so then it wouldn't be closed under scalar multiplcation
#help-36
blissful meadow
The scalar multiplication is defined to be Ku=(0,Ku^2)
marble cliff
@everyone if you need math questions i have igcese level
tranquil pine
The scalar multiplication is defined to be Ku=(0,Ku^2)
yes
so it is like the law of that space
marble cliff
@everyone if you need math questions i have igcese level
tranquil pine
Like the scalar multiplication you're used to doesn't exist
yes
so lets say we are in another universe where ku = (0, ku2)
blissful meadow
@everyone if you need math questions i have igcese level
Can you not? This Channel is occupied and this server is not for self promotion.
tranquil pine
this is a law that cannot be broken
marble cliff
@everyone if you need math questions i have igcese level
marble cliff
Can you not? This Channel is occupied and this server is not for self promotion.
sorry then wich channel
blissful meadow
this is a law that cannot be broken
Yeah the point is to check that under this multiplication, you still have an ordered pair.
blissful meadow
sorry then wich channel
None.
marble cliff
ok
tranquil pine
Yeah the point is to check that under this multiplication, you still have an ord...
yes. ok i see.
vector spaces forces me to think differently hahah
i saw a vector space where 0=1
blissful meadow
Yeah it does make things quite abstract in some cases
solar pelican
Find real numbers x and y that verify the equality
final saddleBOT
Find real numbers x and y that verify the equality
tired walrus
!noping
final saddleBOT
Please do not ping individual helpers unprompted.
solar pelican
Damn
Alright. Ive got to complete to square, but Im stuck doing that
I broke it down into two separate terms
2x² -2x and 2y² + 2y
<@&286206848099549185>
final saddleBOT
@solar pelican Has your question been resolved?
final saddleBOT
@solar pelican Has your question been resolved?
solar pelican
@magic sparrow
vital crag
Alright. Ive got to complete to square, but Im stuck doing that
,tex .cts
soft zealotBOT
riemann
vital crag
<@779131727338930196>
!noping
final saddleBOT
<@779131727338930196>
Please do not ping individual helpers unprompted.
vital crag
He helped me yesterday
Yea that doesn't matter
solar pelican
Mk.
wide tartan
final saddleBOT
next onyx
when we say rate we mean what
wide tartan
when we say rate we mean what
the rate at which something is changing?
wide tartan
yeah like distance/time
next onyx
mmmm. lemme try
wide tartan
(300^2 + 250^2)^1/2 = 390.51
huh? isnt that just the pythagorean theorem
what does it have to do with related rates
so where is the derivates used?
im so confused
so you dont need related rates to solve that?
next onyx
im so confused
well why i think it is pythagoras theorem is because the diagram is in the form of a triangle and i do not see x, that should be enough
if there was something like 3x+4 then it will be derivative
wide tartan
a is distance between plane a and the destination b is distance between plane b and the destination so c is distance between the planes
wide tartan
so why did my thing didnt work
isnt it not a constant though?
isnt 250 the rate of change between plane a and the airport?
250 miles/hour
okay ill think about it thx
.close
sacred lynx
One dot is the the dot that all the bisectors go thro the other dot is the dot that all medians go thru i need to find the distace between them
final saddleBOT
One dot is the the dot that all the bisectors go thro the other dot is the dot t...
sacred lynx
incenter yes
what is a centroid do you mind exlpaining?
incenter is 3 alr found that
but centroid throws me off
so what do i do?
what dose hight have to do with it
ok what do i do?
<@&286206848099549185>
.close
pearl ivy
i need to continue this to show that kd is the gcd, not just a common divisor. what do i do next
final saddleBOT
i need to continue this to show that kd is the gcd, not just a common divisor. w...
desert mantle
you can for example use bezout
wraith crater
^ if you can't use that, then consider the prime factorisations of a,b and k.
grim nebula
^ if you can't use that, then consider the prime factorisations of a,b and k.
actually don't need to use either 
wraith crater
wraith crater
then what do i do from there
Write down what you have first.
pearl ivy
final saddleBOT
@pearl ivy Has your question been resolved?
pearl ivy
or is this only true if kd is the gcd(ka, kb)
final saddleBOT
@pearl ivy Has your question been resolved?
final saddleBOT
@pearl ivy Has your question been resolved?
final saddleBOT
@pearl ivy Has your question been resolved?
pearl ivy
.close
torpid blade
how do you take the derivative of an absolute value?
final saddleBOT
how do you take the derivative of an absolute value?
torpid blade
final saddleBOT
@torpid blade Has your question been resolved?
half salmon
@torpid blade I don't remember exactly, but you split it into a piecewise function
for when x >_ 5 and x < 5
it's just two linear equations or semi-linear equations, but each side has a slope of it's own
torpid blade
oh so like 5 - x - 5 and 5 - x + 5? @half salmon
half salmon
5 + x - 5
5 - x + 5 I believe
civic cargo
could someone help me through these problems?
final saddleBOT
could someone help me through these problems?
civic cargo
(its a physics problem but i find it hard to get real answers in the physics discord)
im pretty sure i solved part a (just need someone to check my answers and work) but im stuck on b
misty torrent
Have you studied series and parallel connections of resistors?
Also do you know that the current divides in inverse ratio of resistances?
misty torrent
Also do you know that the current divides in inverse ratio of resistances?
You know this?
civic cargo
wait can u check my work for a real quick
misty torrent
Sure
civic cargo
okay so
to find the total resistance, you find the resistance of the parallel resistors then treat it as if it is in series with the first resistor
so the resistance of the parallel resistors is 1/r1 + 1/r2 = 1/rt
or 1/4 + 1/6 = 1/rt
so rt = 12/5
and then you add it cuz u simplified it as a series circuit so 2 + 12/5 = 22/5 is the total resistance of the whole circuit
so to find the current at a1 you just take the voltage/resistance = 2V/22/5 ohm = 0.45...
and then to find a2 and a3 you use kirchoffs rule
for a2 it would be
2 - 0.45(2) - a_2 (4) = 0
= 1.1 - 4(a_2) = 0
A_2 = 0.275A
and for a3 it would be
2 - 0.45(2)-A_3(6) = 0
so 1.1 - a_3(6)=0
so a_3 = 0.183A
misty torrent
Indeed
misty torrent
Yes
civic cargo
okay so for part b
to find the current at a1
you just simplify and redraw the circuits until you get the total resistance
so you first simplify the resistors in parallel w resistance 10 and 20
and get a parallel resistor with 20/3 ohm and 5 ohm
misty torrent
You can fast forward things I'll understand
civic cargo
okay
so u get that resistor has the resistance of 20/7
and the first one has a resistance of 1
and the current at a1 is 2/(27/7) = 14/27
and i was kinda confused from here for finding a2 and a3
but i know that current going in a junction = going out
so does that mean that I_t = V1/R1 + V2/R2
misty torrent
Well as its same resistance current just divides in half
civic cargo
ohh okay so v1 = v2 and r1 = r2
misty torrent
Yes
civic cargo
and the current would be 14/54 A ?
misty torrent
Yep
civic cargo
okay
misty torrent
In A2,A3
misty torrent
Well they have diff resistances so not really
civic cargo
current in = current out
misty torrent
You remember the result abt current divides in inverse of res then why don't you use it
misty torrent
2 - 0.45(2) - a_2 (4) = 0
Here too you could use that
civic cargo
ohh
so would i simplify the circuit for that section and just make it 1 parallel resistor
and then find a4+5 and a6
and then split a4 and a5 and solve
misty torrent
so would i simplify the circuit for that section and just make it 1 parallel res...
Yeah first parallel 10 and 20 and then distribute current in inverse of resistances
misty torrent
2 - 0.45(2) - a_2 (4) = 0
For A part also
i1 = (1/R1) / (1/R1 + 1/R2)
civic cargo
wait can you explain the current distribution again
misty torrent
Alr
Let 2 resistors in parallel R1, R2
Total current = I
Current in 1 = (1/R1) / (1/R1 + 1/R2) * I
civic cargo
For A part also
i1 = (1/R1) / (1/R1 + 1/R2)
wait why
misty torrent
Current in 2 = (1/R2)/(1/R1 + 1/R2) * i
civic cargo
wait i did not learn this
misty torrent
I 1 : I 2 = 1/R1 : 1/R2 right?
You understand this?
misty torrent
It is 1/R1
civic cargo
wait what
civic cargo
okok thanks
misty torrent
civic cargo
oh its cuz current is inversely proportional to resistance
misty torrent
Yes
civic cargo
wait
misty torrent
This might clear concept too
civic cargo
so the current in 1
i get the concept but i dont understand why
like how did you get that 1/r1 is on top
wouldnt that already be the current at i1
oh wait u get the resistance of all of the parallel circuits whjich is equal to I_total and it cancels right
bc 1/r1 + 1/r2 + 1/r3 + 1/r4 = I
misty torrent
i1 : i2 = 1/R1 : 1/R2
i1 + i2 = I
Now solve for i1
i1 = k/R1 and i2 = K/R2
Find K by i1+i2 and then put k in i1
You will see how formula came
civic cargo
i1 = (r2 * i2)/r1
misty torrent
Don't simplify
civic cargo
(1/r1 * i2)/(1/r1)
misty torrent
k = I /(1/R1 + 1/R2)
misty torrent
(1/r1 * i2)/(1/r1)
?? denominator
civic cargo
(1/r1 * i2)/(1/r2)
oops
so
(1/r1 * 1/r2)/(1/r1)
oh and u just flip it
wait
what
misty torrent
Bro send your work 💀
civic cargo
im on pc i cant send but ill type it LMAO
so (i1)/(i2) = (1/r1)/(1/r2) right
solving for i1
you move i2 to the other side
and you get
(i1)= ((1/r1) * i2)/(1/r2)
misty torrent
Look mate we want i1 and i2 independently
misty torrent
civic cargo
ohhhhhhhhhh
okay that makes sense
okay so for part b
u can just use that concept
and you get
a6 = (3/20)/(3/20 + 1/5) * (14/27)
which is 2/9 ? @misty torrent
final saddleBOT
@civic cargo Has your question been resolved?
reef canyon
final saddleBOT
reef canyon
where did I mess up?
rocky tusk
how did u get to the last line from the line above it
final saddleBOT
@reef canyon Has your question been resolved?
reef canyon
how did u get to the last line from the line above it
found the integrals of each term seperatley
final saddleBOT
@reef canyon Has your question been resolved?
final saddleBOT
@reef canyon Has your question been resolved?
soft zealotBOT
lament geode
All of your work previous to this is correct
final saddleBOT
@reef canyon Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
tranquil pine
idk how to solve this as it doesnt seem to be a GP or an AP
tranquil pine
.close
halcyon saffron
Use a definite integral to define a function F(x) having derivative $\frac{sinx}{1+x^2}$ for all x and satisfying F(17) = 0
final saddleBOT
Use a definite integral to define a function F(x) having derivative $\frac{sinx}...
soft zealotBOT
ЖѲTЇЇC
halcyon saffron
Is it $\int_{17}^{x} \frac{sinx}{1+x^2} dx$
soft zealotBOT
ЖѲTЇЇC
halcyon saffron
Cause F(x) - F(17) = F(x)
soft zealotBOT
Ann
tired walrus
also \sin btw
halcyon saffron
Could you help with another?
tired walrus
fifty fifty
halcyon saffron
It’s diff to that
tired walrus
can't tell without seeing the other problem
halcyon saffron
halcyon saffron
$\lim_{n \to \infty}\sum_{i=1}^{n} \frac{n}{n^2 + i^2}$
soft zealotBOT
ЖѲTЇЇC
rare beacon
Maybe I am a bit stupid but how would I start to calculate V0?
final saddleBOT
Maybe I am a bit stupid but how would I start to calculate V0?
amber junco
do KCL
and no current goes into the opamp so you know that the current v1 and the node is equal to the node and vo
rare beacon
If we start at the leftbottom at v2, v2/r1 is the current right? From the voltage source 2.
Then because the amps are ideal, the current flows to the feedback resistor above
is that a correct reasoning with KCL?
amber junco
is that a correct reasoning with KCL?
yea
rare beacon
Do we need to calculate anything with how much these opamps amplify the voltage or just use KCL?
rare beacon
that node is grounded so 0 volts in that node
Which node is grounded? The one going into the first opamp? (left bottom)?
rare beacon
ideal opamp has the same voltage in v- and v+
So the voltage would stay the same afterwards?
After going through an amp?
rare beacon
Yes
amber junco
that means the voltage at v- is also 0
cus its grounded
If you know that you can just use KCL alot and get v0
rare beacon
that means the voltage at v- is also 0
Don't you mean v+?
rare beacon
But v- won't be 0 voltage
amber junco
no current goes into opamp and that means basically i = v+- v-/r = 0
rare beacon
Or am I seeing it wrong
amber junco
if no current goes into opamp that means the numerator must be 0
rare beacon
So infinite resistance in the "first resistor in the amp". So no current so no voltage
Okay
So use KCL and the formulas to calculate the gain of these opamps
Is that the right method? @amber junco ? Thanks a lot btw
amber junco
yeah thats it basically
left sigil
Does inf / inf = 1?
final saddleBOT
Does inf / inf = 1?
left sigil
when x = inf, x / x = 1?
left sigil
It is most definitely not 1.
However that isn't because it is something else.
The fact is that it is not defined, and that in turn is because:
A) infinity isn't a number. It is a concept and has different meanings in different contexts.
B) there is no way to define it that does not result in overturning the rules of arithmetic that apply normally.
For example, what would ∞∗0
be? If it is ∞
then infinity divided by infinity would be zero. If it is zero then zero divided by zero would have to be infinity.
It is far better just to accept that we can't define it in such a way that normal arithmetic applies.
manic flame
we're talking limits here, as you said infinity is not a number
infinity * 0 is also indeterminate
left sigil
ys
if inf * 0 is indeterminate, then how come inf / inf is determinate?
as equal to 1?
manic flame
inf/inf is indeterminate
wary blade
the way i was taught is that in limits, infinity is only a tendency
left sigil
<:nameitpls:1072040910289899590>
lol
left sigil
when x approaches inf in x/x it's 1
but why
wary blade
so if both terms grow at the same rate, they will be the same
left sigil
umm
cobalt lodge
when x = inf, x / x = 1?
x = inf is an invalid statement. infinity is not a value you can let just some variable be
left sigil
when x approaches inf in x/x it's 1
as u mention if substitude inf into x, then ur proving inf / inf = 1?
manic flame
no
left sigil
x = inf is an invalid statement. infinity is not a value you can let just some v...
Ok, so les just say x -> inf
soft zealotBOT
Combustion
manic flame
apply l'hopital on it
you'll see that it will approach infinity
that doesn't mean "inf/inf = inf"
manic flame
yes
left sigil
x approches inf, doesn't mean x is equal to inf
manic flame
yeah
left sigil
and that means x is not equal to inf?
manic flame
you already said that
cobalt lodge
still needs to be non zero 
marsh valley
The answer is B. The only thing I know about this question is that a=90 degrees but I’m not even sure about it
final saddleBOT
The answer is B. The only thing I know about this question is that a=90 degrees ...
west berry
Find coordinates of points A B and C
And then use the area of triangle formula using coordinates, in case you know that
Else find the lengths of each side
marsh valley
Ah I see, thanks
obtuse rock
x^x^4 = 64
final saddleBOT
x^x^4 = 64
west berry
$x^{x^4}$?
soft zealotBOT
Lorentz
soft zealotBOT
Haider
obtuse rock
This
west berry
I see
obtuse rock
$x^{x^4} = 8^2$
soft zealotBOT
Haider
$x^{x^4} = 8^2$
obtuse rock
I can just think of this step
final saddleBOT
@obtuse rock Has your question been resolved?
wind bough
Is this proof correct?
final saddleBOT
Is this proof correct?
tranquil pine
**Haider**
```latex
$x^{x^4} = 8^2$
```
Lambert w function
tranquil pine
$x^{x^4} = 8^2$
final saddleBOT
@wind bough Has your question been resolved?
wind bough
no
final saddleBOT
@wind bough Has your question been resolved?
gray skiff
Is this proof correct?
look at 1 and 3
- you got what's given wrong
- it's literally what's given to you
you don't need no.3 btw
the hypotenuses being congruent are given
it's sufficient to show that the legs are congruent, which you already did by the reflexive property
wind bough
1. you got what's given wrong
3. it's literally what's given to you
oops it was a typo thank you for pointing it out 🙂
wind bough
you don't need no.3 btw
ohh okay thank you, my teacher always wants me to be very detailed and takes off points for it so i just included extra 😭
.close
tranquil pine
here to double check if this is correct
final saddleBOT
here to double check if this is correct
storm void
it is correct , but there's a better way for this one , in the original limit you can split n into n 1s and rearrange terms like x-1 , x²-1 , x³-1 , x⁴-1 .... Which factors out the x-1 from numerator at once
wintry mantle
This is not solvable in Z/3Z, right?
final saddleBOT
This is not solvable in Z/3Z, right?
desert mantle
why not
tender dome
This is not solvable in Z/3Z, right?
what do you mean
wintry mantle
- x=1+y
- in 2): 2(1+y) + y = 1 <==> 2 = 1
desert mantle
yes
wintry mantle
so unsolvable?
wraith crater
Yeah
final saddleBOT
tranquil pine
it is solvable
wintry mantle
.reopen
final saddleBOT
✅
tranquil pine
we get answer as x=1/3
wraith crater
tranquil pine
the answer at final are same
wraith crater
Z/3Z
wintry mantle
.close
final saddleBOT
tender dome
This is not solvable in Z/3Z, right?
add two equations,
3x+3y=2
subtract x+y from both sides of both equations, y=1-x-y and x=1-x-y, so x=y
so x+y=2/3 and x=y
so x=y=1/3
wraith crater
@tender dome it’s integer mod 3
final saddleBOT
Available help channel!
main sphinx
Erm idk how to do y intercepts and I'm failing math 🤡
final saddleBOT
Erm idk how to do y intercepts and I'm failing math 🤡
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.
@main sphinx Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
young ore
could someone help me with this question?
final saddleBOT
could someone help me with this question?
@young ore Has your question been resolved?
Available help channel!
next pagoda
Calc 2. I'm confused by what was done here with the integrals of the cos's. It looks like they treated cos(theta) like you would with the integral of like x...
final saddleBOT
Calc 2. I'm confused by what was done here with the integrals of the cos's. It l...
next pagoda
I've never seen this before
formal trail
shorthand for:
rewrite
sin³(θ) = sin(θ)(sin²θ)=sin(θ)(1-cos²(θ)
u-sub u = cosθ
lethal estuary
yeah, I see that now. interesting
final saddleBOT
@next pagoda Has your question been resolved?
final saddleBOT
final saddleBOT
@thorny folio Has your question been resolved?
final saddleBOT
@thorny folio Has your question been resolved?
final saddleBOT
@thorny folio Has your question been resolved?
tranquil pine
Hey
tranquil flume
,rccw
soft zealotBOT
tranquil pine
HELLO??
paper perch
hello! this doesn't seem too bad tbh since it's explicitly allowing you to use a calculator
what kind of calculator do you have? @tranquil pine
tranquil pine
Texas instrument
tranquil pine
30
paper perch
A
never mind, but the concept should be the same
you could also use desmos if that doesn't work
wait is that a scientific calculator??
okay that explains a lot, i assumed you had a graphing calculator which would very much trivialize these problems but scientific calculator works too since pretty much all you have to do is set these guys equal to each other and then solve
squints. it's been a while since i did any trigonometry but there should be an inverse tan function for getting rid of those pesky tans, and then you just have to add pi to the radian result
paper perch
then ask me more specific questions
we're not allowed to do your hw for you
like i can't sit here and explain the entirety of inverse trig functions to you, we'd be here all day and you still wouldn't learn anything LMAO
tranquil pine
we're not allowed to do your hw for you
Tell me where I asked u to do my homework buddy
Get ur faxs straught
paper perch
okay...?
tranquil pine
.close
chilly umbra
I dont really know where to get started on this, could someone please help me?
chilly umbra
.close
final saddleBOT
Available help channel!
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
@tranquil pine Has your question been resolved?
tight star
What is the LCM of 24 and 18
final saddleBOT
What is the LCM of 24 and 18
@tight star Has your question been resolved?
Available help channel!
vocal coral
final saddleBOT
vocal coral
im so confused how to go about with b
final saddleBOT
@vocal coral Has your question been resolved?
lyric summit
im so confused how to go about with b
multiply both sides of the given inequality by 5, and next move 2x+3 to the left side, then you will be able to use factoring
lament wagon
need help with this please
lament wagon
.close
final saddleBOT
Available help channel!
pulsar wadi
Would this be A?
final saddleBOT
Would this be A?
pulsar wadi
I know at first derivative a y value of 0 indicates weather there is a minimum or maximum so thtas what my answer is being based off of
radiant laurel
I know at first derivative a y value of 0 indicates weather there is a minimum o...
you marked global minima even though you're correct
radiant laurel
i see its graph of f'(x)
pulsar wadi
yes
radiant laurel
try to predict second order derivative of the given curve where y=0 if it's -ve then maxima and vice versa
pulsar wadi
E?
radiant laurel
probably
tranquil pine
My friend who’s dumb
final saddleBOT
My friend who’s dumb
tranquil pine
Is wondering
How to this question
And I’m to lazy to help her cause I’m doing a test
I’m handing the phone to her now
!close
.close
final saddleBOT
final saddleBOT
Available help channel!
spice fable
HELLO
final saddleBOT
@spice fable Has your question been resolved?
spice fable
.close
final saddleBOT
Available help channel!
sacred lynx
the rombus diagnals +diagnal eaquels m and the area is s i need to find the sides
final saddleBOT
the rombus diagnals +diagnal eaquels m and the area is s i need to find the side...
hybrid heath
the rombus diagnals +diagnal eaquels m and the area is s i need to find the side...
diagnals +diagnal
what
hybrid heath
the sum of the diagonals?
final saddleBOT
@sacred lynx Has your question been resolved?
sacred lynx
ok so what do i do?
hybrid heath
wait, it is not the same, my mistake
sacred lynx
what do i doo
cerulean horizon
sacred lynx
yes
dull spear
the rombus diagnals +diagnal eaquels m and the area is s i need to find the side...
you have to describe the sides interms of "m"?
sacred lynx
sum of the diagnals is m
area is s
from that info i need to find what the sides would be
dull spear
yes, but i think you wont get a numerical value
dull spear
if L is the side then L = Sqrt( m^2 -4S)/2
final saddleBOT
@sacred lynx Has your question been resolved?
small stump
h(x) = sqrt(3-x) * (2x^2-2)
im asked to get the domain and image of h(x)
for domain i did just
sqrt(3-x) >= 0
x <= 3
not sure how to approach image tho
final saddleBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
wintry kindle
h(x) = sqrt(3-x) * (2x^2-2)
im asked to get the domain and image of h(x)
for d...
consider the min and max value you can construct from h(x)
small stump
so to do that i would get the min and max of (2x^2-2) and just multiply both by sqrt(3-x) right
(2x^2-2) doesn't have a max
wintry kindle
hm a bit careful though
the minimum of sqrt(3-x) is clearly 0
however that won't yield the minimum value of h(x)
because you can construct a lower value by increasing sqrt(3-x) and having a negative value for (2x²-2)
let's first consider the max value since it's easier
sqrt(3-x) is strictly positive
and 2x²-2 is strictly positive for x<-1
now, as x becomes larger in the negative
you don't reach any limit
it grows indefinitely
depending on which subject you're in
you can either use the lim[x->-inf.]
to show that limx->-inf. = inf.
or you can take any y>0
and show that there exists an x such that h(x)=y (as well as h(x-a)>y for a>0)
@small stump so far clear?
at this point you'd know the domain is [ ? , inf. )
.-.
small stump
ok yeah
wintry kindle
k, one boundary completed
now for min value
here it's a bit tricky since we know that sqrt(x-3) is strictly positive and 2x²-2 is only negative on the interval (-1, 1)
however it doesn't tend to -inf. within that interval
since it's a polynomial
therefore the lower bound of the domain is a constant
and you can retrieve it using the derivative
by setting h'(x)=0, you find all extrema of h(x)
and among those extrema is the lowest possible value for h(x) which we seek.
@small stump do you know how to calc the derivative of h(x)
small stump
im realizing that i don't think i was supposed to do this exercice
lol
but out of curiosity continue
derivate was that getting the slope or area of a graph?
i mix them up
wintry kindle
derivate was that getting the slope or area of a graph?
derivative is the slope yes
small stump
ah ok ok
wintry kindle
I don't know how much you know of derivatives of course
small stump
they way im taught to do it is subsistute x with infinity
which is not good cause that's not part of real numbers
but whatever it works
small stump
ok ok
wintry kindle
shall I continue with the derivative or do you want me to elaborate on the limit
wintry kindle
kk, all rules you'd need for the derivative btw are Chain Rule & Product Rule & Derivative of a Polynomial
k, you can consider the function sqrt(x)
now when you want to calculate limx->inf. for instance
that expression's meaning is "what is the value of sqrt(x) as x tends to infinity"
and not "what is sqrt(inf.)"
if you'd instead try to ask that, look at a function like x²-x
if you "insert" infinity for x, you get inf.²-inf., which I suppose is something like infinity again? 0? undefined?
either way it's incorrect due to what you mentioned, infinity is not a number. at least not in R (real numbers)
the actual definition of the limit as x tends to infinity would be
that I could give you any value c
and you can find a starting point, let's call it x0
and from that point x0 and onwards all values of that function are greater than c
small stump
ah yeah i've seen something like that
wintry kindle
simple example: f(x) = x
I'll throw you a value: c = 10
can you give me a x0
such that f(x) > c for all x >= x0
wintry kindle
:D maybe I've written it too formally, but can you think of a starting value x0 from which point onwards all values are greater or equal to c
e.g. f(10) = 10 of course
but f(11) = 11 > c
and f(12) = 12 > c
f(13) = 13 > c
and so on
so if you'd give me x0 = 11
then every value of f(x) will be greater than c if you start from x0
wintry kindle
if you're more of a visual learner I'll try to vis it, a sec
small stump
it's meant to be inferior no?
wintry kindle
this would be the graph of f(x) = x
now I'll set the boundary c = 10
you can quickly see that f(x) is smaller than that boundary on the left
but f(x) is greater than the boundary on the right
exactly after x=10
right?
small stump
yeah
wintry kindle
kk, so x0 = 11 would be valid for the starting value
wintry kindle
yes, I'll attempt not to lean towards formulating it too formally since that's what one is used to at university :)
now I could throw you c = 20
and you could give me x0 = 21
small stump
ye
wintry kindle
meaning that f(x) > 20 for all x >= x0
I could throw you c = 100 and you can give me a x0
c = 1000
small stump
makes sense
small stump
loll
wintry kindle
you can always throw a x0 back
small stump
yeah for this particular function x0>c right
wintry kindle
and f(x) will rise above c from that point onwards
small stump
ok im following
wintry kindle
and that's the essence you need to think of when looking at a limit
I can throw you a greater c indefinitely
but you can always give back a x0 that works
that proves that f(x) = x tends to infinity as x goes to infinity.
this is not just an approach which works for this function f(x) = x, this is how the limit for x->inf. is defined
let's look at a function which makes that a little bit harder:
f(x) = sqrt(x)
Now if I give you a c, e.g. c = 5
it doesn't become directly clear what x0 you should give me
since sqrt(x) isn't just a linear function
small stump
yeah
wintry kindle
here this would be f(x) = sqrt(x) with the boundary c = 5
I mean yeah it seems to rise above the boundary, but how could we actually show that there exists such a x0?
small stump
by saying that we can always plug in f(c + n), n being a positive number?
small stump
oh shoot c is a y value
ok then get the x of c
like the intersection
so sqrt(x) = c
small stump
intersection is c^2
wintry kindle
yup!
wintry kindle
exactly! we'll just choose some arbitrary n, e.g. 1
so if I throw you a c
you can throw back x0 = c²+1
and it will be above the boundary
small stump
yeah that's logical
wintry kindle
let's verify that briefly: we had c=5, so x0 would be c²+1=26
yup, on the right where the green line is, there is 26
and the red graph is just above the boundary c=5
now with what you just used
solving for f(x) = c
and thereby always finding a x0
you've shown that the limit of f(x) = sqrt(x) is infinity as x goes to infinity 🦇
small stump
coool B)
wintry kindle
in theory to complete the proof we'd also need to briefly show that sqrt(x) strictly grows
but that should be fairly obvious
since sqrt(a) > sqrt(b) if a > b
small stump
ah yeah that would be very rigorous
wintry kindle
neat, now we already know the limit of the linear function f(x) = x and the root function f(x) = sqrt(x)
now, what I want you to notice is that
if we shift the graph to the left or right by some number
let's say by n
do you think the limit remains the same as x goes to infinity?
wintry kindle
absolutely, so what that means is that e.g. f(x) = x-3 also has infinity as the limit for x->inf.
or f(x) = x+999
or f(x) = x-239232
small stump
yes
wintry kindle
we can also shift f(x) = sqrt(x)
if we shift it one to the left it's
f(x+1) = sqrt(x+1)
that function still has the same limit
what that means is that we've shown that the limit remains the same for this large group of functions
which are all the same, just shifted to the left or right
so instead of just knowing the limit for f(x) = x and f(x) = sqrt(x), we now already know it for an infinite number of functions!
ok, what about moving it up or down?
if I move the graph up by 1
small stump
also wouldn't change
wintry kindle
yup
small stump
so f(x+h) + k has the same limit as f(x)
wintry kindle
you can take any function, move it somewhere else in the coordinate plane and the limit remains the same
hm, what about multiplication? If I multiply a function by some constant, does the limit remain the same?
wintry kindle
exactly, just wanted to point that out :D
now you can not only move the graph to anywhere, you can also stretch it as you wish
and the limit in infinity remains the same
small stump
i see oki
wintry kindle
I'll have to close it off here as I have to go, but I hope I could intrigue you a bit about it :)
from that point onwards one can show more fundamental properties
such as the sum of two functions (ANY functions), the limits add up
wintry kindle
such as the sum of two functions (ANY functions), the limits add up
likewise for product and division
the proofs for those are a bit longer than what we had so far though
but once one knows these
you can quickly show what the limit of your original function
f(x) = sqrt(3-x)*(2x²-2) is
nywys, cya 🙂
small stump
.close
daring pendant
If we arent uses sum of forces = ma, would this be an energy equation?
minor patio
I missed an entire unit on functions so I don’t know what most of this means
If anyone could help it would really be appreciated
daring pendant
If anyone could help it would really be appreciated
use a different channel please
minor patio
Mb
daring pendant
np
im a little confused on what the system would be here
im thinking that
Initial energy = eleastic potential energy + kinetic energy
Final energy = gravitational potential energy
This is what I have but I don’t think it really fits a system
<@&286206848099549185>
final saddleBOT
@daring pendant Has your question been resolved?
daring pendant
<@&286206848099549185> :D
daring pendant
.close
tranquil pine
final saddleBOT
tranquil pine
Do I use taylor expansion?
eager junco
hi @rapid sky
final saddleBOT
hi <@537482727557955597>
eager junco
im assuming the answer for this is c
random forge
looks right
first and third both seem to satisfy the 3 properties
i assume you checked though
eager junco
i checked for (I) and i kinda know that (II) is not a subspace but have a few doubts
so for (I) its clear to write the augmented matrix thru the given variables
but for (II) how would i write the augmented matrix given that there is a "3"?
random forge
i have no idea what you are writing augmental matrices for, but you can tell its not a subspace because it doesnt contain the zero vector
eager junco
i was told by Chai to do it the augmented matrix way
random forge
oh perhaps wait until the return of chai
eager junco
but how come (II) is not a zero vector?
i mean i kinda see it but wanted to ask anyway
wraith crater
The second entry will always be 3 so you can’t get the zero vector for any a,b,c
eager junco
i have no idea what you are writing augmental matrices for, but you can tell its...
and for the (I) and (III) how would you check the other two properties?
random forge
closure under addition and closure under scalar multiplication
pick two arbitary vectors and add them and see if they are still in the space
and pick some arbitrary scalar and do that
when i say "pick" i mean use variables, not actually pick
eager junco
right, so i was doing that, and added the two vectors then i had
[1 1 0 | 3
2 0 -3| 8
0 1 5 | 2]
thats how i found the augmented matrix
random forge
i have to go (my league game started lol) but i would just do the vector with abc then another with def and then when you add it should just work out
rapid sky
right, so i was doing that, and added the two vectors then i had
[1 1 0 | 3
2 0...
That can only help in proving it for two specific vectors rather than two arbitrary vectors.
Like [a - b, 3, b + 3c] is an arbitrary vector.
You can fill in whatever you want for a, b, and c.
Also, anything you fill in for them will be a vector in that subset.
Because that's the definition of that subset.
eager junco
oh about that, how did you find such good vectors as examples?
rapid sky
Which vectors?
eager junco
when i chose mine, they would always result in decimal values
eager junco
no like before
rapid sky
I'm not sure what you mean.
rapid sky
Oh, I got lucky, I guess.
eager junco
b/c when im doing this as a test, i would have a hard time thinking of good vectors as examples
Personally i find subspaces a bit stupid. I dont really get them besides (II) being not a subspace
rapid sky
Do you know what a vector space is?
eager junco
like on graphs?
rapid sky
No, I mean something else.
You're taking linear algebra, I think.
And the point of linear algebra is vector spaces.
Well, a major point is.
eager junco
like its easy to eye ball the subspaces for the zero vector
only if it were the same for the other 2 properties
rapid sky
It is easy to do the other two properties for these kinds of vectors with variables in them.
You don't fill in the variables.
You do algebra instead.
eager junco
yea but the solution of the variables doesnt really make sense to me
rapid sky
Which solution?
eager junco
the ones we did earlier
rapid sky
Like solving for the values of the variables?
eager junco
like when you did a1 + a2 etc
eager junco
yea
rapid sky
All the vectors will be a variable minus a second variable for the first part, a three for the second part, and the second variable plus 3 times a third variable.
So, you could have like [x - y, 3, y + 3z] or something.
As long as you have the first variable minus the second one, 3, and the second one plus 3 times the third variable.
You could also have [1 - 2, 3, 2 + 3(3)].
You could also have [(a1 + a2) - (b1 + b2), 3, (b1 + b2) + 3(c1 + c2)].
That has the same pattern to it.
Do you see the pattern?
eager junco
a little
rapid sky
You can have anything minus anything in the first part. You can have 3 in the second part. You can have the thing you subtracted before plus 3 times another thing in the third part.
So, like you can have [259 - 56, 3, 56 + 3(8282)] or something.
That's the pattern.
eager junco
what are you trying to say through this?
rapid sky
I'm trying to get across the pattern.
eager junco
ok
rapid sky
When you have like a - b, you can put anything you want for a and anything you want for b.
When you have b + c, the bs have to match and the c can be anything you want.
So, you can fill in anything you want for the variables as long as the letters that are the same have the same values.
Does that make sense?
eager junco
yea
rapid sky
OK, so if I have a - b and b + c, I can fill in a with x + 5, b with y + z + 6, and c with 7.
Because I said earlier I can fill them with anything I want.
So, I can fill them with those.
As long as all the as are the same, all the bs are the same, and all the cs are the same.
Does that make sense?
rapid sky
What do you mean by testing it out?
rapid sky
But you don't know how to prove additive closure for a subspace yet.
eager junco
i do, ive learned it in class, its just that i have yet to practise is
for us we just say u + v we dont say additive closure tho
rapid sky
No, additive closure doesn't mean u + v.
It means that when you add any two vectors, you get another vector in that same subset.
rapid sky
OK.
eager junco
.close
final saddleBOT
final saddleBOT
Available help channel!
haughty moth
final saddleBOT
barren hound
there are a few techniques, including row reducing the augmented matrix
haughty moth
they tell me to use PLU or QR or cholesky
on python
A = PLU and there's an indication that says that P² = P^(-1)
@barren hound
.close
flint fable
i need help solving this problem please
final saddleBOT
i need help solving this problem please
west berry
Did you differentiate it
flint fable
im not sure how to do that
brittle gazelle
is there a name for the these sorts of geometric objects?
final saddleBOT
is there a name for the these sorts of geometric objects?