I have to sleep soon pleasd
#help-41
simple drum
simple drum
Hi! It is to find the dimension and basis of CnU when C is the space of columns of A and U is space of rows of A
I have found the bases for C and U I can't seem to find CnU
river verge
Give me a second.
simple drum
Yes
I wrote a general member of C and U and put the conditions of C on U and wrote the members of C accordingly
My explanation is a bit unclear
But like I got for C=(a,-10a,b,c)
U=(13x,13y,13z,-44x+45y-3z)
river verge
Where is the computation of vectors that lie in both row and column space?
simple drum
How to do that ?
river verge
But like I got for C=(a,-10a,b,c)
U=(13x,13y,13z,-44x+45y-3z)
Well the formula I can conduct from this would be dim(UnC) = rank(A) - (dim(U) + Dim(C)) + Dim(A)
river verge
But like I got for C=(a,-10a,b,c)
U=(13x,13y,13z,-44x+45y-3z)
no steps for vectors exist in both spaces?
simple drum
I tried to write U actually in terms of C but the answer was wrong
The dimension was 2 which was correct
simple drum
Well the formula I can conduct from this would be dim(UnC) = rank(A) - (dim(U) +...
I've never heard of this haha
river verge
The dimension was 2 which was correct
yes this is incorrect
simple drum
O
river verge
I've never heard of this haha
This is dimensionality check formula
river verge
The dimension was 2 which was correct
The final answer should be 0
amber waspBOT
@simple drum Has your question been resolved?
amber waspBOT
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noble plume
Is there any good websites/videos where I could easily learn to factor a quadratic
noble plume
All of the videos I watched so far are so bad
rigid viper
Is there any good websites/videos where I could easily learn to factor a quadrat...
Like quadratic using what method
amber waspBOT
grizzled pagodaBOT
w
silver maple
no
note that F’=f and thus F’’=f’
what can you say since f is strictly decreasing (i.e. f’<0)?
huh
you mean F?
👍
amber waspBOT
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real delta
Why isn't number 3 answer D and number 4 answer c?
amber waspBOT
Why isn't number 3 answer D and number 4 answer c?
silver maple
Why isn't number 3 answer D and number 4 answer c?
$f:[-5,\infty)\mapsto\bR$
grizzled pagodaBOT
;(
real delta
What
silver maple
so $x\geq -5$ in other words
grizzled pagodaBOT
;(
silver maple
now take 1/2 of that and subtract 2
real delta
Oh ok
silver maple
you get your lower bound for h
silver maple
What
its more..."advanced" notation (its just saying the domain is from [5,inf) and its codomain (range) is in the reals)
primal briar
isn't it add 2?
primal briar
because it's af(kx-d) + c
real delta
I thought that I would do this:
2(-5) + 2 =-8
primal briar
d is +2 here
real delta
Wait isn’t it y = af[(1/b)(x-c)]+d
primal briar
yeah you're right i think
real delta
But my answer key says the answer is C but it feels off
Like to get C, you have to:
(-5 + 2) * 2 =-6
The order of operations is messed up
silver maple
??
real delta
:0
silver maple
we're solving an inequality, guys...
real delta
I’m so confused 😵💫
real delta
Yeah
silver maple
you're shitfing to the right 2 units so its adding 2 units
silver maple
grizzled pagodaBOT
;(
real delta
But dawg the answers are diff
What I do wrong
Choking myself on my pencil cause I can’t find why D ain’t the answer
What the fuck, I give up. I know I’m right, stupid ass answer key
.close
amber waspBOT
celest cove
youre not
real delta
.reopen
celest cove
1/2(-8)-2 isnt -5 as a simple check
amber waspBOT
✅
amber waspBOT
potent stump
145
gritty lion
sigh
amber waspBOT
@vivid sandal Has your question been resolved?
shell talon
amber waspBOT
shell talon
What's the mistake
pallid canopy
,w int 1/ sqrt(x^2+2x+3)
grizzled pagodaBOT
shell talon
Where did I go wrong
pallid canopy
4^2 is 16
shell talon
Wait lemme check
pallid canopy
,calc 4^2
grizzled pagodaBOT
Result:
16
fierce edge
if a radian is defined to be the ratio of the arc length cut off by some angle theta with the radius (assuming we're working on the unit circle) then why are negative radians "defined"?
fierce edge
aren't we admitting that arc length can be negative then
cuz clearly the radius is not negative
vocal isle
The value of a radian is defined using arc lengths.
fierce edge
how does that contradict anything i've said
fresh ocean
length of an arc is one of the interpretations of angle
fresh ocean
You can also define an angle as orientation of a vector, and for that clockwise/anticlockwise is importatn
vocal isle
oh
fresh ocean
angle is just a dimensionless vector
radians are measure of angles, and thus they are only measuring the magnitude
If you look at it from a physics perspective, its easier to understand
vocal isle
just like displacements in physics
amber waspBOT
@fierce edge Has your question been resolved?
split sail
Your question is more about why choose a definition, but I think you meant a definition for angles, not radians necessarily
amber waspBOT
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river brook
for a positive integer n, let f(n) denote the smallest positive integer which neither divides n nor n + 1. Which values can f(n) take as n varies
indigo cloud
have you done some examples? eg f(1),f(2),f(3) up to f(20)?
river brook
yes
they're all of the form p^n or 2p^n
where p is a prime
i was able to prove that the former is when f(n) does not divide (n)(n+1)
i thought maybe i could form a case where f(n) does divide (n)(n+1) but made no progress
<@&286206848099549185>
<@&286206848099549185>
amber waspBOT
@river brook Has your question been resolved?
unkempt holly
hi
unkempt holly
can somone explain me the 1+1=2
patent raptor
<@&268886789983436800>
unkempt holly
why
scenic dagger
Please don't use these channels for troll questions.
unkempt holly
Please don't use these channels for troll questions.
im serious
unkempt holly
Please don't use these channels for troll questions.
the mathematic explaination
scenic dagger
At the level of set theory we have a handful of axioms that tell us we can construct objects where this holds.
There's more than one system of arithmetic that have objects which work like 1 and 2.
So a better explanation probably requires specifying more info.
unkempt holly
i dont get it
indigo cloud
in the peano axioms the symbol 2 is by definition the successor of the symbol 1 and "+1" is by definition "take the successor", so 1+1 means take the successor of 1, aka 2
unkempt holly
So a better explanation probably requires specifying more info.
allr
scenic dagger
To add to that, the axioms of set theory tell us an empty set exists, that we can union sets and that we can construct sets containing other known sets. So if we let 0 be the empty set and define A+1=AU{A} we can reconstruct all the peano axioms in plain old set theory
unkempt holly
in the peano axioms the symbol 2 is by definition the successor of the symbol 1 ...
ooow thats make sens
unkempt holly
To add to that, the axioms of set theory tell us an empty set exists, that we ca...
i undersatnd
.close
amber waspBOT
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tough mica
amber waspBOT
tough mica
amber waspBOT
@tough mica Has your question been resolved?
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tardy egret
AD = BD. I just can't see how to prove that these 4 points are concyclic...
amber waspBOT
AD = BD. I just can't see how to prove that these 4 points are concyclic...
@tardy egret Has your question been resolved?
amber waspBOT
@tardy egret Has your question been resolved?
night snow
AD = BD. I just can't see how to prove that these 4 points are concyclic...
is AB the diameter?
tardy egret
is AB the diameter?
I don't think it is, unless that can be proven. At least it is not stated so
night snow
what else is given?
tardy egret
what else is given?
nothing, well the gray angles are equal. That's it
night snow
if we can prove angleACB = ngle ADB, we r done
amber waspBOT
@tardy egret Has your question been resolved?
humble grotto
Fact: if the opposite angles of the quadrilateral sum to 180 degrees, then the points are concyclic
So far, I have deduced that the black angle should be equal to the base angle of the equilateral triangle for this to work. Just rying to prove that now
humble grotto
if we can prove angleACB = ngle ADB, we r done
This also gives me that the two angles in question mmust be equivalent. But I see no obvious way to show that the angles ar equivalent.
amber waspBOT
@tardy egret Has your question been resolved?
night snow
This also gives me that the two angles in question mmust be equivalent. But I se...
ye same even i was trying
night snow
So far, I have deduced that the black angle should be equal to the base angle of...
wait u mean the isosceles triangle right?
humble grotto
yeah i labelled the isoceles triangles base angle as alpha, and the black angle as beta
haughty siren
helpp
haughty siren
.close
amber waspBOT
sterile swan
!da2a
amber waspBOT
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/
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pallid iris
I tried sin(61)x54 but it was wrong
amber waspBOT
I tried sin(61)x54 but it was wrong
sterile swan
try making a drawing
celest cove
doesnt sound like it should be wrong
sterile swan
,calc sin(61)*54
grizzled pagodaBOT
Result:
-52.170359580453
that calc does radians
pallid iris
how can it be a minus?
celest cove
,calc sin(61 * pi/180) * 54
grizzled pagodaBOT
Result:
47.229464185527
thanks!
sterile swan
,w sin(61 degrees)*54
grizzled pagodaBOT
pallid iris
.close
amber waspBOT
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iron rapids
Can someone help me i just wanna learn math in depth from arithmetic to algebraic topology and real analysis. How can I teach myself on the internet. Any recommendations?
vocal wagon
My prof is doing a PhD in algebraic topology
iron rapids
My prof is doing a PhD in algebraic topology
Ye but how can I teach myself
vocal wagon
Idk
outer hull
Can someone help me i just wanna learn math in depth from arithmetic to algebrai...
can you be more specific what do you mean by "arithmetic"
amber waspBOT
@iron rapids Has your question been resolved?
elder mountain
hi, i have problems on finding the electrostatic flow of a semisphere, i know it is the same as the flow of the circle closing it but i am trying to find the flow of the semisphere. i got a surface r
(θ,ϕ)=R(cosθcosϕ,cosθsinϕ,sinθ) and a ds= (R^2 * sinθcosθ)dθdϕ on -k versor (the electric fild is just acting on k versor)
elder mountain
i am integrating
silver maple
i am integrating
have you considered swapping bounds
(btw, i will only deal with the integral portion, as i have no experience in electrostatics)
silver maple
i am integrating
or just use the fact that $\sin(x)\cos(x)=\frac12\sin(2x)$
grizzled pagodaBOT
;(
elder mountain
the problem is that that integral (dθ) is 0
does not have sens, it is the area of half sphere, it should be 2 pi* R^2
it has to be something wonrg in the integral or in my parameterization of the surface
can´t be 0
silver maple
oh, are you trying to prove the volume of a hemisphere?
elder mountain
no, it´s area
silver maple
@elder mountain there is a different method unless you are required to do this
silver maple
no, it´s area
oh surface area mb
how did u get 40 anyways?
also, i don’t think the integral dtheta is 0.
,w integral 0 to pi sin(x)cos(x)
grizzled pagodaBOT
silver maple
nvm
elder mountain
it´s the flow, so it´s the integral of the region s of ds, thats the whole region s, so it´s the area of the semisphere
na, 40 is the value of the electric field acting on k versor, is a constant
silver maple
how did u get 40 anyways?
this seems random tho
silver maple
na, 40 is the value of the electric field acting on k versor, is a constant
oh
well i have no clue then
😅
elder mountain
is this, E and ds are both just on k versor
E is constant and ds is a diferential of the surface
so it´s just E*
E=40 and i am having problem whit the integral
<@&286206848099549185>
amber waspBOT
@elder mountain Has your question been resolved?
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lime jasper
How is |x-4| >= -1 infinite solutions
amber waspBOT
How is |x-4| >= -1 infinite solutions
gloomy tide
why wouldn't it be 
lime jasper
I don’t know why it is
pallid canopy
I don’t know why it is
Think of one solution, then think of a larger one
quick ridge
I don’t know why it is
|anything| >= 0
amber waspBOT
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polar cedar
how can i simplify this further
amber waspBOT
how can i simplify this further
amber waspBOT
@polar cedar Has your question been resolved?
polar cedar
.close
amber waspBOT
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nocturne plaza
Hi! :) How do I find the critical points of V'(x)=(5/2)-3x^2? I got 0 by multiplying the bottom "2" by positive 3x^2, but that feels wrong. @~@
shrewd elm
you want the critical points of V(x) or of V'(x)?
nocturne plaza
pretty sure I'm doing it for the derivative? I suck at word problems tho T~T
shrewd elm
!original
amber waspBOT
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.
nocturne plaza
my teacher basically confirmed up to the V(x)=1/6(15x-6x^3) thing, but I probs messed up after that? Idk
shrewd elm
what is x here? and does V(x) represent the volume?
nocturne plaza
x is just like the length of a side. And I think V(x) should be volume? He didn't explicitly say
shrewd elm
and you took the derivative and found it was 5/2 - 3x^2
so you just need to set that equal to zero and solve for x
shrewd elm
wait you solved
0 = 5/2 - 3x^2
and got x=0?
plug x=0 into that equation, you get 0 = 5/2, so you know that can't be right
shrewd elm
haha has nothing to do with smart, you probably just made an arithmetic mistake somewhere
nocturne plaza
am I solving for x or plugging in 0 for the x vals
nocturne plaza
😅 I'm so sleep deprived, I'd probably mess up 10+5
nocturne plaza
I did it REALLY weirdly :')
and then I checked it, but this makes no sense, so did I enter it wrong ?
nocturne plaza
free our minds for the harder stuff
hehe very true
shrewd elm
ah yea you want to find the critical points of your original V(x)
not of V'(x)
so it's a terminology issue
the critical points of V(x) are the roots of V'(x)
if you tell the calculator to find the roots of 5/2 - 3x^2, that will give you what you want
nocturne plaza
not of V'(x)
but I'm solving for 0 on the V'(x) D:
shrewd elm
yes
that means you're finding the roots
finding critical points of a function means: take the derivative of that function and then set it equal to 0 and solve for x
if you say "find the critical points of V'(x)" then the calculator is gonna take another derivative
but you already did that part
shrewd elm
yea but isn't that the calculator telling you that?
nocturne plaza
I also did it by hand, just badly
shrewd elm
if you solve 5/2 - 3x^2 = 0 yourself, you don't get x=0 do you?
nocturne plaza
apparently xD
nocturne plaza
aight, lemme fail at drawing on a computer rq
nocturne plaza
ok writing it out again, I did mess that up very obviously 😭
divided the 3x^2 and it was alrdy wrong by that point lmao
nocturne plaza
or you can take a picture/screenshot and upload if you like
my phone hates me :<
shrewd elm
ah you want to set it equal to zero
5/2 - 3x^2 = 0
solve for x
so it becomes:
5/2 = 3x^2
nocturne plaza
... a h
ok that definitely simplifies stuff, but the fraction makes me way too nervous for a student in CALC 2. I am NOT prepared. :')
like normally I'd probably divide by the 3 and try to sqrt the whole thing, but that seems very incorrect here, haha
shrewd elm
no that's exactly what you want to do
shrewd elm
yep
nocturne plaza
I AM A GENIUS
shrewd elm
so that gives you the x value that makes the volume maximum
they want the volume itself
so now you need to plug that into the formula for the volume
nocturne plaza
.... I did something wrong lmao
shrewd elm
what did you get
shrewd elm
yea i get the same
nocturne plaza
what did you get
sorry that wasn't me brushing you off xD
oh :o
that is
such a tiny volume
Ig it makes sense tho, $15 isn't a lot 🤔
nocturne plaza
xD true
shrewd elm
oh no,, they said feet nvm
nocturne plaza
ikr, like what am I gonna fit in there >.>
nocturne plaza
😭 maybe like a few beans or somethin
also, this section was amazingly worded and is going into my notes. Thank you, it gives me some perspective that I was lacking 😅
shrewd elm
nice
nocturne plaza
oki I must disappear to grind out the next math problems, but they should be a bit easier. Would not be surprised if I needed more help but :')
for now, I shall close this channel. Thank you again :D
shrewd elm
sounds good, fire up another channel if you need help again
amber waspBOT
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hollow ginkgo
Find all such pairs of natural numbers m, n such that 7^m - 3*2^n = 1
sage bough
amber waspBOT
@hollow ginkgo Has your question been resolved?
winged panther
Consider modulo 9
7^m is {1, 4, 7} mod 9, 3•2^n is {3, 6} mod 9
Build on that, maybe you'll get somewhere
hollow ginkgo
Nothing, Because we need to achieve a limit on m or n from above, mod x won't help
amber waspBOT
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amber waspBOT
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thick canyon
$\Find {dy/dx} if \ (cos(x)^y = cos(y)^x$
amber waspBOT
$\Find {dy/dx} if \ \(cos(x)^y = cos(y)^x$
patent raptor
Find ( \frac{dy}{dx} ) if ( \cos(x)^y = \cos(y)^x)
grizzled pagodaBOT
bagelguy3
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
grizzled pagodaBOT
anti-algebraist 𝔸dωn𝓲²s
Find \( \frac{dy}{dx} \) if \( \cos(x)^y = \cos(y)^x\)
winged panther
Find $\dv{y}{x}$ if $\cos^y x = \cos^x y$
thick canyon
yes
patent raptor
So you know chain rule and x = e^ln(x)
winged panther
That?
patent raptor
( e^{y \ln \cos x} = e^{x \ln \cos y} )
thick canyon
yes
grizzled pagodaBOT
anti-algebraist 𝔸dωn𝓲²s
thick canyon
yes
patent raptor
Actually this assumes cos(x), cos(y) > 0
thick canyon
yes
patent raptor
So you chain rule as in ( e^{u(x)} = e^{v(x)} )
grizzled pagodaBOT
anti-algebraist 𝔸dωn𝓲²s
thick canyon
wait what
patent raptor
Yes
thick canyon
v(x)?
patent raptor
Now you differentiate both sides
patent raptor
yes
patent raptor
\( e^{y \ln \cos x} = e^{x \ln \cos y} \)
( e^{y \ln \cos x} \frac{d}{dx} \left ( y \ln \cos x \right ) = e^{x \ln \cos y} \frac{d}{dx} \left ( x \ln \cos y \right ) )
grizzled pagodaBOT
anti-algebraist 𝔸dωn𝓲²s
patent raptor
you need product + chain rule
patent raptor
it's correct
patent raptor
,texsp ( e^{y \ln \cos x} \cdot \left ( y' \ln \cos x - y \tan x \right ) = e^{x \ln \cos y} \cdot \left ( \ln \cos y - x \tan y \cdot y' \right ) )
grizzled pagodaBOT
anti-algebraist 𝔸dωn𝓲²s
patent raptor
What is the derivative of y
thick canyon
What is the derivative of y
y'
patent raptor
You forgot it
thick canyon
oh right
i cant just
RIGHT
damn it i didn't forget it
I didn't consider it to begin with
lemme redo
how bout this
@patent raptor
patent raptor
yes
patent raptor
**anti\-algebraist 𝔸dωn𝓲²s**
Group the terms, factor y' and solve for it
Bring on one side every thing that has y'
thick canyon
i get that part
its just that
it seems impossible...
thats why i came here to begin with
patent raptor
the algebra is simpler than the differentiation
thick canyon
?
patent raptor
[ e^{y \ln \cos x} \cdot \left ( y' \ln \cos x - y \tan x \right ) = e^{x \ln \cos y} \cdot \left ( \ln \cos y - x \tan y \cdot y' \right ) ]
[ a(y'b-c) = d(e-y'f) ]
How would you isolate y' below
grizzled pagodaBOT
anti-algebraist 𝔸dωn𝓲²s
thick canyon
but we will have y' in terms of x and y
patent raptor
but it's just the terms looking funny
thick canyon
not in terms of x only
patent raptor
but we will have y' in terms of x and y
yes
thick canyon
mhm
patent raptor
not in terms of x only
Well technically speaking y' = y'(x,y(x))
But thats to no surprise, if you implicitly differentiate thats very common
patent raptor
You can still compute the slope at a point (x,y)
thick canyon
sqrt(1+x²) divided by sqrt(1-x²)
patent raptor
Maybe there is some trig sub, not sure tho
thick canyon
k
patent raptor
,, \int \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}} : dx = \int \frac{\sqrt{1+\sin^2\theta}}{\sqrt{1-\sin^2\theta}} \cdot \cos \theta : d\theta
grizzled pagodaBOT
anti-algebraist 𝔸dωn𝓲²s
patent raptor
x = sinθ
thick canyon
try x = sec(theta)
patent raptor
,, \int \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}} : dx = \int \frac{\sqrt{1+\tan^2\theta}}{\sqrt{1-\tan^2\theta}} \cdot \sec^2\theta : d\theta
grizzled pagodaBOT
anti-algebraist 𝔸dωn𝓲²s
patent raptor
I think the denominator is a problem tho
thick canyon
yes
i think sin(theta) might just be the best
cuz
the sqrt(
cancels with cos(theta)
but how to integrate sqrt(1+sin²x) dx
patent raptor
,w Integrate[Sqrt[1+sin^2θ]]
grizzled pagodaBOT
patent raptor
Yea good luck buddy
patent raptor
,w Integrate[Sqrt[1+x^2]/Sqrt[1-x^2]]
grizzled pagodaBOT
thick canyon
mhm
patent raptor
😂
patent raptor
E is probably not what you looking for
patent raptor
Some non-elementary solution
thick canyon
definite function
damn
I tried to integrate
I mean
I tried to find the length of sine wave
so when I ended up with sqrt ( 1 + cos²x) dx
I used u = cosx
and I ended up where I am
amber waspBOT
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.
thick canyon
,w Integrate sin(cos(tan(secx))) dx
grizzled pagodaBOT
thick canyon
damn
patent raptor
Try out every function composition of trig
patent raptor
,w Integrate sin(cos(tan(secx))) dx
Thats what you tried lmao
thick canyon
ok
so uh
thx for ur help
the 12 graded sitting next to me had the cosx cosy problem
I wnated to see the answer
anyway
I got 2 more years for that I got bio to do today cuz i got exam tmrw
so uh bye
thx again
raw stirrup
amber waspBOT
thx again
If you are done with this channel, please mark your problem as solved by typing .close
@thick canyon Has your question been resolved?
amber waspBOT
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fierce edge
amber waspBOT
fierce edge
stupid question but <ACB could be less than 90 or greater than 90 right?
Idk why i just forgot my geoemtry lol
tulip ruin
more
fierce edge
correct
tulip ruin
I your case:
imagine C is infinitely close the centre of your circle
then <ACB should be infinitely close 180degrees. Because it is esentialy a line segment.
and if C is super far away, meaning a really tall triangle
then <ACB aproaches zero
fierce edge
s there an actual proof without heuristics
i worked on a proof rn
do you know something more elementary?
fierce edge
I your case:
imagine C is infinitely close the centre of your circle
then <ACB ...
also not sure 😭 if that's a justification unless i'm missing something
fierce edge
that arc angle of AB should've been 180
whoops
but yep the red angle is (90 + x)/2
and x > 0
but yeah that's my little proof, i'm sure there's a simpler justification
tulip ruin
hmmmm,
a wild idea
maybe you could go with 3:4:5 line legth ratios...
the hypotenuse is fixed so last is 5. and changing the sum of 3 and 4 makes it no longer a right triangle ,
sum is bigger angle smaller, else <ACB bigger
meaning you are inside or outside of circle.
i wonder if it coresponds to moving in a certain direction in the plot of all right triagles.
i saw in one 3b1b video xd
split sail
ye
fierce edge
ye
can u annotate your drawing
as in can you briefly describe what i'm supposed to be looking at
on your image?
fierce edge
fine
fierce edge
okay
amber waspBOT
@fierce edge Has your question been resolved?
violet blaze
hollow vessel
Who can solve this?
amber waspBOT
amber waspBOT
Available help channel!
hollow heart
how does this work?
amber waspBOT
how does this work?
marble thicket
is there somone who understand frensh
sterile swan
hollow heart
i mean i tried like
|a x b| = |a||b|sin(C) right
wait no that doesnt work
on god i have no clue
amber waspBOT
@hollow heart Has your question been resolved?
crude peak
welcome
hollow heart
.close
stoic solar
Hi, is there a way to show that $\frac{2 \sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}} < 2 \sqrt{k+1}
stoic solar
$\frac{2 \sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}} < 2 \sqrt{k+1}$
grizzled pagodaBOT
Minnie
patent raptor
Is k > 0?
stoic solar
yes
gloomy tide
i mean i hope so if you're taking its square root
stoic solar
My thoughts so far are like
patent raptor
I think I have one way
$0 < 2 \sqrt{k+1} - \frac{2 \sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}} \eqqcolon f(k)$
grizzled pagodaBOT
anti-algebraist 𝔸dωn𝓲²s
patent raptor
Try to show that f is strictly decreasing and that f(k) -> 0 for k -> inf
f(0) = 1 so it is bounded from above
stoic solar
Sorry for responding slow the very important call I had finally picked up
I'll be back asap
patent raptor
Maybe I made matters worse actually
stoic solar
Try to show that f is strictly decreasing and that f(k) -> 0 for k -> inf
Would there be an easier way to do this? Because this is part of an induction proof and I was hoping uhh
I could sort of "bind it"
patent raptor
Is k supposed to be an non-negative integer?
stoic solar
yes
patent raptor
ahh
stoic solar
k is a natural k >= 1
patent raptor
Ok induction might be the way indeed
stoic solar
grizzled pagodaBOT
hayley, who shakes the world
gloomy tide
isolate the sqrt and square both sides
grizzled pagodaBOT
hayley, who shakes the world
grizzled pagodaBOT
Minnie
gloomy tide
$\sqrt{k^2+k} > k$
i mean that's true but it doesn't help you much
gloomy tide
**hayley, who shakes the world**
we can square both sides of this to make it easier to prove
grizzled pagodaBOT
Minnie
gloomy tide
confused why that's true?
stoic solar
how do i get from the left side to the right side?
Because I know that $\sqrt{k^2+k} > k$ yet $1>\frac{1}{2}$
grizzled pagodaBOT
Minnie
gloomy tide
ok let's continue to prove it then
we want to show that $\sqrt{k^2+k} < k + \f12$, right?
grizzled pagodaBOT
hayley, who shakes the world
stoic solar
Well its more that I don't understand how to just come up with RHS given the left hand side
gloomy tide
i mean both the RHS and the LHS were derived from the original thing you were trying to prove
i took this
and multiplied both sides by sqrt(k+1)
stoic solar
mhm I know
gloomy tide
ok. so do you understand how we got to the point of having $\sqrt{k^2+k} < k + \f12$ as a goal?
grizzled pagodaBOT
hayley, who shakes the world
stoic solar
yes
stoic solar
Wait let me retype my question
gloomy tide
everything is positive so that won't change the truth value of the statement
stoic solar
cuz i feel like im confusing u
I'm doing an induction problem. Since it is prove by induction I already know $2\sqrt{k} + \frac{1}{\sqrt{k+1}} < 2 \sqrt{k+1}$ holds true.
grizzled pagodaBOT
Minnie
stoic solar
I just have a hard time understand how I actually show the LHS eventually gets to the RHS
gloomy tide
wdym "gets to"
stoic solar
I have to prove $p(k+1)$ holds true from $p(k)$ right?
grizzled pagodaBOT
Minnie
gloomy tide
yes
i can't even tell what p(k) is though
or what it is that you're trying to show
stoic solar
I broke this question into two parts, the upper and lower bound
I am looking at the upper bound right now
the 2 root n part
stoic solar
**Minnie**
I get to this
gloomy tide
you get to that? or you want to prove that that's true?
stoic solar
I want to prove that is true
gloomy tide
then don't say "i already know _______________ holds true"
if you don't know it's true
stoic solar
But the question suggests it has to be true
gloomy tide
you expect it to be true
stoic solar
that is a better wording yes
gloomy tide
we still need to prove it
stoic solar
yup
gloomy tide
which, i think, is what we've been trying to do
stoic solar
you are right
gloomy tide
and we saw that if we can just prove $\sqrt{k^2+k} < k + \f12$, then we will know the target inequality is true.
grizzled pagodaBOT
hayley, who shakes the world
stoic solar
You are right
Would i just work this out on my own
and then skip to the conclusion
gloomy tide
no, i can show you how to structure it
stoic solar
When I see professors / textbooks do examples of induction
they sort of just skip
so I assume they have a magical trick of just knowing
stoic solar
**hayley, who shakes the world**
this is true
gloomy tide
that's because they work stuff out on their own then write it upside down
stoic solar
oh.
gloomy tide
so as mentioned, squaring both sides will make this very tractable
stoic solar
So they just guess if its true, work it out and see that its true, then just write it in one go
gloomy tide
giving us $k^2 + k < k^2 + k + \f14$
grizzled pagodaBOT
hayley, who shakes the world
gloomy tide
which is quite clearly true
stoic solar
100%
but the general process is that
the profs write all this out on some rough paper
and just don't include it in the proof
gloomy tide
yes
stoic solar
and it seems like they just magically know?
gloomy tide
you should generally do the same
stoic solar
I see
gloomy tide
so here you would, say, start with 0 < 1/4
stoic solar
I think you have clarified my confusion
gloomy tide
and then build it up in the opposite order
until you get something where you can apply p(k)
stoic solar
I thought there was a method they could just jump
gloomy tide
and show that p(k) implies p(k+1)
amber waspBOT
robust isle
**Minnie**
maybe starting from here theres another way to get at the inequality
cuz like
,, 2\sqrt{n + 1} - 2\sqrt n = \f 2 {\sqrt {n +1} + \sqrt n}
grizzled pagodaBOT
stoic solar
interesting! This sort of looks like the type where you compare the denominator
I will try it as well thank you for your suggestion
amber waspBOT
Available help channel!
tough mica
amber waspBOT
tough mica
\textbf{2.-} Let $f: \mathbb{R}^4 \to \mathbb{R}^3$ be the linear transformation such that
[
M_{EB}(f) =
\begin{pmatrix}
1 & 1 & -2 & a \
1 & a & -5 & 1+a \
1 & 1 & a & 0
\end{pmatrix},
]
where $B = {(1,-1,1), (0,1,-2), (1,0,0)}$ is a basis of $\mathbb{R}^3$. Find all values of $a \in \mathbb{R}$ for which $\dim(\ker(f)) = 2$, and for each of these values, provide a basis of $\operatorname{Im}(f)$.
worthy remnant
one online teacher told me that the third is irrelevant
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
fast badger
one online teacher told me that the third is irrelevant
!occupied
amber waspBOT
one online teacher told me that the third is irrelevant
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
tough mica
dim(Ker(f)) + dim(Im(f)) = dim(codomain)
dim(ker(f))=2
2 + dim(Im(f)) = dim(R^3)
2 = dim(Im(f)) = 3
dim(Im(f))=1
cunning birch
dim(Ker(f)) + dim(Im(f)) = dim(codomain)
dim(ker(f))=2
2 + dim(Im(f)) = dim(R^3)...
not codomain
tough mica
rank nullity theoerem
formal harbor
it's the dimension of the domain, not the codomain, i thought
cunning birch
cunning birch
rank nullity theoerem
that's what the rank nullity theorem says
tough mica
oh my
tough mica
it's the dimension of the domain, not the codomain, i thought
my baad
tough mica
2 + dim(Im) = dim(R^4)
dim(Im) = 2
f(e1) = (1,1,1)_B
f(e2) = (1,a,1)_B
f(e3) = (-2,-5,a)_B
f(e4) = (a, 1 + a, 0)_B
tough mica
that's what the rank nullity theorem says
can I get some hints
we can find the image
problem is
it will be in terms of a
how to find the image? find the image of f under canonical basis
the basis of those images is the image
tough mica
it will be in terms of a
^
(1,1,1)_B= (1,-1,1)+(0,1,-2)+(1,0,0)
(1,1,1)_B = (2,0,-1)
(1,a,1)_B=(1,-1,1)+(0,a,-2a)+(1,0,0)
(1,a,1)_B = (2,a-1,1-2a)
(-2,-5,a)_B= -2(1,-1,1)-5(0,1,-2)+(a,0,0)
(-2,-5,a)_B= (-2,2,-2)+(0,-5,10)+(a,0,0)
(-2,-5,a)_B= (a-2,2-5,10-2)
(-2,-5,a)_B= (a-2,-3,8)
cunning birch
(1,a,1)_B=(1,-1,1)+(0,a,-2a)+(1,0,0)
missed a -2a
tough mica
?
tough mica
missed a -2a
can u elaborate
tough mica
missed a -2a
ohh i see it now
tough mica
missed a -2a
(1,a,1)_B = (2,a-1,1-2a)
(a,1+a,0)_B= (a,-a,a) + (0,1+a, -2-2a)
(a,1+a,0)_B= (a,1+a-a,a-2-2a)
(a,1+a,0)_B= (a,1,-2-a)
f(e1) = (1,1,1)_B
f(e2) = (1,a,1)_B
f(e3) = (-2,-5,a)_B
f(e4) = (a, 1 + a, 0)_B
(1,1,1)_B = (2,0,-1)
(1,a,1)_B = (2,a-1,1-2a)
(-2,-5,a)_B= (a-2,-3,8)
(a,1+a,0)_B= (a,1,-2-a)
f(e1) = (1,1,1)_B = (2,0,-1)
f(e2) = (1,a,1)_B = (2,a-1,1-2a)
f(e3) = (-2,-5,a)_B= (a-2,-3,8)
f(e4) = (a,1+a,0)_B= (a,1,-2-a)
Im(f) = <(2,0,-1),(2,a-1,1-2a),(a-2,-3,8),(a,1,-2-a)>
tough mica
missed a -2a
dim(Im(f))=2
tough mica
Im(f) = <(2,0,-1),(2,a-1,1-2a),(a-2,-3,8),(a,1,-2-a)>
now the question gets reduced to, reduce this basis of Im(f) to a two vector basis
tough mica
missed a -2a
we can use det = 0
cunning birch
we can use det = 0
that will be so cumbersome
grizzled pagodaBOT
cunning birch
only one det
which one?
tough mica
,w det {{2,a-1,1-2a},{a-2,-3,8},{a,1,-2-a}} = 0
,w det {{2,a-1,1-2a},{a-2,-3,8},{a,1,-2-a}} = 0
?????
what happened with WA
tough mica
,w det {{2,a-1,1-2a},{a-2,-3,8},{a,1,-2-a}} = 0
grizzled pagodaBOT
tough mica
Im(f) = <(2,0,-1),(2,a-1,1-2a),(a-2,-3,8),(a,1,-2-a)>
if I plug a = 1 I get this REF
tough mica
which one?
this is two pivots
dim(Im(f)) = 2 when a = 1?
,w rref {{2,a-1,1-2a},{a-2,-3,8},{a,1,-2-a}}^T when a = 1
tough mica
the matrix I REFed was this
grizzled pagodaBOT
tough mica
Im(f) = <(2,0,-1),(2,a-1,1-2a),(a-2,-3,8),(a,1,-2-a)>
basically this as columns and plugging a = 1
tough mica
which one?
sir
Im(f) = <(2,0,-1),(2,a-1,1-2a),(a-2,-3,8),(a,1,-2-a)>
when a = 1
Im(f) = <(2,0,-1),(2,1-1,1-2),(1-2,-3,8),(1,1,-2-1)>
,w ref {{2,0,-1},{2,1-1,1-2},{1-2,-3,8},{1,1,-2-1}}^T
grizzled pagodaBOT
tough mica
,w rref {{2,0,-1},{2,1-1,1-2},{1-2,-3,8},{1,1,-2-1}}^T
grizzled pagodaBOT
amber waspBOT
@tough mica Has your question been resolved?
amber waspBOT
@tough mica Has your question been resolved?
amber waspBOT
@tough mica Has your question been resolved?
tough mica
Im(f) = <(2,0,-1),(2,a-1,1-2a),(a-2,-3,8),(a,1,-2-a)>
when a = 1
Im(f) = <(2,0,-1),(2,1-1,1-2),(1-2,-3,8),(1,1,-2-1)>
,w rank {{2,0,-1},{2,1-1,1-2},{1-2,-3,8},{1,1,-2-1}}^T
grizzled pagodaBOT
tough mica
,w rref {{2,0,-1},{2,1-1,1-2},{1-2,-3,8},{1,1,-2-1}}^T
grizzled pagodaBOT
tough mica
.solved
paper ibex
For part b, ik that it's related to part a but I couldn't figure out how to make use of the ans I got in part a so I just ignored it and went the hard way ig...could someone explain how I'd relate the 2 parts
paper ibex
Also for this question, part d, I got a as 1/4 but the ans is 4, I'm confused because a is a vertical stretch so shouldn't it originally be 1/4 but when we work with it we use the reciprocal such that it becomes 4?
mint sparrow
yall better help this person im interested too damn...
amber waspBOT
@paper ibex Has your question been resolved?
cursive jay
For part b, ik that it's related to part a but I couldn't figure out how to make...
For this one, you should do a substitution and let y = x + 20 degrees
See if that helps
mint sparrow
i think you could try saying
sinx = 2sqrt3 cosx
and then substitute???
So then when u expand in part b then you could end up with all cos?
i dunno
cursive jay
No no you're overthinking it
cursive jay
Also for this question, part d, I got a as 1/4 but the ans is 4, I'm confused be...
For this one you should just work through the algebra carefully. You've gone backwards
paper ibex
But I just don't get it....I watched a video abt it and it didn't make sense to me
cursive jay
Okay so let y = x + 20
Or x = y - 20, it's all equivalent
Then substitute y into that equation instead of x
What do you get?
paper ibex
For this one you should just work through the algebra carefully. You've gone bac...
If A was 1/4 originally wouldn't we have to take the reciprocal of it? Wouldn't it then become 4 and when we multiply by 5 we get 20 which is our new y coordinate for the vertex?
I understood it this way
cursive jay
I'll be honest I don't really understand where you're coming from lol sorry
But I do think you're overthinking that one too
if f(x) = 5 and af(x) = 20, solve for a
sorry no no
Solve for a
That's what it's asking
paper ibex
Or x = y - 20, it's all equivalent
Let me try this 😭😭 I'm actually very sleepy rn it's like 2 am. I'm gonna attempt it tomorrow again cuz I can't focus. Is it possible to keep this channel open or will it automatically close 😭😭
cursive jay
No it'll close in like 15 mins but you can reopen one
paper ibex
That's what it's asking
Check my working in the pic I sent, its at the bottom
paper ibex
No it'll close in like 15 mins but you can reopen one
Ah damn
mint sparrow
For this one, you should do a substitution and let y = x + 20 degrees
Oh my god i just understood this Thats
Wtf
I could have never imagined that
cursive jay
What's that 3rd symbol? Sorry I'm also pretty tired haha
I'm fairly certain that you've gone backwards though
cursive jay
if f(x) = 5 and af(x) = 20, solve for a
Do you see where this came from?
paper ibex
What's that 3rd symbol? Sorry I'm also pretty tired haha
5*x=20 so x=4
cursive jay
Oh that's multiplication in the middle okay
paper ibex
I'm fairly certain that you've gone backwards though
Wait let me explain to u whats going in my mind
So for stretch
We have vertical and horizontal right
paper ibex
A horizontal would be smthing like f(7x), so our factor is 7. In vertical though it would be f7(x) but our factor is going to be the reciprocal of that and so will become 1/7
cursive jay
Ah I think I see
paper ibex
Now here if the question was given to us but reversed, asking us for example to find the new coordinate of the vertex
cursive jay
This is just one of those things you need to think over in your head
f(7x) is f(x) but squished by a factor of 7 horizontally
7f(x) is f(x) but stretched by a factor of 7 vertically
paper ibex
f(7x) is f(x) but squished by a factor of 7 horizontally
How's this squished, wouldn't we multiply the 7 with the x coordinate and so we get a larger value so it spreads
cursive jay
It's kinda hard to explain in words
But imagine our function looks like this
Now let's try to compute f(7x) at this point
First we multiply x by 7
So f(7x) = f(z), where z is this point here (approximately)
Then we plot that point back at x=x
So in other words it gets squished in by a factor of 7
Does that make any more sense?
paper ibex
I actually don't get it, I'm so sorry😭😭😭😭 maybe cuz it's too late but I'm actually trying to understand
