Okay neither of you are acting in good faith. Take some time off
#help-42
stark sparrow
reef solar
this isn't fair
elfin dragon
WHY ME
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royal raft
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royal raft
does my proof make sense?
torpid canopy
yes
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woven light
How many four letter words with or without meaning can be formed by the letters of the word
'EXAMINATION'
woven light
Ping when replying
winter elbow
Ping when replying
Tried anything yet?
woven light
Tried anything yet?
I don't know where to start
strong knoll
How many possible letters are there?
strong knoll
Ah, so like you can't use more than 1 E, or 2 As?
woven light
Yeah
short salmon
perhaps important to ask: can letters be reused more times than they appear in the original word?
woven light
perhaps important to ask: can letters be reused more times than they appear in t...
No
strong knoll
Ugh I can't remember if there's a formula for this
short salmon
I don't know where to start
first, have you heard of permutations?
let's say every letter of examination is indeed distinct. then, what would the answer be?
woven light
first, have you heard of permutations?
Yes
strong knoll
Oh pre-uni? Okay probably not a formula then
woven light
let's say every letter of examination is indeed distinct. then, what would the a...
8C4*4!
strong knoll
Yeah
short salmon
yeah
strong knoll
8!/4! would also work
short salmon
8!/4! would also work
yeah this is the more straightforward answer imo lol
calm coralBOT
@woven light Has your question been resolved?
coarse zephyr
how do i do the 2nd one?
ocean ice
velvet osprey
this feels like there isn't enough info... we know the major axis (one way or another) but how are we supposed to find the minor axis
also what's up with the numbers in option d
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@coarse zephyr Has your question been resolved?
tough shadow
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tough shadow
Hello I'm currently doing c) and want to confirm if when finding if g is increasing or decreasing I just check if the gradient is negative or positive ?
velvet osprey
you want to check whether g'(x) is always >0 or always <0, yeah
tough shadow
you want to check whether g'(x) is always >0 or always <0, yeah
okay thank you
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✅ Original question: #help-42 message
tough shadow
If there is no solution for x then would it just be g(x)=3x ? and then the function would just be increasing as 3 is positive ?
velvet osprey
... what is k. ????
tough shadow
also I just realised I shouldnt have written x=k ? don't know why I wrote that
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dull spruce
ab and ba are 2 digit natural numbers that satisfy ab - ba = a² - b²
how many different (a,b) pairs is it possible to write?
my initial thought was to just write it as 9a - 9b = (a-b)(a+b) then cancel out a-b's for both sides since they don't have any root that would make it 0 in this case and then got the equation 9 = a + b
then I wrote down
(1, 8)
...
(8, 1)
and got 8 different pairs
however the correct answer is supposed to be 17, how is that?
velvet osprey
it's precisely those pairs where a-b=0
ie a=b
which, numbers-wise, works out as 11, 22, 33, ..., 99
those are all the ones you missed
dull spruce
oh wait
glass heart
whenever you cancel something out, take a second to think about whether that thing is 0
dull spruce
so there's a root i missed?
glass heart
very important habit
dull spruce
ahhh
glass heart
you cancelled out a-b from 9(a-b)=(a+b)(a-b)
dull spruce
I thought a + b = 9 shouldn't have any a = b case but yeah that's post cancelling
got it ty
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ivory flume
,rotate
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ivory flume
Question 7 I don’t understand how reduction to echelon form can be used to find lambda where it has no unique solution
It has no unique solution when the determinant is 0
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upper sparrow
If you decide to reduce the augmented matrix, you'd either have a row which is all zeros apart from the last entry (which would correspond to stating "0 = 1"), and/or you'd have a row of all zeros (provided the previous point doesn't hold, you'd have solutions but not unique) 
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lilac rock
Is this correct?
stoic oyster
Is this correct?
looks good yes
lilac rock
The paper I'm reading has this though:
I didn't know papers published in top venues could have such mistakes
stoic oyster
what's the paper?
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unborn stone
i can choose 0 to 1 as the first intgegral too right? and still get same answer
unkempt drift
i can choose 0 to 1 as the first intgegral too right? and still get same answer
yep, the order of dx, dy doesn't matter here
just make sure you swap the order of (0, 1) and (0, 2)
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blazing coyote
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blazing coyote
so I suppose I start by finnding the marginals?
$F_X(x) = \int_{y-2}^{2} \frac{6x}{7} dx$
potent lotusBOT
wai
blazing coyote
and $F_{Y}(y) = \int_{x-2}^{2} \frac{6x}{7} dy$?
potent lotusBOT
wai
velvet osprey
so I suppose I start by finnding the marginals?
idt those will be helpful to you unless you want to suffer through 18 pages of boring and errorprone integration
rather think about what region on the plane the inequality max(x,y) > 1 defines
well ok maybe this can be done with evaluating the joint cdf (if you can even get the balls to calculate that) at one specific pt
blazing coyote
rather think about what region on the plane the inequality `max(x,y) > 1` define...
the region outside a unit square of side 1
I think I should do this tomorrow
just saw the time
,ti
potent lotusBOT
The current time for math_rocks is 12:06 AM (IST) on Sun, 26/10/2025.
unique heath
hey guys i have a question for this equation here im only trying to find a,h,b,k if we have no value for h what’s it gonna be? and for h we always flip the sign right? and b we flip the fraction? or do we keep it as it is
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desert nymph
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idle geode
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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
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7. None of the above
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idle geode
nice job
atomic notch
misclick!
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upbeat crypt
Why is it (sin((4x+3)^5)) and not what i wrote?
calm coralBOT
Why is it (sin((4x+3)^5)) and not what i wrote?
jolly pilot
because the 5th power is not of the sin but of the angle
I believe someone did tell you yesterday something along the lines that $\sin^5(x)$ is different from $\sin(x^5)$
Its the second case in the problem, while what you did is according to the first case, which makes it wrong
potent lotusBOT
βαχτϵρ10Φρ4γ
jolly pilot
$\sin^5(x) \neq \sin(x^5)$
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βαχτϵρ10Φρ4γ
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unreal cloud
im currently learning logarithms. and i stuck on those problems. i dont understand the square of log up there(^2). can anyone help
rustic osprey
im currently learning logarithms. and i stuck on those problems. i dont understa...
$\log_3^2 9=\left(\log_3 9 \right)^2$. Similar for 4.012 and 4.014.
potent lotusBOT
Civil Service Pigeon
velvet osprey
well photomath understands the notation correctly
the exponent means you first calculate the log and then raise that to the power it says
unreal cloud
i still dont understand
dull wagon
which part
unreal cloud
so this two is equivalent?
dull wagon
bad notation
$$(\log_3 9)^2$$
to indicate the whole log expression being squared, and not the 9
potent lotusBOT
ραμOmeganato5
unreal cloud
ohh
unreal cloud
$$(\log_3 9)^2$$
to indicate the whole log expression being squared, and not the...
but does it make difference?
dull wagon
yes, they're not the same
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blazing coyote
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blazing coyote
I started off by constructing a basis of $M^0$. Call it ${e_1,e_2,\dots}$. Then extended that to a basis of $M'$, call it ${e_1,e_2,\dots, f_1,f_2,\dots}$. Finally , this can be extended to form a basis of $V'$, call it ${e_1,e_2,\dots, f_1,f_2,\dots, g_1,g_2,\dots}$.
A basis of $V'/ M^0 is {M^0,f_1+M^0,f_2+M^0,\dot g_1+M^0,g_2+M^0 }$
potent lotusBOT
wai
blazing coyote
The issue is establishing a mapping b/w M' and V'/M^0
now f_i+M^0-> f_i does do some of the job
but I wonder if I'm even on the right track
ivory latch
**wai**
Maybe you could do something like this, but I think it would be easier to use the first isomorphism theorem
blazing coyote
which is?
ivory latch
If $\varphi : V \to W$ is a linear map, then $V/\text{ker}(\varphi) \cong \text{Im}(\varphi)$
potent lotusBOT
ucheo
remote mural
Yooo
remote mural
yooo
How would you describe this set? Would you say “its not a vector space” or would you say “its not a vector space in R^2” or what would you describe it as exactly considering its in R^2 but the definition changes for scalar multiplication
sour monolith
If you are trying to describe the structure of the set with the two operations, then this structure is not a vector space
There is no rewaon relating it R^2 as it doesn't have the same operations as R^2
calm coralBOT
@remote mural Has your question been resolved?
calm coralBOT
@remote mural Has your question been resolved?
outer sedge
if I have f(x) = x + 5, for 2<=x<=3, is x a parameter here?
calm coralBOT
if I have f(x) = x + 5, for 2<=x<=3, is x a parameter here?
outer sedge
what's the difference between a parameter and a variable
pearl skiff
I read your question and searched online a bit, found this explains well:
https://math.stackexchange.com/questions/1290373/difference-between-variables-parameters-and-constants
calm coralBOT
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patent kayak
i need help understanding this
calm coralBOT
i need help understanding this
patent kayak
violet turret
What have you tried
patent kayak
all, but i just got more confused since I've never encountered a problem like that
violet turret
The first problem seems to be repeated application of chain and product rule
patent kayak
do i need to derive e^-2x and cos3x separately?
patent kayak
yes
violet turret
What do we get if we apply that
patent kayak
3e^-2x? (I don't know the trigo one)
but if we drive cos it's equivalent to -sin right?
violet turret
We get cos3x•[e^-2x]' + [cos3x]'•e^-2x
violet turret
Now we can differentiate the expressions
[f(x)•g(x)]' = f'(x)g(x) +f(x)g'(x)
Basically the product rule
patent kayak
what's next after that?
violet turret
We get cos3x•[e^-2x]' + [cos3x]'•e^-2x
Now we differentiate e^-2x and cos3x which still remain in the expression
violet turret
It will be -3sin3x
violet turret
Yes
patent kayak
will f(x) be f'(x) = 3e^-ex or will it stay the same?
violet turret
I am not quite sure what you are trying to say here
hexed nebula
won't cos3x be -sin3x? or I'm wrong
it is but then you multiply by d/dx (3x)
u treat cos(3x) as f[g(x)]
tall moon
how does plugging (sqrta,sqrtb,1,sqrtab) get that?
calm coralBOT
how does plugging (sqrta,sqrtb,1,sqrtab) get that?
tall moon
shouldnt it get
$$\frac{f(\sqrt{a})^2+f(\sqrt{b})^2}{f(\sqrt{ab}^2)+f(1^2)}=\frac{\sqrt{a}^2+\sqrt{b}^2}{1^2+\sqrt{ab}^2}$$
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glacial tapir
Given f(x) = ax^2 +bx + c. Suppose that f(x) = x doesn't have sol, prove that f(f(x)) = x also doesn't have sols
desert nymph
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3. I got an answer but I was told that it's wrong.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
also "doesn't have solution" - do you mean in R? in Z? in N?
glacial tapir
the question says so 🤷♂️
desert nymph
okay, so I think we can assume it means in the real numbers
glacial tapir
if we have f(x) = x has no sols, which means that (b-1)^2 - 4ac < 0
desert nymph
if it has no real solutions, then it never crosses the x-axis. it's either always > 0 or always < 0
in fact, it never crosses the line y=x
glacial tapir
a wild idea: does proof by contradiction work?
desert nymph
could try it 🙂
glacial tapir
a*(f(x))^2 + b f(x) + c = x
desert nymph
I think the issue is you would run into solving a quartic which is....ew
trail glacier
cant you just say
if f(x) has no x in reals such that f(x) = 0
when you do f(f(x)) you are just plugging in a subset of the reals
glacial tapir
a f(x)^2 - ax^2 + b f(x)^2 - bx + f(x) - x = 0
glacial tapir
what did you read
trail glacier
hold on
I think the same reasoning applies
just with f(x) = x
if there is no x in reals that satisfies this
then there won't be one in the subset of the reals either
wait no
🤦♂️
ignore me sorry
desert nymph
no, because you're given that the pair (x, x) doesn't exist - not that the pair (f(x), x) doesn't exist 🙂
glacial tapir
a f(x)^2 - ax^2 + b f(x)^2 - bx + f(x) - x = 0
so from this, I have (f(x) - x) (a f(x) + ax + b + 1) = 0 (still has sols)
desert nymph
if it has no real solutions, then it never crosses the x-axis. it's either always > 0 or always < 0
in fact, it never crosses the line y=x
if f(x) is always > x, then what does this say about f(f(x))?
glacial tapir
also > x
desert nymph
if that's too much of a jump:
given that "f(x) = x no sols ---> f(x) always greater than x", what's the equivalent statement about "f(y) = x no sols where y=f(x)"?
* I say "always greater than" - you can do the exact same logic with "always less than"
glacial tapir
if that's too much of a jump:
given that "f(x) = x no sols ---> f(x) always grea...
where does y come from?
wait i just saw y = f(x)
desert nymph
where does y come from?
I just wanted to make it clearer that when I said "f(f(x)) = x no sols", I was trying to show how it mirrors f(x) = x no sols
but yeah - if f(x) is always > x, then f(f(x)) must be always > f(x)
if you let y = f(x), then it becomes more obvious: f(y) > y
this is the same as our original statement about x, but about y (which is f(x))
glacial tapir
i kinda understand
desert nymph
I can try to explain more if there's something not clear
but the basic list is:
- we can say something about every x in f(x)
- obviously if we just change which letter we use then it remains the same - so let's say every y in f(y)
- if we say y = f(x), then our statement becomes about every f(x) in f(f(x))
if f(x) > x, and f(f(x)) > f(x), then we can combine them: f(f(x)) > f(x) > x for all x
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dim plover
@zealous flume
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old forge
Hello evyone please give me sum tips to calculate fast in math if it is possible 😁
sour monolith
Sorry man but this one is closed
old forge
ok
uneven shell
I saw it 😂
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simple musk
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simple musk
how do I find the partial derivative of this wrt x?
pallid halo
would you know how to do it if you had, say 5 instead of y?
simple musk
you mean, for like, single variable?
pallid halo
yes
simple musk
,, f(x) = \frac{1}{x^2 + 5} \ \pdv{f}{x} = -(x^2+5)^{-2} \cdot 2x
potent lotusBOT
Renato
pallid halo
yea good, it works the same way when you do the partial derivative with respect to x
just treat y as a constant (number)
pallid halo
i can do a similar one for you if it will help
simple musk
yeah that would help me a ton
pallid halo
say $f(x,y) = \sin(x^2 + y^2 + 1)$
potent lotusBOT
Bungo
pallid halo
then $\pdv{f}{x} = \cos(x^2 + y^2 + 1)\pdv{(x^2 + y^2 + 1)}{x} = \cos(x^2 + y^2 + 1)(2x)$
potent lotusBOT
Bungo
simple musk
I see
yeah let me do this
,, f(x,y) = \frac{1}{x^2 + y^2 + 1} \ \pdv{f}{x} = -(x^2+y^2+1)^{-2} \cdot \pdv{(x^2+y^2+1)}{x} \ \pdv{f}{x} = -(x^2+y^2+1)^{-2} \cdot 2x
potent lotusBOT
Renato
pallid halo
yep good
simple musk
I appreciate it
pallid halo
yw
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hardy spindle
i dont understand b)
calm coralBOT
i dont understand b)
sonic rose
Please try to do translation work on this one
turbid violet
I think I understood sorta
hardy spindle
When a freezer breaks, its internal temperature rises, which can cause the loss of its contents.
The rule 𝑓(𝑡)=19/5 sqrt (𝑡−10) makes it possible to calculate the internal temperature f(t) (in °C) of a freezer as a function of the time elapsed
t (in hours) since the breakdown, until the temperature of the freezer reaches that of the surrounding air.
b) What is the rule of the function that makes it possible to calculate the elapsed time (in hours) since the breakdown as a function of the internal temperature (in °C) of the freezer?
turbid violet
You already have f(t) which calculated temperature in terms of time, and b) is asking you to find the function that calculates the time in terms of the temperature
hardy spindle
yeah
turbid violet
Consider f(t) = T and put it in the function rule
or C instead of T to make it different
And isolate t in the right hand side
turbid violet
And isolate t in the right hand side
In other words, try to make $C=\frac{19}{5}\sqrt{t}-10$ into $t=\text{something with C}$
I'm pretty sure the result will be a quadratic function
hardy spindle
ive obtained 50/19 = sqrt(t)
C = 19/5 sqrt(t) - 10
+10 both sides
10 = 19/5 sqrt(t)
potent lotusBOT
Mathlympian / Gab
hardy spindle
multiply by 5 both sides?
turbid violet
10 = 19/5 sqrt(t)
The C needs to be kept in lhs
hardy spindle
oh
turbid violet
So 10 + C = 19/5 sqrt(t)
turbid violet
Ye
hardy spindle
divide by 19?
turbid violet
Yes
turbid violet
(50/19 + 5c/19) ^2 = t
It's better if you write it as (50 + 5c)/19 instead of 50/19 + 5c/19
hardy spindle
but how will i put a ^2 then
turbid violet
The ^2 will apply for both numerator and denominator
And so just use (a + b)² = a² + 2ab + b² for numerator, and 19² = 361 for denominator
hardy spindle
2500/361 + 25/361 = t
hardy spindle
ok what
turbid violet
2500/361 + 25/361 = t
The C didn't appear again
hardy spindle
25c/361
turbid violet
And also you did (a + b)² = a² + b² only, but not 2ab
hardy spindle
theres 1 fraction
and another
both powered by 2
i dont get where a 3rd fraction comes from
turbid violet
It's because there are 3 terms, a², 2ab and b²
hardy spindle
okk and how do i know that
hardy spindle
average quebecer book
this book is stupid
turbid violet
Or you can think (a + b)² = (a + b)(a + b) = a² + ab + ab + b² = a² + 2ab + b² (by distributive property)
tiny bolt
i hated those books as a teen
hardy spindle
even my teacher said it i knew i wasnt imagining things
but my teacher is also equally stupid
tiny bolt
yeah i mean hs teachers are not always the smartest of the bunch
hardy spindle
this guy speedran notions and here i am and my whole class learning alone
we aren't doing jack for 3 weeks now
tiny bolt
happens
hardy spindle
Or you can think (a + b)² = (a + b)(a + b) = a² + ab + ab + b² = a² + 2ab + b² (...
ok
tiny bolt
just remind urself its going to be over at some point 😅
hardy spindle
Or you can think (a + b)² = (a + b)(a + b) = a² + ab + ab + b² = a² + 2ab + b² (...
(a+b)^ 2 = a2 +2ab + b^2
?
turbid violet
where did c come from
hardy spindle
oh b
hardy spindle
wait but we cant just keep it like how it is already?
turbid violet
You can
hardy spindle
like 2500/361 + 25c/361 = (t)
turbid violet
Bcuz it's asking for t in terms of C
hardy spindle
yeah
turbid violet
like 2500/361 + 25c/361 = (t)
- 500C²/361
hardy spindle
i didnt really learn that a2 + b thing whatever this year that was last year
turbid violet
Understood
hardy spindle
ok but where did 500C^2/361 come from
hardy spindle
2(2500)(25)
hardy spindle
yes
turbid violet
And 25c²/361
hardy spindle
why
turbid violet
The c is squared too
hardy spindle
oh right
turbid violet
So 2500/361 + 500c/361 + 25c²/361 = t
turbid violet
You can leave it like that or combine the fractions
turbid violet
yeah i get it
Exactly
hardy spindle
alright but we isolated t now what
hardy spindle
ohh ok thanks dude
turbid violet
yw!
hardy spindle
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turbid violet
glad to have helped
hardy spindle
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✅ Original question: #help-42 message
hardy spindle
@turbid violet sorry my man 1 last question if u can
hardy spindle
it says we need to avoid the temperature to be above 4Celcius. How much time at maximum the food can be conserved?
so since t = to those fractions
4 = 2500/361 + 500c/361 + 25c²/361
?
hardy spindle
ah
turbid violet
And then you'll get the t
hardy spindle
thanks again
turbid violet
Yw!
hardy spindle
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lost mantle
(sec(x) - 1) / (1 - cos(x)) = sec(x) what is the domain restriction
lost mantle
for domain restrictions for trig identity proofs, do you look at both sides or only the side your solving for
<@&286206848099549185>
dusky vapor
both
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quasi delta
need help with part b
calm coralBOT
need help with part b
quasi delta
part b: given A is a 2x2 matrix. Prove that A^k = 0 (k > 2) <-> A^2 = 0
sour monolith
quasi delta
A^2 = 0 leading to A^k = 0 is easy
sour monolith
Yes
quasi delta
how do i do the opposite direction
remote mural
yes
exotic cosmos
how do i do the opposite direction
consider the minimal polynomial of A
quasi delta
so stuff like matrix addition, multiplication, determinants, inverse matrix, matrix rank and gauss elimination
and that is it
quasi delta
did you learn about eigenvalues?
nope
glass heart
char poly and cayley hamilton?
quasi delta
never heard of it
glass heart
then wtf did you even do in part a)
novel spindle
do you know about nilpotent matrix at least ?
quasi delta
then wtf did you even do in part a)
just evaluate A^2 and plug everything in
glass heart
ok
so the quadratic polynomial on the left is the char poly of the matrix
and cayley hamilton says that if you plug a matrix into its char poly then you get 0
quasi delta
that's good to know
but i dont think my prof will accept that
so i do need another solution
glass heart
I'm not saying thats the proof
divide x^k by the polynomial in a)
what happens if you plug A into the resulting equation
something along those lines should work
quasi delta
lemme try later
calm coralBOT
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quasi delta
$x^k = [x^2 - (a+d)x + ad-bc] \cdot x^{k-2} + (a+d)x^{k-1} - (ad-bc)x^{k-2}$
potent lotusBOT
(De)Carbonized
quasi delta
divide x^k by the polynomial in a)
like this? or do i need to expand a few more terms?
seems like im just getting back to where i started
sour monolith
You know A^k=0, so it has determinant and trace equal to 0, and the polynomial in part a becomes x^2
quasi delta
det(A^k) = 0
there's det(AB) = det(A)*det(B)
oh
so det(A) = ad-bc = 0
so we got down to A^2 = (a+d)A
hmm
wait a sec
$A^k = (a+d)^{k-1} \cdot A = 0$
potent lotusBOT
(De)Carbonized
quasi delta
oh wow
if a + d is not 0 then A = 0 and then A^2 = 0
if a+d = 0 then we just plug into the characteristic polynomial in a) and get A^2 = 0
thanks everyone
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calm coralBOT
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viscid path
can someone explain to me this?
calm coralBOT
can someone explain to me this?
teal drift
What part is unclear?
viscid path
What part is unclear?
from step 2 onwards
teal drift
If alpha (here 60°) is a solution, then also 360° - alpha is.
Nothing more than this 🤔
viscid path
If alpha (here 60°) is a solution, then also 360° - alpha is.
Nothing more than ...
so its always 360 degrees - alpha?
but how do we identify which Q the angles are in?
teal drift
but how do we identify which Q the angles are in?
Well, you should know them 😅
Q1 is for angles between 0 and 90°, Q2 from 90° to 180° and so on and so forth
teal drift
so its always 360 degrees - alpha?
Always referring to what? Trig equations in general?
viscid path
Always referring to what? Trig equations in general?
for solving trig eqn in linear form
viscid path
Q1 is for angles between 0 and 90°, Q2 from 90° to 180° and so on and so forth
if tan is positve it'll be in quad 3 right?
calm coralBOT
@viscid path Has your question been resolved?
teal drift
for solving trig eqn in linear form
Well, of course not 😅
teal drift
if tan is positve it'll be in quad 3 right?
1 or 3
Since tangent being positive means sin and cos having the same sign
I suggest you review the unit circle
viscid path
bro for this how they get 2cos x = -sqrt 3 which becomes cos x = -sqrt 3/ 2?
teal drift
They've just divided both sides by 2...
mental raven
calm coralBOT
mental raven
so basically Fi/2 < F(i-1)
1, 1, 2, 3, 5, 8, 13, 21, 34, ...
idk
how to "show that"
though
calm coralBOT
@mental raven Has your question been resolved?
potent lotusBOT
Rbit
atomic notch
And for S>3 you can use the inequality to deduce it
and note that the first two terms are already summing to 2
visual viper
And for S>3 you can use the inequality to deduce it
how so?
atomic notch
$\sum_{i=4}^{\infty} \frac{1}{F_i} > \frac{1}{2} \sum{i=4}^{\infty} \frac{1}{f_{i-1}} = \frac{1}{2} \sum_{j=3}^{\infty} \frac{1}{F_j}
= \frac{1}{2} T$
potent lotusBOT
Rbit
calm coralBOT
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Available help channel!
mental raven
S = 2+T>2+1=3
calm coralBOT
mental raven
it can be shown that T>1
ohhh ok
i figured out how
its by induction
i didnt realise because it said "show that" when it was a proof
and i felt it should say "prove that"? idk
mental raven
anyways here
idk because if you expand that
wouldnt you get z^n + az^n-1 + bz^n-2 + ... - 1 or something
idk
i dont fully understand
sour monolith
Show is used in the same way as prove
mental raven
Show is used in the same way as prove
fe
sour monolith
Both are monic polynomials of degree n
Show they have the same roots
And then you are done
mental raven
Show they have the same roots
wdym by "same roots"
thick grail
alr
sour monolith
i need help related chemistry
Open a new channel #❓how-to-get-help
thick grail
but it contain math
mental raven
but it contain math
yeah then go to another unused help channel
not one where im currently on trying to solve a question
thick grail
alr thx
mental raven
Root = a solution to p(x)=0
wait but with the roots
that doesnt necessarily mean the equations are the same
like (x-1)(x-2) is not the same equation as 3(x-1)(x-2) for example
but i guess you can infer they have the same scale factor
because z^n coefficient is 1
sour monolith
Yes! But the difference is the leading coefficient
sour monolith
because z^n coefficient is 1
Right, the polynomials in your question are monic, which means they have leading coefficient 1
mental raven
Right, the polynomials in your question are monic, which means they have leading...
yeah
i understand
ok
ill end the channel if i dont get stuck at the rest of the question
hold on
calm coralBOT
@mental raven Has your question been resolved?
calm coralBOT
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Available help channel!
night lion
How do I solve it
calm coralBOT
How do I solve it
night lion
I worked on it as I attached work in picture
<@&286206848099549185>
<@&286206848099549185>
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calm coralBOT
desert relic
@night lion maybe try finding the x for which your condition applies, then apply it into F = -kx
calm coralBOT
Available help channel!
night lion
.
night lion
<@1219929477925179424> maybe try finding the x for which your condition applies,...
It says E = 1/2K + 1/2U
So We have K_max = 1/2kx² and U is also 1/2kx²
right?
desert relic
Well U I'm sure about
daring haven
PE is kx^2/2, and E is kA^2/2, so find A in terms of x. I'm not sure though this will work
desert relic
Wait KE = PE too
night lion
Wait KE = PE too
Yes that
desert relic
There was a formula for Energy of a spring yeah?
night lion
Yes that is $ 1/2 k (x_m)²$
desert relic
whoops
blissful maple
Yes look, the place where potential energy and kinetic energy is half is
When the object is at half the distance of the amplitude
The 1/4th part
calm coralBOT
Available help channel!
visual sigil
calm coralBOT
visual sigil
not sure how to do part c
mortal wren
not sure how to do part c
show ur diagram
visual sigil
show ur diagram
i don't think we need one
NB is just OB - ON
ON we already have
OB = b
round widget
You should still draw on the question to visualize all the givens. Then you can start marking all the other stuff proven from the givens
visual sigil
You should still draw on the question to visualize all the givens. Then you can ...
doesn't really help here
anyways, let k = lambda
i get:
NB = (2 - 1.5k)a + (1 - 0.5k)b
ON = (2 - 1.5k)a + 0.5kb
idk how this is 2:1?
umbral elk
@visual sigil your previous solutions will be helpful, share if you can
oh lol you sent that same time i did
visual sigil
<@1417933317537398824> your previous solutions will be helpful, share if you can
CM = 0.5b - 1.5a, ON is in the question
calm coralBOT
@visual sigil Has your question been resolved?
calm coralBOT
@visual sigil Has your question been resolved?
rustic osprey
anyways, let k = lambda
i get:
NB = (2 - 1.5k)a + (1 - 0.5k)b
ON = (2 - 1.5k)a...
I wouldn't use the same scalars for $\overrightarrow{NB}$ and $\overrightarrow{ON}$ since things get confusing otherwise
potent lotusBOT
Civil Service Pigeon
calm coralBOT
rustic osprey
.reopen
calm coralBOT
✅ Original question: #help-42 message
rustic osprey
oh wait you didn't start part c
ok
$\overrightarrow{ON} = (2-1.5 \lambda) \mathbf{a}+0.5 \lambda \mathbf{b}$
potent lotusBOT
Civil Service Pigeon
rustic osprey
My hint for you is to consider scalar multiples, specifically
,texsp ||$\overrightarrow{ON}$ is a scalar multiple of $\overrightarrow{OB}$||
potent lotusBOT
Civil Service Pigeon
rustic osprey
**Civil Service Pigeon**
now from this, consider what equation you can set up
especially ||regarding the "coefficient" of a||
calm coralBOT
@visual sigil Has your question been resolved?
dawn plover
Hey, how can I derivate X^x...?
calm coralBOT
Hey, how can I derivate X^x...?
glad parrot
x^x = e^[xln(x)] and you can use chain rule on it
dawn plover
x^x = e^[xln(x)] and you can use chain rule on it
Why "e**^[xln(x)]**", I've seen it A LOT. I asked ChatGPT and Copilot and they do the same, also Photomath does it... But in a weirdest way, I don't know how to converse it into ln.
glad parrot
u = e^ln(u)
Since e and ln are inverse functions
Let u = x^x and apply log proprety
dawn plover
Is there any name for that rule or method? Just to search further.
glad parrot
i think its just exp propreties or smth
dawn plover
I see, I found that there are multiple methods to do it, I asked to my professor about it and he told me that this can be derivated by this formula:
The 12 one.
glad parrot
,rcw
potent lotusBOT
glad parrot
Ok well yes
dawn plover
Is it possible to solve it with the 12 one?
glad parrot
It is
dawn plover
I barely understood that since ln appeared in most of the AIs and Photomath, and they are my unique saviors before asking on this server. 😭
glad parrot
It does make a ln appear there also
dawn plover
Can you help me to solve it using the formula?
dawn plover
I'll take notes, give me a sec.
@glad parrot So, there are 2 ways of making it, th one with the e^[xln(x)] and the other one using the formula, right?
fair dock
hello
dawn plover
So; 𝑦 = 𝑥^𝑥 ⇒ 𝑦 = 𝑒^𝑥ln(𝑥)
Here's the chain rule
Then the derivate, but what happened to e & x?
leaden thunder
e^ln(stuff) = stuff
leaden thunder
u = e^ln(u)
.
leaden thunder
Then the derivate, but what happened to e & x?
They're doing the derivative of the other part separately.
calm coralBOT
@dawn plover Has your question been resolved?
dawn plover
They're doing the derivative of the other part separately.
wha
dawn plover
Here's the chain rule
Okay, so here's the chain rule, so they derivated the right part?
d/dx xln(x) = (1) ln(1)?
leaden thunder
Okay, so here's the chain rule, so they derivated the right part?
the word is "differentiated"
potent lotusBOT
leaden thunder
Then the derivate, but what happened to e & x?
this was correct
dawn plover
this was correct
d/dx xln(x) = ln(x) + 1?
leaden thunder
yes
leaden thunder
that's exactly what this says
dawn plover
that's exactly what this says
So, we got this now:
dawn plover
Sorry, wrong image...
I meant to send this one, mbmb sorry.
What here? Do I just go back to the original expression?
If I go to the original expression, the answer will be d/dx x^x = x^x . [ln(x) + 1]
Is this correct?
potent lotusBOT
dawn plover
Yes
I see, thank you! Just before I go, can you help me with the same derivate but now doing it with this formula? #help-42 message (The XII one)
rustic osprey
I see, thank you! Just before I go, can you help me with the same derivate but n...
This is mainly just plug and chug
Did you identify what u and v should be
dawn plover
Did you identify what u and v should be
I believe that x is u and ^x is v, right?
rustic osprey
v is just x, not ^x
rustic osprey
Why are you trying to put exponents without a base in the functions?
$u=x$ and $v=x$
potent lotusBOT
Civil Service Pigeon
rustic osprey
$x^x=u^v$
potent lotusBOT
Civil Service Pigeon
dawn plover
Well, you have a point, at the end of the day they are the same. U & V = X
I'll try to solve it by my own, thanks a lot to everyone who helped.
.close
calm coralBOT
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Available help channel!
simple musk
calm coralBOT
marsh valley
!status
calm coralBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
Sure what do you have to share?
simple musk
Sure what do you have to share?
do you have any idea on how to do this ?
teal drift
If you have already done something, why not sending it?
simple musk
the progress i done is finding the intersections between the boundaries of the restrictions
x = -1 , x^2 + 4y^2 = 4 ,
(-1)^2 + 4y^2 = 4
y^2 = 3/4
y = +- sqrt(3)/2
@teal drift
@here
marsh valley
That seems good.
simple musk
That seems good.
anyways, do you have any idea on how to find the extrema for the frontier of the region
marsh valley
Well the boundary is made of two pieces.
x=-1 and x^2 + 4y^2 = 4
With those you should be able to investigate extrema of the function restricted to either of those separately.
simple musk
With those you should be able to investigate extrema of the function restricted ...
whats your plan?
marsh valley
Those problems usually have a pretty straightforward approach.
Find critical points on the interior with the gradient, find critical points on the boundary (that includes weird intersection points), compute the function at those points and see which is the maximum/minimum.
For the boundary critical points you can either substitute in when it's simple enough, or use Lagrange multipliers if necessary.
simple musk
Those problems usually have a pretty straightforward approach.
Find critical poi...
how do you plan to find the critical points on the boundary
marsh valley
With those you should be able to investigate extrema of the function restricted ...
.
simple musk
For the boundary critical points you can either substitute in when it's simple e...
do you mind if we do both, first lagrange then parametrizing each curve from the restriction
simple musk
Those problems usually have a pretty straightforward approach.
Find critical poi...
are u here?
marsh valley
I am doing other stuff at the same time. I can help but you can also go ahead and work on it no?
simple musk
I am stuck at the lagrange multipliers step
marsh valley
Ok how come are you stuck?
simple musk
well first of all, when I try to apply lagrange multipliers for the first restriction
I get that I want to apply lagrange to x = -1 right?
well x = -1 is not even a function
because for same x it gives different y
@marsh valley
marsh valley
g(x,y) = x is a function
That piece of the boundary is given by the level set g = -1
simple musk
alright
simple musk
g(x,y) = x is a function
alright so let me try to apply lagrange multipliers
tried to apply lagrange multipliers method but I am getting a weird system
maybe I messed up? can you confirm this is correct? @marsh valley
I just set the two gradients equal and multiplied by lambda the gradient of g
marsh valley
That looks okay you can continue..
The boundary is fairly simple so in particular if you use the fact that x=-1 you should be able to find a value for y fairly quickly
marsh valley
Okay. So we got one possible point on that piece of the boundary. Is that point actually on our boundary though?
(-1,-3/2)
marsh valley
In any case it's not in the actual region we're interested in
simple musk
In any case it's not in the actual region we're interested in
did u checked?
potent lotusBOT
marsh valley
Yes you can graph it it's on x=-1 but too low
simple musk
Yes you can graph it it's on x=-1 but too low
you mean is outside of the elliptic disk?
marsh valley
Yeah if you want
So it worked but this point is of no use.
So you can go for the second piece of the boundary
marsh valley
Yes
simple musk
So you can go for the second piece of the boundary
but did we found all of the solutions to the system?
marsh valley
We found that there are no relevant points to investigate on the boundary where x=-1
Except the intersection points, say
Those you always check
marsh valley
Okay
simple musk
ok so basically I got this system
and I am asking myself what happens when x = 0 or y = 0 so I can then divide by x and divide by y in the equations to solve for lambda and since I dont care about lambda I will set the lambdas equal to each other
after I set the lambdas equal to each other I find the values of x and y, but I got that (0,0) is a candidate so far
marsh valley
Well (0,0) is not on your boundary so something went wrong.
Those points still have to satisfy the equation x^2 + 4y^2 = 4
simple musk
right
so x = 0 is impossible
no wait
x = 0 => (0,-1) and (0,1)
but if x = 0 the first equation of the system is a contradiction
2x - 12y = 2lambda x
because if x = 0 then y = 0
so basically we are finding that x = 0 is not possible
@marsh valley
marsh valley
Yes essentially
simple musk
ok
we have the system
i) 2x - 12y = 2λx
ii) 8y - 12x = 8λy
iii) x^2 + 4y^2 - 12xy = 4
if x = 0 then y = 0 by i), but then 0 = 4 by iii)
so x is never 0
if y = 0, then x = 0 by ii), but then 0 = 4 by iii) so y is never 0
@marsh valley
now that we now that x ≠ 0 and y ≠ 0 we can proceed with dividing by x and dividing by y
by i) we get that λ = (2x - 12y)/(2x)
by ii) we get that λ = (8y-12x)/(8y)
we set them equal we get
(2x)(8y-12x)=(8y)(2x-12y)
16xy - 24x^2 = 16xy - 96y^2
24x^2 = 96y^2
x^2 = 4y^2 basically
simple musk
Well start with x^2 =4y^2 and use that in your constraint
2x^2 - 12xy = 4
marsh valley
The constraint is x^2 + 4y^2 = 4
simple musk
The constraint is x^2 + 4y^2 = 4
I did it
marsh valley
You got x^2 = 4y^2
simple musk
The constraint is x^2 + 4y^2 = 4
oh mybd
marsh valley
You need to plug that in the constraint, not the objective function
simple musk
2x^2 = 4
x = ±sqrt(2)
now we go back here x^2 = 4y^2
we get y^2 = 1/2 and consequently y = ± 1/sqrt(2)
marsh valley
Good so that gives you 4 possible points. Check if those are in the region we're looking into
simple musk
,w x >=1 and x^2 +4y^2 <=4 with (x,y)=(sqrt(2), 1/sqrt(2))
potent lotusBOT
simple musk
,w x >=1 and x^2 +4y^2 <=4 with (x,y)=(sqrt(2), -1/sqrt(2))
potent lotusBOT
simple musk
,w x >=1 and x^2 +4y^2 <=4 with (x,y)=(-sqrt(2), 1/sqrt(2))
potent lotusBOT
simple musk
,w x >=1 and x^2 +4y^2 <=4 with (x,y)=(-sqrt(2), -1/sqrt(2))
potent lotusBOT
simple musk
Good so that gives you 4 possible points. Check if those are in the region we're...
so we got 2 new candidates to the bag
marsh valley
Yep
So all in all in the x=-1 piece of the boundary we found nothing interesting.
In the x^2 + 4y^2 = 4 part of the boundary we found 2 possible points.
We also have the intersection of the two pieces of the boundary that we need to keep track of.
It remains to check whether the interior has any critical points though
simple musk
It remains to check whether the interior has any critical points though
alright I set the gradient of f equal to zero, check the candidates and we are mostly done
marsh valley
Yep
Then you'll have all interesting points and you can check all of them (not that many) to see which is the biggests and which is the lowest.
You'll only be able to conclude if the region is compact (closed and bounded) though, which it is for this case and for those problems in general.
simple musk
You'll only be able to conclude if the region is compact (closed and bounded) th...
even though the first restriction is unbounded the region talks about the intersection of both restrictions, so like, the elliptic disk is clearly bounded and even though the first x >= -1 line is unbounded the region is bounded
marsh valley
Yes yes that's fine. The point is the domain we're investigating has to be bounded and closed. Whether it's made up of pointwise bits of other constraints doesn't really matter.
In particular it guarantees by the extreme value theorem that the function has an absolute minimum and an absolute maximum in this region, which is necessary to conclude that you can indeed reduce your search to those candidates you found with the multipliers/in the interior.
simple musk
In particular it guarantees by the extreme value theorem that the function has a...
yes
simple musk
this is not a solid proof but a geometrical intuition as to why the region is compact
marsh valley
It is the intersection of two closed regions, so it is closed.
It is the intersection of one bounded region with some other region, so it is bounded.
There we go
simple musk
recall that the region x >= -1 => x ∈ [-1, ∞) is not closed
nor bounded
but... we don't need to get into the nitty gritty of compactness
marsh valley
recall that the region x >= -1 => x ∈ [-1, ∞) is not closed
This is closed.
Although your implication doesn't hold
{x,y : x>=-1} is closed
So you intersect two closed regions that always gives a closed region
Then for boundedness we know that the ellipse is bounded, so taking the intersection with anything will still yield a bounded region
Anyways
Point is it's compact.
simple musk
{x,y : x>=-1} is closed
care to elaborate how this set is closed?
marsh valley
Its complement {x,y : x<-1} is open.
simple musk
Point is it's compact.
yes, luckily all of this exercises that I have been given we have compact regions, so we can apply weirstrass/evt
marsh valley
Well {x,y : x>=-1} is certainly not open since for any point on the line x=-1 there are no open balls around that point that are contained in the set
simple musk
Its complement {x,y : x<-1} is open.
how does there exists some open balls for this
any examples that come to mind. of open balls for subsets of this set
marsh valley
Any point on the left of the line x=-1 can be written as (-1-r, y) for some r >0 and y in R.
Then you can draw an open ball of radius r around (-1-r,y)
It's still in the set {x,y : x < -1}
It should be fairly easy to see that if you take any point on the left you can just draw a small enough ball around it that stays within the set.
simple musk
Well {x,y : x>=-1} is certainly not open since for any point on the line x=-1 th...
I was thinking like: for any point on the right of the line x = -1 can be written as (-1+r,y) for some r > 0 and y ∈ R
but I don't think I can draw open balls that are still in this set because that would pressume r = 0 at some point or another. correct or incorrect?
marsh valley
Well that one is closed specifically because {x,y : x<-1} is open (closed iff complement is open).
marsh valley
but I don't think I can draw open balls that are still in this set because that ...
Well it's just that this is true only for points strictly to the right of the line x=-1.
But {x,y : x>= -1} also contains those points on the line, so the argument doesn't help product a ball around those points since r=0.
And in fact any open ball you draw around a point (-1, y) won't be contained in {x,y : x >= -1}
simple musk
And in fact any open ball you draw around a point (-1, y) won't be contained in ...
care to elaborate?
simple musk
I just noticed i just drawed x = -1 not x >= -1
marsh valley
Well you drew the intersection of both constraints
Which is the region we're looking at
marsh valley
what kind of math is this? anal?
If anything it's just Euclidean geometry, maybe some very basic point set topology?
But it's also borderline just set theory
marsh valley
care to elaborate?
Take any point on the line.
Draw an open ball around that point
However small you draw the ball, it will stick outside the set {x,y : x>= -1}
simple musk
Well you drew the intersection of both constraints
yes
simple musk
However small you draw the ball, it will stick outside the set {x,y : x>= -1}
yeah
I get it now
anyways back to the exercise
we had the candidates from the intersection
marsh valley
Yes that's good just missing the ones for the interior
simple musk
alright
simple musk
2x-12y = 8y - 12x = 0
2x - 12y = 0
x - 6y = 0
x = 6y
2y - 3x = 0
y = (3/2)x
only possibility is when x=y=0
otherwise this is nonsense
marsh valley
Indeed
simple musk
alright
and the origin is inside the region
@marsh valley
I think I only like math, the rest of the things are not as cool
marsh valley
OK so you have points $$(\sqrt{2}, \frac{\sqrt{2}}{2}), (\sqrt{2}, -\frac{\sqrt{2}}{2}), (-1, \frac{\sqrt{3}}{2}), (-1, -\frac{\sqrt{3}}{2}), (0,0)$$
simple musk
but I am bad at it, any tips on improvement?
potent lotusBOT
Azyrashacorki
simple musk
**Azyrashacorki**
we just need to find the minimum and maximum of these and we are done
marsh valley
we just need to find the minimum and maximum of these and we are done
Yes
simple musk
,w x^2 + 4y^2 -12xy at (x,y) = (sqrt(2), sqrt(2)/2)
,w x^2 + 4y^2 -12xy at (x,y) = (sqrt(2), -sqrt(2)/2)
,w x^2 + 4y^2 -12xy at (x,y) = (-1, sqrt(3)/2)
,w x^2 + 4y^2 -12xy at (x,y) = (-1, -sqrt(3)/2)
marsh valley
Whoawhoa
We already know those points are on the boundary
It's the objective function you want to evaluate
potent lotusBOT
marsh valley
That's what you're trying to optimize
marsh valley
They do haha
simple musk
,w x^2 + 4y^2 -12xy at (x,y) = (0,0)
potent lotusBOT
simple musk
max is this
simple musk
min is this