#help-19
weary flame
if q was 2011^2+2012^2 it would be a whole number
wait is sqrt(2011^2 + 2012^2) the same as sqrt(2011^2) + sqrt(2012^2)?
or 2011 + 2012
steep mantle
no
weary flame
I thought so
steep mantle
if q was 2011^2+2012^2 it would be a whole number
btw do you wanna let q be this or the original question
weary flame
original question
odd edgeBOT
odd edgeBOT
Available help channel!
solid rain
Let P be a polynomial of degree n,n>=2. Knowing that the set of real solutions of P(f(x)) =0 has 3n-1 elements, prove that P(-1)*P'(1/e)!=0.
solid rain
I know P(-1) !=0 since 1/e is the min value
In the answer there was this statement: The equation has 3n-1 solutions if P has n real distinct roots and n-1 of them being from the interval (0,1/e) and a root 1/e
sullen ferry
this seems very inductiony
solid rain
how would I use induction here
solid rain
In the answer there was this statement: The equation has 3n-1 solutions if P has...
i understand that the roots are from that interval but I dont get the part about the 3n-1 solutions
!status
odd edgeBOT
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
5
odd edgeBOT
@solid rain Has your question been resolved?
solid rain
.close
odd edgeBOT
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Available help channel!
fallow wedge
Calculate the line integral (\int_C \mathbf{F} \cdot d\mathbf{r}) for the vector field (\mathbf{F}(x, y) = (2xy, x^2 - y^2)) along the curve (C), which is the straight line segment from ((0, 0)) to ((1, 1)).
clever fjordBOT
lovely💐
merry finch
what have you tried?
merry finch
what do you know about line integrals on conservative vector fields?
fallow wedge
That the vector field which is conservative only depends on the endpoints of the curve
@merry finch
merry finch
and you're given the end points
also
the vector field doesn't depend on the endpoints
a line integral over a conservative vector field only depends on the end points
and how does it depend on the end points?
fallow wedge
That the path is not important
merry finch
That the path is not important
that's what it means for the line integral to only depend on the end points
merry finch
and how does it depend on the end points?
what about this?
fallow wedge
I am not sure
merry finch
it's sort of the fundamental theorem of calculus
fallow wedge
I remember seeing something about having a phi function
merry finch
take the potential scalar function (usually phi)
fallow wedge
But I didn’t understand it
merry finch
and just do phi(end point) - phi(start point)
Let $F$ be a conservative vector field with scalar potential function $\phi$, that is, $F(x, y, z) = (\partial_x\phi(x, y, z), \partial_y\phi(x, y, z),\partial_z\phi(x, y, z))$.
Then for any points $a, b \in \mathbb R^3$, the line integral of any curve $C$ starting at $a$, ending at $b$ over $F$,
[\int_C F\cdot d\vec r = \phi(b) - \phi(a)]
clever fjordBOT
frosst
fallow wedge
So do I calculate phi (b) wrt the partials x, y, and z?
merry finch
you need to find the scalar potential function phi
you do that by integrating each component of F wrt the partial
see if this phi exists
if it doesn't exist then F is not conservative
odd edgeBOT
@fallow wedge Has your question been resolved?
winter breach
**lovely💐**
Since $C$ is a straight line segment from $(0,0)$ to $(1,1)$, we can parameterize it using a parameter $t$ such that:
$$
r (t)=(t, t), \quad \text { where } 0 \leq t \leq 1
$$
$$r (t)=(t, t) \Longrightarrow \frac{d r }{d t}=(1,1) \Longrightarrow d r =(1,1) d t$$
$$F (x, y)=\left(2 x y, x^2-y^2\right) \Longrightarrow F (t, t)=\left(2 t \cdot t, t^2-t^2\right)=\left(2 t^2, 0\right)$$
clever fjordBOT
dghf
winter breach
$$F (t, t) \cdot \frac{d r }{d t} d t=\left(2 t^2, 0\right) \cdot(1,1) d t=2 t^2 \cdot 1+0 \cdot 1 d t=2 t^2 d t$$
$$\int_0^1 2 t^2 d t=2 \int_0^1 t^2 d t=2\left[\frac{t^3}{3}\right]_0^1=2\left(\frac{1^3}{3}-\frac{0^3}{3}\right)=2 \cdot \frac{1}{3}=\frac{2}{3}$$
clever fjordBOT
dghf
merry finch
!nosols @winter breach
odd edgeBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
Available help channel!
analog lichen
odd edgeBOT
tawdry cave
You can rationalise the expression to achieve a stronger conviction of that.
odd edgeBOT
@analog lichen Has your question been resolved?
analog lichen
You can rationalise the expression to achieve a stronger conviction of that.
Yes i did that too
1/√n+1+√n
tawdry cave
Where will that converge?
odd edgeBOT
@analog lichen Has your question been resolved?
analog lichen
Where will that converge?
N tends to infinity 0
tawdry cave
N tends to infinity 0
Yeah. As n tends to infinity, 1/{{infinity tending}} goes to zero.
analog lichen
Yeah so?
It converges to 0
I am mixing some concepts
Is it not necessary condition?
@tawdry cave
analog lichen
.close
subtle iron
Can someone help me with this question, I don’t understand what this means.
mystic saffron
$16^{12} * 16^{-16}$
clever fjordBOT
Cnidarian
mystic saffron
$16^{12}$ means 16 is being multiplied by itself 12 times
clever fjordBOT
Cnidarian
subtle iron
So is 16^-16
16x(-16)x(-16)like that?
mystic saffron
nope
subtle iron
What does it mean
mystic saffron
$16^{-16}$ can be rewritten as $1/16^{16}$
clever fjordBOT
Cnidarian
subtle iron
So it’s
16^15 basically?
subtle iron
I just don’t get it when there is a negative symbol
mystic saffron
what about 1 / 16^16
mystic saffron
I just don’t get it when there is a negative symbol
it means that the term is in the denominator
subtle iron
(1/16)x16 for 16 times
subtle iron
Alr
mystic saffron
lol
odd edgeBOT
If you are done with this channel, please mark your problem as solved by typing .close
subtle iron
.close
odd edgeBOT
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Available help channel!
wide drift
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wide drift
Im not really sure what to do with this one
Cause I assume u = sin(t)
(Not me googling this)
tulip mica
Or you can use the trigo formula
wide drift
Or you can use the trigo formula
I need to google that too 💀
tulip mica
I need to google that too 💀
Nah I think u got this
tulip mica
Write sin^3 t = (sin t)(sin^2 t)
Then use formula and u substitution
orchid torrent
I just realised I misread it as sin^2 rather than sin^3
wide drift
Sin(t)(1 - cos^2(t))
tulip mica
Ok u corrected it
tulip mica
Nooo
wide drift
no?
tulip mica
Use u substitution
tulip mica
Yess
tulip mica
Yup now integrate ez
tulip mica
Yup
tulip mica
Yayaayy
tulip mica
Np!
wide drift
Oh wait
I wanna try the other way
Cause I need to learn all of these formulas QQ
Oh
wide drift
Hahaha oops
tulip mica
I think they misread it
tulip mica
Have fun integrating!
wide drift
🤗
tulip mica
Byee
odd edgeBOT
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Available help channel!
quaint bolt
hyia! I have a question
odd edgeBOT
hyia! I have a question
quaint bolt
one, i don't know what RR is
oh, and a and b are rational and I have to prove that x is rational
so firstly can someone help me understand what RR is
then if i still cant solve it, i'll ask for a hint
north badger
I think it's a mistype for $\mathbb R$, i.e. the real numbers. So $x\in\mathbb R \setminus{0}$
clever fjordBOT
π=√g
north badger
What is the whole question?
quaint bolt
That translates to "Consider numbers a and b, where x belongs to that RR. Knowing that the numbers a nd b are rational, prove that x is rational"
north badger
Yeah that is correct, remember the rationals are a subset of the reals
The question asks that if a and b are rational, then prove x is rational too
north badger
Yeah, or if you can find x = ... where the dots only have a's and b's, then it will show x is rational
Like if you could get x = (6a+2) / (a + 4b), then x would be rational
Gtg now tho.. have fun
quaint bolt
Like if you could get x = (6a+2) / (a + 4b), then x would be rational
that's pretty hard though
<@&286206848099549185> i've gotten down to px^2 + (p-q)x + p = 0
is that enough to prove that x is rational
because to me it looks like it
odd edgeBOT
@quaint bolt Has your question been resolved?
quaint bolt
<@&286206848099549185>
odd edgeBOT
@quaint bolt Has your question been resolved?
odd edgeBOT
@quaint bolt Has your question been resolved?
quaint bolt
<@&286206848099549185>
north badger
<@&286206848099549185> i've gotten down to px^2 + (p-q)x + p = 0
Hey im back, that doesn't look very correct as p=1 and q=4 does not give a rational x, (and p=1, q=2 gives an imaginary x)
Ill have a go at the problem
north badger
<@&286206848099549185>
Had a look but can't solve it, you might first have to find conditions on a and b (or c=1-1/a, d=-1-b to make things simpler) for which a solution x exists, and then show that x is rational...
north badger
Had a look but can't solve it, you might first have to find conditions on a and ...
Cuz the two equations imply $$x=\frac{-c \pm \sqrt{5c^2 + 4d - 1}}{2(c^2 + d - 1)}$$ which isn't rational for \textit{all} $c$ and $d$
clever fjordBOT
π=√g
north badger
Oh, actually i just missed an extra step
It gives $x=\frac{c^2 + d}{c(2-c^2-d)}$ which is rational!
clever fjordBOT
π=√g
odd edgeBOT
@quaint bolt Has your question been resolved?
tepid cave
Hello, just a quick question about the topic of big O notation. I was asking a programming community about it and got the answer 'experience bruh'. Quite literally. And I've seen a few articles on it that seems to break it down better, but what bemuses me is how a number which is a specific quantity, like log(n), or n^2, looks like it should be calculatable by some set of steps. But that's rarely what I have gotten referred to in the past. And now, when I do have some freetime and some resources at hand, maybe I sorta nail my understanding for it. I do understand the premise of it at least.
But in the end I'm still curious, why is there so much vague info about big O notation? They don't seem like they're... wrong, these tutorials I often stumble upon are probably written by skilled people, but there's still this weird... lack of care with whatever is written about it. Why?
tepid cave
Another tendancy is that explanations of big O notation tend to consist solely of 3 examples. A notation of
-n steps
-n^2 steps
-log2(n) steps
Which feels... lacking. I can't do anything with that.
odd edgeBOT
@tepid cave Has your question been resolved?
south plume
Another tendancy is that explanations of big O notation tend to consist solely o...
O(n!) is the complexity for bogosort xd
n*log(n) is something i see sometimes too
tawdry cave
Hello, just a quick question about the topic of big O notation. I was asking a p...
You are right. Typical programmer blogs or tutorials on the topic won't go into deep details as they are often written with the purpose of serving as a quick how-to guide for programmers sitting for interviews where interviewer will ask cliche complexity questions. That is why there are also a lot of tricks for finding complexity instead of a rigorous way. Of course, those tricks are accurate but we rarely see any insight into how they work.
As about the definition, it's not vague really. It's just that while figuring out the complexity, people often don't employ the definition and move ahead with what seems to be the case, which more often than not, is the case.
tepid cave
You are right. Typical programmer blogs or tutorials on the topic won't go into ...
Huh, that's pretty interesting. Yeah, I feel like I'm about to sort this out at some point. Thank you a lot for your input, I hope you have a good rest of your evening/day.
static totem
i don't think it's true?
like if you mean the definition somehow tells you how to get an exact number
the definition as far as i know is like, "if g(x) is always less than cf(x) starting from some x, for some c, then g is in O(f)"
so 3x² is in O(x³) by definition
it doesn't even require that it's the "smallest" function, and it wouldn't help if it did
odd edgeBOT
@tepid cave Has your question been resolved?
subtle canyon
odd edgeBOT
subtle canyon
Is this correct
nimble blaze
no
odd edgeBOT
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Available help channel!
bright cliff
got this question in an assignment, and managed to do everything up to the final part about the median. I got my value for M as root(9/2), which doesnt fit the equation. I even tried asking chatgpt to see if i'd missed something basic, but it also arrived at the same answer and got stumped. for reference k=4
nocturne belfry
got this question in an assignment, and managed to do everything up to the final...
are you sure k = 4?
this is a pdf right
i get that F(x) = (k-1)/9
bright cliff
are you sure k = 4?
confirmed that with chat too yeah
clever fjordBOT
jan Niku
bright cliff
yeah exactly
nocturne belfry
,w integrate 2/27 x (k-1) with respect to x
clever fjordBOT
nocturne belfry
so $\frac{9(k-1)}{27}$
clever fjordBOT
jan Niku
nocturne belfry
or $\frac{k-1}{9}$
clever fjordBOT
jan Niku
wicked kestrel
9/27 is not 1/9 jan niku
wicked kestrel
good morning
nocturne belfry
thanks
@wicked kestrel maybe you should help ike finish the problem
if u are not busy
wicked kestrel
man
nocturne belfry
well okay so $P(x) = \int _0 ^x f(t) \dd t$
clever fjordBOT
jan Niku
nocturne belfry
which should be ... $\qty( \frac 23 )^3 \frac{x^2}{2}$?
clever fjordBOT
jan Niku
nocturne belfry
got this question in an assignment, and managed to do everything up to the final...
but then i get a different answer too, i think?
,calc (3/2)^(3/2)
clever fjordBOT
Result:
1.8371173070874
,calc (9/2)^(1/2)
clever fjordBOT
Result:
2.1213203435596
mines a bit lower
bright cliff
goddamn
nocturne belfry
whats up
bright cliff
decided i'd try find the roots of the equation directly and apparently they give M to be this
nocturne belfry
,w integrate 2/27 (4-1) t from t=0 to t=x
clever fjordBOT
nocturne belfry
sure
but then
man it was a mistake to jump into a help channel right when i woke up sorry
im movin slow
bright cliff
LMAO no worries
nocturne belfry
lemme do this again more slowly
bright cliff
substituted the goofy surd one back in to confirm and it checks out
the question might just be cooked ngl
clever fjordBOT
jan Niku
nocturne belfry
then $x = \frac{3}{\sqrt 2}$
clever fjordBOT
jan Niku
nocturne belfry
which is what you got 
bright cliff
exactly LMAO
clever fjordBOT
Result:
2.1213203435596
as in i tried to satisfy the equation and it didnt work
nocturne belfry
hmm well
itd be the x such that $\int _0 ^x f(t) \dd t = \int _x ^3 f(t) \dd t$ right
clever fjordBOT
jan Niku
nocturne belfry
we could solve this instead and see if we get the same
we know its $\frac{x^2}{9} = \int _x ^3 f(t) \dd t$
clever fjordBOT
jan Niku
wicked kestrel
the question might just be cooked ngl
i'm also of the opinion that something seems to have gone wrong in the question if my opinion means anything
nocturne belfry
hello again steak 
wicked kestrel
good evening
bright cliff
how's lunch
wicked kestrel
,w solve 4x^3 - 30x^2 + 81=0
clever fjordBOT
nocturne belfry
**jan Niku**
this was dumb to try my bad
wicked kestrel
,w integrate 2x/9 from 0 to sqrt(9/2)
clever fjordBOT
nocturne belfry
wicked kestrel
this says the median is 9/2 + sqrt(3)/2
nocturne belfry
i see
wicked kestrel
indeed
also there are 2 question cs
also this wording makes my teeth itch
i hate this question
bright cliff
yeah might just call it a day
i'll confirm with the other guys if it is indeed chalked
thanks for the help
nocturne belfry
theres some typo in that last equation
id bet a dollar on it
I dont see the other integer solution though
which is strange
so my guess is that they were working with the wrong M
or made some otherwise stupid algebra mistake like me
its a stupid part c anyways
because either it doesnt work or its arithmetic
wicked kestrel
they are both the stupid part c
bright cliff
LMAO
wicked kestrel
jan niku how could you suggest i solve such a dorky question
nocturne belfry
well i was worried for a second
but then i realized its a stupid problem
you always have an out with stupid problems
you can just go well of course i got it wrong the problem is smoked or cranked or whatever the kids say
wicked kestrel
i'll confirm with the other guys if it is indeed chalked
chalked
odd edgeBOT
odd edgeBOT
Available help channel!
elder fjord
trying to do part one
odd edgeBOT
trying to do part one
elder fjord
F(x) we got to pick our own one
so i choose 2x + 6
but i need help my teacher doesnt explain good and idk if im wrong
(ping me when u can help me please and thank you)
elder fjord
trying to do part one
for part 2 of this ^i have this
( havent compressed it by 3 or down shift it by 10 yet )
open knot
for part 2 of this ^i have this
not sure why you've shown the one before then
you use f(x)=2x+3 for Part 2 yh, but where did you use f(x)=2x+6
elder fjord
not sure why you've shown the one before then
shi i forgot to change it mb let me fit it rn
elder fjord
you use f(x)=2x+3 for Part 2 yh, but where did you use f(x)=2x+6
is part one right
elder fjord
trying to do part one
^?
open knot
kk, we seek a function that first increases the value by 10
and then triples it
so we could construct f(x) = (x+10)*3
since you can insert any x, it'll increase its value by 10 and then multiply the value by 3
not sure why they worded it as "move point upwards and stretch vertically", but I'm probably missing additional context
then for Part 2 they want you to find the according inverse function, that undos the operation of f(x)
open knot
ah yh k
well then what I mentioned above
the inverse function in part 2 would perform the same steps, but in reverse:
so divide by 3, then decrease by 10:
f^-1(x) = x/3 - 10
elder fjord
yes but it says i need to find f(x)
elder fjord
you use f(x)=2x+3 for Part 2 yh, but where did you use f(x)=2x+6
so 2x+6 +10
2x + 16
3(2x +16 )
6x + 42
f(x) = 6x + 42 ?
open knot
so 2x+6 +10
2x + 16
3(2x +16 )
6x + 42
not sure what you intended to calculate
elder fjord
not sure what you intended to calculate
is that right ?
i was just showing the steps
open knot
well then no, I don't quite get why you start with 2x+6+10
it asks us to first start finding a function f(x)
which performs the +10 and then the *3
so f(x) = (x+10)*3
elder fjord
it asks us to first start finding a function f(x)
F(x) = 2x + 6
Y = 2x + 6
x = 2y + 6
x – 6 = 2y
(x –6=2y )/2
(x-6)/2=y
1/2 x-3+10
3(1/2 x+7)
3 *1/2 x + 21
open knot
you don't get to "choose" f(x), you're restricted to the demands of the tasks
open knot
so f(x) = 2x+6 wouldn't satisfy the conditions of
"move by 10 and stretch by factor of 3"
f(x) = (x+10)*3 would however
or what remains unclear?
elder fjord
so im a little confused
i have to find f(x) then do + 10 then * 3
how do i find my f(x)
open knot
i have to find f(x) then do + 10 then * 3
nop
you find f(x) *that itself performs +10 and 3
f(x) is the function that does these operations
so for any input x, f(x) shall first add 10 to it, so x+10
and then multiply the result of that by a factor of 3: (x+10)*3
the tasks states that f(x) shall fulfill these requirements
meaning f(x) = (x+10)*3
gtg hope another passenger will take over :)
elder fjord
trying to do part one
okay so for my part one ^ i write down
F(x) + 10
3(F(x) + 10)
3*F(x) + 30
elder fjord
gtg hope another passenger will take over `:)`
oh damn okay ty for trying to help me
open knot
okay so for my part one ^ i write down
F(x) + 10
3(F(x) + 10)
3*F(x) + 30
No f(x) = 3x + 30 :D
why do you try to insert f(x) into the formula
elder fjord
why do you try to insert f(x) into the formula
the way my teacher explains it
open knot
the way my teacher explains it
this looks like pretty horrible notation
okay let's take one step back instead then
elder fjord
without the mark ups
open knot
f(x) first of all would mean that you have a function called f that takes one input, which is called x
open knot
without the mark ups
yeah no still horrible 😄
open knot
f(x) first of all would mean that you have a function called `f` that takes one ...
that is clear right @elder fjord
elder fjord
f(x) first of all would mean that you have a function called `f` that takes one ...
eys
yes
open knot
k, and if we want to define that function, the notation is f(x) = ...
and some expression would be on the right side
for instance f(x) = 5
then this function f always yields the output 5 regardless of the input x
whereas if I'd write f(x) = x then that function yields the input as output
e.g. f(7) spits out 7
now, if we have a task where we shall find a function
that satisfies specific conditions
like "this function should first add 10 to the input and then multiple that with 3"
then the function f(x) = (x+10)*3 fulfills these conditions
since we can take any input x and it'll add 10 to that input and then multiply the result by 3
the iterative process outlined in the slide you've sent above
is to iteratively change the function matching the conditions
here your teacher started with
f(x) = x
which he/she for some odd reason denoted as "Y = f(x)"
then he changes the function to suit the restriction "the function should add 10":
f(x) = x + 10
and then he changes it so suit the restriction "the function should then multiply by 3":
f(x) = (x + 10)*3
mostly clear? @elder fjord
elder fjord
i understand it but the way he explains it makes my head explode idk what to do
i understand what u are saying but the way my teacher says it idk
odd edgeBOT
@elder fjord Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
mystic saffron
odd edgeBOT
mystic saffron
im not getting what they mean by |g(x)| < M
dawn tiger
M is just some finite number since they can conclude that g is bounded near a for the limit there to exist
they use it to simplify the inequality
mystic saffron
M is just some finite number since they can conclude that g is bounded near a fo...
so M is like g(x) where x approaching a ?
dawn tiger
M isn't necessarily the limit of g(x) as x approaches a
for example, if g(x) = x^2 and a was 0, then the limit at a is 0, but they're saying in a neighborhood around 0, g(x) is bounded by some finite value M
mystic saffron
ahhhhh i get it
dawn tiger
M isn't 0 in this case since clearly |g(x)| < 0 isn't true
but just some other number
like 1
mystic saffron
like 1
OHHHH now it makes sense like limit exists at a for g(x) so it will have some value M around that neighborhood
(x-a, x+a)
dawn tiger
not that it has some value M, but that its size is always finite
the magnitude of the output never exceeds M in that neighborhood
mystic saffron
the magnitude of the output never exceeds M in that neighborhood
oh!! so its like
maximum value
in that neighborhood?
dawn tiger
sure, maybe not exactly the maximum
since it's strictly less than M
in that neighborhood
but it's like an excluded upper bound
mystic saffron
that makes sense
dawn tiger
it's all just a fancy way of saying your function doesn't blow up to +/- infinity in a neighborhood of a
mystic saffron
it's all just a fancy way of saying your function doesn't blow up to +/- infinit...
we are putting a upper bound on it got it
this might help in other proofs too
so thats Cool
i learnt a new thing
@dawn tiger thank you
it may become mroe and more clearer
as time pass
i have "some" idea of it now
.close
odd edgeBOT
mystic saffron
it's all just a fancy way of saying your function doesn't blow up to +/- infinit...
i was thinking about what if its going negative the value of g but then i realize we are taking absolute value
its making much more
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mystic saffron
sense
odd edgeBOT
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glacial timber
is this correct?
odd edgeBOT
is this correct?
glacial timber
prove gcd(a + b, b) = gcd(a, b)
let gcd(a, b) = x, clearly x divides a and b,
let a = xc, b = xd
gcd(xc + xd, xd) = gcd(x(c+d), xd) = x
for this to be true, gcd(c+d, d) = 1, which is true
since gcd(a, b) = x = gcd(xc + xd)
divide by x to get
1 = gcd(c, d) = gcd(c + d, d)
odd edgeBOT
@glacial timber Has your question been resolved?
slow sphinx
why did you divide by x at the end
since gcd(a,b)=x=gcd(xc+xd, xd)
and gcd(xc+xd,xd)=gcd(a+b,b)
=> gcd(a,b)=gcd(a+b,b)
you need to conclude with the statement you're trying to prove
also mention that c and d are coprimes, thats what makes this true
glacial timber
also mention that c and d are coprimes, thats what makes this true
thanks for the help btw
actually nvm i get it
i was trying to say c d are coprimes by the last few lines of the proof
slow sphinx
why are c, d coprimes?
you said a=xc, b=xd
and gcd(a,b)=x
slow sphinx
np
glacial timber
.close
heady plover
$\text{suppose}\mid x\mid\neq0\text,\text{slove}\lim_{n\to\infty}\cos\frac x2\cos\frac x4\cdotp\cdotp\cdotp\cos\frac x{2^n}$
clever fjordBOT
nino
dawn tiger
this might be useful
odd edgeBOT
@heady plover Has your question been resolved?
heady plover
this might be useful
what's this? where did you know this? Seems very useful
meager juniper
This is just repeated application of the double angle formula
sin(2x) = 2 sin x cos x
Solve for cos x
@heady plover
mystic saffron
damn so many formulas that i need to mermorize<:angerysad:602329540131553290>
You don't need to memorize anything just practice more and use them more everything will get obvious in no time
odd edgeBOT
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rich dust
Hi, are Physics questions allowed here?
odd edgeBOT
Hi, are Physics questions allowed here?
static totem
yeah
rich dust
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mystic saffron
odd edgeBOT
mystic saffron
i can simplify the rxthesum - the sum
but i cant seem to make the jump to find a result for Sn
pls help
summer cradle
but i cant seem to make the jump to find a result for Sn
what did you get after simplifying it?
mystic saffron
r^n - 1
i think
bc rSn is the sum from i=1 to n
and Sn is from 0 to n-1
it takes 1 off
and r^n is left
but idk how to get a formula for Sn
i assume it has r in the denominator but
idk
summer cradle
so we have $rS_n - S_n = r^n - 1$
clever fjordBOT
sharp #1 simp
mystic saffron
yep
oh
im
stupid
ggs
take a factor of sn out
and divide by
r-1
how
tysm
summer cradle
yep
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glad shell
$$f(x) = k(x-a)(x-b)(x-c)(x-d)(x-e)(x-f)$$
$$f'(x) = 6(x^5-8x^4+24x^3+Ax^2+Bx+c)$$
if $ e\leq k_1 $ and $f \geq k_2$
$$a<b<c<d<e<f$$
find $k_1+k_2$
clever fjordBOT
xd_senBugha
odd edgeBOT
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glad shell
<@&286206848099549185>
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@glad shell Has your question been resolved?
elder fjord
.reopen
odd edgeBOT
@glad shell Has your question been resolved?
glad shell
<@&286206848099549185>
lapis linden
what is 8+8
glad shell
<@&286206848099549185>
glad shell
Any help?
odd edgeBOT
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hoary yacht
Hi
cold sage
its the kth entry of the point
like if i have (1,2,3,4,5)
4 is x_4 or the 4th entry (k=4)
hoary yacht
Kth?
hoary yacht
Yes
cold sage
(x_1,...,x_n) has n entries
hoary yacht
Yes
cold sage
x_k represents the kth entry along for (x_1,...,x_(k-1), x_k, x_(k+1), ..., x_n)
hoary yacht
Lost you
cold sage
if k was 3, then it would be the third coordinate
hoary yacht
Kth?
cold sage
like first second, ..., thirtieth
kth
K (natural number) th, coordinate K
x_1 (k=1), x_2 (k=2), ..., x_n (k=n)
first, second,... nth
hoary yacht
So k is just a coordinate? What are they trying to get across
cold sage
its just a way to denote each individual coordinate entry you have
hoary yacht
When do i use k like that
cold sage
say you have two vectors, and youre doing the dot product of them, its much easier to denote this in sigma notation if you have notation for each entry
its more useful when youre doing things with matrices as well, there we write x_ij for the ith row and jth column
makes it a lot easier to write things out
hoary yacht
Im just starting with the book linear albert done right, i dont know what a sigma notation or a dot product is 😅
Is it okay if i only understood your matrice explenation?
Also do we define k later or we just say for any k from 1 to n
cold sage
k is just in the set of natural numbers from 1 to n
written here as k in (1,...,n)
cold sage
x_1 (k=1), x_2 (k=2), ..., x_n (k=n)
this is all they really mean
odd edgeBOT
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granite delta
i dont get this. the area of the shape should be 2400 cubic feet. and when we try to find the answer to the question i have to only get the surface area of the top and bottom and multiply that by 10 right? and then subtract it from 2400 right?
granite delta
so i got 2400 in the first question
then i do 12x4 for the area of the bottom face
i get 48
and i multiply that by 2 for 96 because theres two faces not being used
and then multiply it by ten bc there is ten boxes
i get 960
then i subtract 960 from 2400 to get 1440
that was my thought process
vapid quartz
youre trying to get surface area
granite delta
what did i do wrong?
vapid quartz
which is all the areas added together
2400 is the volume (ft^3) and you need surface area (ft^2)
try getting all the visible faces instead of basing off of the volume
granite delta
ohhhh
that makes sense
wait but im still not getting it
so now
i get every side add them together and multiply that by ten
so 12x5= 60
and then 60 again for the other side
wait
nevermind
im stupid
silly mistake
okay thanks
alot
.close
odd edgeBOT
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grim coral
would proving that ${\frac{n}{4^n}}{n=1}$ is decreasing and that $\lim\limits{n\to\infty}\frac{n}{4^n}=0$ be sufficient to show that $\sum\limits_{n=1}^{\infty}\frac{n}{4^n}$ converges?
lyric dust
maybe im misunderstanding what u are trying to say but the summation does not decrease
if u are beginning for positive x and its an exponentional as the denominator, there is no negative terms and hence ur summation will not decrease
grim coral
let me edit that
clever fjordBOT
lILi
grim coral
I meant to refer to the sequence of the series
i'm trying to find the convergence of the series but i'm not sure how to go about it for this one
I know all of the convergence tests apart from the root test
but i've seen my professor use the fact that the sequence is decreasing and that the limit of it as it goes to infinity is zero, as proof that the series containing that sequence converges
but it confuses me because I know that 1/n diverges
lyric dust
convergence and diverence isnt part of highschool syllabus so im not too well versed but i believe its sufficient n/4^n < 1 as n approaches infinite
grim coral
if it doesnt equal 0 as it goes to infinity then it must diverge
orchid torrent
would proving that $\{\frac{n}{4^n}\}_{n=1}$ is decreasing and that $\lim\limits...
If you're asking if $a_n$ being a decreasing sequence that tends to zero is sufficient to show that $\sum a_n$ converges, then no
clever fjordBOT
Civil Service Pigeon
orchid torrent
an easy counterexample is 1/n
grim coral
that's what I thought
lyric dust
oh sorry i meant to write that (n+1)/4^n+1 / (n)/(4^n)
grim coral
so i'm not sure why my homework assumes that it is a valid proof
oh, ratio test
i'm not sure why I hadn't thought to use that
lyric dust
it should evaluate to 1/4 as n approaches infinite and i think that is sufficient to show it converges
grim coral
i'll work through it myself but yeah that would be proof
again idk why i hadn't thought to use ratio test
Well, thank you
.close
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winter spruce
I have to solve this integral. Here's what's going through my mind: √(x-1) could maybe be transformed in such a way as to be the derivative of a trigonometric function. √(x^3) could be simplified as √(x) * x. Maybe I could switch the numerator and the denominator by saying that the whole thing is ^(-1/2). That could get me closer to an immediate integral.
Am I on the right path?
winter spruce
winged rune
In the end it should be divided by x not multiplied
winter spruce
right
I don't think going with substitution would help me. Maybe by parts?
i think im onto something
i came to this conclusion but it's wrong. How so?
winged rune
You're supposed to pick u and v from the integral right?
So you picked v as √((x-1)/(x)) and du as 1dx... And what about the 1/x ?
odd edgeBOT
@winter spruce Has your question been resolved?
winter spruce
I'm looking at my teacher's solution to help me through this and he apparently shifts the ^2 from the denominator to the nominator. Must be a mistake right?
.close
odd edgeBOT
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slow thorn
I have a solution for this problem. Can anyone say if it is right or if it is wrong? If it's wrong, please tell me my mistake.
slow thorn
Step 1: Simplify Factorials
Simplify the factorials in the equation:
$\left(n^{2}+1\right)! = (n^{2}+1)(n^{2})(n^{2}-1)!$,
$\left(n^{2}\right)! = (n^{2})(n^{2}-1)!$,
$\left(n^{2}-1\right)! = (n^{2}-1)(n^{2}-2)!$.
Step 2: Substitute Factorials
Substitute the simplified factorials back into the equation:
$\frac{(n^{2}+1)(n^{2})(n^{2}-1)! + (n^{2})(n^{2}-1)! + (n^{2}-1)(n^{2}-2)!}{(n+1)!} = (n+3)^{2}$.
Step 3: Cancel Common Terms
Cancel out common terms in the numerator and denominator:
$\frac{(n^{2}+1)(n^{2}) + (n^{2}) + (n^{2}-1)}{(n+1)!} = (n+3)^{2}$.
Step 4: Expand and Simplify
Expand and simplify the numerator:
$\frac{n^{4} + n^{2} + n^{2} + 1 + n^{2} - 1}{(n+1)!} = (n+3)^{2}$,
$\frac{n^{4} + 3n^{2}}{(n+1)!} = (n+3)^{2}$.
Step 5: Simplify Further
Simplify the expression:
$\frac{n^{2}(n^{2} + 3)}{(n+1)!} = (n+3)^{2}$.
Step 6: Rewrite Factorial
Rewrite the factorial as $(n+1)! = (n+1)n!$.
Step 7: Cancel Terms
Cancel out common terms:
$\frac{n^{2} + 3}{n+1} = (n+3)^{2}$.
Step 8: Solve for n
Solve the equation for $n$:
$n^{2} + 3 = (n+1)(n+3)$,
$n^{2} + 3 = n^{2} + 4n + 3$,
$4n = 0$,
$n = 0$.
clever fjordBOT
shlok-ggg
worn sandal
ChatGPT ahh solution
icy kindle
you definitely can't cancel like that....
those terms just vanished, with no actual cancelation
worn sandal
@slow thorn You can also see pretty easily that n=0 isn't a solution by plugging it in. Do the problem yourself instead of asking AI.
odd edgeBOT
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slow thorn
.close
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thick pebble
so far i have sin 2x = +-1
odd edgeBOT
so far i have sin 2x = +-1
sand horizon
so far i have sin 2x = +-1
What if you let X = 2x ?
thick pebble
sinx = +-1
sand horizon
Yes and solve for X now
sand horizon
i get pi/2?
And ?
thick pebble
- 2pin
sand horizon
Its for 1, what about -1 ?
thick pebble
3pi/2 + 2pin
sand horizon
Exact so those two are equal to X, but X = 2x so we have ?
thick pebble
pi/4 + pin, 3pi/4 + pin
sand horizon
Gg wp
thick pebble
its asking for degrees though
thick pebble
0 <= x <= 360
sand horizon
Yeah so x is in [0;2pi]
So we have 4 solutions
x= pi/4
x= 3pi/4
x=5pi/4
x = 7pi/4
Now i guess you convert them into degrees
Since pi/4 is 45°
odd edgeBOT
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bright topaz
Hello, I'm in Calculus 3 and need help with my homework assignment on double integrals. Would anyone be willing to help walk me through things?
odd edgeBOT
Hello, I'm in Calculus 3 and need help with my homework assignment on double int...
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/
sacred stirrup
don't ask to ask, just ask
bright topaz
Sounds good, Here is my first question
sacred stirrup
were you able to sketch this?
if yes, your job is basically done. proceed with fubini
sinful grove
if yes, your job is basically done. proceed with fubini
they are specifically asked to calculate it geometrically
bright topaz
no i havent yet. Im going to be honest with you, I am really confused with this stuff. I am taking an 8 week college calc 3 course and we are going really fast. I also have a full time job so I am trying to learn as much as I can. Would you be willing to walk me through while challenging me with questions along the way?
sacred stirrup
they are specifically asked to calculate it geometrically
mb didn't check it
bright topaz
in this case are the bounds on the outer integral -2 and 2 or is it 0 and 4?
sacred stirrup
no i havent yet. Im going to be honest with you, I am really confused with this ...
pretty hard, https://www.youtube.com/playlist?list=PL4C4C8A7D06566F38 this is pretty goated though so follow it. I feel like you haven't studied this topic and hence you're not able to solve this because this problem is a rather elementary problem in multivariable calculus [LECTURE 16-18]
bright topaz
Yeah we just started this today, and this assingment is due wednesday. So I am just starting on double integrals
sacred stirrup
yeah do it, he explains it really well.
odd edgeBOT
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shell stirrup
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shell stirrup
so the question asked was "prove no a value in the line family is tangent to the circle c"
no when D > 0 I know it intersects at two points with the cirkel, I also know D = 0 means only one intersect between line and cirkel and when D is below 0 it never intersects
my answer is "correct", I just dont understand why
i can conclude from the d = 36 - 0 that a pretty much has no impact, no matter what the value is?
oh wait, since D > 0 it means its intersects twice with any a value
therefore is never tangent
Idk why I didnt think of that... sorry
!close
.close
odd edgeBOT
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odd elk
Wanted to see if these are correct.
odd edgeBOT
Wanted to see if these are correct.
sand horizon
Wanted to see if these are correct.
Why did you put -1 twice ?
odd elk
Why did you put -1 twice ?
Oh whoops. Is it just or-1, 1
sly loom
Hi, I have the equation: Factor the perfect square trinomial.
x^2 − 4x + 4 i got (2x+4)^2, I know the anwser is wrong im confused on where I went wrong
latent scaffold
If you have a perfect square, then $x^2 + bx + c = (x+\frac{b}{2})^2$.
clever fjordBOT
Azyrashacorki
odd edgeBOT
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quartz pier
odd edgeBOT
quartz pier
I dont understand why we cant use master method here
surely its not leaf heavy or balanced. but why isnt it root heavy?
odd edgeBOT
@quartz pier Has your question been resolved?
quartz pier
<@&286206848099549185>
odd edgeBOT
@quartz pier Has your question been resolved?
odd edgeBOT
@quartz pier Has your question been resolved?
warped grove
It's because for it to be root heavy, n log n needs to grow faster than n * n^epsilon for some small number epsilon
But log n grows slower than any n^epsilon, no matter how small epsilon is
even if it's n^0.000001 or something, log n still grows slower in the long run
so therefore n log n does not grow fast enough for it to be root heavy
does that answer your question @quartz pier
quartz pier
does that answer your question <@925451734380318801>
I have another quesrion regarding master theorem
I dont get this whole process.
- how does, 2^m = n imply m = log(n)?
shouldnt it have been: 2^m = n => mlog(2) = log(n) => m = log(n)/log(2)
-
How did they get rid of the 2 inside T?
-
and finally how do they even come up with a value for the subsituiton? is it just guessing?
warped grove
I dont get this whole process.
1) how does, 2^m = n imply m = log(n)?
shouldnt...
2^m = n means that m = log_2 n by definition
and then they didn't write the base of the log explicitly bc it doesn't rly matter for asymptotic analysis
quartz pier
2^m = n means that m = log_2 n by definition
oh because 2^log_2 will cancel out ?
warped grove
also, log(n)/log(2) (the thing you got) is the same as log_2(n)
quartz pier
warped grove
I dont get this whole process.
1) how does, 2^m = n imply m = log(n)?
shouldnt...
what is S?
odd edgeBOT
warped grove
.reopen
odd edgeBOT
✅
quartz pier
what is S?
idk either
warped grove
I dont get this whole process.
1) how does, 2^m = n imply m = log(n)?
shouldnt...
I think this is a typo, it should be
T(2^m) = 2T(2^(m/2)) + m
and then just define
S(m) to be T(2^m) and you get the equation for S
(because n is 2^m, so T(n) is T(2^m), not T(m))
to answer 3), the motivation for defining m to be log n is just simplifying the original equation to not have log n anymore
quartz pier
I think this is a typo, it should be
T(2^m) = 2T(2^(m/2)) + m
and then just defi...
so they subbed m instead of 2^m? Im a bit confused here
warped grove
so they subbed m instead of 2^m? Im a bit confused here
which part are you confused by
quartz pier
im confused how exacrly they got the equation for S(m)
warped grove
are you good with T(2^m) = 2T(2^(m/2)) + m
quartz pier
bow they transformed T(2^m) to S(m)
quartz pier
are you good with T(2^m) = 2T(2^(m/2)) + m
yep
warped grove
okay, now define a new function S
which is related to T by S(m) = T(2^m)
is that okay?
quartz pier
that part gets me confused. how exactly would one even do that
so are we saying that
quartz pier
S(m) = 2T(2^(m/2))+ m = T(2^m)?
warped grove
sure, but the point is that S is a new letter that we're introducing to make the problem easier
quartz pier
alright
warped grove
because T(2^m) is S(m) and T(2^(m/2)) is S(m/2)
quartz pier
oh
warped grove
so now we get the recurrence S(m) = 2S(m/2) + m
which is way simpler than the thing we started with
and now we can solve using the master thm
warped grove
you're welcome!
quartz pier
.close
odd edgeBOT
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minor cove
Can anyone help with this.
trying to find cos(B)
odd edgeBOT
Can anyone help with this.
trying to find cos(B)
agile crag
Can anyone help with this.
trying to find cos(B)
Yeah
So cos(B) would be the side adjacent from it over the hypotenuse
We can use the pythagorean theorem:
c^2 = 2^2 + 7^2 = 4 + 49 = 53
minor cove
ohh okay
agile crag
👍
minor cove
then I would solve that out right
agile crag
Tell me what you get for cos(B) so I can check it
agile crag
Yup, I got that for the hypotenuse as well
minor cove
Okay great
agile crag
now all you need is the side adjacent from angle B and setup the adj/hyp ratio
agile crag
That would be opposite of the angle
minor cove
Ahh so BC?
agile crag
Yup
minor cove
Okay
so 7 over root53
agile crag
Yeah
And if your teacher/professor wants you too, you can simplify that to 7*sqrt(53)/53
agile crag
You're welcome
minor cove
is there one more thing you can help me with? real quikc
agile crag
Sure
agile crag
So the part next to the x would be 102/tan(70)
minor cove
ohh okay
agile crag
Do you know anything like the law of sines or cosines?
minor cove
I think so
agile crag
Cool
minor cove
So i got 37.124
agile crag
For x, or for the green side of the pciture below:
agile crag
Okay
One thing I'd recommend is leaving numbers in intermediate calculations with 4 or more decimal places
minor cove
ohh okay got it
agile crag
So we'll leave it as 37.1250 (the fourth unrounded digit was a 9)
minor cove
okay got it that makes sense
agile crag
We can now use the pythagorean theorem to find the yellow side
agile crag
Tell me what you get for the yellow side and I'll confirm it
minor cove
okay great
agile crag
Now, we need to find the red and purple angles in the image below
minor cove
okay how would we get those
agile crag
We know that the sum of all angles in a triangle is 180
So given two angles of a triangle, we can solve for the third
agile crag
The red angle isn't 110
minor cove
oh shoot im trippin
agile crag
It's cool
agile crag
40 is close
agile crag
Yeah
agile crag
Here's the info we have
minor cove
okay
agile crag
Do you know the law of sines?
minor cove
ohh I knew that but i forgot now i learned it a while ago
agile crag
We know that sin(32)/108.5461 = sin(38)/x by the law of sines
And now we can solve for x
agile crag
Let me check mine real quick
minor cove
Okay
agile crag
Yeah, I get the same
minor cove
okay let me check if it worked
Oh yup it worked thank you so much!
Really appreciate it 🙏
agile crag
No problem! This was the first geometry problem I've done in a while, so this was really fun for me
agile crag
You're welcome!
You can type .close when you're done reviewing anything so the help channel can be re-used
odd edgeBOT
@minor cove Has your question been resolved?
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lime swift
Yea I must’ve messed up chain rule ∨ smthnt
odd edgeBOT
Yea I must’ve messed up chain rule ∨ smthnt
lime swift
Idk how though
(This is self study, and the book did not provide step by step for this solution)
inland rover
Idk how though
(This is self study, and the book did not provide step by step fo...
do you know differentiability rules?
inland rover
DQ/DT = DQ/Du * Du/DT
This is what I’m doing (u is u0 on the paper)
you dont ned to do this yk
lime swift
Yea Ik I can just f’(g(x))*g’(x), but I wanted to practice more on that way
I wanted to see what I did wrong this way
inland rover
DQ/DT = DQ/Du * Du/DT
This is what I’m doing (u is u0 on the paper)
if you want to do like this you have to define 'U'
lime swift
I did, wait I’m writing it more formal
inland rover
ok ill wait
lime swift
inland rover
oh a mistake is there in second line
lime swift
?
inland rover
you have to write T in terms of U if you want to do this