which entails just taking the highest possible value in the range instead of lowest
#help-33
wooden shard
why wouldnt it be minimum
limber wigeon
because the mean of A is larger than the mean of B
and you are trying to find the smallest difference between them
wooden shard
ok i see
limber wigeon
because the mean of A is larger than the mean of B
you can see this just by looking at the histograms
they have the same frequencies, but B has them over the intervals with lower values
so everything is smaller in B
limber wigeon
then the difference would be?
marsh citrusBOT
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gentle patrol
What the heck does (cf)(x) = cf(x) mean and how does that suffice to prove closure under scalar multiplication?
marsh citrusBOT
What the heck does (cf)(x) = cf(x) mean and how does that suffice to prove closu...
proud prism
well, (cf) and f are both functions
gentle patrol
well, (cf) and f are both functions
How?
proud prism
cf: D to R,
(cf) (x) := c* f(x)
in the right side we have pointwise real multiplication
and in the left it’s scalar multiplication to f not necessarily usual real number muliplication
f itself should be viewed as a single object
gentle patrol
and in the left it’s scalar multiplication to f not necessarily usual real numbe...
so c doesn't have to be in R for it to out to R when multiplied by f?
proud prism
yes, the right side to should make sense
whole thorn
so c doesn't have to be in R for it to out to R when multiplied by f?
c is in R, for our current purpose
but i think dotdoc meant f is not in R, so cf is not usual real number multiplication
proud prism
if you just wrote
(cf)(x) = c* f(x), it completly depends on how you interpret.
A standard setting is c is in R, and f: D to R.
the key distinction is right side the multiplication operation is pointwise on the values of f
and in left it’s on f itself.
proud prism
What the heck does (cf)(x) = cf(x) mean and how does that suffice to prove closu...
also what’s the original question?
gentle patrol
also what’s the original question?
It is in the context of vector spaces
{ f : I -> F }
function spaces
proud prism
What’s structure you have in T?
gentle patrol
What’s structure you have in T?
F is an arbitrary field
proud prism
right, so c should come crom F
proud prism
{ f : I -> F }
This is exactly what I meant by should look at function as a single entity
cf should also be another object in the collection.
gentle patrol
I don't get what (cf)(x) = c * f(x) really means
Maybe I would like to start from knowing what (cf)(x) is
proud prism
quick question,
How do you know check if two function f, g are the same?
gentle patrol
quick question,
How do you know check if two function f, g are the same?
f = g
?
No
delicate prairie
given a function f and scalar c, this is the way to define another function cf, and (cf)(x) denotes cf evaluted at x
gentle patrol
quick question,
How do you know check if two function f, g are the same?
Iff for any x in I, f(x) = g(x)
Yes
gentle patrol
given a function f and scalar c, this is the way to define another function cf, ...
mm
proud prism
if F is a field you can always do this
it’s matter of whether you can afford to define it
marsh citrusBOT
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coarse nova
Calc 2
- Find the length of the loop of the curve x=3t-t^3 y=3t^2
How I attempted to solve: I assumed the loop of the curve meant the length between 2 t values that both had the exact same (x,y) output. For the y output to be the same, t0=-t1. So I set x(t)=x(-t) and using a calculator found an intersection at sqrt(3) and 0. I assumed that the intersection at 0 didn't matter since it seemed more like obvious. So I did the integral from -sqrt(3) to sqrt(3) of sqrt((dx/dt)^2 - (dy/dt)^2)dt and got about 20.785
My main question is can the parametric go negative or should I have done the integral from just 0 to sqrt(3) instead. Or is there something else I'm missing.
marsh citrusBOT
Calc 2
12. Find the length of the loop of the curve x=3t-t^3 y=3t^2
How I atte...
@coarse nova Has your question been resolved?
wind hare
Calc 2
12. Find the length of the loop of the curve x=3t-t^3 y=3t^2
How I atte...
It doesn't matter if t is negative
marsh citrusBOT
@coarse nova Has your question been resolved?
tight furnace
There's no intersection of the curve with itself at t=0
You looked for pairs t, -t with the same x-value, since you identified that was required for the y-values to match
all that getting t=0 told you was that the position is the same at t=0 as it is at t=0
coarse nova
all that getting t=0 told you was that the position is the same at t=0 as it is ...
Ah
So t=0 is irrelevant
Does that mean the limits of integration are just -sqrt(3) and sqrt(3) then?
tight furnace
Yeah
spark halo
marsh citrusBOT
spark halo
by balancing the force along the tangent to the particle, i obtained the result $q=\frac{mg}{vB}$ . I didnt quite get what "force of interaction" essentially meant, i assumed it would be any normal reaction force. Intuitively along the radial direction, forces cannot be balanced because the particle has some centripetal acceleration so i chose the tangential direction. Now how do i find "r"?
elfin berryBOT
atharvxd
marsh citrusBOT
@spark halo Has your question been resolved?
spark halo
<@&286206848099549185>
marsh citrusBOT
@spark halo Has your question been resolved?
spark halo
yeah sure
topaz escarp
“no force of interaction”
does not mean “no centripetal force”
it means the ring should not need to push or pull on the particle at all
means all forces needed for the particle’s motion must come from external fields only
gravity + magnetic force
you’re right about the tangential direction
now how to get r
don’t balance forces radially
insteaduhh
spark halo
yeah that doesnt work
topaz escarp
the particle is moving in a circle of radius r right
that means it needs centripetal force
spark halo
even if i write the law of motion of the particle in the radial direction to equate to the centripetal force however the parameter for angular position cannot be eliminated
topaz escarp
that means it needs centripetal force
mb^2/r
spark halo
thats correct
topaz escarp
even if i write the law of motion of the particle in the radial direction to equ...
you’re overthinking
topaz escarp
mb^2/r
and since the ring cannot push on it
the ONLY force allowed to provide this is
the magnetic force
qvB
so just set them equal
can you do that rn
spark halo
r=mv/qb
topaz escarp
no bro
spark halo
wait a min
spark halo
but
topaz escarp
spark halo
that is the correct answer
however the problem is
doesnt the particle also have a velocity as a consequence of pure rolling?
topaz escarp
yes but not an extra velocity
pure rolling determines how the direction of velocity changes not how big the velocity is
spark halo
wont you write $\vec{v_p}= \vec{v}+ \vec{\omega} \cross \vec{r}$
elfin berryBOT
atharvxd
topaz escarp
yes
topaz escarp
it doesn’t change the final force balance
bc the magnetic force uses the total particle velocity
topaz escarp
**atharvxd**
writing that is correct like
but treating those as separate physical speeds is no
the mag force sees only the resultant speed
that speed is constant
so that’s valid
but yeah write in full kinematics
spark halo
the magnitude of the resultant velocity changes with angular position irght
and im sorry the answer is not r=mv/qb
topaz escarp
yes
spark halo
its v^2/g
topaz escarp
yes
spark halo
so uh
topaz escarp
tell me
spark halo
yeah i got r=mv/qb
the angular position parameter got eliminated
i just realised i got r=mv/qb
topaz escarp
nice you closed the loop now
spark halo
ok yeah i got the answer to match
but how did you predict by intuition that the angular component wont be involved in the lorentz force equation
i didnt understand that
topaz escarp
so
if the only forces acting come from uniform fields
and the system has no angular asymmetry
then any angular parameter you introduce must cancel out
spark halo
what does uniform field mean precisely?
topaz escarp
its magnitude and direction are the same at every point in the region of motion
a field is uniform if that checks
spark halo
does this hold true for non conservative fields?
topaz escarp
yess
uniformity hasnothing to do with being conservative
what matters is spacial uniformity
not whether a potential exists
spark halo
and uh
i still didnt get how you equated qvb= mv^2/r
now its true the net force acting on the particle is qvb
topaz escarp
newton’s 2nd law
spark halo
because of symmetry
topaz escarp
in the normal direction
topaz escarp
lorentz force magnitude: qvB
it (force) is perpendicular to velocity
perpendicular force = normal acceleration
normal acceleration for curved motion v^2/r
topaz escarp
i still didnt get how you equated qvb= mv^2/r
so this
topaz escarp
spark halo
ill attempt it once before tho
topaz escarp
okie dokie
topaz escarp
shortest way i can think of
is to see what velocities matter for lorentz first
then kill the w * r part
using symmetry
then take the motion
and treat it like a charge particle with charge Q
and ig then bring in rolling constraint if needed
topaz escarp
nothing from rotation enters the radius
spark halo
wdym
topaz escarp
yeah you’re right
spark halo
????
topaz escarp
like you’re right
topaz escarp
the radius of the center of mass is to be found
correct
and nothing from the rolling or angular motion changes the cm radius
spark halo
the sphere would keep pure rolling while its center of mass performs uniform circular motion
the radius wouldnt simply be mv/qb
topaz escarp
yes
topaz escarp
the radius wouldnt simply be mv/qb
it would be
the center of mass is what moves in a circle
the only net external force causing that curvature is the magnetic force
so all rotational/rolling details affect torque, not the net force on the cm
spark halo
the center of mass needs to move in a circle while the sphere rolls purely
spark halo
the only net external force causing that curvature is the magnetic force
friction?
topaz escarp
write newts law only for the center of mass
static friction adjusts itself to satisfy the rolling constraint
it is not fixed in direction or magnitude
when the cm that starts curving:
the required torque to keep rolling changes
friction supplies that torque
but radial acceleration of the CM is already fixed by the magnetic force
there is no kinematic requirement that friction contribute to radial acceleration
if it did, then the CM speed would change, or rolling would break
spark halo
the torque equation can be written as $\frac{2}{5}MR^2 \omega \Omega = f.R + \frac{1}{5} QR^2 \omega B$ where $\Omega$ is the angular velocity of the circular motion of the center of mass and \omega is the angular velocity of pure rolling.
elfin berryBOT
atharvxd
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
topaz escarp
the torque equation only fixes the friction f needed to keep pure rolling. it does not alter the force balance on the CM, which is still
‘’’QvB=\frac{mv^2}{r_{\text{CM}}};\Rightarrow; r_{\text{CM}}=\frac{mv}{QB}’’’
can you copy paste this into here
latex is mad at me
we broke up last year apparently
spark halo
$QvB=\frac{mv^2}{r{\text{CM}}};\Rightarrow; r{\text{CM}}=\frac{mv}{QB}$
elfin berryBOT
atharvxd
spark halo
the torque equation will fix friction but why will the friction not alter the motion of cm?
topaz escarp
bc static friction only enforces the rolling constraint and its line of action passes through the instantaneous center of rotation
it contributes literally 0 net work and no required radial acceleration of the cm
the only force that can curve the cm path is the magnetic force
so friction cannot change the cm motion
spark halo
if friction passes through the instantaneous center of rotation then it should indeed be involved in the body's law of motion right?
i get that it does no work because there is no displacement of the point of contact
topaz escarp
because the static friction force is purely tangential to the cm trajectory, while only a radial force can change the cm’s circular path, friction cannot alter the cm motion or its radius
spark halo
oh yeah ive assumed friction to be radially outward
however
if lorentz force acts radially inward there must be a force acting radially outward to keepthe center of instantaneous rotation static?
topaz escarp
no no
nothing needs to act radially outward
the instantaneous center of rotation is a kinematic concept
not a force balance point
and the inward lorentz force by itself supplies the required centripetal acceleration of the cm
marsh citrusBOT
@spark halo Has your question been resolved?
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shadow field
I have a matrix in Unity called: localToWorldMatrix which describes the transformation from a local object space to a world space.
This matrix is represented by a basis (with each column describing the scaled right vector (x-axis), up vector (y-axis), and forward vector (z-axis)) and a position alongside a dummy 4th row to allow for correct multiplication:
x y z
[1, 0, 0, pos.x]
[0, 1, 0, pos.y]
[0, 0, 1, pos.z]
[0, 0, 0, 1 ] (dummy 4th row)```
I am attempting to convert this to a *Transform3D* matrix in Godot, which is identical to the above except for the forward direction of the z axis being interpreted by the engine as opposite. In other words moving along the z-axis in the positive direction in Unity moves in the negative direction in Godot.
My first thought was to simply invert the x and y components of the z axis and the z position as follows:
INVERT_Z =
[1, 0, 0, 0]
[0, 1, 0, 0]
[0, 0, -1, 0]
[0, 0, 0, 1]```
INVERTED_LOCAL_TO_WORLD_MATRIX = INVERT_Z * localToWorldMatrix * INVERT_Z
However this resulted in the weird phenomena where some rotations were off, the Euclidean representation of z-axis rotations seemed to show negative theta (degrees) instead of positive theta (degrees) and vice versa, but it seemed to be only in some cases, and it seemed to be in cases where the rotation was quite minimal, with other seemingly random cases being flipped along the y axis leading to the z-axis rotation being correct.
Would anyone have any idea why this might be happening and how I can correctly reverse the direction of the z-axis in this matrix?
marsh citrusBOT
I have a matrix in Unity called: *localToWorldMatrix* which describes the transf...
frigid topaz
the sandwich multiplication is correct in principle
the wrong rotations are probably an artifact of comparing euler angles across two engines that use different euler decomposition orders
unity uses ZXY and Godot uses YXZ
verify the basis vectors directly instead of comparing extracted angles
frigid topaz
I have a matrix in Unity called: *localToWorldMatrix* which describes the transf...
.
marsh citrusBOT
@shadow field Has your question been resolved?
shadow field
yeah one sec typing
okay so the transform in unity of some object A is:
[2.2311, 0.175, 0, 155.8315]
[-0.302, 1.1157, 0, 57.1153 ]
[0, 0, 0.7471, 2.1879 ]
[0, 0, 0, 1 ]```
in godot it is:
[2.231005, 0.17558391, 0, 155.8315]
[-0.09068846, 1.1523061, 0, 57.11531]
[0, 0, 0.7470923, -2.18791]
[0, 0, 0, 1 ]```
which is incorrect, the desired transform is:
[2.231005, -0.17558391, 0, 155.8315]
[0.09068846, 1.1523061, 0, 57.11531]
[0, 0, 0.7470923, -2.18791]
[0, 0, 0, 1 ]```or incorrect is probably the wrong word, but you get what i mean
<@&268886789983436800>
shadow field
I also noticed a bunch of transforms had their scales flip inside out.
These seemed to be the ones which had been rotated correctly.
marsh citrusBOT
@shadow field Has your question been resolved?
marsh citrusBOT
@shadow field Has your question been resolved?
bold ice
okay so the transform in unity of some object A is:
```
[2.2311, 0.175, 0, ...
I have no Idea about game development engines, and I also don't understand the description of the math involved. I tried reading documentation to understand your problem, but it didn't help much
bold ice
okay so the transform in unity of some object A is:
```
[2.2311, 0.175, 0, ...
Also if you only did z -> -z that -0.302 -> -0.9 seems weird
shadow field
Also if you only did z -> -z that -0.302 -> -0.9 seems weird
oh so i'm not just negating the z, imagine a plane across the z axis, i'm reflecting it over that plane
still temple
hello there
shadow field
hi
marsh citrusBOT
@shadow field Has your question been resolved?
marsh citrusBOT
@shadow field Has your question been resolved?
marsh citrusBOT
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Available help channel!
novel juniper
marsh citrusBOT
quaint elm
there are only eight possible outcomes for 3 coins this takes 30 seconds the bruteforce way
marsh citrusBOT
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gentle patrol
Definition0.3.21. A relation R on a set 𝐴 is said to be
(i) Reflexiveif 𝑎 R 𝑎 for all 𝑎∈𝐴.
(ii) Symmetric if 𝑎 R 𝑏 implies 𝑏 R 𝑎.
(iii) Transitive if 𝑎 R 𝑏 and 𝑏 R 𝑐 implies 𝑎 R 𝑐.
Definition 0.3.23. Let 𝐴 be a set and R an equivalence relation. An equivalence class of
𝑎∈𝐴, oftendenotedby[𝑎],is the set {𝑥∈𝐴: 𝑎R 𝑥}.
For example, given the relation ‘★’ above, there are two equivalence classes, [1]=[2]=
{1,2} and [3]={3}.
Question:
What does a R a mean? a in A? How does that relate to "An equivalence class of
𝑎∈𝐴, often denoted by[𝑎],is the set {𝑥∈𝐴: 𝑎R 𝑥}."?
What really is an equivalence class?
stoic saddle
R is a binary relation
you can think of it informally as a "function" which takes two arguments from set A and returns a boolean
when we say "a R a" you can think of it as meaning "R(a,a) = TRUE"
gentle patrol
I see R as a subset of A x A
proud ice
Definition0.3.21. A relation R on a set 𝐴 is said to be
(i) Reflexiveif 𝑎 R 𝑎...
$R$ represents a relation between two values in a set. A good example is the equality relation $=$ in the set of real numbers $\bR$. For any real number $a\in\bR$, the relation $=$ satisfies $a=a$ and is therefore reflexive. Also, if $a=b$, then $b=a$, so $=$ is symmetric. Finally, if $a=b$ and $b=c$, then $a=c$, so $=$ is also transitive
elfin berryBOT
SWR
proud ice
I see R as a subset of A x A
This is the definition of a relation
gentle patrol
then `a R a` means (a,a) ∈ R
Yes
stoic saddle
What really is an equivalence class?
an equivalence class is a subset of A in which all elements within are related to each other by R but nothing outside it is related by R to anything inside
gentle patrol
> What really is an equivalence class?
an equivalence class is a subset of A in ...
But, does that mean that, if there exists a relation a R b, then c can't have a relation with a or b (since c is outside the relation a R b)?
stoic saddle
i have... no idea what you're talking about.
a R b does not by itself imply that (∀c ∈ A \ {a,b})[not(a R c) and not (b R c)] .
gentle patrol
a R b **does not** by itself imply that (∀c ∈ A \ {a,b})[not(a R c) and not (b R...
Okay
I am sorry
I used tuples instead of sets
they are subsets of A 
Thanks, I get it now
.close
marsh citrusBOT
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echo scaffold
Could somebody check these for me since my teacher did not post an answer key since it’s optional
marsh citrusBOT
Could somebody check these for me since my teacher did not post an answer key si...
@echo scaffold Has your question been resolved?
shy harness
marsh citrusBOT
shy harness
how do i solve this??
noble geyser
!status
marsh citrusBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
knotty trellis
<@&268886789983436800>
shy harness
<@&286206848099549185>
whole sleet
Both circles travel the same amount of "circumference distance", if that makes any sense
ionic ferry
yeah I guess it's just the ratio of circumferences, but the distance between circles has a factor too
eh or does it though, something tells me the distance between them doesn't matter, not sure though
marsh citrusBOT
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shy harness
.close
marsh citrusBOT
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devout mauve
<@&268886789983436800>
elfin badger
marsh citrusBOT
elfin badger
wHy wouldn't it be the bit in red?
main idol
why do you think you can't do it your way in red?
elfin badger
becus its done in a different way but I guess my way would work
main idol
if you get the same answer then it's unlikely that it's wrong
untold spruce
How is $G(x) -\alpha A(x)$ concave?
marsh citrusBOT
How is $G(x) -\alpha A(x)$ concave?
elfin berryBOT
LXDL
untold spruce
Are they using first order condition/test?
main idol
there's probably multiple ways to prove G(x) - alpha * A(x) is concave. you can try using the definition on f(x) = G(x) - alpha * A(x) directly since G(x) is concave and A(x) is linear
marsh citrusBOT
@untold spruce Has your question been resolved?
marsh citrusBOT
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white rivet
If you have thirteen items that each have a 50/50 of being true or false, and their outcomes are irrespective of each other is 13 trues equally likely as 6 trues and 7 falses?
marsh citrusBOT
If you have thirteen items that each have a 50/50 of being true or false, and th...
white rivet
Personal question, not for school
devout mauve
13 trues is equally likely as eg "the first 6 are true, the rest is false"
however, if you just say "6 are true, 7 are false", then those can be distributed in lots of different ways
which gives you way more options
obtuse umbra
We use uhm
white rivet
however, if you just say "6 are true, 7 are false", then those can be distribute...
i meant like this
devout mauve
then you have your answer
obtuse umbra
however, if you just say "6 are true, 7 are false", then those can be distribute...
$\binom{13}{6}$ to find how many cases of this can happen
elfin berryBOT
1 divided by 0 equals Infinity
devout mauve
(this is called a binomial distribution if you want to look it up)
obtuse umbra
And there are only 1 case where there are 13 trues
obtuse umbra
(this is called a binomial distribution if you want to look it up)
Oh thanks for the name
obtuse umbra
how do I end?
!done
marsh citrusBOT
how do I end?
If you are done with this channel, please mark your problem as solved by typing .close
devout mauve
.close
fickle nova
Suppose we have a sum S_n = 1 - 2 + 3 - 4 - 5 + 6 - 7 - 8 - 9
where between every positive term there's k negative ones, and k increases by 1 every time
Calculate S_435
marsh citrusBOT
Suppose we have a sum S_n = 1 - 2 + 3 - 4 - 5 + 6 - 7 - 8 - 9
where between ever...
fickle nova
Just had an exam with this and i thought i was a genius and wanted to know if i did it right
polar dew
Start with +1
Then 1 negative
Then +3
Then 2 negatives
Then +6
Then 3 negatives
Then +10
Then 4 negatives
etc.
So:
• The positive terms are
1, 3, 6, 10, 15, 21, ...
These are triangular numbers.
The m-th positive term is:
𝑇𝑚=𝑚(𝑚+1)2Tm
=
2
m(m+1)
Now look at how many terms are used up to the m-th positive block.
Each stage contributes:
1 positive
m negatives
So total terms after m blocks
1+(1+1)+(1+2)+(1+3)+⋯+(1+𝑚)
1+(1+1)+(1+2)+(1+3)+⋯+(1+m)
idk how to make the T and V signs in here
fickle nova
i had a different approach
polar dew
What did u use?
fickle nova
I noticed that the positive terms all follow the same pattern where it's (n n+2) (binomial distribution)
So i converted the entire sum into:
$2 \cdot \sum_{i=0}^m \binom {n + 2}{n} - \sum_{i=1}^n i$
elfin berryBOT
Saarker
fickle nova
where n is 435
and m is the number of these terms
which you find by verifying whether 435 itself is one of them
by solving $435 = \frac{m \cdot (m + 1)}{2}$
elfin berryBOT
Saarker
fickle nova
then you verify if the m is an integer and if it is it means that the last term of the sum is positive
$2 \cdot \sum{i=0}^m \binom {n + 2}{n} = \sum_{i=1}^m (i^2 + i)$
elfin berryBOT
Saarker
polar dew
Yeah i see what you did
polar dew
im not sure beacuse i didnt do it like that
jagged relic
I noticed that the positive terms all follow the same pattern where it's (n n+2)...
These are the triangular numbers
fickle nova
$2 \cdot \sum_{i=0}^m \binom {n + 2}{n} = \sum_{i=1}^m (i^2 + i)$
fickle nova
These are the triangular numbers
but they follow this pattern dont they?
elfin berryBOT
Saarker
jagged relic
,, \binom{n+2}{n} = \binom{n+2}{2} = \frac{(n+1)(n+2)}{2} = T_{n+1}
elfin berryBOT
Nel
fickle nova
,, \binom{n+2}{n} = \binom{n+2}{2} = \frac{(n+1)(n+2)}{2} = T_{n+1}
im worried i might have to just bruteforce this
i cant trust what i wrote down
but i think the result was right
i got -85600 or something along that
worst case scenario i just forgot to change up some indexes or whatever so its a bit off
jagged relic
S_n = -T_n + 2U where U is the sum of the positive terms
fickle nova
yup thats what i did
jagged relic
The terms in U are in T_n, and 435 is a triangular number
fickle nova
but U has a different index
fickle nova
The terms in U are in T_n, and 435 is a triangular number
so i found the index that gives 435
which is like 29
jagged relic
,calc 29*30/2
elfin berryBOT
Result:
435
Right
fickle nova
could this mean i got the hardest question on the exam right
jagged relic
So you can just take the sum of T_n from 1 to 29
fickle nova
So you can just take the sum of T_n from 1 to 29
i switched up the index of U to get i ( i + 1 ) to just use m instead of m - 1
jagged relic
Not sure what you mean by that
,, 2U = 2 \sum_{n=1}^{29} \frac{n(n+1)}{2} = \sum_{n=1}^{29} n(n+1)
elfin berryBOT
Nel
fickle nova
originally i didnt have n = 1
i had n = 0 because i was referring to the binomial coefficient
so i made n = 1 so it would be n(n+1)
jagged relic
Ok sure
fickle nova
i expanded it and it was like $\sum_{i=1}^m (i^2 + i) = \frac{m(m+1)(2m+1)}{6} + \frac{m(m+1)}{2}$
elfin berryBOT
Saarker
jagged relic
jagged relic
Right so that's m(m+1)(m+2)/3
fickle nova
awesomesauce
jagged relic
,calc 435436/2 - 2930*31/3
elfin berryBOT
Result:
85840
If you don't have any more question, sure
marsh citrusBOT
marsh citrusBOT
Available help channel!
marsh citrusBOT
Available help channel!
vague jay
marsh citrusBOT
tight furnace
Have you done <= yet?
vague jay
Suppose $\alpha = p(\lambda)$ for some eigenvalue $\lambda$ of $T$. We know that there exists a nonzero $v \in V: Tv = \lambda v$. Thus, $p(T)v = p(\lambda)v = \alpha v$, showing that $\alpha$ is an eigenvalue of $p(T)$.
elfin berryBOT
holo_morph
vague jay
This idea was already mentioned in LADR but the $p(T)v = p(\lambda)v$ equivalence follows from the fact that $T^kv = \lambda^k v$
elfin berryBOT
holo_morph
tight furnace
Nice
vague jay
yeah but I need the other direction 😭
tight furnace
Well you might suspect this already but it has to do with the field being specifically C
You might have a guess as to which property of C is needed here
vague jay
Conversely, suppose $\alpha$ is an eigenvalue of $p(T)$. Then for some nonzero $v \in V$,
\begin{align*}
&p(T)v = \alpha v\
&\Rightarrow (p(T) - \alpha)v = 0
\end{align*}
By the fundamental theorem of algebra, we have that
\begin{align*}
&p(z)-\alpha = c(z-\lambda_1)\cdots(z-\lambda_m)\
&\Rightarrow p(T)-\alpha = c(T-\lambda_1)\cdots(T-\lambda_m)\
&\Rightarrow (p(T)-\alpha)v = c(T-\lambda_1)\cdots(T-\lambda_m)v = 0.
\end{align*}
The above equation implies that some $\lambda_j$ is an eigenvalue of $T$, and since $\lambda_j$ is a root of $p(z) - \alpha$, we have that $p(\lambda_j) - \alpha = 0 \implies \alpha = p(\lambda_j)$, as desired.
elfin berryBOT
holo_morph
vague jay
did u do all of ladr?
tight furnace
nah
I feel like Ive seen this problem before
But probably like posted in a discord server before lol
Actually my first instinct was to nuke it with Jordan normal form lol
but that's not the LADR way 
Well also ig we don't have finite dimensionality lol
steady fjord
\textbf{Axiom - Pairing Axiom:} $\forall x\forall y \exists z \forall w ( w \in z \iff (w = x$ or $w = y))$
This means "for any sets x and y, there exists a set z such that the elements of z is either x or y
marsh citrusBOT
\textbf{Axiom - Pairing Axiom:} $\forall x\forall y \exists z \forall w ( w \in ...
elfin berryBOT
toast
steady fjord
or im trying to understand this axiom
so its just saying theres some set z that contains either x or y or both?
fervent rampart
it means "w is an element of z if and only if it is equal to either x or y" (i.e. the elements of z are x and y)
so it's saying that given x and y exist, z = {x,y} is a set that also exists
steady fjord
so i am curious
why cant the axiom just be given x exists, z = {x} is a set that also exists?
i think you can prove it from the pairing axiom
but why cant we just set z = {x} as an axiom itself? is it just hard to proof stuff with just this?
fervent rampart
this allows for constructions other than just that, for example constructing the ordered pair/tuple (x,y) = {{x},{x,y}}
steady fjord
this allows for constructions other than just that, for example constructing the...
hmm im actually a little confused by this notation
why do we say (x,y) = {{x}{x,y}} and not {{x},{y}}?
fervent rampart
tuples (x,y) is ordered, so you need to be able to tell which element is first and which element is second, but sets are not ordered, so {{x},{y}} = {{y},{x}}, which would mean (x,y) = (y,x) using that definition, whereas {{x},{x,y}} = {{x,y},{x}} retains the information about which element is first (the one in the singleton) in both orders
steady fjord
tuples (x,y) is ordered, so you need to be able to tell which element is first a...
ohhh
tysm!!!!
this makes alot of sense
.close
marsh citrusBOT
vague jay
this allows for constructions other than just that, for example constructing the...
how do u write an ordered 3-tuple using sets, would it be (x,y,z) = {x, {x, {y, {y, z}}}}
marsh citrusBOT
Available help channel!
stuck tendon
help i dont know how to do it
marsh citrusBOT
help i dont know how to do it
stuck tendon
so far i found derivative or slope of the first equation
serene bramble
Can you then find the tangent at (4, -4)?
stuck tendon
the tangent is when i plug in x = 4 into my derivative equation right?
serene bramble
(I mean, this is a two-parter question, the first part literally asking you to find the equation of the tangent — so I'd think to do that in any case)
serene bramble
the tangent is when i plug in x = 4 into my derivative equation right?
That will give you the gradient of the tangent at that point, yes
[important to get that conceptual phrasing down correctly]
stuck tendon
ok i got y = -x
stuck tendon
That will give you the **gradient of the tangent at that point**, yes
is gradient just slope
serene bramble
yes - but I'm pointing out that "gradient" != "tangent"
stuck tendon
oh yeah
serene bramble
"gradient" is the value that represents the rise-over-run
"tangent" is a line
So they're different objects, hence the need to make the wording a little clearer there
exotic valley
some people actually think that the first derivative's equation is the equation of the tangent
stuck tendon
but thats not the equation of the tanggent
exotic valley
so yeah best to clarify
stuck tendon
some people actually think that the first derivative's equation is the equation ...
first derivative is the gradient of the original equation
exotic valley
first derivative is the **gradient** of the original equation
yes I know
I'm saying it's a common mistake
stuck tendon
oh yeah
opaque agate
Yes i clicked on the channel because of the name
stuck tendon
ok
exotic valley
sorry to interrupt
stuck tendon
i was just wondering
if i was right
so im sorry
but anyways
to find the equation
i will
serene bramble
The first derivative gives you a function to find the gradient of a tangent at a point which you can plug in, yh
stuck tendon
just plug in for y= mx + b?:
serene bramble
ok i got y = -x
But yh, that sounds correct
stuck tendon
ok good
serene bramble
Now what's the second part of the question asking?
stuck tendon
find the coordinate of the tangent where it meets again
wait but are we done the first part of the qustion?
is it just y = - x
serene bramble
ye
marsh citrusBOT
marsh citrusBOT
Available help channel!
vivid beacon
how to do this
marsh citrusBOT
how to do this
sharp vigil
hint: ||factor theorem||
exotic valley
heh they give you space to type working? wow
vivid beacon
hint: ||factor theorem||
idk what that is
vivid beacon
heh they give you space to type working? wow
yh
vivid beacon
what is a poly
vivid beacon
so how do we do it
i was thinking complete the square
but seems wrong
or qudratic formula
knotty trellis
That's unnecessarily complicated
sharp vigil
or vieta
knotty trellis
Either write it in the factorized form, or use vieta's
vivid beacon
idk what vieta is
vivid beacon
Either write it in the factorized form, or use vieta's
how can i factorise if i dont know what p and q is
exotic valley
you have the two roots and thus the two factors!
distant turtle
one simple way is forming 2 linear equations in 2 variables by plugging in 5 - (3)^1/2 and then 5 + (3)^1/2
u can then solve the linear equations
to find p and q
vivid beacon
subtituating the soluton into the equation for both + and minus?
jagged relic
There is really no reason to form a system of equations
exotic valley
but idk wht the numbers p and q are
consider that you have the two factors. then after writing the quadratic in factored form using the two factors, you can expand the factors and thus compare coefficients
that's the most straightforward way
that I can think of at least
jagged relic
If a polynomial $ax^2 + bx + c$ has roots $r_1$ and $r_2$, then it equals $a(x-r_1)(x-r_2)$
elfin berryBOT
Nel
marsh citrusBOT
marsh citrusBOT
Available help channel!
warm pebble
Let $R=\mathbb{R}/\mathbb{Z}$ denote the ring of real numbers mod $1$. Given an integer matrix $A$ with non-zero determinant and vector $b\in R^n$, I want to show that $Ax=b$ has exactly $|\det(A)|$ solutions in $R^n$. It is easy to see when $A$ is diagonal, as $0, 1/k, 2/k, ..., (k-1)/k$ all vanish in $R$ when multiplied by $k$. Also, I know that the number of solutions is invariant with left-multiplication by an invertible integer matrix
marsh citrusBOT
Let $R=\mathbb{R}/\mathbb{Z}$ denote the ring of real numbers mod $1$. Given an ...
elfin berryBOT
Hanno
marsh citrusBOT
@warm pebble Has your question been resolved?
marsh citrusBOT
@warm pebble Has your question been resolved?
marsh citrusBOT
@warm pebble Has your question been resolved?
night relic
Let $R=\mathbb{R}/\mathbb{Z}$ denote the ring of real numbers mod $1$. Given an ...
Have you considered decomposing the matrix so that you can get a diagonal matrix and some left multiplication of invertible matrices? ||LUP decomposition of the form PA = LU|| seems promising (||start with A, argue that left-multiplication with P is invariant on solutions, then examine LU instead||); of course, you would have to prove that everything required for this decomposition works under R/Z.
The other one that sprang to mind is ||diagonalizing,|| but i'm not 100% you can do that without right-multiplication. (you'd have ||AP = PD||, the LHS could cause problems)
Try those and see what you get?
marsh citrusBOT
@warm pebble Has your question been resolved?
marsh citrusBOT
marsh citrusBOT
Available help channel!
stone fulcrum
Can someone please explain this to me?
marsh citrusBOT
Can someone please explain this to me?
stone fulcrum
I was doing a bunch of problems where you do up forces=down forces, f=ma and it was about splitting things into components, but the model answer for this question shows barely any working
paper kelp
Which part would you like help understanding?
stone fulcrum
Why didn’t we need to find theta and divide it into components? Ty
wispy jungle
It was originally in equalibrium
high pine
since these three forces cause the point to be in equilibrium, it means that each force balances the other two, in particular the force of 35 N balances the other two
stone fulcrum
I think I somewhat understood? So if i were to find the resultant force between 12 and 37 it would also be 35?
high pine
Exactly
marsh citrusBOT
high pine
Of course you will have to apply it in the opposite direction to achieve balance
But the magnitude would be the same
marsh citrusBOT
Available help channel!
compact flower
9.14
wintry gale
is bro casually flexing the happy valentine's day gift?
compact flower
is bro casually flexing the happy valentine's day gift?
LMAO my bad
obtuse umbra
is bro casually flexing the happy valentine's day gift?
exotic valley
for future helpers, might wanna briefly explain what you've tried and where you're stuck on!
obtuse umbra
is bro casually flexing the happy valentine's day gift?
bro ain't single
compact flower
I’m stuck on it conceptually, and how I can actually use any of 9.12 to show this
Conceptually, it’s really simple, but I can’t figure out the starting step even though I did 9.12
I currently have |a^n+1/a^n| < epsilon
And even then I don’t think that’s a good starting point
obtuse umbra
is bro casually flexing the happy valentine's day gift?
!redir btw
marsh citrusBOT
is bro casually flexing the happy valentine's day gift?
This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
exotic valley
!redir btw
you also lol
obtuse umbra
you also lol
lol
wintry gale
mb
odd agate
i needed to talk to you for a bit <:KEK:586240877358350341>
!redir?
marsh citrusBOT
i needed to talk to you for a bit <:KEK:586240877358350341>
This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
knotty trellis
I currently have |a^n+1/a^n| < epsilon
which part are you even doing?
out of the 3 cases
compact flower
which part are you even doing?
The first
knotty trellis
So if |a| <= 1, we need to bound |a^n / n^p - 0|. How did you even get this to the |a^n + 1/a^n| < epsilon?
I'd just bound it by |a^n / n^p| <= |1 / n^p| and then make that < epsilon
compact flower
Ok now the 2nd
I actually just got the first
Can I assume n/(n+1) converges to 1
knotty trellis
Can I assume n/(n+1) converges to 1
Well, have you ever proved something like that?
n/(n+1) = 1 - 1/(n+1)
so if you have proved some theorems about arithmetic of limits and that 1/n -> 0, then you should be fine
knotty trellis
n/(n+1) = 1 - 1/(n+1)
you could explain it by saying this and referencing theorems about arithmetic of limits and 1/(n+1) -> 0
compact flower
That was the hardest part for me so I think I have the q down now, thank you
Like figuring out the setup and cancelling things
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
gaunt lynx
hello, can you somehow determine if a extreme point is a minima or maxima without pluging it into the 2nd derivative? my teacher does it somehow but i lowkey dont listen and i forgot how he did it
marsh citrusBOT
hello, can you somehow determine if a extreme point is a minima or maxima withou...
devout mauve
you can plug nearby points into f(x)
knotty trellis
or inspect the derivative to see if it's negative -> 0 -> positive or positive -> 0 -> negative
gaunt lynx
or inspect the derivative to see if it's negative -> 0 -> positive or positive -...
wdym -> 0
wait hold on
devout mauve
left of the extreme point its negative, at the extreme point its zero, right of the point its positive
gaunt lynx
yeah
i remmeber now
he said something like
like after a minimum there must be a maximum yk
but generally its always better to plug it into f''(x)?
gritty schooner
hello, can you somehow determine if a extreme point is a minima or maxima withou...
yes, plug nearby points in f'(x) and see if it goes from p-n or n-p
gaunt lynx
yes, plug nearby points in f'(x) and see if it goes from p-n or n-p
so n-p is a minimum and p-n is maximum?
gritty schooner
so n-p is a minimum and p-n is maximum?
yes
gaunt lynx
okay also
this is another topic but
How do i know if i have to do a case thingy when parameters are involved?
its so confusing , sometimes you have to do it, and sometimes you dont
stuck tendon
How do i know if i have to do a case thingy when parameters are involved?
practice my friend
lots and lots of practice and you will see differently
you will find trends and patterns at an instant of a glance
stuck tendon
that is why it is called intuition
gaunt lynx
for telling when?
gaunt lynx
yeah i will thanks
stuck tendon
tulip galleon
marsh citrusBOT
dire basalt
Fr smth feels off innit
marsh citrusBOT
@tulip galleon Has your question been resolved?
wind root
prove f'(x)=x
tulip galleon
and why is a a function of x instead of a constant
'a' isnt a constant
marsh citrusBOT
'a' isnt a constant
tulip galleon
a is an integral and g(x) is an integral with the upper bound as an integral...
if that makes sense
lavish pelican
'a' isnt a constant
Yes that's what riemann said
marsh citrusBOT
@tulip galleon Has your question been resolved?
oak marsh
hi
oak marsh
Hello
Salutations
I ran into a small error. For some reason it keeps saying i'm wrong but I checked with desmos and it has the same points
cunning fiber
I ran into a small error. For some reason it keeps saying i'm wrong but I checke...
(In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.)
cunning fiber
I ran into a small error. For some reason it keeps saying i'm wrong but I checke...
perhaps this is a "do as I want, not as I say" moment?
try taking out the +0
oak marsh
try taking out the +0
Tried that too, same error
oak marsh
(In the future, please make the first message you send the actual question since...
Ohhh, silly me lol
jagged relic
it has the same points
No it doesn't
marsh citrusBOT
If you are done with this channel, please mark your problem as solved by typing .close
oak marsh
Question, How do I sketch this graph, more so. Where do I start?
marsh citrusBOT
Question, How do I sketch this graph, more so. Where do I start?
oak marsh
lethal kite
ig a good starting point is to find the vertex
and figure out which way it is pointing
lethal kite
u could if u want but it wont be rly usefull. Just notice that the max of -3x^2 is 0, so what can we say about the vertex?
oak marsh
I'm not understanding, could you elaborate a bit more please
lethal kite
so we know that x^2>=0, right
oak marsh
yes
oak marsh
So the vertex would be (0, -2)?
lethal kite
yes
oak marsh
And I owuld just plug in random number till I find a second point with whole numbers?
lethal kite
well if you plug in a whole number for x the output is also a whole number, right?
oak marsh
yes
lethal kite
so ig you could do that..
but theres a better way
do you know parabola tranformations?
if not its ok
oak marsh
No, but i am interested in learning it if thats ok with you
Maybe i do, might not ring a bell
oak marsh
Yeah, its where the vertex is (0, 0)
oak marsh
OHHHH, its where the graph has a stetcha dn yeah, now it rings a bell
lethal kite
so in general ax^2 is "thinner" that x^2
oak marsh
ohhh, that makes sense now
oak marsh
Oh yes, my favorite tool
lethal kite
i reccomend you play around wih the graphing calculator with parabolas to get a fell on how much it "thins"
lethal kite
np!
oak marsh
.close
marsh citrusBOT
lethal kite
wait a sec
just a quick tip
if you were to graph it on paper, i would reccomend you to plot points because it is more accurate, and the transformation if only good for a rough sketch of more complicated polynominals
but yea nice job
!
marsh citrusBOT
Available help channel!
gaunt lynx
hello did i do this correctly
marsh citrusBOT
hello did i do this correctly
marsh citrusBOT
@gaunt lynx Has your question been resolved?
gaunt lynx
yeah its correc
slim lark
two externally disjoint circles o1 and o2. find locus of centers of circles that at orthogonal to o1 and externally tangent to o2
marsh citrusBOT
two externally disjoint circles o1 and o2. find locus of centers of circles that...
slim lark
i think it might be a parabola but i tave no idea how to prove it
marsh citrusBOT
@slim lark Has your question been resolved?
frail obsidian
i think it might be a parabola but i tave no idea how to prove it
branch of a hyperbola
have you made any progress yet
slim lark
i havent
slim lark
branch of a hyperbola
what with it
slim lark
do you have full solution? english isnt my first language and especially in math it makes learning harder
frail obsidian
well yeah, but we're not supposed to share full solutions in this channel
slim lark
is it okay with rules for me to ask you to send me that in dm?
im gonna translate it and analyse it step by step
frail obsidian
oh that sort of full proof 😭
I'm sorry, don't have my phone on me rn
the proof is quite messy the way I have it
slim lark
ive just started learning bout elipses and parabolas and ummm its not going well
slim lark
the proof is quite messy the way I have it
can i have a tip for now?
slim lark
i think
frail obsidian
so by pythagorean theorem, (distance between centers)^2 = (radius 1)^2 + (radius 2)^2
that's your first equation
slim lark
in my lannguage its caled perpendicular circles 😭
frail obsidian
in my lannguage its caled perpendicular circles 😭
yeah its the same lol
orthogonality is js fancy math speech for perpendicularity 💀
2nd equation comes from condition for tangency
distance between centers = sum of radii
slim lark
so by pythagorean theorem, (distance between centers)^2 = (radius 1)^2 + (radius...
yea thats where my ideas stop
frail obsidian
distance between centers = sum of radii
for external tangency, this thing must hold
slim lark
distance between centers = sum of radii
yea but those circles have diff radius
frail obsidian
yea but those circles have diff radius
yeah that's ok
let radius of O1 be r1
radius of O2 be r2
radius of the locus circle would be equal in both equations right?
let that be r
so you rewrite both equations as r = something 1, r = something 2
set those somethings equal
and you'll have your locus
frail obsidian
i need to find the focus and the directrix
you'll get that from the equation
ok, I'll type out a specific proof here, you can try generalizing it for arbitrary disjoint circles ok?
frail obsidian
is o1 the focus? do i look in the right place?
you don't need to worry about focus or directrix
you'll straight up get the equation of the curve
give me a few mins to latex
slim lark
you don't need to worry about focus or directrix
i dont think my teacher wolud accept that
frail obsidian
no like you can extract those details too
I'm js saying that its not required for the solution itself
frail obsidian
Let the two circles be:
- $O_1:$ Center $C_1$ = (-a,0), radius $r_1\n$
- $O_2:$ Center $C_2$ = (+a,0), radius $r_2\n$
- Locus point P(x,y) (center of third circle having radius r\n)
Condition 1: Orthogonality to $O_1\n$
Distance between centers squared is same as sum of squares of radii$\n$
$(x+a)^2 + y^2 = r^2 + r_1^2\n$
i.e $r^2 = (x+a)^2 + y^2 - r_1^2\n$
Condition 2: External tangent to $O_2\n$
Distance between centers is sum of radii $\n$
$\sqrt{(x-a)^2 + y^2} = r+r_2\n$
i.e $r = \sqrt{(x-a)^2+y^2}-r_2\n$
Substitute eqn 2 to eqn 1 $\n$
$(x+a)^2 + y^2 - r_1^2 = (\sqrt{(x-a)^2+y^2}-r_2)^2\n$
$(x+a)^2 + y^2 - r_1^2 = (x-a)^2 + y^2 + r_2^2 -2r_2\sqrt{(x-a)^2+y^2}\n$
$\frac{r_1^2+r_2^2-4ax}{2r_2}=\sqrt{(x-a)^2+y^2}\n$
square both sides
oh my god
how do you format this 😭
elfin berryBOT
donkey
Compile Error! Click the reaction for more information.
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tulip galleon
uh
tulip galleon
where did my doubt go
knotty trellis
#help-33 message
This?
marsh citrusBOT
@tulip galleon Has your question been resolved?
tulip galleon
https://discord.com/channels/268882317391429632/907022703230345237/1474822692355...
yes🥲
this is pre university calculus btw
im in the 12th grade
marsh citrusBOT
@tulip galleon Has your question been resolved?
shell tulip
I've big issues i don't understand what's limited development and what's his goal?
marsh citrusBOT
I've big issues i don't understand what's limited development and what's his goa...
azure ivy
!xy
marsh citrusBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
sinful thistle
what lol
shell tulip
"Hey! I'm struggling to understand the concept of Taylor expansion at infinity (limit development). Could someone explain how the change of variable $X = 1/x$ works and what's the point of it? Thanks!"
elfin berryBOT
tintinnono
shell tulip
.close
marsh citrusBOT
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Available help channel!
gaunt lynx
marsh citrusBOT
gaunt lynx
Determine, for each extremum and for each inflection point of the function, the coordinates as functions of the parameter and the equation of its locus.
hello what did i do wrong
marsh citrusBOT
@gaunt lynx Has your question been resolved?
gaunt lynx
<@&286206848099549185>
void pine
hello what did i do wrong
Given what you've done you still haven't answered the question
gaunt lynx
Given what you've done you still haven't answered the question
yes so
im basically stuck
thats what i meant to say
gaunt lynx
so uhh
void pine
Why did you calculate the derivative
gaunt lynx
bascially use the pq formula
gaunt lynx
Why did you calculate the derivative
for the extremes
void pine
Ok yes how do you link extrema with the derivative
gaunt lynx
Ok yes how do you link extrema with the derivative
the zeropoints of the derivative
void pine
Yeah the extrema are zero points of the derivative
So i see you've calculated the zeros of the derivative
void pine
Do you have any idea what to do next
gaunt lynx
i did something wrong i think
void pine
Don't you zgree that x1/2 are zeros of thederivative
gaunt lynx
Don't you zgree that x1/2 are zeros of thederivative
they are but
it seems too much
like
void pine
You think you lessed up that
Ok lemme check
Did you calculat ethe discriminant?
I mean
Whatever you call it
Delta
void pine
So you did multiply by 2t before and this works okay
void pine
And it does give a different result for the discriminant but the solution should be the same at the end
void pine
Because now you changed your a, b, c coefficient of your pilynomial
So you have to use the new values for the formula
void pine
No
gaunt lynx
the stuff inside the root isnt D?
void pine
This is the formula
It is tye stuff inside the root, i just think you got the formula wrong
void pine
So it would give 16t²-12t²
void pine
t
Oh okay so we just haven't seen the exact same formula
So p and q arethe2 last coefficients right?
void pine
So what is p
gaunt lynx
and q is 3t
gaunt lynx
yeah
void pine
So you just miscalculated the stuff by forgetting a square i think
void pine
Okay i see the problem
Again what is p
p is 4t not just 4
T is a coefficient
You're not solving for t
You're solving for x
Because it's f(x), not f(t)
void pine
Happens
gaunt lynx
wait so we can just pull the root now ? it cancels the ^2?
void pine
So yeah your problem seems to be a small calculation but your overall strategy is going in the right direction
void pine
wait so we can just pull the root now ? it cancels the ^2?
As long as the stuff inside is positive
void pine
Root of square turns everything inside positive
gaunt lynx
so its better to do a case differentiaation here?
void pine
For exemple root of square of -1 is 1 not -1
void pine
Yes you have to do a case differentiation