn't factor, you would keep it as one big curve in the $(p, s)$ diagram. This curve is often called the "Bifurcation Locus" or the "Fold Curve" (since the SN/fold is the most generic $\lambda=0$ event).2. Differentiate the Bifurcation Types (The Hard Part)Since the $\det(J)=0$ curve contains all three $\lambda=0$ events, you have to use a secondary condition to figure out which part of the curve is which.A. Look for the Transcritical/Pitchfork (TC/PF) EventsThe TC/PF bifurcations always happen at an equilibrium that exists for all parameter values (like the trivial $\mathbf{E}_0$ in your system).Find the Collision Point: Find where the two solution branches must cross. For TC/PF, this happens when the non-trivial equilibrium $\mathbf{E}^$ coincides with the trivial equilibrium $\mathbf{E}_0$.Substitute $\mathbf{E}_0$ into $\mathbf{E}^$:First, find the formula for $\mathbf{E}^$ (the non-trivial fixed point) by setting $\dot{n}_A = 0$ and $\dot{n}_B = 0$.Then, find the parameter values $(p_c, s_c)$ where $\mathbf{E}^ = \mathbf{E}_0$.The points where this happens must lie on the $\det(J)=0$ curve. This isolates the TC/PF points along the curve.B. The Rest Are Saddle-Nodes (SN)Anywhere else on the non-factorable $\det(J)=0$ curve where the TC/PF condition is not met must be a Saddle-Node bifurcation.Saddle-Nodes are the generic $\lambda=0$ bifurcation. They are the "default" event, occurring whenever the fixed point equation $f(\mathbf{x}, \mu) = 0$ is tangent to the $\mathbf{x}$-axis in a higher dimension.
-
Find the LocusDefine the full bifurcation curve.$\det(J) = 0$ (non-factorable)
-
Locate TC/PF Identify the special collision points. $\mathbf{E}^ = \mathbf{E}_0$*
-
Identify SNThe rest of the curve must be SN.$\det(J)=0$ AND $\mathbf{E}^* \neq \mathbf{E}_0$
wraith daggerBOT
papaop
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l.49 ...k).For your previous problem (where A=1−
s and p,q are parameters):...
You may provide a definition with```
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mossy pivot
I need help, i dont even know how to start, ive been watching videos trying to learn and get an idea but idk , can i get pointed to the first step pls
granite ether
I need help, i dont even know how to start, ive been watching videos trying to l...
for the first one id use l'hospitals probably
it looks like you could do it explaining how it would approach but if you want a more formal proof l'hospitals
cedar kilnBOT
@mossy pivot Has your question been resolved?
grizzled vale
okay there's a couple of ways of determining the limit of a sequence, if you've learnt the monotone convergence theorem you can use that, but l'hopitals rule also works nicely
just note that strictly, you need to "embed" the sequence in a continuous function to use l'hopitals rule as the domain of the sequence is limited to natural numbers (therefore a derivative does not exist for them)
you can just say "let $f(x)=(func)$ where $x\in\Bbb R$"
but ignore this if you weren't taught it
wraith daggerBOT
lucas
mossy pivot
just note that strictly, you need to "embed" the sequence in a continuous functi...
Uh ion really understand what you’re saying here tbh
Does monotone mean monotonic and for that isn’t it easier to see when graphed
dusk goblet
Does monotone mean monotonic and for that isn’t it easier to see when graphed
you can ignore that
it will almost never be relevant in these calc 2 problems
what do you think the limit is for the first one
$a_n = \frac{n^3}{e^{2n}}$
wraith daggerBOT
knief
mossy pivot
Oh i didn’t think of writing it that way
dusk goblet
generally a good idea
you should remember that exponentials grow far faster than polynomials
mossy pivot
The bottom is increasing a lot quicker then the top i think
dusk goblet
yes
mossy pivot
So its getting smaller fast
dusk goblet
which says what about the limit
mossy pivot
So is it reaching 0?
mossy pivot
Tbh my brain saw negative exponent and e and it shut off
dusk goblet
or if you can just say that exponentials >> polynomials
dusk goblet
Tbh my brain saw negative exponent and e and it shut off
whenever you see a negative exponent just rewrite it using a positive one
and flip it of course
mossy pivot
Looking at it rewritten made it easier
dusk goblet
can you try part b?
mossy pivot
So to justify
dusk goblet
do you know the limit of arctan?
dusk goblet
So to justify
uhh yea you can just write e^{2n} >> n^3 or if your teacher is really petty you can say by lhopitals rule
mossy pivot
Ohh okay
dusk goblet
or you could squeeze it
use squeeze theorem
$0 \leq \frac{n^3}{e^{2n}} \leq \frac{n^3}{n^4}$
wraith daggerBOT
knief
dusk goblet
for appropriate n
where you use the fact that e^(2n) >= n^4 eventually
but probably easier to just say by lhopitals rule
mossy pivot
Isnt this also like power rule or something
dusk goblet
Isnt this also like power rule or something
what is?
mossy pivot
Question a😶 or am i wrong
dusk goblet
but where?
you mean when doing lhopitals rule?
to take the derivative of the numerator you would need the power rule
mossy pivot
uhh yea you can just write e^{2n} >> n^3 or if your teacher is really petty you ...
no when look at it like this
dusk goblet
if thats what you meant
dusk goblet
no when look at it like this
uhh there is no power rule there
🤔
power rule is for derivatives
dusk goblet
you mean like comparing the degree
like saying n^4 > n^3
it doesn't apply here since e^(2n) isn't a polynomial
mossy pivot
no
Pi/2 according to my searches
So if the limit is pi/2 for arc tan and the bottom can just infinitely get bigger isnt this like the first problem where bottom like dominates the problem
dusk goblet
Pi/2 according to my searches
yes
dusk goblet
So if the limit is pi/2 for arc tan and the bottom can just infinitely get bigge...
yes
mossy pivot
So is this reaching 0 too
Does the last one have no limit
Because its just going back and forth like i see the pattern but itll never reach anything
Not monotonic ?
dusk goblet
Because its just going back and forth like i see the pattern but itll never reac...
perfect
dusk goblet
Not monotonic ?
well it doesn't have to be monotone to have a limit
consider $a_n = \frac{(-1)^n}{n}$
wraith daggerBOT
knief
dusk goblet
even though we aren't always increasing or decreasing the terms do approach 0
mossy pivot
Oh so how do i justify no limit
Oh
Wait i just explain
Actually idk
Am i supposed to justify using some theorem or just explain the pattern
dusk goblet
yea im guessing they dont want a formal proof
dusk goblet
Am i supposed to justify using some theorem or just explain the pattern
doubt it
you could say that you can take two subsequences with one consisting of only the 0 terms and another consisting only of the 1 terms and say their limits dont agree
that would be a theorem
and is probably the best explanation youll get
mossy pivot
ok👍
dusk goblet
by subsequence i just mean a sequence within the sequence essentially. so we take whatever terms we want and make sure we keep the same order of the terms
like imagine i have
$a_1, a_2, a_3, \dots,$
wraith daggerBOT
knief
dusk goblet
i might take a subsequence consisting of only the even index terms
so
$a_2, a_4, a_6, \dots$
wraith daggerBOT
knief
dusk goblet
or the odd terms of the multiples of 5 or whatever i want
so long as the new sequence i take maintains the order that the sequence members appear in the original sequence
dusk goblet
oh right we skipped c lol
dusk goblet
now for the ln problem 😼
use log rules
mossy pivot
So rewrite as ln(2n^2+7/n+3)
dusk goblet
So rewrite as ln(2n^2+7/n+3)
yes
mossy pivot
So here the numerator is getting larger faster then the denominator because of the square and then we apply ln which is just division? But since were going to infinity the num is just gonna get bigger
dusk goblet
So here the numerator is getting larger faster then the denominator because of t...
wdym just division
the limit of the inside is what?
you are correct that the numerator grows faster
mossy pivot
Like its opposite of exponential or e whatever the name for that is
mossy pivot
the limit of the inside is what?
Infinity?
dusk goblet
Infinity?
yes
then if the input of ln is going to infinity
what is the entire thing going to
mossy pivot
Infinity
dusk goblet
nice
dusk goblet
yea
if you want to be more formal you might divide both the numerator and denominator by n
$\frac{2n^2 + 7}{n + 3} \cdot \frac{\frac{1}{n}}{\frac{1}{n}} = \frac{2n + \frac{7}{n}}{1 + \frac{3}{n}}$
then take n -> inf
mossy pivot
I thought when u do that you do it with the highest power
wraith daggerBOT
knief
dusk goblet
I thought when u do that you do it with the highest power
well no
if we did that then we would get 2/0 which still works
but
i don't do that
dusk goblet
if we did that then we would get 2/0 which still works
since this goes to inf
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Available help channel!
wispy nebula
but thats not the question i believe
cedar kilnBOT
but thats not the question i believe
dusk goblet
isnt n = 1
what
wdym isn’t n = 1?
this is true regardless of the value of n so long as the expression is defined
🤔
it’s an identity
long swan
i am very not smart
Luckily you don't have to he smart to do math. Just look at Newton. He died poor and with no bitches
wispy nebula
Luckily you don't have to he smart to do math. Just look at Newton. He died poor...
wow
wispy nebula
it does dw
dreamy vale
you can do what ever you want,do not inspirating negative @wispy nebula
wispy nebula
you can do what ever you want,do not inspirating negative <@759863736734318592>
what does this mean
dreamy vale
you are smart enough
wispy nebula
ohh ty
wicked mantle
I need help, i dont even know how to start, ive been watching videos trying to l...
I believe this is the original question(?
latent palm
I believe this is the original question(?
Yeah but she closed the ticket
cedar kilnBOT
@wispy nebula Has your question been resolved?
hollow trail
dreamy vale
thank you
hexed vortex
whats the difference from continuity and a limit in this definition?
hexed vortex
I know that a limit doesnt require the function to be defined at x0
while continuity does
but where is that said in this definition
hallow pelican
continuity just means limit agrees with the real function
pastel vault
but where is that said in this definition
$\epsilon > 0, \delta > 0$ implies those can't be zero
wraith daggerBOT
south
pastel vault
if $\epsilon = 0$, then the absolute value is always non-negative, so you'd end up with $f(x) - f(x_0) = 0$
wraith daggerBOT
south
cerulean sail
(The fact that in general limits, you usually have "0 < |x - x0| ...")
hexed vortex
oh
pastel vault
similarly you can't have x = x0
hexed vortex
I see that makes sense, so |x - x0| can be 0 hmm but then wouldn't x = x0 in that case but we cant have that?
pastel vault
**south**
no, you can't have this either
you have $f(x_0) - \epsilon < f(x) < f(x_0) + \epsilon$ in general
wraith daggerBOT
south
cerulean sail
I see that makes sense, so |x - x0| can be 0 hmm but then wouldn't x = x0 in tha...
(In general, you can't, they do that so that you don't need to worry about the function being defined at x0, but for continuity you do want the function to be defined at x0)
pastel vault
oh wait that didn't answer the question
cedar kilnBOT
dusk goblet
(In general, you can't, they do that so that you don't need to worry about the f...
its not so much about it being defined but that we dont care about what happens at x_0
the information the limit gives us is about the behavior in deleted neighborhoods around x_0
cedar kilnBOT
Available help channel!
dusk goblet
as reflected by the 0 < |x - x_0| < delta
cedar kilnBOT
as reflected by the 0 < |x - x_0| < delta
dusk goblet
it could be the case that the function is defined there but doesn't agree with the limit
and is thus not continuous
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
hexed vortex
does for any epsilon > 0, there exists a delta mean that each epsilon has a delta basically?
hexed vortex
where the delta can be the same or not
flint plinth
yep
tropic oxide
different epsilons may be "served by" different deltas yes
flint plinth
you can think of delta as a function of epsilon
hollow trail
usually we come up with some formula which gives delta as a function of epsilon
hexed vortex
ok makes sense thanks
cedar kilnBOT
@hexed vortex Has your question been resolved?
hexed vortex
I'm proving the left direction first and my scratch work is above, can someone check my understanding of this? I just want to make sure I'm getting a better sense of how this all works intuitively
Were given that f is continuous at $x_0$, and by the definition of continuity for any distance of how close we want our function values to be to $f(x_0)$, we can find some distance for which we can make our x values close enough to $x_0$ where our function values satisfy that given distance with $f(x_0)$. So for any given sequence of our domain values that converges to $x_0$ we can go far enough in our sequences such that the distance between $x_n$ and $x_0$ is less than the distance needed for our function values to be within any given distance which implies function applied to the sequence values would also satisfy that epsilon.
I tried to write it out abstractly but I think I just confused myself with all the distances
wraith daggerBOT
Branshi
hexed vortex
I remember knief mentioning of wanting to think of these things in terms of distance
I think thats a good way to look at it also just not sure if im looking at it right lol
hexed vortex
like I cant say I for sure understand why in this image here were allowed to use the epsilon from the continuity definition to prove the last statement I wrote, well im not sure if were allowed to but I think we are but cant say for sure why
dusk goblet
like I cant say I for sure understand why in this image here were allowed to use...
wdym?
for all x within a distance delta from x_0, the distance from f(x) to f(x_0) is less than epsilon. you can guarantee that for sufficiently large n all x_n will be within that distance from x_0 and hence their images will be within epsilon from f(x_0)
hexed vortex
Ok that makes sense I think I was just over complicating things in my head a bit
dusk goblet
probably
hexed vortex
ill give the other direction a go now
dusk goblet
@hexed vortex any luck?
dusk goblet
what have you been trying so far
hexed vortex
dusk goblet
you do that thing again where you write x_n > M but i assume you're having trouble writing what delta would be
have you tried any other approaches?
dusk goblet
oh
hexed vortex
because we’re looking at the actual sequence values right
dusk goblet
well f(x_n) is a sequence but its still dependent on the index n?
for example, take f(x) = 1/x and just x_n = 1/2^n or something then you're saying there is some M such that if 1/2^n > M then .... but this shouldnt make sense because x_n is a decreasing sequence
its dependent on the index n coming from the sequence x_n
f(x_1), f(x_2), ...
its the same idea we really
the n corresponds to the f(x_n)
so when we say f(x_n) is getting closer to f(x_0) its capturing that same idea as it was with just x_n
each n corresponded to some sequence member
remember that x_n is itself a map from N into R so f(x_n) is a map from N into R as well
its like a composition
dusk goblet
remember that x_n is itself a map from N into R so f(x_n) is a map from N into R...
ok just assume f is real valued actually
if thats what notabot was going to point out
🤔
the point is the domain though
N
hexed vortex
hmm I see why f is a composition from N to R, but not sure what's wrong with x_n > M yet
because the limit to infinity definition
dusk goblet
well consider my example?
hexed vortex
requires x values to be greater than M right
dusk goblet
if you take x_n to be a decreasing sequence here like 1/2^n then you only ever have finitely many values x_n > M
hexed vortex
hmm yeah that makes sense
dusk goblet
since 1/2^n converges to 0
so it wouldnt make sense to speak of convergence for the sequence f(x_n) like that if the tail is just finitely many values
hexed vortex
ok that makes sense
cedar kilnBOT
@hexed vortex Has your question been resolved?
lament hemlock
could i get help
cedar kilnBOT
could i get help
lament hemlock
part ii
i know i need to apply part i recursively
but i cant telescope properly
wraith daggerBOT
south
pastel vault
so from $n = 1$ and given $I_0$, you can write out what $I_1, I_2, I_3 \cdots$ must be to give you some of a start
wraith daggerBOT
south
pastel vault
then the actual proof will be by induction
cedar kilnBOT
@lament hemlock Has your question been resolved?
pastel vault
I'm not sure actually whether you need to do it by induction
lament hemlock
I don’t understand the answers but I know you can just apply recursively
Like write out the first 4 terms
And then justify
pastel vault
I don’t understand the answers but I know you can just apply recursively
oh can you send out the solution
pastel vault
ah that's clever
so they multiplied top and bottom by (2n) * (2n - 2) * (2n - 4)...
cause now the denominator will be (2n + 1)!
lament hemlock
Oh to fill in those terms
pastel vault
so okay, you have $[(2n)(2n - 2)(2n - 4) \cdots]^2$
wraith daggerBOT
south
pastel vault
there are n factors of 2 before squaring
so then after squaring, you can take 2^(2n) out
lament hemlock
On so it just so happens the missing terms in the denominator are the ones in the numerator
pastel vault
and then you are left with $[n(n - 1)(n - 2) \cdots]^2$
wraith daggerBOT
south
pastel vault
so they multiplied top and bottom by (2n) * (2n - 2) * (2n - 4)...
yep
tidal fable
Guys can someone tell me how i should study for 3-4 hours daily
covert violet
Guys can someone tell me how i should study for 3-4 hours daily
divide your time properly
id spend an hour or so learning new concepts, and the rest on application since practice is really important
and maybe dont spend all that 3-4 hours at one go, its important to take breaks too
wicked mantle
vague halo
Guys can someone tell me how i should study for 3-4 hours daily
It's not just about the hours we study, it is about the understanding
rocky harness
I mean just start studying and study as much as you see fit. Then you can change your judgement depending on what results you get
cedar kilnBOT
@tidal fable Has your question been resolved?
cerulean sail
(this is more for #study-discussion btw 
)
dreamy cairn
hi
wicked mantle
hi
dreamy cairn
I'm testing an AI and I need some very human style algebra tests to see if it can create abstract thinking
could anyone give me a few, easy medium difficult and extremely difficult ones so I can see if it can solve them
I'm trying to build a.... new AI that will smash the mathematics benchmarks
cedar kilnBOT
@dreamy cairn Has your question been resolved?
pliant atlas
could anyone give me a few, easy medium difficult and extremely difficult ones s...
chatgpt it
im kidding
there should be many in google
just search it up 😭...
dreamy cairn
Im not sure how I should take this answer, If I wanted to be trolled I'd go to reddit math.
I genuinely wanted help.
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
craggy heart
wtf am i suppose to do
cedar kilnBOT
wtf am i suppose to do
teal wedge
you can replace |x| with f(x)
teal wedge
and the minimum of f(x) is 0
(for range)
teal wedge
what did i get from that
idk but the questions wants you to do that
teal wedge
idk
ask your teacher
probably because the other questions does that too
(6): 2f(x); domain:(-∞,∞), range:[0,∞); idk what is monotonicity (if it means 'one-to-one', probably not); it is even since g(-x)=2|-x|=2|x|=g(x); x=0
probably like this
this is just an example, do the other ones by yourself
cedar kilnBOT
cedar kilnBOT
Available help channel!
marble solstice
cedar kilnBOT
marble solstice
Is this impossible?
I habve been trying to prove it for a while but I can't
From a truth table it is not the same thing too
Via proof theory ^
cedar kilnBOT
@marble solstice Has your question been resolved?
marble solstice
that is how it is written i did not write it like that
dreamy vale
what " | "means?
if you define " | " I can help most probably
if you read above please pin me
marble solstice
if you define " | " I can help most probably
The pipeline and the hpyhen mean semantically entails
|—
cedar kilnBOT
@marble solstice Has your question been resolved?
dreamy vale
so left and right phrase has same value ( <=>) ? or its one way correct like ( => )
marble solstice
yes exactly
so that thing pretty much means =>
whenever LHS is true the RHS must be true
but any other case does not matter since that is what I am trying to prove
does that make sense?
I am trying to show that from LHS, I can get to RHS
dreamy vale
right phrase is necessary for left one ,,, if im correct tell me to begin
cedar kilnBOT
cedar kilnBOT
Available help channel!
cedar kilnBOT
wispy marten
😔
dreamy vale
@marble solstice
cedar kilnBOT
<@715295051458150542>
dreamy vale
urban coral
is this supposed to be a question?
urban coral
.close
dreamy vale
.close
cedar kilnBOT
urban coral
also probably don't directly send a solution (if it is)
cedar kilnBOT
Available help channel!
tight widget
I started coordinate geometry and came across section formula.
But I cannot understand/visualise the meaning of a point (x,y) dividing a line segment externally. Any help please.
zinc quarry
Could I get some help with this?
cedar kilnBOT
Could I get some help with this?
zinc quarry
Does that just mean plugging in a/b for r? I dont get the second and third sentence of the hint
tropic oxide
yes plug in a/b for r
then multiply both sides by b^3
do nothing else besides this for now. stop there and show me what you get
zinc quarry
a^3 + ab^2 + b^3 = 0
tropic oxide
wonderful, ok
zinc quarry
So now both a and b turn into 2*k to check theyre even or odd?
tropic oxide
not like that, no.
now there are 4 cases based on the parities of a and b, namely:
- a even, b even.
- a even, b odd.
- a odd, b even.
- a odd, b odd.
before you proceed with any more algebra, here is a question i am gonna require you to think about:
one of these cases automatically rules itself out based on info already written in the hint.
which one, and why?
dusk goblet
i mean why not just use rational root theorem
tropic oxide
i mean why not just use rational root theorem
im having op work thru what the hint says
and she might not even have access to RRT yet
dusk goblet
yea i would suggest learning it, it’s pretty straightforward
tropic oxide
again,
dusk goblet
and makes this problem take 2 steps
zinc quarry
is it something about the multiplying?
zinc quarry
ah if theyre both even they can be divided by two?
tropic oxide
yet
tropic oxide
ah if theyre both even they can be divided by two?
if a and b are both even then a/b isn't in lowest terms because a 2 can be cancelled out, yes.
ok so yes you got the important thing right.
zinc quarry
yeah that makes sense
tropic oxide
case 1 is the one that's auto eliminated.
the other 3 cases you'll have to look at in more detail.
if you can tell me immediately the parity of a^3 + ab^2 + b^3 in each one, then do so.
otherwise, write a and b according to the defns of even and/or odd number as applicable.
zinc quarry
if you can tell me immediately the parity of a^3 + ab^2 + b^3 in each one, then ...
im not sure what that means 😂
tropic oxide
careful not to use the same letter for each -- if you try to write a = 2k+1, b = 2k+1 in case 4, you will be asserting a=b, which is wrong
tropic oxide
im not sure what that means 😂
the word "parity" means the property of being even or odd.
im saying that if you can reach an immediate verdict on that, maybe using your knowledge on how parity behaves under multiplication and addition, then you should do that
if you don't, then work it out the "honest way".
zinc quarry
ok just to be safe im gonna do it the honest way
zinc quarry
careful **not** to use the same letter for each -- if you try to write `a = 2k+1...
what is the +1 for?
tropic oxide
i used case 4 as a way to showcase the pitfall im warning you against.
zinc quarry
ah
tropic oxide
that case says "a odd, b odd"
do you understand why translating that into specifically a = 2k+1, b = 2k+1 is incorrect?
zinc quarry
yeah because thatll give the same value for both of them when they could be different values in reality
tropic oxide
yes, ok, you understand it.
now go and work it all out the honest way.
it may take you a little while.
zinc quarry
now there are 4 cases based on the parities of a and b, namely:
1. a even, b ev...
case 2- a=2k, b=2j+1
case 3- a=2k+1, b=2j
case 4- a=2k+1, b=2j+1
like that?
tropic oxide
would appreciate commas between those eqs
yes those are the correct translations
now work out what happens with a^3 + ab^2 + b^3
zinc quarry
just go case by case?
tropic oxide
yes go case by case.
i do think that it would be easier to recall properties such as even*odd = even but like, you do you.
zinc quarry
not sure if this is how I was supposed to solve it but:
case 2 - even + even + odd = odd
case 3 - odd + even + even = odd
case 4 - odd + odd + odd = odd
tropic oxide
if a and b are odd then a×b×b is even?
zinc quarry
ah wait i mixed up that squared as multiplied by 2
zinc quarry
no
ah so thats the proof by contradiction?
wait where did we contradict the statement
tropic oxide
we supposed that a lowest-terms fraction a/b exists that will make the eq a^3+ab^2+b^3=0 true
it then turned out that this equation is impossible to satisfy
zinc quarry
ah that makes sense
ok sweet that definitely helped. im probably still gonna be stuck on proofs for a while but each practice definitely helps me understand it better. thanks for all your help
tropic oxide
ok sweet that definitely helped. im probably still gonna be stuck on proofs for ...
there's not really a royal road to proofs tbh, don't worry about it too much
everyone sucks at proofs for the first thousand or so
gaunt plaza
I dont really understand why this step is important
cedar kilnBOT
I dont really understand why this step is important
lyric sphinx
just cut the log both sides lol
gaunt plaza
I did that once and my teacher deducted points because of it😭
dire geode
It's necessary if you don't know log is injective
That's a question for your teacher then
lyric sphinx
I did that once and my teacher deducted points because of it😭
well if steps matter then yeah use the property where a*loga(b)=b
gaunt plaza
That's a question for your teacher then
What is your pfp? Some kind of 3d integral?
tender patio
could anyone explain me why result its 32/45. is it wrong? cuz in work says that
tender patio
could anyone explain me why result its 32/45. is it wrong? cuz in work says that
i got other result and i asked ai to solve and got another result than ds
upper laurel
could anyone explain me why result its 32/45. is it wrong? cuz in work says that
you here?
tender patio
yea
upper laurel
couldnt you just put this into a calculator?
to verify that 32/45 should be close to the answer
sacred anchor
could anyone explain me why result its 32/45. is it wrong? cuz in work says that
Does $0, \hat{8}$ mean $0.888\ldots$?
wraith daggerBOT
brittle berry
cedar kilnBOT
@tender patio Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
lethal vapor
How do I solve this question. I didn't understand it which is why I dont have any working to show
violet flume
no solutions 
lethal vapor
ye
violet flume
How do I solve this question. I didn't understand it which is why I dont have an...
try this
$2|3x+4y-2| = - 3 \sqrt{25-5x+2y}$
wraith daggerBOT
jan Niku
violet flume
do you see the issue?
2: Always positive
|3x+4y-2|: never negative
-3: Always negative
Sqrt: Never negative
lethal vapor
so
violet flume
so if we have any luck of this ever being true
what must happen
if we could just force these 4 factors to be whatever we want following these rules, how could we make the equation true?
lethal vapor
I'm new to modulus
violet flume
is this a complex question
lethal vapor
this is one of my first questions could you please elaborate
violet flume
sorry i mean
lethal vapor
is this a complex question <:thonk:406575732563902485>
for me yeah
violet flume
complex numbers
lethal vapor
nooo
violet flume
not complicated
lethal vapor
it does not use complex
violet flume
this is one of my first questions could you please elaborate
sure so, you see what i mean by 4 factors
I'm talking about the 4 pieces here
the two numbers, the absolute value, and the square root
Lets assume the equation is true for some values
then both sides must be the same sign, yea?
they have both be positive, or both be negative, or both be zero
lethal vapor
yea
violet flume
but we have some problems
-3 * sqrt(something) can never be positive
you see why?
lethal vapor
yes
violet flume
and 2 * absolute value something can never be negative
so, both sides cant be positive
and both sides cant be negative
if we have any hope at all, they must both be zero
lethal vapor
ok
violet flume
how do you make $2|3x+4y-2|=0$
wraith daggerBOT
jan Niku
violet flume
can you solve it?
lethal vapor
yeah
violet flume
its a line, yea?
lethal vapor
yes
wraith daggerBOT
jan Niku
lethal vapor
yes
lethal vapor
ok
violet flume
,w solve 2|3x+4y-2|+3 sqrt(25-5x+2y) = 0
wraith daggerBOT
cedar kilnBOT
cedar kilnBOT
Available help channel!
civic sentinel
Im genuinely so sorry but i got this homework assigned and I couldn't find any tutorials on it, my classmates asked chatgpt (i dont support ai, i avoid it since it feels pointless when IM the one who needs to learn, not the ai) and photomath and got different results, can anyone walk me through the process of ANY of these? I asked my teacher to explain this topic before and he didnt even try to explain it he just counted it but i do NOT get anything, can anyone tell me the process or steps?? I tried looking for tutorials but none explained these specific like topics, only a piece of them which didnt help
cedar kilnBOT
Im genuinely so sorry but i got this homework assigned and I couldn't find any t...
@civic sentinel Has your question been resolved?
warm oriole
so much radicals 😔
warm oriole
Im genuinely so sorry but i got this homework assigned and I couldn't find any t...
some ideas to search up explanations would be “simplifying radical expressions” but these seem more difficult since there’s a lot of fractions and radicals and whatnot
it is pretty ugly to look at tho i’ll say 
if it is any advice, you should always work from inner-out
like for example 1) if you wanna begin simplifying you wanna start at the numerator and combine like terms, then subtract the powers of the numerator and denominator to get it all under “one” variable
civic sentinel
it is pretty ugly to look at tho i’ll say <:bleak:877093280905592842>
I know ☹️ no one in our class knows i tried asking around if anyone can tutor me
civic sentinel
some ideas to search up explanations would be “simplifying radical expressions” ...
thank you!! ill try searching it up
warm oriole
I know ☹️ no one in our class knows i tried asking around if anyone can tutor me
i feel ur pain and i remember searching this up aswell when i was taking it but theres only very easy examples
civic sentinel
aaaarghh
warm oriole
goodluck bro 😔😔
civic sentinel
thank youu
haughty yoke
Im genuinely so sorry but i got this homework assigned and I couldn't find any t...
I feel like i got it but idk the order of the equations
if i can simplify the c first then
civic sentinel
id appreciate even the tiniest attempt of a help trust me 🥹
warm oriole
u reached the right conclusion but 1/6 is meant to be positive
haughty yoke
after that we can change the root to power
oh yea
mb
the mistake i always repeat smh
so, if power meets multiplication and the thing that has the formula is the same, we can just add the power
civic sentinel
ohh
haughty yoke
wait
do you get it?
can you see it?
yea there
then we get this
after that we can change the root to an exponent
then we get this
since it has "()" means that we'll multiply the -3 with the 1/5
then finally
haughty yoke
then finally
and negative exponent can be changed to posotive by this way
@civic sentinel do you get it?
civic sentinel
let me see
haughty yoke
just reply the pic that you're confused with
civic sentinel
why does divide turn into minus
haughty yoke
why does divide turn into minus
so, if multiplication with exponent becomes addition then division with exponent is subtraction, but remember that this works because it has the same thing under the exponent (idk the name of it)
in this case its C
civic sentinel
OHH
haughty yoke
thats why
civic sentinel
then we get this
can i ask why it turns to a like this thing now, is it because it was a fraction
but you explain it so well thank you
civic sentinel
OHH right thank you
civic sentinel
then we get this
thank you im a bit confused 🥹 like i get fraction turns into division but then whats the division here
haughty yoke
then we get this
then we get this
you can try the other questions by yourself so you can understand more, if you have trouble you can ask me, i might go off cuz i need to get some rest, sorry
civic sentinel
YEYE thank you so much! im trying to count some myself, you really explained it SO well, thank you so much and have a nice rest
haughty yoke
do you get it now?
civic sentinel
yess i do get that now but now i dont get one more thing like
the red thing like how did it appear
haughty yoke
Im genuinely so sorry but i got this homework assigned and I couldn't find any t...
from the question, we did it step by step
i mean, we can just write the whole thing but it could make it messy
civic sentinel
OHH okay thank you
haughty yoke
you see the brackets right?
civic sentinel
yes yes i forgot they are there but i see now
like i forgot we were counting inside of them at first, thank you so much enjoy your rest
appreciate it
haughty yoke
ight then, ima go, just remember if you're having trouble ask someone, dont be shy to ask ight.
cya
civic sentinel
cya!!
fossil dawn
hey folks, first time being on the other side of the helping!
the question is in the first image, my answer is in the second.
is my answer justifiable? if it is, is it too verbose? short? if it is not, what else should I have mentioned?
I was particularly thinking about whether I should have brought up the prime factorization of 2^p and 3^q, considering that I let p, q \in Z already from the start.
(please ping on initial reply!)
void sand
seems fine to me 
gritty galleon
yeah
flint cape
yep
gritty galleon
much better than anything i'd write lol
fossil dawn
icic, I was quite afraid I messed up the prime factorization part by including it in the first place
void sand
also, your handwriting is really nice 
fossil dawn
halfway through writing those two sections I was like
"should I have written them....?"
also, is the conclusion supposed to be "no such rational r exists", or "r cannot be rational"?
flint cape
Either is fine
fossil dawn
I feel like these two sentences mean different things
flint cape
Our assumption is:
So if we want to be very explicit that we end up with a contradiction, negate this statement
"There is no rational r such that..." or "No such rational r exists"
If this weren't the goal, say if you wanted to find a good approximation somehow, and you're whittling down options for r e.g. by finding out that it has to be irrational, then you could preferably write instead "So r is not rational"
fossil dawn
also, about the prime factorization part, could I just have left that out and went directly to saying that 2^p (and 3^q) are integers, since we let p, q be integers?
fossil dawn
"There is no rational r such that..." or "No such rational r exists"
got it, thanks!
fossil dawn
also, about the prime factorization part, could I just have left that out and we...
add: not just integers, but even/odd
flint cape
also, about the prime factorization part, could I just have left that out and we...
This is actually a helpfully stronger argument tbh
fossil dawn
to confirm, you mean that including the bit about prime factorizations strengthens the argument?
flint cape
yeah
Because you can amend this argument e.g. to show that, "For any two prime numbers p, q, there is no rational number r such that p^r = q"
fossil dawn
right, so it can turn into a general argument about any two primes
flint cape
That is to say, your proof can be generalised neatly, yh
fossil dawn
gotcha, thank you guys a ton!
I may be back in the future with more dumb questions, so a quick yoroshiku
.close
cedar kilnBOT
cobalt saffron
I need help to understand probability
fossil dawn
I need help to understand probability
sorry fam, this channel is still occupied
and I see you have a channel alrd
cobalt saffron
sorry fam, this channel is still occupied
Ayein wbat?
I can’t understand
I just asked a question
urban coral
this channel is not yet open
cobalt saffron
So where i can ask?
urban coral
though since you already have a channel, this doesn't matter
cobalt saffron
B i and ii
cobalt saffron
though since you already have a channel, this doesn't matter
What are channels?
I dont know how to use discord
urban coral
these things to the left (you can also read #❓how-to-get-help)
cobalt saffron
What to do now?
cedar kilnBOT
Available help channel!
quaint valve
can someone explain this i dont quite get why the loci looks like this
pastel vault
can someone explain this i dont quite get why the loci looks like this
I think it's pretty clear from the explanation
- z - 2j represents the vector connecting z and 2j, and the same goes for z + 3 = z - (-3) connecting z and -3
- arg(a/b) = arg a - arg b
so then you just think about a circle centred at the origin, for when the difference between the arguments = angles is constant
I guess if you translate the entire diagram by (-z) units
-3 goes to -3 - z, 2i goes to 2i - z, and z goes to the origin
quaint valve
why is there nothing in the second quadrat
pastel vault
why is there nothing in the second quadrat
play around with this interactive yourself
quaint valve
alright cheers
pastel vault
cheers
quaint valve
ive done this next question without quite knowing whats going on by monkey see monkey do
pastel vault
at some point you'll realise the angle between two lines doesn't change depending on where they are
you can have those lines intersect at the origin and it should be clear
just imagine moving the lines around so they intersect at the origin
I can't put it any better way to you
quaint valve
this isnt right apparently
well in their answer z=1 , z= 3+2i lies on the circumfereance of the circle
cedar kilnBOT
@quaint valve Has your question been resolved?
blissful glade
cedar kilnBOT
blissful glade
i have no clue what to do
surreal cave
blissful glade
the regular formula
the M + kI = 0
where k is the eiegn value
then multiplying it with an arbitrary 3x1 but i just get a = ic
surreal cave
the M + kI = 0
isn't the formula (A-kI)v=0 
surreal cave
what is b exactly?
blissful glade
(a b c r the values within the eigenvector)
blissful glade
you end up with -x + iz= 0 or x +iz = 0
its still weird
because the seond equation contradicts
with the correct way
you end up with x+iz = 0 and -ix + z = 0
which dont work
surreal cave
why wouldn't that work?
that's the same equation just rewritten 
in fact if x=-iz then note that:
-i(-iz)+z=(-i)(-i)z+z=-z+z=0
blissful glade
o yeah
would the other 2 eigenvalues be i and -i
i think thats right
and when you normalise this one the normalisation constant is 1/ root 2
blissful glade
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
crimson sedge
i am completely lost
crimson sedge
i am completely lost
<@&286206848099549185> i'd appreciate some help 🙏
winged umbra
؟
crimson sedge
discrete math is making me pull my hair out
cedar kilnBOT
@crimson sedge Has your question been resolved?
cedar kilnBOT
@crimson sedge Has your question been resolved?
zealous root
i am completely lost
a,b,c or d?
Basically what they are saying is that $(a,b)\in R$ if and only if there is an arrow from $a$ to $b$ and then you have to check if its a poset (reflexive, transitive and anti-symmetric)
wraith daggerBOT
VincentBH
cedar kilnBOT
@crimson sedge Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
dense jewel
need help with these
cedar kilnBOT
cedar kilnBOT
Available help channel!
slender holly
how do i graph and write piece wise functions and equations
cedar kilnBOT
how do i graph and write piece wise functions and equations
slender holly
coral jewel
consider each equation in the piecewise function, identify their domains and draw the corresponding equation in each domain
slender holly
to be honest i don’t know how to read those equations
don’t give up on me yet..
i’m going to end it
pastel vault
i’m going to end it
please don't make self-harm or suicide jokes
teal rivet
@slender holly get a piece of scratch graph paper, draw a straight line from the left to right side at y=4 until x=-2. then another straight y=-1 from x=-2 to wherever your graph ends
that should be helpful
pastel vault
for q2, it's like if you draw the graph of $(x + 4)^2$ starting from the very left of the graph
but then since the condition is "x less or equal to -3", you stop drawing once you hit $x = -3$
wraith daggerBOT
south
pastel vault
but how do i graph (x+4)^2
start from x^2
now (x + 4)^2 tells you that you need to move the graph 4 units left
humble rivet
Pls , help me to solve this math of A inverse 1
pastel vault
so the vertex is now (-4, 0)
pastel vault
Pls , help me to solve this math of A inverse 1
!help
cedar kilnBOT
Pls , help me to solve this math of A inverse 1
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
slender holly
Pls , help me to solve this math of A inverse 1
brooo get ur own ticket
ts just confused me
pls delete ts
slender holly
so it basically 0
pastel vault
so it basically 0
wdym?
slender holly
it’s basically just x as 0
pastel vault
it’s basically just x as 0
you need to be much more clear about what you say
slender holly
or does the parabola matter
pastel vault
now (x + 4)^2 tells you that you need to move the graph 4 units *left*
now read this again
pastel vault
yes, that's where the bottom point ends up