viral dagger
OHHH
#help-49
twilit field
midnight plankBOT
twilit field
Would this jsutification be enough : $\bigcup_{i \in \N} \R = \R$ and $\bigcup_{i \in \N} [i, i+1] = [1,\infty)$ as proven earlier
grand pondBOT
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runic hamlet
why should that be enough
limber panther
Lol isnt that just stating the answer
twilit field
Why is that not enough
I've proven $\bigcup{i \in \N} [i, i+1] = [1,\infty)$ just a while ago
grand pondBOT
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runic hamlet
ok sure but how is that related to the problem. you need to show how it is related and stuff
you cant just state some other facts and claim thats enough
in general $\bigcup_{n\in \bN} A_n \times B_n = \left(\bigcup_{n\in \bN} A_n\right)\times \left(\bigcup_{n\in\bN} B_n\right)$ is false
grand pondBOT
Denascite
runic hamlet
and it smells like you want to use that
limber panther
What is X?
runic hamlet
cartesian product
limber panther
and it smells like you want to use that
Is [1, infinity) the answer tho?
runic hamlet
no. and no one claimed that it is
limber panther
Oh I see
runic hamlet
there is no need to try to be smart to skip two lines of double inclusion
twilit field
let $A= {(x,y) : x \in \R , y \in [1,\infty)} $
\
Then $A \subseteq \bigcup_{i \in \N} \R \times [i, i+1]$ as the first element of every set in any element of $A$ is real and the second element lie in an interval $i\leq x <i+1$ by the archemidian property. The same is true for any element in $\bigcup_{i \in \N} \R \times [i, i+1]$.
grand pondBOT
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twilit field
How does this look
We now prove that $\bigcup_{i \in \N} \R \times [i , i+1] \subseteq A$. This follows again, as any element in $\bigcup_{i \in \N} \R \times [i , i+1] $ has a real number as its first element , and a number between two integers as its second
grand pondBOT
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runic hamlet
you are skipping over steps
nowhere in your proof have you used explicitly that you are dealing with a union
which is very important
twilit field
nowhere in your proof have you used explicitly that you are dealing with a union
in my second inclusion proof?
runic hamlet
in both
twilit field
I'm not sure what to do in that case
midnight plankBOT
@twilit field Has your question been resolved?
runic hamlet
how and where would your proof fail if this was about an intersection instead of a union
then rewrite those parts to more explicitly actually use that you have a union
twilit field
hmm
twilit field
how and where would your proof fail if this was about an intersection instead of...
If this were an intersection problem it would fail as $\cap \R \times [i, i+1]$ would be $\varnothing$
grand pondBOT
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twilit field
<@&286206848099549185>
We consider $A_i = \R \times [ i, i+1] \cup A_j = \R \times [j,j+1]$ We now prove that $ A_i \cup A_j = \R \times ( [i,i+1] \cup [j,j+1])$
grand pondBOT
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twilit field
Consider $(x,y) \in A_i \cup A_j$, by definition $x \in \R$ as the first element in both$A_i \text { and } A_j$ belongs to $\R, y \in [i,i+1] \lor [j, j+1]$, which is the definition of union, so $A_i \cup A_j \subseteq \R \times ([i,i+1] \cup [j,j+1])$
grand pondBOT
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midnight plankBOT
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last slate
midnight plankBOT
last slate
3/4
What are the bounds ?
also for 2 can theta have a negative number as one of the bounds?
like -pi/2 to pi/2
@gray widget
shit
sorry
<@&286206848099549185>
.close
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sturdy tendon
Is there a way possible I can calculate sample size on my ti-89 titanium calculator
midnight plankBOT
Is there a way possible I can calculate sample size on my ti-89 titanium calcula...
@sturdy tendon Has your question been resolved?
sturdy tendon
<@&286206848099549185>
midnight plankBOT
@sturdy tendon Has your question been resolved?
agile crystal
Is there a way possible I can calculate sample size on my ti-89 titanium calcula...
Yes u can
agile crystal
How then
Use the calculator home screen and add ur Enter the equation directly by substituting values for z , p , E...ECT and u calculator will return it to sample size
sturdy tendon
Okay
@agile crystal do you know when yto use a specific formulas for n value with trying to figure out the mean sizes or the population sizes
agile crystal
<@1004497619017551993> do you know when yto use a specific formulas for n value ...
It depends on the context and type of deta u are working with
This pretty much wt Ik ^^"
agile crystal
Thank you then
Anytime ^^
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short hill
can someone explain how F(4) becomes 0? wouldnt u try to integrate f(t) and do F(4) - F(x)?
midnight plankBOT
can someone explain how F(4) becomes 0? wouldnt u try to integrate f(t) and do F...
slender walrus
F(4) is a constant
the derivative of that is 0
short hill
how do we know F(4) is constant if we didnt integrate f(t)?
slender walrus
F represents an antiderivative function
F(4) is that function evaluated at 4
the result of that is some number independent of variables
short hill
d/dx F(4) = f(4) = (4^3 + 5)^1/2?
slender walrus
no
short hill
how not
slender walrus
$\dv{x} F(a) \redneq \left[\dv{x} F(x)\right]\eval_a$
grand pondBOT
ℝαμOmeganato5
short hill
is that 4 from f(x) or f(t) o.o
OH
it must be f(x)
oh
wait no its f(t)
nvm then
.close
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safe onyx
midnight plankBOT
safe onyx
I got the graph right, but for the area I messed up and I'm not sure why
I did 9 - (x^3)/3 as the antiderivative
dawn dagger
I did 9 - (x^3)/3 as the antiderivative
and the 9?
safe onyx
oh it needs to be 9x?
dawn dagger
yup
safe onyx
That's where I messed up tysm!
dawn dagger
np!
midnight plankBOT
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short viper
I need help with this
midnight plankBOT
I need help with this
wicked trench
I need help with this
with what?
lavish venture
with this
wicked trench
There's nothing
lavish venture
what have you tried so far
midnight plankBOT
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wide shell
Hello I’m working on trig equations and I’m having trouble with the equation, I have solved a few on my own but I find this one to be confusing because there is no constant to move to the right side.
Tan(x)-cot(x)=0
midnight plankBOT
Hello I’m working on trig equations and I’m having trouble with the equation, I ...
small jasper
Hello I’m working on trig equations and I’m having trouble with the equation, I ...
Multiply both sides by $\tan(x)$
grand pondBOT
Civil Service Pigeon
wide shell
Multiply both sides by $\tan(x)$
I’m still lost would that look like
(Tan(x)) Tan(x)-cot(x) = tan(x)
And would i distribute?
small jasper
Uh ig, but write your parentheses properly
Or just use $\tan x \cot x=1$ from the out
grand pondBOT
Civil Service Pigeon
wide shell
Sorry I don’t know how to type out equations
worthy wing
?
worthy wing
(tan x)(tan x - cot x) = tan x
What
lavish venture
tan(x) * 0 = tan(x)
small jasper
Oh bruh
lavish venture
🔥
small jasper
@wide shell ^
Yeah that too
So fix your right hand side and don’t drop the parentheses on the left
wide shell
Oof lol
worthy wing
worthy wing
You can also factor
Tanx - cotx = tanx - 1/tanx
Add fractions multiply by tanx
And factor tan^2x-1
You can then keep factoring
lavish venture
probably easiest to just do tanx = cotx imo
worthy wing
(Tanx-1)(tanx+1) = -tanx(cotx-1)(cotx+1)
worthy wing
probably easiest to just do tanx = cotx imo
We are here to do some math
Hold my beer
midnight plankBOT
@wide shell Has your question been resolved?
wide shell
This made me realize I suck at factoring
Where would I factor from here?
That’s If I understood it right
dusky harbor
Where would I factor from here?
How do you factor x^2 - 1
wide shell
So once it’s factored would I solve the numerator first by setting the terms in the parentheses to equal 0 and What happens to to the denominator?
obtuse willow
How do I handle dividing a rational function by a non-rational number? 2/3 divided by 3?
midnight plankBOT
How do I handle dividing a rational function by a non-rational number? 2/3 divid...
obtuse willow
I don't full remember/know how to do that.
Plus.
How do I remember/memorize the Unit Circle?
And the values of Sin and Cosine?
severe basin
How do I remember/memorize the Unit Circle?
hi
obtuse willow
Hallo.
severe basin
with the unit circle, you first have to remember that it goes cos, sin
severe basin
The way I remember this is that sin is secnod
severe basin
lol
anyway it's 1/2
so what does that make cos 30
(using these trig functions a lot or memorizing the value is very helpful in memorizing the unit cricle)
obtuse willow
square root of 3 over 2?
obtuse willow
I see.
obtuse willow
Cos - Sin +.
severe basin
yes
so you would do the same thing, just following the x and y signs
so what is sin 120
and cos 120
midnight plankBOT
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obtuse willow
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midnight plankBOT
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twilit field
midnight plankBOT
twilit field
I was thinking of contrapositive
If $T -\sqrt{2}I$ is not injective, then for some $v \in V$, $\norm{Tv} > \norm{v}$.
This tells us
$\norm{T(v-u)} = \norm{\sqrt{2} (v-u)}$
\
\
As $T -\sqrt{2}I$ is not injective, for some $v,u; v \neq u$; $T(v-u) = \sqrt2({v-u})$
\
Thus $\norm{T(v-u)} \geq \norm {v-u}$
\
As $T -\sqrt{2}I$ is not injective, for some $v,u; v \neq u$; $T(v-u) = \sqrt2({v-u})$
midnight plankBOT
@twilit field Has your question been resolved?
twilit field
<@&286206848099549185>
If $T -\sqrt{2}I$ is not injective, then for some $v \in V$, $\norm{Tv} > \norm{v}$.
This tells us
$\norm{T(v-u)} = \norm{\sqrt{2} (v-u)}$
\
\
As $T -\sqrt{2}I$ is not injective, for some $v,u; v \neq u$; $T(v-u) = \sqrt2({v-u})$
\
Thus $\norm{T(v-u)} \geq \norm {v-u}$
\
We now want to assure ourselves that this isn't always an equality
grand pondBOT
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midnight plankBOT
midnight plankBOT
✅
twilit field
If $T -\sqrt{2}I$ is not injective, then for some $v \in V$, $\norm{Tv} > \norm{v}$.
This tells us
$\norm{T(v-u)} = \norm{\sqrt{2} (v-u)}$
\
\
As $T -\sqrt{2}I$ is not injective, for some $v,u; v \neq u$; $T(v-u) = \sqrt2({v-u})$
\
Thus $\norm{T(v-u)} \geq \norm {v-u}$
\
We now want to assure ourselves that this isn't always an equality
\
to do so we look at the definition
$\sqrt{\ang{T(v-u),T(v-u)} }\geq \sqrt{\ang {v-u,v-u}}$
This gives us $\ang{\ang{T(v-u),T(v-u)} \geq \ang {v-u,v-u}$
grand pondBOT
A dense set(Ping when reply)
twilit field
If $T -\sqrt{2}I$ is not injective, then for some $v \in V$, $\norm{Tv} > \norm{v}$.
This tells us
$\norm{T(v-u)} = \norm{\sqrt{2} (v-u)}$
\
\
As $T -\sqrt{2}I$ is not injective, for some $v,u; v \neq u$; $T(v-u) = \sqrt2({v-u})$
\
Thus $\norm{T(v-u)} \geq \norm {v-u}$
\
We now want to assure ourselves that this isn't always an equality
\
to do so we look at the definition
$\sqrt{\ang{T(v-u),T(v-u)} }\geq \sqrt{\ang {v-u,v-u}}$
\
This gives us $\ang{\ang{T(v-u),T(v-u)} \geq \ang {v-u,v-u}$
grand pondBOT
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twilit field
If ( T - \sqrt{2}I ) is not injective, then for some ( v \in V ), ( | T v | > | v | ). This tells us
[
| T(v - u) | = | \sqrt{2} (v - u) |
]
As ( T - \sqrt{2}I ) is not injective, for some ( v, u ) where ( v \neq u ), we have ( T(v - u) = \sqrt{2} (v - u) ).
Thus,
[
| T(v - u) | \geq | v - u |.
]
We now want to assure ourselves that this isn't always an equality.
To do so, we look at the definition:
[
\sqrt{\langle T(v - u), T(v - u) \rangle} \geq \sqrt{\langle v - u, v - u \rangle}.
]
This gives us
[
\langle T(v - u), T(v - u) \rangle \geq \langle v - u, v - u \rangle.
]
grand pondBOT
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twilit field
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midnight plankBOT
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jolly roost
Find the angular velocity of an engine on a jet aircraft that turns at 12500 pm in exact radians per second?
how do i do the exact value in radians per second
i got that ω = 1250/3
but thats not per second
midnight plankBOT
Find the angular velocity of an engine on a jet aircraft that turns at 12500 pm ...
jolly roost
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regal barn
velvet oar
Could you translate the question to English? I tried using Google lens to translate but it didn't do very well
regal barn
In the given diagram TQ is bisector of PTR angle then TS will be?
velvet oar
@regal barn Have you tried anything yet? Do you have a guess which one might be right?
regal barn
I feel like there must be a theorem when it touches the circle and one is passed through it@velvet oar
velvet oar
Do you know the inscribed angle theorem?
It also applies when one of the segments is just touching
for example, angle P = (1/2) arc TR, and also angle STR = (1/2) arc TR
@regal barn by using ideas like this, you can find an isosceles triangle
regal barn
But it is asking for TS
velvet oar
It's asking which of those choices is equal to TS
You can use the angles to find an isosceles triangles, meaning that TS is equal to one of those choices
You can't find the actual length of TS, since no lengths are given
But you can show that it's equal to SQ, or TQ, or one of those other choices
,rcw
grand pondBOT
velvet oar
angle P = angle STR, since both angles subtend arc TR. I've marked them both as x
angle QRT = angle PTU, since both angles subtend arc TP. I've marked them both as y
@regal barn
Using the fact that TQ bisects angle PTR, you can find more equivalent angles
regal barn
Let me understand it thanks btw
velvet oar
np 👍
I added the point U, btw
feel free to ping me if you have any questions, I'll be up for a little bit still
midnight plankBOT
@regal barn Has your question been resolved?
twilit field
midnight plankBOT
twilit field
$\ang {u,u} - \ang {v,u}=0 \implies \ang{u-v,u} =0$ and $\ang{v,v} -\ang{u,v} = \ang{v-u,v}=0$ . Thus we have , on adding the two previosu equations $\ang{v-u, v-u}=0 \implies v-u=0 \implies v=u$
too handwavey?
spiral osprey
looks like circular reasoning. in the first step you're assuming that u-v = 0, or u=v. that is what you're trying to prove @twilit field
oh wait i misread
quiet hinge
Its fine
Just justify why <v-u,v>=0 and <u-v,u>=0 => <u-v,u-v>=0, because ||<v-u,v>=<u-v,-v> so 0=<v-u,v>+<u-v,u>=<u-v,-v>+<u-v,u>=<u-v,u-v>||
twilit field
Will do , thanks
grand pondBOT
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midnight plankBOT
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twilit field
midnight plankBOT
twilit field
We proceed by contrapositive
\
$$\sqrt{1-\norm{u}^2} \sqrt{1- \norm{v}^2} > 1- \abs{\ang{u,v}} $$
\
So
\
$$1 - \norm{u}^2)(1-\norm{v}^2)> 1 +\abs{\ang{u,v}}^2- 2 \abs{\ang{u,v}} $$
\
Thus
\
$$1 - \norm{v}^2-\norm{u}^2+ \norm{u}^2\norm{v}^2 > 1 +\abs{\ang{v,u}}^2 - 2 \abs{\ang{u,v}}$$
\
$$- \norm{v}^2-\norm{u}^2+ \norm{u}^2\norm{v}^2 > \abs{\ang{v,u}}^2 - 2 \abs{\ang{u,v}}$$
grand pondBOT
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west iron
Yeah it's probably not necessary
twilit field
Yeah, definitely a bad idea
west iron
Imagine fixing the values of ||u|| and ||v||
twilit field
$\norm{u}$ exists btw
grand pondBOT
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west iron
how can we then make the right side as small as possible
twilit field
I've already shown in my rough that $\sqrt{1-\norm{u}^2}\sqrt{1-\norm{v}^2} \leq 1$
grand pondBOT
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twilit field
I'll have to constrain that a bit more I guess
west iron
that's trivial isn't it?
twilit field
yeah, but squaring this allows us to exploit cauchy-scrhawrtz
west iron
**A dense set(Ping when reply)**
This is fine except the sign should just be flipped is all
since you're not doing contrapositive
twilit field
$(1-\norm{u}^2)(1-\norm{v}^2) = 1 - \norm{v}^2-\norm{u}^2+ \norm{u}^2\norm{v}^2 \geq 1 - \norm{v]^2 -\norm{u}^2 + \abs{\ang{v,u}}^2$
grand pondBOT
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west iron
random square bracket in there
$(1-\norm{u}^2)(1-\norm{v}^2) = 1 - \norm{v}^2-\norm{u}^2+ \norm{u}^2\norm{v}^2 \geq 1 - \norm{v}^2 -\norm{u}^2 + \abs{\ang{v,u}}^2$
grand pondBOT
Dreyuk
$(1-\norm{u}^2)(1-\norm{v}^2) = 1 - \norm{v}^2-\norm{u}^2+ \norm{u}^2\norm{v}^2 \geq 1 - \norm{v}^2 -\norm{u}^2 + \abs{\ang{v,u}}^2$
```Compilation error:```! Undefined control sequence.
<argument> \left \lvert {\ang
{v,u}}\right \rvert
l.59 ...- \norm{v}^2 -\norm{u}^2 + \abs{\ang{v,u}}
^2$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```
twilit field
oop
sorry
$(1-\norm{u}^2)(1-\norm{v}^2) = 1 - \norm{v}^2-\norm{u}^2+ \norm{u}^2\norm{v}^2 \geq 1 - \norm{v}^2 -\norm{u}^2 + \abs{\ang{v,u}}^2$
grand pondBOT
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west iron
I was gonna say no I think I just don't have \ang lol
twilit field
This tells us $1 - \abs{\ang{v,u}}^2 \geq 1-\norm{v}^2\norm{u}^2$
grand pondBOT
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grand pondBOT
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twilit field
$\rangle$
grand pondBOT
A dense set(Ping when reply)
west iron
ah whatever
hard umbra
I was gonna say no I think I just don't have \ang lol
prolly have inner
grand pondBOT
Dreyuk
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west iron
:kekehands:
twilit field
why get rid of the `||u|| and ||v||`?
They cancel out, no?
twilit field
with each other in the inequality, yes
west iron
Where did they ever enter the other side
I kinda missed they just magically appeared there without reason
twilit field
$1 - \norm{v}^2-\norm{u}^2+ \norm{u}^2\norm{v}^2 \geq 1 - \norm{v}^2 -\norm{u}^2 + \abs{\ang{v,u}}^2$
grand pondBOT
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twilit field
okay, so I have this
twilit field
Cauchy-schwartz
west iron
Why would you want to make the right side smaller
That just makes it so that proving this statement no longer proves the original
twilit field
I thought I can prove the square root terms are even smaller
twilit field
huh?
west iron
**A dense set(Ping when reply)**
Here?
west iron
The sign is flipped from here
twilit field
so I have $\sqrt{1-\norm{u}^2} \sqrt{1-\norm{v}^2} \leq 1- \norm{v}\norm{u}$
to prove instead
west iron
≤ you mean?
hard umbra
≤
grand pondBOT
A dense set(Ping when reply)
twilit field
Which looks easier to prove
west iron
Ok yeah this was basically my sol
2||u||^2||v||^2 ≤ ||u||^2+||v||^2
not hard to show
You can literally just move everything to one side and write it as a square lol
twilit field
Yeah
twilit field
Well, the only thing is I have to use the fact that the norm is less than or equal to 1
I guess the square root takes care of that
west iron
Yeah that's how we know the thing is smaller
west iron
but you can write something smaller as a square
^
twilit field
**A dense set(Ping when reply)**
I was actually thinking , this is equivalent to $1+\norm{v}^2\norm{u}^2 - \norm{v}^2 -\norm{u^2} \leq 1+\norm{v}^2\norm{u}^2-2\norm{v}\norm{u}$
which is always true
so we're done
west iron
I have no idea why we're randomly complicating the inequality by adding things to both sides but as long as you're convinced ig
twilit field
I have no idea why we're randomly complicating the inequality by adding things t...
This gives us what you mentioned
$0 \leq (\norm{v} -\norm{u})^2$
grand pondBOT
A dense set(Ping when reply)
west iron
I promise you this inequality doesn't simplify to that
grand pondBOT
A dense set(Ping when reply)
twilit field
**A dense set(Ping when reply)**
This does, doesn't it
hard umbra
using CS is a bit quicker 
,, \s {1-\norm u^2} \s {1 - \norm v^2} + \norm u \norm v \le 1
grand pondBOT
west iron
Doesn't quite look like it
twilit field
so I have $\sqrt{1-\norm{u}^2} \sqrt{1-\norm{v}^2} \leq 1- \norm{v}\norm{u}$
grand pondBOT
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twilit field
yes?
hard umbra
that's what you need to prove, not what you have
twilit field
yes
hard umbra
you prove it in 1 application of CS
twilit field
**A dense set(Ping when reply)**
I just applied cauchy schawrtz to get this
hard umbra
yes but then you prove that inequality using 1 application of CS
twilit field
Why is squaring both sides invalid
I'm dropping the norm sign for convience for now
$(1-u^2)(1-v^2) \leq 1-2vu+v^2u^2$
grand pondBOT
A dense set(Ping when reply)
twilit field
$1-v^2-u^2+v^2u^2 \leq 1-2vu+v^2u^2$
grand pondBOT
A dense set(Ping when reply)
twilit field
$0 \leq v^2+u^2-2vu$
grand pondBOT
A dense set(Ping when reply)
twilit field
which is always true
hard umbra
sure but to write the proof you need to run this the other way
probably not the best way to do this though
hard umbra
as i said it's literally just CS
hard umbra
,, \s {1-\norm u^2} \s {1 - \norm v^2} + \norm u \norm v \le 1
this inequality follows directly from CS
twilit field
yes
hard umbra
i'm not saying it follows from the original inequality in the problem statement
that's not true
i'm saying you prove what i wrote in that picture using CS
,, \s {1-\norm u^2} \s {1 - \norm v^2} + \norm u \norm v \overset{\text{by CS}}\le 1
grand pondBOT
feral sedge
simply inequality the inequality
twilit field
hmm
west iron
Try to write the left side as a dot product
hollow oyster
How tf do you think of this
snow is just that guy always with very clear concise proofs
hollow oyster
hi convergence
hello wai 
hollow oyster
simply inequality the inequality
hello Mqnic :3
west iron
Gauss pilled
hollow oyster
snow prolly felt rudin was too expository
twilit field
I suppose I could start by multiplying both sides by $|u||v|$
grand pondBOT
A dense set(Ping when reply)
west iron
Don't
west iron
Luckily the dot product induces one
twilit field
$\ang {(u\sqrt{1-{v}^2} , v \sqrt{1 - u^2}}$
grand pondBOT
A dense set(Ping when reply)
twilit field
I think something close to this
west iron
Second hint: it should not be a dot product of one dimensional vectors. That is just called a product
twilit field
$(v+\sqrt{1-u^2}) )(u+ \sqrt{1-v^2}$
grand pondBOT
A dense set(Ping when reply)
twilit field
No
not this either
I'm not sure
I think I'll do this instead : $0 \leq (\norm{v} - \norm{u})^2 \implies 2\norm{v} \norm{u} \ \norm{u}^2+ \norm{v}^2 \implies 1-\norm{v}^2 - \norm{u}^2 + \norm{v}^2\normPu}^2 \leq 1 -2 \norm{v} \norm{u} + \norm{v}^2 \norm{u}^2 \implies (1- \norm{u}^2)(1- \norm{v})^2 \leq (1- \norm{v} \norm{u})^2 \implies \sqrt{1- \norm{v}^2} \sqrt{1- \norm{u}^2} \leq 1- \norm{v} \norm{u}$
oh boy
yeah
this works
grand pondBOT
A dense set(Ping when reply)
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hard umbra
this is just reproving CS
twilit field
Yeah,but I'm not able to think of a CS that will yield what you suggested
hard umbra
theres only 1 sensible way to write this as a dot product
and that 1 sensible way is the way that works
twilit field
Well, the issue is , I can't seem to think of it
hard umbra
write out the definition of the dot product
twilit field
$(u,\sqrt{1-u^2})(v,\sqrt{1-v^2})$
grand pondBOT
A dense set(Ping when reply)
hard umbra
okay now use CS on those two vectors
twilit field
$uv+ \sqrt{1-u^2} \sqrt{1-v^2} \leq u^2v^2 +(1-u^2v^2$
grand pondBOT
A dense set(Ping when reply)
hard umbra
and thats the end 
midnight plankBOT
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slim eagle
100!=2^m
grand pondBOT
A dense set(Ping when reply)
twilit field
no of zeroes at the end of 100! and m=?
m isn't an integer
I can tell you that much
opaque fjord
no of zeroes at the end of 100! and m=?
ya know how to find the exponent of a prime in a factorial?
i dunno if its generally taught, but im pretty sure its taught at the start of combinatorics
and im assuming 100! = 2^m means finding the exponent of 2 in the prime factorisation of 100!
slim eagle
and im assuming 100! = 2^m means finding the exponent of 2 in the prime factoris...
yea
slim eagle
ya know how to find the exponent of a prime in a factorial?
i have the shortcut trick but forgot how it works
opaque fjord
yeah just try to use that
slim eagle
i was wondering about how it works
midnight plankBOT
@slim eagle Has your question been resolved?
midnight plankBOT
100!=2^m
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
slim eagle
!original
Find the value of 𝑚 and the number of trailing zeros in 100!, given that 100 ! = 2 ^𝑚 × 𝐼, where 𝐼 is an odd integer
opaque fjord
i was wondering about how it works
you can probabl find a derivtion online, or in a textbook
but the quick version of explaining it is
lets say you prime factorize 100!, and you wanna know the exponent of 2
that means you need to find the number of 2s that go into each number up to 100
slim eagle
upto 64 then?
cedar coral
there's more even numbers past 64
opaque fjord
well first we try to divide all of them by 2
cedar coral
no. You're only considering numbers that are a power of two.
slim eagle
oh wait u mean all even numbers
opaque fjord
there are 50 which can be divided by 2
cedar coral
every even number contributes at least a two to the prime factorization
slim eagle
yea
opaque fjord
then, there are still numbers which can be divided by 2 additionally
slim eagle
yess
opaque fjord
so you divide by 2 again, now there are only 25 which can do that
midnight plankBOT
cedar coral
the "simplest" and most stupid method would be brute forcing. Prime factor every number from 1 to 100, and add the exponents of two's
opaque fjord
so on so forth
cedar coral
wtf does the bot mean with timeout?
slim eagle
.reopen
midnight plankBOT
✅
opaque fjord
so its like the number of even numbers + divisible by 4 + divisible by 8...
opaque fjord
how did we infer that
well if there are any more, just repeat the process
te process will stop eventually, since you cant have a numebr which is less than 100 divisible by 128
slim eagle
hmm
opaque fjord
its easier to explain this on paper, but i recommend asking the teacher for a derivation, or finding one online
as for the number of 0s afterwards, you can just find the exponent of 10 and thats your answer
midnight plankBOT
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austere ledge
h
austere ledge
$\int_{0}^{1}\left( \int_{z}^{1}e^{x^2}dx \right)dz$
grand pondBOT
Dome
last slate
still need help ?
last slate
wich one ?
last slate
same
midnight plankBOT
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
oh my bad
last slate
what did u do ?
austere ledge
a friend of mine told me to change the order of integral, but I don't know how that works
last slate
idk too, but if u dont know it yet, u shouldnt use it. Do u know integration by parts ?
austere ledge
mhm
last slate
u can integrate by parts the integral from 0 to 1 of f(z) dz
austere ledge
oh
I did not think this way
I looked at the non-elementary integral and lost my shit
last slate
hm about improper integral i assumed all the conditions were good
midnight plankBOT
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slow thorn
If $a, b, c$ are positive real numbers, then the minimum value of $\frac{a^4+b^4+c^4}{3}$ is?
midnight plankBOT
If $a, b, c$ are positive real numbers, then the minimum value of $\frac{a^4+b^4...
grand pondBOT
slow thorn
no idea what to do
lethal path
If $a, b, c$ are positive real numbers, then the minimum value of $\frac{a^4+b^4...
You can just make a, b, c be as small as possible
Given they are positive ofc
lethal path
Ah what are the options?
slow thorn
A) $a + b + c$\
B) $\left(\frac{a+b+c}{3}\right)^4$\\
C) $\frac{1}{3}[abc(a+b+c)]$\\
D) $(abc)^{\frac{4}{3}}$
grand pondBOT
lethal path
Ah okay just use AM-GM
slow thorn
yeah but im not sure how
slow thorn
and id get D with that
lethal path
Yep
sage helm
what's the correct or fully correct answer then
lethal path
Isn't the quickest way to set a = b = c = k then
And see which of those options equals k^4
sage helm
true
slow thorn
apologize for 50p quality photo
slow thorn
And see which of those options equals k^4
hold on
that kind of makes sense
why didnt i try that
theres still another one
ill try it on that
yeah that practically works
maybe thats what they want us to do?
another one while im here
$\tan\left[\sum_{r=1}^{\infty}\arctan\left(\frac{1}{2r^2}\right)\right]$
grand pondBOT
last slate
hm i think that if u say that this is the sum of argument of a complex number it should work
but i dont remember how to do it
slow thorn
scary
safe to say that method is out of our convention
but im still interested to know
.reopen
is the bot glitched?
midnight plankBOT
@slow thorn Has your question been resolved?
midnight plankBOT
@slow thorn Has your question been resolved?
midnight plankBOT
@slow thorn Has your question been resolved?
queen herald
$\tan\left[\sum_{r=1}^{\infty}\arctan\left(\frac{1}{2r^2}\right)\right]$
$$z=\prod_{r=1}^\infty \left(1+\frac{i}{2r^2}\right).$$
The answer in terms of $z$ is $\frac{\Im[z]}{\Re[z]}$. Have a good day 😇
grand pondBOT
EQUENOS
small jasper
Serious answer: Use the arctan difference formula and it telescopes
midnight plankBOT
@slow thorn Has your question been resolved?
slow thorn
**EQUENOS**
i have no remote idea of how you got that
i have only a basic understanding of complex numbers
slow thorn
Serious answer: Use the arctan difference formula and it telescopes
difference formula?
oh
ill try
yeah it works
$= (\arctan 1 - \arctan \frac{1}{3}) + (\arctan \frac{1}{3} - \arctan \frac{1}{5}) + (\arctan \frac{1}{5} - \arctan \frac{1}{7})... = \arctan 1$
grand pondBOT
slow thorn
thanks
one final question
and im done
this one is a bit scary
If $\frac{1^4}{1\cdot 3} + \frac{2^4}{3\cdot 5} + \frac{3^4}{5\cdot 7} ... + \frac{n^4}{(2n-1)\cdot (2n+1)} = \frac{1}{48}f(n) + \frac{n}{16(2n+1)}$, and if $f(10) = k$, then the value of $\frac{k}{1550}$ is?
grand pondBOT
small jasper
If $\frac{1^4}{1\cdot 3} + \frac{2^4}{3\cdot 5} + \frac{3^4}{5\cdot 7} ... + \fr...
Can't you just do polynomial division?
,w divide \frac{x^4}{(2x-1)(2x+1)}
slow thorn
seems like it
slow thorn
whoa
small jasper
slow thorn
that is quite a hideous form
small jasper
not rlly tbh
slow thorn
but how can you reach that expression normally
small jasper
but how can you reach that expression normally
wdym
small jasper
The answer that I should give: https://www.youtube.com/watch?v=_FSXJmESFmQ
small jasper
The answer that I should give: https://www.youtube.com/watch?v=_FSXJmESFmQ
What I'd actually do: $$\frac{1}{4} \cdot \frac{4x^4}{4x^2-1}=\frac{1}{4} \cdot \frac{4x^4-x^2}{4x^2-1}+\frac{1}{4} \cdot \frac{x^2}{4x^2-1}=\frac{1}{4} x^2+\frac{1}{16} \cdot \frac{4x^2}{4x^2-1}$$
grand pondBOT
Civil Service Pigeon
small jasper
fill in the rest yourself because it doesn't fit
which is basically the same thing but
easier to do in your head
midnight plankBOT
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Available help channel!
fair kayak
guys, i don’t understand the yellow highlighted part
midnight plankBOT
guys, i don’t understand the yellow highlighted part
median gust
Houses in the style of A and D will be on the left group
Houses in the style of B and C and E will be in the right group
Refer to the diagram
Also this looks like an A level question
midnight plankBOT
@fair kayak Has your question been resolved?
fair kayak
Houses in the style of A and D will be on the left group
yes but how to solve it
median gust
yes but how to solve it
The right hand side
fair kayak
Also this looks like an A level question
yeah i’m in a cambridge curriculum
fair kayak
You understand part a right
yess
median gust
yess
Apply the same thing to part b
This time, you are doing it twice, with 2 smaller groups
fair kayak
yesyess
fair kayak
thank you so much
fair kayak
guys, i don’t understand the yellow highlighted part
btw do you understand the c part?
median gust
btw do you understand the c part?
Yes
fair kayak
1 house in B, 1 house in C, and what are the other 2 tho?
median gust
1 house in B, 1 house in C, and what are the other 2 tho?
Other two are notB and notC
So you have 2A, 4D, 1E
Total of 7
Choose two out of seven, 7C2
fair kayak
OHHH
midnight plankBOT
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twilit field
I'm nots sure what's being asked of me here
midnight plankBOT
I'm nots sure what's being asked of me here
runic hamlet
what is unclear
twilit field
I have to prove if $p=2$ then this is an inner product space with this norm?
grand pondBOT
A dense set(Ping when reply)
keen saddle
a^n + b^n = c^n only exist for n=2
runic hamlet
if $p=2$ then there is an inner product $\langle \cdot,\cdot\rangle$ such that the norm given by $\norm{(x,y)} = \sqrt{\langle (x,y),(x,y)\rangle}$ has the formula $\left(|x|^2 + |y|^2\right)^{1/2}$
grand pondBOT
Denascite
twilit field
Ah, okay
runic hamlet
and if p is not two, then there is no inner product such that the norm has the relevant formula
twilit field
**Denascite**
so I basically have to prove that $x^2+y^2$ is an inner product?
grand pondBOT
A dense set(Ping when reply)
runic hamlet
thats cause it is
twilit field
This would mean that $\ang {(x,y),(x,y)} \cdot \ang {(x,y),(x,y)} = x^2+y^2$
yes
no?
runic hamlet
which is not what you wrote earlier
runic hamlet
also you are fucking up the meaning of x and y
(x,y) is a point in R^2
its <(x,y),(x,y)> etc
grand pondBOT
A dense set(Ping when reply)
twilit field
where
hard umbra
,, \inner {(x, y)} {(x, y)} = x^2 + y^2
grand pondBOT
hard umbra
assuming dot product
runic hamlet
you only need to square $\sqrt{\langle (x,y),(x,y)\rangle} = \left(x^2+y^2\right)^{1/2}$
twilit field
ah right
grand pondBOT
Denascite
twilit field
I now need to prove that $\ang{(x,y),(x,y)}$ is an inner product then?
grand pondBOT
A dense set(Ping when reply)
runic hamlet
no
you need to show that an inner product exists which has this formula if you plug the same vector into both sides
and this really is the easy direction of the exercise
twilit field
Okay, so I just have to show it satisfies the inner product axioms, no?
the formula
$x^2+y^2 = \ang{(x,y),(x,y)}$$
grand pondBOT
A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
hard umbra
well first of all that formula doesnt define an inner product
it only tells you how to evaluate it on two of the same vector
and second of all, you're only required to show that there is an inner product which gives you that formula
and there is an obvious one which you already know which does precisely that
twilit field
the dot product
hard umbra
and that is all there is to it
twilit field
I see
okay
now I have to show that if the norm associated with an inner product is given by $\norm{(x,y)}= (\abs{x}^p+ \abs{y}^p})^ {\frac{1}{p}} $, then $p=2$
now I have to show that if the norm associated with an inner product is given by $\norm{(x,y)}= (\abs{x}^p+ \abs{y}^p})^{\frac{1}{p}}$, then p=2
grand pondBOT
A dense set(Ping when reply)
now I have to show that if the norm associated with an inner product is given by $\norm{(x,y)}= (\abs{x}^p+ \abs{y}^p})^{\frac{1}{p}}$, then p=2
```Compilation error:```! Extra }, or forgotten $.
l.1422 ...by $\norm{(x,y)}= (\abs{x}^p+ \abs{y}^p}
)^{\frac{1}{p}}$, then p=2
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```
twilit field
Now I have to show that if the norm associated with an inner product is given by
[
\norm{(x,y)} = \left( \abs{x}^p + \abs{y}^p \right)^{\frac{1}{p}},
]
then ( p = 2 ).
grand pondBOT
A dense set(Ping when reply)
twilit field
right
We start with the fact that this means that $\ang{(x,y),(x,y)} = \left( \abs{x}^p + \abs{y}^p \right)^{\frac{2}{p}},$
grand pondBOT
A dense set(Ping when reply)
twilit field
We start with positivity
We start with positivity
$\ang{(x,y),(x,y)} = \left( \abs{x}^p + \abs{y}^p \right)^{\frac{2}{p}}$, which is true as squaring any real quantity ensures it's positive.
We now cover definitness
$p$ must be positive, for if not, the inner product of any vector of the form (x,-x), with itself would be 0, which would violate definiteness.
grand pondBOT
A dense set(Ping when reply)
midnight plankBOT
runic hamlet
you might want to look at a few results for norms from an inner product
midnight plankBOT
Available help channel!
gaunt jetty
Hello can I get some help finding the Fourier series representation of this
midnight plankBOT
Hello can I get some help finding the Fourier series representation of this
gaunt jetty
I tried finding the coefficients as follows but im not sure if i did it correctly. j is the imaginary number
midnight plankBOT
@gaunt jetty Has your question been resolved?
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@gaunt jetty Has your question been resolved?
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@gaunt jetty Has your question been resolved?
plucky atlas
i try wrap my head around conversion of plane equations. is this correct?
midnight plankBOT
i try wrap my head around conversion of plane equations. is this correct?
@plucky atlas Has your question been resolved?
plucky atlas
.close
midnight plankBOT
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gaunt ibex
What does this show as a matrix
midnight plankBOT
What does this show as a matrix
gaunt ibex
fallow scarab
What does this show as a matrix
Rref gives the inverse
jolly roost
A group of 12 people are going out to a concert on Saturday night. The group will take three cars with four people in each car. If they distribute themselves at random, what is the probability that Elim and Vimalan will be in the same car (NOTE: we won’t consider the cars as “distinctive”, just concern yourself with who are sitting together.)?
can someone explain why this is not the answer:
(2C2)(10C2)(8C4)(4C4)/(12C4)(8C4)(4C4)
the (2C2) accounts for picking them toghether and then 10C2 for picking the remaining two as there are 4 seats in a car. then 8C4 for picking 4 for the other car, and 4C4 for the last car. the bottom is quite self explanatory. what is wrong here?
midnight plankBOT
A group of 12 people are going out to a concert on Saturday night. The group wi...
woeful cloak
i think you add instead of multipling for the different cars
midnight plankBOT
@jolly roost Has your question been resolved?
midnight plankBOT
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worthy agate
why do the sides change when the corners change
midnight plankBOT
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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
alpine mist
why do the sides change when the corners change
Because trigonometry shows the relationships of angles and sides in a triangle. Of course if one changes the other will change as well
worthy agate
sorry my mistake
alpine mist
sorry my mistake
No problem
tacit rose
but sin a and be are different
It's in perspective of the angle
worthy agate
could you explain further
tacit rose
You know that sin theta = Opp/Hyp, correct?
tacit rose
Basically the info from here
worthy agate
yea but english is not the languege we learn in so i dont know wat sin theta is
grand pondBOT
CaptainNova22
worthy agate
oh ok
tacit rose
So here, you are asked to find $\sin \alpha$
grand pondBOT
CaptainNova22
tacit rose
What side is opp to that angle?
worthy agate
of theta
tacit rose
Of a, in your image
worthy agate
ca
worthy agate
bottom right corner
worthy agate
bottom left
tacit rose
No
worthy agate
is it not c
worthy agate
oh so cb is opposite
worthy agate
ba
tacit rose
Yes, now what's the sin ratio for a?
worthy agate
bc/ab
worthy agate
ac/ab
tacit rose
Now if you plug in the values for those ratios, what do you get?
worthy agate
wiche one
tacit rose
Do sin a first
worthy agate
0.01
tacit rose
The ratio
worthy agate
4/5
tacit rose
Do you see how the ratios for sin a and sin b is based on the perpective of the angle?
worthy agate
yea
tacit rose
Does that clear up your confusion now?
tacit rose
That's where the right angle is, in that image there's no ratios for that corner
worthy agate
so this wont work for corner c
worthy agate
sin c in corner c cant exist
tacit rose
Because the hypotenuse is defined as the longest side, which is across the right angle, so if you did sin a, then the opp side equals the hypotenuse which would be AB/AB
worthy agate
oh ok
wait i have a question
but wait a minute please
how would i got about to solve this
are they bot cosin
both
tacit rose
are they bot cosin
No
Remember what I said about the perspective of the angle
If you are at angle j, is 10 the opposite, adjacent, or hypotenuse?
worthy agate
adjacent
tacit rose
And is 12 opposite, adjacent, or hypotenuse?
worthy agate
hypotenuse
tacit rose
What ratio uses adj and hyp?
worthy agate
cos
tacit rose
Yes and what ratio uses opp and hyp?
worthy agate
sin
tacit rose
There you go
worthy agate
thx
omg thank you so much
you explained it better then my teacher
well that was everything i wanted to ask thank you
.close
midnight plankBOT
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soft canopy
ik this a physics problem but i need someone to verify my math
midnight plankBOT
ik this a physics problem but i need someone to verify my math
soft canopy
here's what i've done
last slate
Idk what any of that means but it looks good
soft canopy
Idk what any of that means but it looks good
lmao i feel u
even i think its correct cuz its basic integration but im not sure about the limits being correct
midnight plankBOT
@soft canopy Has your question been resolved?
last slate
even i think its correct cuz its basic integration but im not sure about the lim...
basic?!
grand pondBOT
Blitz
last slate
its just the integral of $t^{-2}$ it looks scary but it isn't
I mean I have not a clue what that means, apologies
soft canopy
I mean I have not a clue what that means, apologies
ah its alr u'll get there mate
last slate
ah its alr u'll get there mate
I’ll try 🤣
soft canopy
.close
midnight plankBOT
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hearty nexus
midnight plankBOT
ripe gulch
Where have you
Gotten so far
ripe gulch
Well you should at least see the first step
hearty nexus
i can do this but not the other ones
hearty nexus
Well you should at least see the first step
wdym?
ripe gulch
What can you immediately factor out of the expression
hearty nexus
im sorry i have literally no idea
hearty nexus
huhhh
ripe gulch
We’re talking about this right