fading narwhal
sweet momma damn thank god im done with this
#help-38
grim crown
based on the svd, how do i compute low rank approximation
wraith hinge
Hi! What do you mean how?
To obtain a low-rank approximation, keep only the k largest singular values in Σ, and set the rest to zero, where k is the desired rank.
Multiply the modified U, modified Σ, and V^T matrices back together to get the low-rank approximation. The rank of the approximation will be equal to the value of k.
trim joltBOT
@grim crown Has your question been resolved?
grim crown
grim crown
Hi! What do you mean how?
like this?
trim joltBOT
@grim crown Has your question been resolved?
wise pecan
The author plans to charge $8 per book. About how many should be printed to make a profit?
wise pecan
i dont even understand the question
wait y value is cost
but if the author prints one book
,calc 120+4/1
solid kilnBOT
Result:
124
Result:
192
,calc 120+96
solid kilnBOT
Result:
216
,calc 216/24
solid kilnBOT
Result:
9
wait im confused
solid kilnBOT
Result:
30
ivory gull
The line y = 8 intersects somewhere around 30
wise pecan
The line y = 8 intersects somewhere around 30
but how do we know he is making a profit
there
ivory gull
Anything above 30 he makes a profit
wise pecan
Anything above 30 he makes a profit
how do u know?
ivory gull
Cause he’s charging 8 per and the cost per book decreases below $8 for every book after 30
wise pecan
Cause he’s charging 8 per and the cost per book decreases below $8 for every boo...
im not understand this at all
ivory gull
So the intersection point is (30,8)
That’s the break even point
If he was selling them for $8 each
wise pecan
So the intersection point is (30,8)
how do u know that?
ivory gull
,calc 120/4
^
ivory gull
What unit is it
In real world terms, the author wants to sell the book for $8 each, so they would need the book to cost $8 or less per book to turn a profit
wise pecan
In real world terms, the author wants to sell the book for $8 each, so they woul...
but how can someone by looking at the graph see that $8 makes a profit
ivory gull
Oh wait
Forgot about the price of the book
But starting from the beginning, the function is decreasing
wise pecan
yea
ivory gull
And the author wants to sell for $8 a piece
wise pecan
mhm
ivory gull
So they cost needs to be at least $8 or lower to make a profit
wise pecan
So they cost needs to be at least $8 or lower to make a profit
but why tho?
if they charged $16 for a book
wouldnt that make them more money?
ivory gull
It’s a math problem they could i guess but that’s not the point
Y axis means the cost for the author
While $8 is the price they want to sell at
wise pecan
Y axis means the cost for the author
but how do we know that the cost needs to be $8 or less than 8?
ivory gull
Because that’s what’s needed to make money if the author wants to sell for $8
(30,8) isn’t the break even point but it is the point where they start making money on each book they sell
I was wrong about that before that’s my bad
wise pecan
(30,8) isn’t the break even point but it is the point where they start making mo...
what if it was 16? do i do 16x=120+4(16)/16?
ivory gull
You could do 16
But the questions asking for the point to start making a profit
So at (30,8) total cost would’ve been $240
wise pecan
So at (30,8) total cost would’ve been $240
so they lost 240
wise pecan
But the questions asking for what point they make a profit
so when do they make a profit looking at the graph
wise pecan
This could be solved with calculus
thats wht they are trying to do
alg 2 which is build up to calc
but how do i solve it with basic alg 2?
wait i think i got it
YEA
wait
let me tell u
lets say the guy chrges 16 for the book
and when does he make a profit
u do
16x=120+4(16)/16
whatever u get is how many books
he needs to sell to break even
right?
,calc 4(16)
solid kilnBOT
Result:
64
,calc 120+64
solid kilnBOT
Result:
184
,calc 184/16
solid kilnBOT
Result:
11.5
wait
ivory gull
Kind of, the question want to know how many books are required not the price
wise pecan
Kind of, the question want to know how many books are required not the price
yea that question i gave u was
The author plans to charge $8 per book. About how many should be printed to make a profit?
so we do
8x=120+4(8)/8
and solve
wait not
@ivory gull how did u do it?
ivory gull
With calculus
ivory gull
Hold on I’m thinking of an algebra way to solve it
wise pecan
shouldnt it be 8x?
wise pecan
@ivory gull ?
wise pecan
how
ivory gull
Yeah I misenterpreted the question
So just find the point where the cost is $8
Everything past that point is profit
ivory gull
So 120 + 4x / x = 8
wise pecan
yea
so my method works, right?
lets say 16 books
sorry 16 is the cost
so we do
16=?
ivory gull
But 16 isn’t what they’re asking for they’re asking for 8
wise pecan
But 16 isn’t what they’re asking for they’re asking for 8
yea ik
ivory gull
Linear algebra
wise pecan
wait so does he need to sell 30
books
or more then 30
to make a profit
just 30, right?
wise pecan
Linear algebra
ur in college?
ivory gull
No every book after 30 is a profit
wise pecan
No every book after 30 is a profit
wait what?
how?
wait so once we solve 8x=120+4x
what is the answer
in
30 what?
30 books sold?
ivory gull
Because 30 is when the cost to make a book equals the price the author is selling it at
So everything after 30 is when they make a profit
X > 30
wise pecan
wait what?
wise pecan
Because 30 is when the cost to make a book equals the price the author is sellin...
so thats true for every prices answer?
ivory gull
No for every pride you would just plug it in differently so instead of c(x) = 8, you would switch out 8 for any number
When you plug it in you’re really looking for the intersection
Because that’s the point that you start to gain money per book
wise pecan
No for every pride you would just plug it in differently so instead of c(x) = 8,...
but why is it after that number u make a profit
wait
so if i sell 31 books
ivory gull
Because at the number your profits are 0
wise pecan
120+4(31)
ivory gull
It’s a break even point
wise pecan
,calc 120+4(31)
solid kilnBOT
Result:
244
,calc 244/31
solid kilnBOT
Result:
7.8709677419355
thats lower than 8
ivory gull
Yeah so 31 makes profit
wise pecan
how
ivory gull
But 30 is exactly equal to 8 so if it was sold for 8 profit would be 0
8-7.87 is the profit
Everything before 30 is negative profit
Everything after 30 is positive profit
30 exactly is 0 profit
wise pecan
But 30 is exactly equal to 8 so if it was sold for 8 profit would be 0
ohh
ok
thx bro
@ivory gull man ur in college?
wise pecan
Nah high school
how u taking lin alg in high?
ivory gull
No problem
quartz plume
are these equivalent
trim joltBOT
are these equivalent
quartz plume
are these equivalent
this means that the input of g^-1 is the inverse function of f
left oriole
this is assuming the inverses exist
quartz plume
then the input of h^-1 is the input of the composition of the inverse of g of f
and yes
this is assuming that they exist
left oriole
the input to g^-1 is the value of f^-1(x), which is the number y such that f(y) = x
trim joltBOT
trim joltBOT
Available help channel!
ripe valley
okay so new question
trim joltBOT
okay so new question
ripe valley
$5 {\pi}^4 x^4 + 2 {\pi}^4 x^3 - 30 {\pi}^2 x^2 - 12 {\pi}^2 x + 48 = 0$
solid kilnBOT
ItzKraken
ripe valley
whats the best way of solving an equation like this
ripe valley
this is not related to the previous problem btw
tiny valve
Newton-Raphson method
ripe valley
Newton-Raphson method
u cant be serious
tiny valve
I'm 100% serious
ripe valley
pi²x²(5pi²x²+2pi²x) - 6(5pi²x²+2pi²x) + 48 = 0
oo I didnt notice that
raw magnet
oo I didnt notice that
How 💀
vocal night
Newton-Raphson method
Lagrange method is better
ripe valley
substitution is still terrible imo
vocal night
Lagrange method is better
Up until 6th degree you can get exact results
ripe valley
quad formula will become messy I am looking for a clean solution
ripe valley
Lagrange method is better
I havent heard of this one
prisma drift
idk probably starts from there
vague citrus
pi²x²(5pi²x²+2pi²x) - 6(5pi²x²+2pi²x) + 48 = 0
good luck on solving that one
(nested radicals are fun)
ripe valley
5pi²x²(x²-6) + 2pi²x(x²-6)+48 = 0
still sorta terrible
any form of substitution is too tough
prisma drift
idk then
ripe valley
Up until 6th degree you can get exact results
Can u explain? google shows Lagrange multipliers so Idk what u talking about
vague citrus
,w roots of 5*(pi^4)(x^4)+2*(pi^4)(x^3)-30*(pi^2)(x^2)-12*(pi^2)(x)+48=0
solid kilnBOT
vocal night
Can u explain? google shows Lagrange multipliers so Idk what u talking about
Mb it was for the ones without the power of three term
ripe valley
Mb it was for the ones without the power of three term
kk
vocal night
The general method is called ferrari method
ripe valley
ferrari 
vague citrus
yeah the roots are pretty nice ones. (defo not irratonal)
vocal night
ferrari <:opencry:586078614865444865>
vagrant prism
yeah the roots are pretty nice ones. (defo not irratonal)
0.69.. is very nice indeed
vague citrus
that is a quartic without a x^3 term
ripe valley
yeah I just saw that too
ripe valley
Btw iirc theres a quartic formula too who wants to try
vocal night
That was in my take away exam
ripe valley
this?
raw magnet
This is so fucked
woven trail
raw magnet
No one needs to know that
vague citrus
exact form
tiny valve
Use bisection method
vocal night
yk what I will just say 'I cOmPUTed iT wHAT WiiLL U dO bOuT it'
It's an hw question lol ?
tiny valve
forget about these complicated methods
vague citrus
don't tell me you are asked this on a no calculator exam
ripe valley
It's an hw question lol ?
nah me doing something else
ripe valley
don't tell me you are asked this on a no calculator exam
U expect me to do that?
vague citrus
nah me doing something else
bro cooking meth
wraith hinge
Would solving this equal to x = 13/9
trim joltBOT
Would solving this equal to x = 13/9
edgy cedar
,w 3x-3 = 4/3
solid kilnBOT
edgy cedar
yea
sudden sentinel
What would a good answer for a) be?
And how would I visualize/interpret b and c?
sudden sentinel
for a) i know that 2w1 = w2
so would I say, "every vector w that is perpendicular to v has a y component that is double the x component and the dot product between w and v is equal to 0"
For b) could I say that all vectors lie on a plane?
granite cove
for a) i know that 2w1 = w2
so would I say, "every vector w that is perpendicula...
You're right, but one could write that down a bit more formally
sudden sentinel
Could you tell me how?
granite cove
Like w = r * (1,2) for an arbitary r
sudden sentinel
Oh wow, alright thanks
Do you know how to do C? I could visualize B but can't do C
granite cove
Are you a bit fimiliar with orthogonal complement and dimensions in linear algebra?
sudden sentinel
😅 Not really, no
I'm still learning the basics
granite cove
Are you fimiliar with dimensions though? Or do I need to rely on school knowledge?
sudden sentinel
I guess not
granite cove
Ok, then it is hard to explain it rigorously without that knowledge. Say the first vector is a the second is b. Since a and b are not parallel, we can say that if c is orthogonal to a,b then it is orthogonal on the plane formed by a and b
Do you know any vectors that are orthogonal on a plane?
sudden sentinel
like any example of a vector that's orthogonal on a plane?
granite cove
like any example of a vector that's orthogonal on a plane?
Do you know about normal vectors?
sudden sentinel
yes
sudden sentinel
ahhh gotcha
granite cove
Can you relate that a bit?
sudden sentinel
So it's gotta be a line?
granite cove
So it's gotta be a line?
Exactly!
granite cove
.close
trim joltBOT
granite cove
thank you so much!
You're welcome
trim joltBOT
Available help channel!
lament stone
trim joltBOT
lament stone
Where do I start?
trim joltBOT
@lament stone Has your question been resolved?
summer matrix
I need to solve the inequality for x and draw interval:
x−|x| < |x+5|−13
I rearanged:
x - |x| - |x+5| < -13
then I looked at each one:
|x + 5| < 13 when x < - 18
|x| < -13 when x < -13
same for the x.
The problem is that they are in dependent of each other so I don't know what to do now from this point on.
summer matrix
so I failed where I rearanged?
wise cipher
you almost got it right actually
U just didn't considered the effects of absolute value on the equation
let's say x = 4
Ah shet
Ok ok
summer matrix
okay so how would I continue from where I a, now ?
wise cipher
I see now
So what I said earlier is only true if x is positive
when it is negative, then it would be
-2x < abs(-x + 5) - 13
or -2x < abs (-(x - 5)) - 13
summer matrix
but you didn't rearange like I did
wise cipher
You need to figure out first what would happen for values of positive x and negative x
Then when you know it already you can rearrange it sorry mb I was in another channel
Ok im back
Let's go back to the start
summer matrix
agreed
wise cipher
Let's assume that x is positive first
so it would just
x - x < x + 5 - 13
or
8 < x
When x is positive
see?
summer matrix
I see
wise cipher
Now lets look what happens when it is negative
-x - x < abs (-x + 5) - 13
obviously the left side will be -2x
but the right side is a bit strange so lets focus on that
instead of thinking it as -x + 5, for me genrally I just think of it like this
abs (-(x-5))
summer matrix
okay
wise cipher
And ask the question, wehn will x-5 > 0 and < 0
wise cipher
Since its absolute value
summer matrix
oh okay, by definition then
wise cipher
Ok ok
Wait lemme explain
The problem here is
The abs(-x+5) would actually generate 2 scenarios here
if x = 3, then
abs (-2)
if x = 4, then abs(-1)
BUT when x = 10 then
abs (5)
or when x = 9, then
abs (4)
and we know that abs(-4) = abs(4)
so this generates 2 scenarios
when -x + 5 >0 and <0
get it?
@summer matrix
summer matrix
2
wise cipher
no its -2
summer matrix
what
wise cipher
yeah but then you have this absolute value
and we have another scenario where we can also get 2
when x = 7
cause -7 + 5 = -2
ahh wait my bad
wtf
nah I got the first wrong
yeah at first you're correct
What I meant to put the -7 + 5 here
which is - 2
yah yah mb
wise cipher
what I want to say is we can have 2 scenarios that have the same answer
That is what I'm trying to explain
when x is negative
cause you can have -2 and 2 but since its absolute value
they're the same
so abs(-7 + 5) = abs (-3 + 5)
that is why we need to figure out what happens when
-x + 5 > 0 and < 0
summer matrix
okay
wise cipher
What I usually do is do something like this
(-(x-5))
cause this is easier to work with
its the same thing as -x + 5 = -(x-5)
summer matrix
why -x + 5 and not x + 5 ?
summer matrix
|x+5| is what we have
wise cipher
it means -1 x ( x - 5)
Yes
But where done when x is positive
So that equation doesn't really work when dealing when x is a negative
that is why earlier when x is postive we just did this
x - x < x + 5 - 13
or 8 < x when simlified
but when x is a negative
the abs (x+5) changes
that is why we change all x with -x
so
-x - abs(-x) < abs(-x + 5) - 13
or - x - x < abs(-x + 5) - 13
which will be simplified as
-2x < abs(-x + 5) - 13
Then we need to figure out how -x + 5 is affected by the absolute value
Here you can try your method or mine but this is what I usually do
abs (-x+5) = abs (-(x-5))
x - 5 >= 0 when x >= 5
x - 5 < 0 when x <5 and x >= 0
we can't have x going to the negative side on this part
since we're accounting when the inside is abs(-x+5)
so let's check for the solution when x >= 5
-2x < abs (-(x-5)) - 13
-2x < x - 5 - 13
summer matrix
doesn't multiplying by negative change the symbol direction?
wise cipher
doesn't multiplying by negative change the symbol direction?
ah no, that is when you multiply one side into a negative
Here we used the distributive property
But its still the same thing
summer matrix
how do you get rid of absolute ?
summer matrix
sorry I am at part where we |-(x-5)| > 0 and |-(x-5)| < 0
wise cipher
ah no erase the absolutes there
summer matrix
they are not needed ?
wise cipher
Its already accounted for
We already have a solution for x when it is positive
now we're only asking when it is negative
And since the right side has abs(-x+5)
it will have two different scenarios
when x>5, the answer in -x + 5 is negative right?
but when x >= 0 and x < 5, then -x + 5 is positive
so you have two outcomes when x is negative
the value inside the absolute symbol either becomes negative or positive
And that is why we have these two
wise cipher
sorry I am at part where we |-(x-5)| > 0 and |-(x-5)| < 0
these two
accounting for either cases
In which case you don't need to the absolute sigs
The only reason we multiply -1 on both sides is because of property of equations
yes go ahead
ah ok
summer matrix
I understand the above yeah
but just to sum up to this point
we first took a look at inequality
then see two cases
x > 0
y < 0
wise cipher
ah no
almost
the only reason we're only looking for different cases
is because we have an absolute value
if this was a normal inequality
then just do basic algebra
the absolute value gives two cases
when x is negative and when it is positive
summer matrix
yeah okay
wise cipher
and then we this other case when x is negative
wise cipher
because we have abs(-x+5)
wise cipher
Can you help me with something easy, it takes like 2 mins
you need another channel
minor thistle
you need another channel
Which
summer matrix
this is so different from what was shown to us to solve such cases, well inequality cases
wise cipher
ah because I think they only shown you the simple ones
you also need to learn about property of equations here and fully understand how it works
since I think you have a misconception earlier that we should multiply the other side by -1
You only do that for cases like this
-16 < - x
when you multiply -x with -1, that -1 was not a part of the equation
summer matrix
no what I am lost is from the third step or something
wise cipher
ah ok ok
summer matrix
- we take alook at inequality
wise cipher
just reply it
summer matrix
- because we have absolute equation on left and right side
we need to compare that and take a look what happens >= 0 and < 0
wise cipher
yeah correct
summer matrix
so we come to first case
wise cipher
it's because of the absolute that we need to account what happens when x is postive and negative
summer matrix
(-(x-5)) >= 0
wise cipher
You can actually also use
-x + 5
summer matrix
and here is where I am not sure what you did
-x >= -5
and
x <= 5 if we multiply it by -1
wise cipher
it's just that -x + 5 = - (x-5)
they're basically the samething
I only use this since I'm more comfortable comparing x-5
since we don't actually care whether it the inside of (x-5) becomes a negative or positive
It's absolute value so it's always going to be positive
the problem is the value it will spit out
wise cipher
Yeah you that is correct
wise cipher
yes
summer matrix
so we have x <= 5 and x > 5
wise cipher
so far so good
you also need to include
you're actually pretty close
let's have this one
wise cipher
-x >= -5
and
x <= 5 if we multiply it by -1
this came rom -x + 5 >= 0 right?
so it means that -x + 5 is positive
summer matrix
yeah and -x + 5 < 0
wise cipher
so this part is
wise cipher
okay, then we take a look at -x + 5 < 0
this part is when -x + 5 will become negative
summer matrix
-x + 5 is positive exactly when x > <= 5 right ?
wise cipher
yeah
summer matrix
so this is all the cases we have for |x+5|
wise cipher
but remember that it should end at 0
since if we substitute
-(-4) + 5
then this would not be accounting for the original inequality
since everything that we done so far is WHEN x is negative
so make sure that it would have this format
-x + 5 when you simplify it
wise cipher
-x + 5 is negative exactly when x > 5
so this part is (ooops this the wrong reply)
wise cipher
-x + 5 is positive exactly when x > <= 5 right ?
when x <= 5 and x > 0
but you got the general idea
wise cipher
-x + 5 is positive exactly when x <= 5
and negative when x > 5 ?
here we have this inequality when x > 5
We're not talking about the -x +5 any more since I believe you already get that part
summer matrix
whywe can't have x as negative?
wise cipher
Because we are already accounting for it on this part
If we actually did a negative x
then This would turn something like this
-2x < abs(-(-x) +5) - 13
or
-2x < abs(x +5) - 13
which is not our original inequality
it should be
-2x < abs (-x + 5)
that's the reason we can't have x as negative values when we are already testing it when it becomes negative
wise cipher
I mean earlier you got it already
just figure out what happens when
-x + 5 > 0 and < 0
you already got that righ
summer matrix
we did that
wise cipher
so let's have -x + 5 when x > 5
-2x < abs(-x + 5) - 13 when x > 5
we know that -x + 5 = -(x-5)
like earlier when -4 + 5 = - (4-5)
or when -10 +5 = - (10 - 5)
so we have
-2x < abs(-(x-5)) - 13
or -2x < x - 5 - 13
and then we will have
18 < 3x
or 6 < x when x > 5
let's have when x < 5
-2x < abs (-x + 5) - 13
or -2x < 5 - x - 13
8 < x when x < 5
so we have this when x is postive
lemme find it
ok lemme just solve it again
0 < x + 5 - 13
8 < x when x is postive
and when it is negative
wise cipher
or 6 < x when x > 5
we have this
summer matrix
so let's have -x + 5 when x > 5
all we do from this point on is for negative x
wise cipher
yeah
since that is the tricky part
you have two scenarios for -x here
THat would be easy if its just abs(-x)
summer matrix
take me again through how it's with x > 0
x - |x| < | x + 5 | - 13
x - x -x < 5 - 13
-x < -8
x > 8
right ?
wise cipher
=yeah
summer matrix
when x is negative we have
x - abs(-(-x)) < abs(-(x - 5)) - 13
correct
correct me if I am wrong at above syntax
not sure about abs and -
wise cipher
when we have negatives
its when -x + 5 > 0 and < 0
cause if -x + 5 turns into 2, its because -x itself is -3
but when -x + 5 turns into -2, its because -x itself is -7
summer matrix
x - abs(-(-x)) < abs(-(x - 5)) - 13
is this correct ?
wise cipher
ok so what is -3 + 5
summer matrix
2
wise cipher
what is -7 + 5
summer matrix
-2
wise cipher
the problem is they're the same under absolute value
wise cipher
x - abs(-(-x)) < abs(-(x - 5)) - 13
you got the syntaxin wrong here for abs(-(-x))
but doesn't really matter since
summer matrix
thought so
wise cipher
that would just be x
wise cipher
x - abs(-(-x)) < abs(-(x - 5)) - 13
and the first x
is wrong
it should be -x - abs
summer matrix
let me retype
wise cipher
since you're replacing all x with -x
wise cipher
yeah
wise cipher
the problem is the -(x-5) part
since -2 and 2 are equal under absolute value
so figure out when -(x-5) is going negative
and when it is going positive
wise cipher
they became the same thing when put under the absolute value
summer matrix
ohhh
wise cipher
abs(-2) = abs (2)
summer matrix
that is just an example you mean
so we are here:
-2x < abs(-(x-5)) - 13
now we figured out that right now we only care what is happening for negative x.
furthermore we see:
-(x-5) >= 0
when
x < 5
and
-(x-5) < 0
when
x > 5
correct so far ?
wise cipher
yeah
wise cipher
so for when x > 5 -> infinity it just get bigger right?
since this is absolute value
But when x < 5 from 0 to 5 gets bigger, the answer becomes smaller
summer matrix
wait
summer matrix
when x > 5 then -(x-5) < 0
summer matrix
lets go slow
summer matrix
when x > 5 then -(x-5) < 0
this
you saying than when x is bigger than 5 and goes towards infinity x gets bigger
wise cipher
no the answer for
summer matrix
but we concluded earlier that when x grows from 5 -(x-5) < 0
wise cipher
abs(-(x-5))
not the x itself
yeah it gets bigger
but what I'm asking is what happens with abs(-(x-5))
summer matrix
that is what we figured out earlier no ?
-(x-5) >= 0
when
x < 5
and
-(x-5) < 0
when
x > 5
wise cipher
no, we're only finding the domain
summer matrix
when x is less than 5 it's bigger or equal to 0
wise cipher
no, we're trying to find what happens with the range
summer matrix
when x is more than 5 it's negative
wise cipher
with respect with the domain
wise cipher
when x is more than 5 it's negative
yeah how about the value
wise cipher
what is abs(-(1-5))
answer these questions first and you will see
summer matrix
yeah how about the value
value being ?
wise cipher
what is abs(-(1-5))
It's here
summer matrix
value is negative
wise cipher
|1-5| = 4
|2-5| = 3
|3-5| = 2
|4-5| = 1
|5-5| = 0
|6 - 5| = 1
|7-5| = 2
|8 - 5| = 3
|9 - 5| = 4
Do you see a pattern
summer matrix
it's growing ?
summer matrix
but this is only when x > 0
we are talking about x < 0
no ?
I am sorry, I just wont be able to understand this.
Thank you for the effort I will give channel to someone else.
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placid valve
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placid valve
having trouble understand this, why would k/2 +2 = 0 lead to 2 basic eigenvectors?
there are no leading 1's in any of the rows, doesn't that mean that there are 4 parameters and therefore 4 basic eigenvectors?
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placid valve
<@&286206848099549185>
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astral mortar
I have two functions f,g from R to R satisfying
- f(0)>0
- g(0)=0
- |f’(x)| <= |g(x)| for all x
- |g’(x)| <= |f(x)| for all x
- f(r) = 0
What’s the smallest positive real number r such that this is true for some function f,g
astral mortar
Anyone knows how to start this?
marble wharf
there are two obvious candidate functions for f and g. maybe you can manipulate them somehow to get other solutions?
do you actually mean smallest real number or do you mean smallest positive real number
astral mortar
Oh yeah the smallest positive real number sorry
astral mortar
there are two obvious candidate functions for f and g. maybe you can manipulate ...
You mean coax and sinx right
marble wharf
yes
astral mortar
That’s what I was thinking as well but not sure how to prove that’s pi/2 is the smallest
marble wharf
I have no idea either rn. gotta go, good luck
astral mortar
Alright thanks
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astral mortar
I have two functions f,g from R to R satisfying
- f(0)>0
- g(0)=0
- |f’(x)| <= |g(x)| for all x
- |g’(x)| <= |f(x)| for all x
- f(r) = 0
What’s the smallest positive real number r such that this is true for some function f,g
sin and cos clearly satisfy this and r = pi/2 works but not sure if you can find a lower r
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novel jay
Let f be the function defined by f(x) = 2x^2+3x. What is the average value of f on the interval [0,2] written in simplest form?
novel jay
how should i solve this problem?
i got 7 which is the incorrect answer
basically i used the average value formula
then simplified it to the integral from 0 to 2 of x^2+(3x/2)dx
then just did f(b)-f(a)
dire kraken
I think you forgot to integrate x^2 + (3x/2)
f(b) - f(a) only comes after integrating
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novel jay
doesnt FTC and f(b)-f(a) find the integral
because solving a defitive jntegral is plug in top minus plug in bottom
oh FTc says F(b)-F(a)
which is after taking the integral of f(x)
ok that clears up so much stuff i thought was random in FTC
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silk jungle
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silk jungle
I guess I could move the 16 over instead
but idk what to do with the -16x and -16y
Unfortunately professor ZOOMED past this section (we crammed everything this week because finals are tomorrow)
and he makes his in-class work/exams really easy so I never got something like this
Oh, this is probably going to involve completing the square... hell
keen moat
Oh, this is probably going to involve completing the square... hell
it already is the square
silk jungle
it already is the square
NNo
keen moat
(2x-4)^2(2y-4)^2=0
silk jungle
We're missing the rest for x
keen moat
just add 16 both sides
silk jungle
what
Sorry, going to close this. I'm short on time and decided to skip it since it's going to be a lot more work than I'd like atm
and I only have 2 hours to complete the rest of my classes's work
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thick cloud
why is the limit 0 if its oscillating?
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why is the limit 0 if its oscillating?
unique minnow
A limit can oscillate and still converge
proven hound
It would oscillate between two values(1 or -1), but the denominator tends to infinity so it becomes 0
The same way lim(sinx/x) tends to 0 as x tends to infinity
cinder swallow
Hello, I am trying to gain insight into differentiating $y=\sqrt[q]{x^3}$
solid kilnBOT
kytsu1
cinder swallow
$y + dy = \sqrt[q]{(x + dx)^3}$
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kytsu1
cinder swallow
We add a little bit of y to y as we add a little bit of x to x.
Or y changes a little bit, as x changes a tiny bit.
This is the result of it with Sympy.
import sympy as sp
x, q = sp.symbols('x q')
y = (x**3)**(1/q)
der_1 = sp.diff(y, x)
print(der_1)
# 3*(x**3)**(1/q)/(q*x)
$\frac{\sqrt[q]{(3x^3)}}{qx}$
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kytsu1
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wraith hinge
$$\text{prove that if } {\epsilon_i}_{i=1}^n \text{is basis for } V^n, \text{then } \det(\epsilon_1, \ldots, \epsilon_n) \neq 0.$$ How can i do that? 🥺
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Maria
boreal dome
if (e_i) is a basis of V^n, then they are linearly independant
marble wharf
only put the $ around the math bits, dont put the text inside \text
like this $\varepsilon$
solid kilnBOT
Denascite
wraith hinge
if (e_i) is a basis of V^n, then they are linearly independant
yes, but what's next move
wraith hinge
only put the \$ around the math bits, dont put the text inside \text
ok, i'll do that next time
marble wharf
what properties of the determinant do you know
wraith hinge
only basic ones 😦 i don't know any property connected with linearly independent combination of vectors
marble wharf
well which basic ones
boreal dome
only basic ones 😦 i don't know any property connected with linearly independent...
do you know that determinant is a multilinear alternate application (idk the name in english for real)
wraith hinge
no, i didn't know that, can you please tell me the statement?
boreal dome
f is multilinear => f(x1, ..., ax_i, ..., x_n) = af(x1, ..., x_i, ..., x_n) (linear for all composants)
alternate : if there is x_i = x_j for i ≠ j, f(x_1, ..., x_n) = 0
which means that if you have two equal composants, f will return 0
we can prove your statement by contraposition i think
for a n-tuples of vectors from V^n, we have det(e_1, ..., e_n) = 0, prove that (e_i) isnt linearly independant
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boreal dome
im getting ignored i guess
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sage jacinth
im getting ignored i guess
🌹
sage jacinth
It appreciates ur effort
boreal dome
It appreciates ur effort
sage jacinth
!
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molten hinge
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molten hinge
trapezoids 🗿
molten hinge
i've gotten 8,7,11,40,39
dapper swift
how?
molten hinge
it has this example thing and i just applied the problem to it
dapper swift
you're supposed to think for yourself you know
dapper swift
ok what's the max value of x^2 on [3, 4]?
molten hinge
16
molten hinge
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glass night
hi
glass night
who can solve this to me please
marble mauve
glass night
is not my homework I have taken the exam today and this was my first question and I couldn't solve because it was so hard so I want to know what is the answer of this equation and sorry for distrub.
marble mauve
you are not asking anything but how to solve it, and no details
wraith hinge
is not my homework I have taken the exam today and this was my first question an...
wouldnt you rather
learn how to solve it
marble mauve
^
wraith hinge
than just know the final number or the equation
marble mauve
learn to fish
marble mauve
the final thing is useless anyways
unless it's homework
glass night
I don't want the result I want the steps please
marble mauve
"how do i solve this equation"
wraith hinge
who can solve this to me please
look
simply everything you can
and re-arrange into proper symmetric looking sides
with y functions on the right or left
and the x on the opposite
glass night
Can you explain more ?
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mossy niche
I’m not really sure how to approach this question if I’m being totally honest. I’ve tried plugging in numbers for each box but then my inverse trig functions to find the angles never match eachother
fathom ivy
you’re definitely not getting an exact solution so i guess just approximate
might want to start by finding a target ratio for your side lengths
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jaunty scarab
I’m not really sure how to approach this question if I’m being totally honest. I...
since you mentioned inverse trig functions i assume you actually know trig.
Let's call x your opposite side (1 digit), y your adjacent side (2 digits), and z your hyppothenuse (sqrt of 3 digits)
you know that your target angle is 10º; Which means that your target tangent, x/y, is 0.17632698
that means that you want to obtain a fraction that is as close to that value as possible, with x being a single digit, and y being one or two digits
i'd recommend going into EXCEL, and making a table of the thousand of fractions that you can get
and then checking which ones are closer to that tangent value
mossy niche
Okay thank you that helps a lot
frank merlin
do I need to find derivative of just ln(t^2)?
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do I need to find derivative of just ln(t^2)?
frank merlin
idk what they want
jaunty scarab
they give you a function. That ugly integral.
They want you to derive that.
frank merlin
they give you a function. That ugly integral.
They want you to derive that.
how would i "derive" it?
wraith hinge
tho i would assume if you want the derivative of something you need to define f(x) first
so maybe try integrating the function
and then simplify it
and then derivate it
hm but that might lead to different answers
odd onyx
how would i "derive" it?
$\int_{1}^{e^{x^{2}}}\ln\left(t^{2}\right)dt=F\left(e^{x^{2}}\right)-F\left(1\right)$ where F is the anti derivative