then just say K does not equal to 0
#help-26
fathom sinew
mellow cypress
but K=0
means its just lim K\theta=KP=> 0=0, you can leave it at that
mellow cypress
because I need to show
|KP-K\theta|->0
mellow cypress
isnt it supposed to be like K*theta
when did I divide K by theta
fathom sinew
when did I divide K by theta
never but u put \
fathom sinew
ok
mellow cypress
(K\theta)
thorny flameBOT
quartz geyser
How to go about solving this?
On 1 and -1, all of them make sense but sec and cosec don't have any values.
acoustic tangle
topaz sinewBOT
@quartz geyser Has your question been resolved?
hollow knoll
How do I find the graph of f(x)=3cos2x-2?
topaz sinewBOT
How do I find the graph of f(x)=3cos2x-2?
modern rock
topaz sinewBOT
@hollow knoll Has your question been resolved?
turbid matrix
my name is beans — Today at 7:14 PM
Of course! Here's a unique, easy, and delicious recipe for Indian household: Mushroom Tikka.
Ingredients:
For the Marinade:
250g button mushrooms, cleaned and halved
1/2 cup thick yogurt
1 tablespoon ginger-garlic paste
1 teaspoon red chili powder
1/2 teaspoon turmeric powder
1 teaspoon garam masala
1 teaspoon cumin powder
1 teaspoon coriander powder
Salt to taste
2 tablespoons vegetable oil
For the Skewers:
Skewer sticks (wooden sticks soaked in water for 30 minutes)
Instructions:
Marinate the Mushrooms:
In a mixing bowl, combine all the marinade ingredients - yogurt, ginger-garlic paste, red chili powder, turmeric powder, garam masala, cumin powder, coriander powder, salt, and vegetable oil. Mix well.
Thread the Mushrooms:
Thread the marinated mushroom halves onto the soaked skewer sticks. Make sure to leave a little space between each mushroom for even cooking.
Grill or Bake:
You can cook the mushroom skewers in two ways:
Grill Method: Preheat your grill to medium-high heat. Place the skewers on the grill and cook for about 8-10 minutes, turning occasionally until the mushrooms are tender and have a slight char. Baste them with some oil or butter while grilling for added flavor.
Oven Method: Preheat your oven to 200°C (400°F). Place the skewers on a baking sheet and bake for about 15-20 minutes, turning them once halfway through, until the mushrooms are cooked and have a slight char.
Serve: Serve the Mushroom Tikka hot with a side of mint chutney and lemon wedges. Enjoy the flavorful and healthy snack or appetizer!
topaz sinewBOT
my name is beans — Today at 7:14 PM
Of course! Here's a unique, easy, and delici...
topaz sinewBOT
@turbid matrix Has your question been resolved?
jagged glen
is my answer correct?
topaz sinewBOT
is my answer correct?
plucky halo
helloguys any one tell me a cube plus b cube formula please
topaz sinewBOT
helloguys any one tell me a cube plus b cube formula please
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
jagged glen
is my answer correct?
My method looks like its correct to me. Can anyone correct it?
Or is the given solution wrong
astral temple
topaz sinewBOT
@jagged glen Has your question been resolved?
neon iron
how to come up with algorithm that given room size outputs how much floor heating pipe is needed
talking about spiral format like this:
topaz sinewBOT
how to come up with algorithm that given room size outputs how much floor heatin...
neon iron
assuming for example that the pipes are 20cm from wall and there is 10cm between pipes
topaz sinewBOT
@neon iron Has your question been resolved?
topaz sinewBOT
@neon iron Has your question been resolved?
topaz sinewBOT
@neon iron Has your question been resolved?
topaz sinewBOT
@neon iron Has your question been resolved?
topaz sinewBOT
@neon iron Has your question been resolved?
neon iron
@neon ironyou can ping @Helpers if you still need help btw
topaz sinewBOT
topaz sinewBOT
Available help channel!
languid surge
Hey so im a bit confused,
So if we let Let X = {a,b, c}. Would the discrete topology just be τ = {∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}} or would it be multiple, such as like τ = {∅,X,{a}}, τ = {∅,X,{a,b},{b},{b, c}}, τ = {∅,X,{a},{b, c}}, τ = {∅,X,{a},{b}}, etc
drifting swift
Would the discrete topology just be τ = {∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}
yes
the discrete topology is the one where everything is clopen
languid surge
the discrete topology is the one where everything is clopen
thanks, thought it was that, but my professor wrote it down pretty weirdly in his textbook
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
timber mirage
topaz sinewBOT
https://cdn.discordapp.com/attachments/1081837589780234241/1165215272093876315/i...
timber mirage
what's going on here?
neon iron
I'd think you have to establish some sort of pattern
customer 1 gets served by rep 1
meanwhile customer 2 appears, rep 2 takes the job
timber mirage
right it can be in reverse order too right?
clever stirrup
So am i racist?
neon iron
customer 3 waits until rep 1 finishes
timber mirage
I honestly didn't bother reading all that description above
timber mirage
So am i racist?
?
timber mirage
customer 3 waits until rep 1 finishes
indeed
neon iron
right it can be in reverse order too right?
you just have to worry about who served customer 1 so I think either of them works
clever stirrup
im so confused about this
neon iron
but yeah I think you can continue like that. customer 4 gets rep 2 as rep 1 is finishing the 3rd
so it can't be customer 4
timber mirage
but yeah I think you can continue like that. customer 4 gets rep 2 as rep 1 is f...
hmm wait i don't get why customer 4 started later when rep 2 was already vacant all along
oh maybe they arrived late?
neon iron
yeah but I think you can continue off with the rest
timber mirage
rep 1 helped 5
timber mirage
yeah but I think you can continue off with the rest
i think i did that but i missed a nuance or too
or i'm just sleep derived
*deprived
lemme see
neon iron
lmaoo
funnily enough I think they just alternate between even/odd
I am pretty sure it's just 9 by the pattern
timber mirage
I am pretty sure it's just 9 by the pattern
😭 it's 10
neon iron
oh darn
timber mirage
1 - 1
2 - 2
3 - 1
4 - 2
5 - 1
6 - 2
7 - 1
8 - 2
9 - 2
10 -1
timber mirage
okay these exercises are just messing with our brain 😭
neon iron
I see now
timber mirage
thanks thanks
neon iron
that's the point of the thingy I see
they just want to bamboozle you by thinking it's even/odd and checking off 9
like I did 
neon iron
thanks thanks
np
timber mirage
im so confused about this
do you want this channel?
actually since you're in my channel ig i can help you
what have you done?
neon iron
they can open their own channel. They can't "take over" your channel
if someone starts like spamming questions in Ur own channel just guide them to open their own
neon iron
in how many ways can 9 teachers be assigned to 3 teaching positions
neon iron
so this is really about knowing whether I'm dealing with combinations or permutations
are questions like these like, ambiguous in nature?
am I expected to assume that the teachers are distinct from each other?
cause if they weren't, I assume it's akin to choosing between a set of, say, {T, T, ..., T} where that set is of cardinality 9
so it can be T, T, T, for any particular combination but that doesn't make sense?
am I misinterpreting something
ivory panther
It’s implied that if a teacher fills a position they have been ‘used up’
neon iron
ohhh I see that's a very good way to think about it thanks
ivory panther
And this is a permutation problem I would say
There is a general formula but this one is small enough that you should be able to figure it out directly
neon iron
it's just product rule yeah
also
just to elaborate
when we consider nCr,
are we treating the r objects as indistinguishable from each other, or the n objects as indistinguishable from each other?
I'm assuming the former?
ivory panther
nCr counts how many different subsets of size r there are from a set of size n
Can you see why this is not the same as your problem?
neon iron
no not really
ivory panther
In the teacher example it’s implied that there is an order, like T1-P1, T2-P2, T3-P3 =/= T2-P1, T3-P2, T1-P3 for example, but these are the same entity in nCr
neon iron
hmm i see yeah that makes sense
if we changed the wording to
in how many ways can 9 teachers be assigned to a 3 people tutoring group
ivory panther
Then that would be nCr yes
neon iron
then it'd be combinations in that case yed
okay
I think I have rationalised this better now
so when the r positions are distinguishable from each other, we are considering permutations
when they are not, it's combinations
it also describes the equality between them
[
\map C{n,r} = \f{\map P{n,r}}{r!}
]
thorny flameBOT
neon iron
by dividing by r!, we are thus treating the r objects all indistinguishable from each other
ivory panther
so when the r positions are distinguishable from each other, we are considering ...
Ah this is what you meant, yes in your original problem the positions are distinct in that once you pick the 3 teachers you have to consider all the permutations within the positions
Yes exactly
neon iron
another example could be something like
In how Many ways can 3 people of 10 get a $1, $5, or $10 bill?
or rephrased
In how Many ways can 3 people of 10 get a $1 bill
the former, the bills are distinct from each other, so we are considering permutations
the latter, the $1 bills for all intents and purposes are indistinguishable
so it's combinations
ivory panther
Yeah you’ve got it
And how you explicitly count nPr is the following
How many options for $1?
How many subsequent options for $5?
How many subsequent options for $10?
And you use multiplication principle
So it is always n(n-1)…(n-r+1)
neon iron
yeah
I get that now thank you
one additional thing though
The number of ways of partitioning a set of $n$ objects into $r$ subsets with $n_1$ elements in the first, $n_2$ elements in the second, and so forth: [
\binom n{n_1, n_2,\hdots, n_k} = \f{n!}{n_1!n_2! \hdots n_k!} ]
where $n_1 + \hdots + n_k = n$
thorny flameBOT
neon iron
how does one prove the above theorem?
neon iron
what do you mean?
ivory panther
How many ways can I pick an n1 subset of n?
neon iron
but wait
I feel like this is a bit circular isn't it
we define combinations by this same theorem (P(n,r)/r!)
neon iron
n choose n1
so using this fact doesn't seem like it says much
ivory panther
nCr = n!/(r!(n-r)!)
neon iron
we define combinations by this same theorem (P(n,r)/r!)
how can that formula be independently derived without using this
ivory panther
In a set of n elements, if I want to pick r elements from it I have n choice for the 1st one, n-1 choices for the 2nd,… etc.
neon iron
okay sure
neon iron
we define combinations by this same theorem (P(n,r)/r!)
but you're still looping back to this are you not
ivory panther
They are equivalent formulae
neon iron
okay I don't think you're understanding my confusion here at the moment, and that's fine. I'll continue searching independently
thank you
.close
polar gyro
Hey! I'm struggling to find how to determine the coordinates on a semi circle based on a vector of which I know the direction but that doesn't cross in the center.
topaz sinewBOT
Hey! I'm struggling to find how to determine the coordinates on a semi circle ba...
@polar gyro Has your question been resolved?
topaz sinewBOT
@polar gyro Has your question been resolved?
polar gyro
<@&286206848099549185> sorry for the ping, still couldn't find something on my end. I tried to simply calculate 180 points for each degree based on origin and currently trying to find closest point when aggregating the vector position as I know its origin. For more context I'm trying to determine the coordinate of a point on the perimeter of a shape with one rectangle and 2 semi circles like such:
I already managed to determined the position on the rectangle, now when y < height/2 I'm trying to do what I described in my original question
topaz sinewBOT
@polar gyro Has your question been resolved?
topaz sinewBOT
@polar gyro Has your question been resolved?
versed cairn
median zenith
to me it says its still open
vernal vale
.
median zenith
yeah but its still occupied
craggy haven
it will still show up in the list for about five minutes
median zenith
ah
craggy haven
don't worry about that, just get another
median zenith
oh I dont need to I just didnt want to hog up a channel when I had no question
civic owl
hi, sorry id like to ask why is the area of triangle 1/2(4-k)(k-1) why can it not be (k-4) is k not >4?
oh sorry i didnt know how asking questions works my bad.
steel violet
We can write those coordinates in form a matrix
civic owl
We can write those coordinates in form a matrix
i just need the area of T x the determinant to = 10, my issue is with finding the area of T
steel violet
|k 1 1|
|4 1 1 |
|4 k 1 |
This will be the matrix
Find its determinant and divide it by 2
Then put it equal to 10
civic owl
Find its determinant and divide it by 2
ye when you do that you get (k-4)(k-1) but the answer is (4-k)(k-1)
topaz sinewBOT
Available help channel!
steel violet
I think u get (k-4)(1-k)
topaz sinewBOT
I think u get (k-4)(1-k)
steel violet
I think u get (k-4)(1-k)
Take minus common from tje both
civic owl
.close
alright i get how kind of but
its because the triangle has the hyptonuse on the left but i still dont get why k cant be >4
topaz sinewBOT
@steel violet Has your question been resolved?
steel violet
.close
neon iron
,tex A box contains 500 envelopes, of which 50 contain $100 in cash, 150 contain $25, and 300 contain $10. An envelope may be purchased for $25. What is the sample space for the different amounts of money? Assign probabilities to the sample points and then find the probability that the first envelope purchased contains less than $100
neon iron
.close
flint adder
yo need help
topaz sinewBOT
yo need help
neon iron
flint adder
maybe hard to see but the question is basically
what happens if the paranthesis of the denominator gors away
a = 12 btw
neon iron
,rccw
thorny flameBOT
topaz sinewBOT
@flint adder Has your question been resolved?
topaz sinewBOT
@flint adder Has your question been resolved?
flint adder
.close
frozen garnet
No matter how I solve it I get 24
topaz sinewBOT
No matter how I solve it I get 24
frozen garnet
How did they get 6
fast yarrow
My guy 1 volume of cylinder = 2×volume of sphere
Not ½×volume of sphere
16pi×r²= ½ × 4/3pi×r³ means cylinder volume is half of sphere
topaz sinewBOT
topaz sinewBOT
Available help channel!
tawny yacht
Is this true? And if yes, why?
topaz sinewBOT
Is this true? And if yes, why?
odd pagoda
hasty sail
topaz sinewBOT
topaz sinewBOT
hasty sail
well i clearly posted it before so ...
loud glen
well i clearly posted it before so ...
bruh
we posted literally same time
dont be a jerk abt it my god
hasty sail
i'm not you pinged me
topaz sinewBOT
Available help channel!
prisma galleon
Is this a valid proof?
topaz sinewBOT
Is this a valid proof?
prisma galleon
its at the top might be hard to read prove log_10 (2) is irrational
if any steps are hard to read i can clarify
hallow depot
..
prisma galleon
the only thing im not sure on is after you get to 2^b = 10^a and you have to prove there are no solutions
I know its pretty obvious that all powers of 2(mod10) are 2, 4, 8, 6 and then repeating i just have to prove
hallow depot
ok it's easy .... like you said you have to prove that $10^{\frac{a}{b}}=2$ has no soluzion
thorny flameBOT
everg
hallow depot
i.e. $10^a=2^b$ where a and b are integers
thorny flameBOT
everg
hallow depot
with b not equal to zero
now LHS is divisible by 5 and RHS isn't
for a not equal to zero
this imply that a=0
so we have to solve $10^0=1=2^b$
thorny flameBOT
everg
carmine marten
now LHS is divisible by 5 and RHS isn't
this only holds for positive a
prisma galleon
Yeah but is what I did wrng though
to just say 2^a (mod10) can never be equal to 10^b (mod10) where a and b are natural number
hallow depot
this only holds for positive *a*
true ...we can assume a>=0 because of 10^a/b=2>1
prisma galleon
specifically the second little proof by induction
hallow depot
to just say 2^a (mod10) can never be equal to 10^b (mod10) where a and b are nat...
i think its true ...if you want more motivation i think you have to follow my steps
prisma galleon
uhhh is the second part right is the only thing im not sure about
the inductive hypothesis -> step part
i feel like it mightve been too trivially easy so i mightve messed up
even for your steps if you wanted to make it complete wouldnt you have to say 10^n (mod5) is always congruent to 0
the picture again
topaz sinewBOT
@prisma galleon Has your question been resolved?
topaz sinewBOT
topaz sinewBOT
Available help channel!
digital carbon
Is there a diffrence between (Grad dot F)f and f(F dot Grad)?
topaz sinewBOT
Is there a diffrence between (Grad dot F)f and f(F dot Grad)?
digital carbon
F is a field, f is a function
I think there was this thing where (F dot grad) was an operator which I dont get
hallow depot
F dot grad isn't it just the divergence ? d/dx1 F1+....+d/dxnFn
digital carbon
yeah it should be the divergence but
no grad dot F is
F dot grad is an operator which again im confused on, and not divergence so like the dot product order matters
like to try and shwo what I mean
there was some example where you had (2xi - 2yj + 2zk dot grad)xyz
hallow depot
dot product is commutative
digital carbon
hmmm, lemme come back with like a photo
the fact i dont know latex is bricking my efforts
i gotta eat something first rq
ill reopen ty so far tho
.close
neon iron
Confirm the identity without evaluating the determinant directly
$\begin{vmatrix} a_1 + b_{1} t & a_2 + b_{2} t & a_3+ b_{3} t \
a_{1} t + b_1 & a_{2} t + b_2 & a_{3} t + b_3 \ c_1 & c_2 & c_3 \end{vmatrix} = (1-t^2) \begin{vmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix}$
topaz sinewBOT
Confirm the identity without evaluating the determinant directly
$\begin{vmatrix...
thorny flameBOT
Derivative
neon iron
anybody know how to do this
im lost
all I know is that if we do cR_i + R_J determinant does not change
cR_i = cdet(A)
hallow depot
try to substitute the second row with second row-t*first row
neon iron
and switch two rows, we add negative sign
hallow depot
then you have delited a's form second row
..now you can delete b's from first row with a similar substitution
now you would have (1+t) factor in the second row ... and you can plug out by multilinearity
neon iron
ok
for now i have this
$\begin{vmatrix} a_1 + b_{1} t & a_2 + b_{2} t & a_3+ b_{3} t \
b_1 - b_1 t^2 & b_2- b_2 t^2 & b_3- b_3 t^2 \ c_1 & c_2 & c_3 \end{vmatrix} = (1-t^2) \begin{vmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix}$
thorny flameBOT
Derivative
hallow depot
ok now you can group (1-t^2) in the second row
neon iron
and then i do -tR_2 + R_1
hallow depot
gg
topaz sinewBOT
topaz sinewBOT
Available help channel!
mellow abyss
.ocupe
golden edge
Trying to settle a debate, I believe a) is 6! and b) is 5! (circular permutation n-1)
My friend says a) 6! b)6!
Can anyone confirm if we're both wrong or one is wrong XD
topaz sinewBOT
Trying to settle a debate, I believe a) is 6! and b) is 5! (circular permutation...
worthy storm
Trying to settle a debate, I believe a) is 6! and b) is 5! (circular permutation...
what is your reasoning for why they are different?
hallow depot
b) 6! is impossible
golden edge
what is your reasoning for why they are different?
In a circular permutation, there are (n-1)! ways to arrange n distinct objects in a circle. Is that not correct?
worthy storm
isn't the dinner table also a circle?
golden edge
isn't the dinner table also a circle?
Hmmm, i suppose since its not specified i assumed not 🤔
I suppose "around" would imply even for a rectangular table its the same situation
worthy storm
seems so to me unless there's some subtlety
is it maybe because a circular ring can be flipped upside down
whereas a dinner table cannot
golden edge
is it maybe because a circular ring can be flipped upside down
That!
That would make sense
I was trying to come up with an explanation but that hits the nail on the head
So in that case a) 6! and b) 5! still applies I assume?
long stirrup
yes the ring has 2 sides, it's like left right symmetry
a) 5! and b) something completely different
golden edge
Oh seriously?
Damm more complex than we'd both first assumed then
Quora and mytutor seem to believe 6 people can be arranged around a circular table 120 different ways (5!)
so saying b = 5!
golden edge
And that's because it can be flipped right?
I've just figured that out as you sent that XD
So table = 5! and Ring can be flipped so 5!/2
long stirrup
i believe so
worthy storm
i agree with both of those
golden edge
So final answers a) 5! = 120 b) 5!/2 = 60 as the ring can be flipped whereas the table cannot
worthy storm
yea, unless there's something about the ring that makes upside down somehow distinguishable from rightside up
golden edge
Ty for clarification on a) idk why i assumed the table was linear and not "circular"
Perfect
Thank you for your help ill probably be back XD
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
topaz sinewBOT
thorny flameBOT
everg
opal vault
this is the definition of the preimage
hallow depot
no its the definition of $f^{-1}(A)$
thorny flameBOT
everg
opal vault
no its the definition of $f^{-1}(A)$
the... preimage of A by f
thorny flameBOT
everg
opal vault
so the only thing that changes is the X is swapped with M
and that's it
yes
and?
idk why it wouldn't be
thorny flameBOT
Madashi
opal vault
**Madashi**
this is the proof that $N\subseteq f^{-1}(f(N))$, we've shown every element of the first set is an element of the other
thorny flameBOT
rafilou2003
hallow depot
why "but okay"?
opal vault
Back to the definition. Let $f:M\to T$. If $A\subseteq M$ and $B\subseteq T$, we define $f(A)={f(x), x\in A}$ and $f^{-1}(B) = {x\in M, f(x)\in B}$
thorny flameBOT
rafilou2003
hallow depot
if you want to prove that $A\subset B$ you have to prove that for all $a\in A$ you get $a\in B$
thorny flameBOT
everg
opal vault
**everg**
indeed, to prove that $N\subseteq f^{-1}(f(N))$, we prove that $n\in N \implies n\in f^{-1}(f(N))$
thorny flameBOT
rafilou2003
opal vault
so let $n\in N$.
thorny flameBOT
rafilou2003
opal vault
By definition of the image of N, $f(n)\in f(N)$.
thorny flameBOT
rafilou2003
opal vault
So $n$ is an element of $M$ such that $f(n)\in f(N)$
thorny flameBOT
rafilou2003
opal vault
So by definition of the preimage of $f(N)$, $n\in f^{-1}(f(N))$
thorny flameBOT
rafilou2003
opal vault
take a constant function
like f(x) = 1 on R
f({0}) = {1}
but the preimage of {1} is...
is 0 the only element such that f(x) = 1?
(remember that f is constant)
$f:\begin{cases}\bR\to\bR \ x\longmapsto 1\end{cases}$
thorny flameBOT
rafilou2003
opal vault
$f(x) = 1$ for all $x\in \bR$
thorny flameBOT
rafilou2003
opal vault
what is $f^{-1}({1})$?
thorny flameBOT
rafilou2003
opal vault
if you're hesitating, go back to the definition of the preimage
so write the definition of the preimage applied to $f^{-1}({1})$
thorny flameBOT
rafilou2003
opal vault
yes, so f(x) = ?
no, f(x) is a number
yes
and what are the x that verify f(x) = 1?
if you go back to the definition of f
?
You know what $\bR$ means right?
thorny flameBOT
rafilou2003
opal vault
ok
opal vault
**rafilou2003**
so when I write this, can you tell me which x verify f(x) = 1?
every real number yes
so $f^{-1}({1}) = ?$
thorny flameBOT
rafilou2003
opal vault
It's not a number, it's a set
it's the set of all numbers that verify f(x) = 1
So it's which set?
Thank you
$f^{-1}({1}) = \bR$
thorny flameBOT
rafilou2003
opal vault
And $f({0}) = ?$
thorny flameBOT
rafilou2003
opal vault
**rafilou2003**
we defined f RIGHT here
so f(0) = ?
yes thank you
So what's the set $f({0}) = ?$
thorny flameBOT
rafilou2003
opal vault
yes
opal vault
**rafilou2003**
so now, using this
thorny flameBOT
rafilou2003
thorny flameBOT
rafilou2003
opal vault
and this : $f^{-1}(f({0}))$
thorny flameBOT
rafilou2003
thorny flameBOT
rafilou2003
opal vault
You tell me, we've literally found $f^{-1}(f({0}))$ two seconds ago
thorny flameBOT
rafilou2003
opal vault
Is ${0} = f^{-1}(f({0}))?$
thorny flameBOT
rafilou2003
jaunty citrus
topaz sinewBOT
jaunty citrus
Is my logic correct that because Ammeter almost has no resistance, and because it's connected in parallel, all the currrent will choose to move through the circuit with the ammeter isntead of the 4 ohm resistance?
long stirrup
yep
jaunty citrus
neat
so
1 is zero because no current will flow ther
2 is gonna just be the voltage divided by the internal resistance so 5A
how would i find the terminal voltage in this case?
like, i was told to find the overall current in the circuit and then multiply it by the resistance not including the internal resistance of the battery....
so in this case would it be 0?
long stirrup
i don't actually know but logically, you;re given 2 ohm as internal battery resistance
so voltage would come from multiplying current by 2 ohm or something similar
jaunty citrus
4 is just 1.67 bc 10V/6 ohm
jaunty citrus
damn ok, everything else look right tho?
long stirrup
like, i was told to find the overall current in the circuit and then multiply it by the resistance not including the internal resistance of the battery....
oh so you do that
jaunty citrus
yeah so like, if the ammeter was connected properly, or if it was just gone. then it'd be like, 10V / 6 Ohm = 1.67 so thats the overall current. And 1.67 x 4 = 6.68 V so that's the terminal voltage
but like
the resistance in the circuit is just, 2 ohm cuz we ignore the 4ohm resistor, so current is 5A
but then, is it just gonna be 5 x 0?
long stirrup
probably
rocky skiff
i need very general help on series and sequences
topaz sinewBOT
i need very general help on series and sequences
rocky skiff
ive watched many videos but i feel like im missing something big but im not sure what im missing
so when ive done 30 for example, i took the limit of the sequences and got 7, and because the limit was 7 that means the series diverges
ok, but what if, like in 31, the limit of the sequence coverges to 0
that means the series converges right
but to what value??
in the videos ive seen they just tell you that it "converges" or "diverges" but how do i figure out to what value?
loud oasis
that means the series converges right
not necessarily. the sequence 1/n converges to 0 but the harmonic series (sum of 1/n) famously doesn't converge
rocky skiff
oh?
loud oasis
also, for many series it's easy to find out what the series converges to but not to find its value
generally any series that converges must have its sequence converge to 0, but not all sequences that converge to 0 have a convergent series
in this case though, 31 is a geometric series so we can find its value with the geometric series formula
rocky skiff
stupid question, but if i take the limit of a sequence and get a number (like not infinity) does that mean it converges
loud oasis
yes, that's the definition of a convergent sequence
rocky skiff
bet ok
rocky skiff
in this case though, 31 is a geometric series so we can find its value with the ...
a/(1-r) i remember thatt
loud oasis
unless we have some special case like a geometric or telescoping series, we generally can't get an exact value for a convergent series. We can estimate it by taking a finite partial sum and finding a bound on the error, though
loud oasis
careful. the geometric sum formula applies for k= 0 -> infinity, but we have a sum from k = 1 -> infinity
loud oasis
it's fine because we can just subtract the 0th term from our answer, but something to keep in mind
loud oasis
no you got it right the first time
rocky skiff
oh wait the 0th term of the series not the sequence
loud oasis
that's the same thing, but the 0th term is 3 * 1, not 1
rocky skiff
could i possibly leave this open for a sec cuz i might have a question on the next problem
rocky skiff
username checks out though
i get that a lot 💀
is 32 geometric even though it doesnt go up at all
it converges at 1/2 so the series diverges so i guess it doesnt matter (?)
loud oasis
yes. in general geometric series converge (and the formula applies) for |r| < 1
rocky skiff
sorry one last question
for 32 the formula wouldn't work because 1/0 right, is that how you (with numbers?) know it diverges?
loud oasis
geometric series converge for |r| < 1 (strictly less than 1), so with r = 1, the series diverges. the divergence test (sequence doesn't converge to 0) also tells us it must diverge
rocky skiff
WHEN R <1
yes
of course
im gonna take a break
im obviously too tired lol
thank you for the help
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
sudden dragon
“One interpretation of a baseball players batting average is as the empirical probability of getting a hit each time the player goes to bat.
If a player with a batting average of 0.204, bats 5 times in a game, and each at-bat is an independent event, what is the probability of the player getting at least one hit in the game?”
topaz sinewBOT
“One interpretation of a baseball players batting average is as the empirical pr...
sudden dragon
i dont know how im supposed to interpret these kinds of questions
topaz sinewBOT
@sudden dragon Has your question been resolved?
grim elk
I don't understand question b) at all
topaz sinewBOT
I don't understand question b) at all
grim elk
the second image answer
vernal vale
so you have to close that vertical gap
if you subtract k
the parabola will move down k units
grim elk
what does the two solutions part mean
vernal vale
you need the vertex to go below the x axis
after which, you have two roots
does that make sense?
like the vertex is below the x axis
so it crosses the axis, goes below, comes back up, crosses again
vernal vale
what are the roots?
roots to f(x) are solutions to f(x)=0
if k is one
you have x^2 -4x +5 -1
or x^2-4x+4
or (x-2)^2
which just barely touches the axis
idk does this help 
grim elk
so how would it work in this case
grim elk
so how would i asnwer a)
vernal vale
think of setting k like moving a horizontal line instead if it helps
where do you have to put a horizontal line on that plot to only cross f(x) one time
grim elk
if its bigger than 2 or less than -2?
vernal vale
if its bigger than 2 or less than -2?
yea
you can think of this too like moving the function up or down by subtracting k
or as placing a line if its easier
grim elk
ahh i see, is question b asking where the horizontal line can go to hit f(x) twice?
vernal vale
yes
jade thunder
topaz sinewBOT
https://cdn.discordapp.com/attachments/903430333096132618/1165459049647702046/24...
jade thunder
So like this then
and so then now i just need to figure out why the first line is true
probably has somethign to do with the definition of the abs sign and the partial derivative
topaz sinewBOT
@jade thunder Has your question been resolved?
topaz sinewBOT
@jade thunder Has your question been resolved?
topaz sinewBOT
@jade thunder Has your question been resolved?
topaz sinewBOT
@jade thunder Has your question been resolved?
opal vault
probably has somethign to do with the definition of the abs sign and the partial...
Well the module of the product is the product of the modules
$|ax^n\partial^\alpha\varphi(x)| = |a| |x|^n |\partial^\alpha\varphi(x)|$
thorny flameBOT
rafilou2003
opal vault
Then we use that $|\partial^\alpha\varphi(x)| \leq$ ...
thorny flameBOT
rafilou2003
jade thunder
well i mean it just asserts this statement without proof
im guessing you can arrive to the above from this definition of what's in S?
so it'd be all the phi's in C^inf such that this finite
for all alpha and betas in Z+
opal vault
Basically the intuition is to say that $p_{\alpha,\beta}(\varphi) = \max(\sup_{x\in \bR^n, |x|<1}|...|,\sup_{x\in\bR^n,|x|\geq 1}|...|)$
thorny flameBOT
rafilou2003
opal vault
We apply the fact that this quantity is finite to $\alpha = (d+1,...,d+1)$
thorny flameBOT
rafilou2003
opal vault
We let $C_{d,\beta} = \max(p_{\alpha,\beta}(\varphi),2^d\sup_{|x|<1}|\partial^\beta\varphi(x)|)$
jade thunder
ok let me digest this
opal vault
Not easy to make the proof on the spot xd
jade thunder
it looks to me like random symbols everywhere
opal vault
But this should work because for |x|<1, we use the term on the right
And for |x|>=1, we use the term on the left
jade thunder
what is this 1 + √n doing
opal vault
To take into account the (1+|x|²)^(d/2)
Which can go up to 2^d on the unit ball
Wait no i'm dumm
thorny flameBOT
rafilou2003
opal vault
Which can go up to 2^d on the unit ball
And edited this
jade thunder
ok hold up
jade thunder
well i mean it just asserts this statement without proof
where does the other side of this inequality even come from
like im hold (\partial^\alpha \varphi(x)) in my hand now what
thorny flameBOT
Frosst
opal vault
where does the other side of this inequality even come from
Don't get me wrong it would have been simpler if the denominator were |x|^d
opal vault
somehow i feel like the 1 is important
It's important because around x=0 this upper bound is useless
jade thunder
ah
that's true too
it'd just say it's bounded by infinity ish
ok but that's besides the point
where did the ≤ sign come from
jade thunder
well i mean it just asserts this statement without proof
in here
jade thunder
so it'd be all the phi's in C^inf such that this finite
this one
opal vault
Yes we start with this one
jade thunder
aha
opal vault
When |x|>=1 we use this supremum
jade thunder
ok i think im folloiwng give me a moment
opal vault
It's not "finite to alpha"
It's "for alpha =..., the quantity p_alpha,beta is finite"
jade thunder
well p_alpha,beta is finite for all alpha betas in Z+
otherwise we would not consider the phi's
right?
jade thunder
ok so from $p_{\alpha,\beta}(\varphi) = \max(\sup_{x\in \bR^n, |x|<1}|x^\alpha\partial^\beta\varphi(x)|,\sup_{x\in\bR^n,|x|\geq 1}|x^\alpha\partial^\beta\varphi(x)|)$
thorny flameBOT
Frosst
jade thunder
we choose alpha = (d+1, d+1, ..., d+1)
opal vault
Oh wait we might need to use a different alpha, one sec
Keep that in mind for the moment
opal vault
Ok i found something
jade thunder
well the first one is always smaller than the second one isn't it?
opal vault
You will need to prove the following intermediate result :
opal vault
**Frosst**
Talking about this ?
jade thunder
yeah
opal vault
Well yeah there's an equality
jade thunder
if |x| < 1 then each of the components of x mustn't be greater than 1
then when i take it to the alphath power
they should all remain under 1
when i multiply all them together
i still get something less than 1
so in the max function you never pick the left side
opal vault
?
jade thunder
**Frosst**
well here we want to pick hte left or right option for the max
wouldn't the left side always be smaller than the right side
since that x^alpha term at the front is always < 1 for all x that satisfies |x| < 1
opal vault
First, when alpha = 0 this is not true, and second of all, phi might decrease faster than x^alpha
jade thunder
well when alpha = 0 it doens't matter what you pick
opal vault
Well it does
jade thunder
and they both have phi so the rest would still be the same no?
opal vault
Take $\varphi(x) = e^{-x^2}$
thorny flameBOT
rafilou2003
opal vault
The maximum is in (0,0,...,0)
jade thunder
First, when alpha = 0 this is not true, and second of all, phi might decrease fa...
that's what you mean
opal vault
Yeah so left one bigger than right one
jade thunder
if phi decreases quickly then a small |x| is better
cos phi of that x
would also be bigger
right i see
opal vault
Ok so intermediate result
If $\varphi \in \mathcal{S}$, then $\forall d\in \bZ_+, \forall \alpha\in \bZ_+^n$, $p_{d,\alpha}^*(\varphi) = \sup_{x\in \bR^n}(|x|^d|\partial^\alpha\varphi(x)|)<+\infty$
thorny flameBOT
rafilou2003
jade thunder
ok that's just saying my thing is finite?
opal vault
Proof: apply the definition to the exponent (d,0,...,0), then to (0,d,0,...,0), etc...
jade thunder
my p is finite?
opal vault
ok that's just saying my thing is finite?
I'm saying this new p* is finite yes
jade thunder
is (|x|^d=\left(|x|^d\right)^n)?
thorny flameBOT
Frosst
opal vault
No, one of them is |x|^d, the other is |x|^(dn)
jade thunder
is that not the same
opal vault
?
jade thunder
oh
opal vault
is 2^2 the same as 2^(2n)?
jade thunder
i have d here as 1 number not the vector
opal vault
i have d here as 1 number not the vector
yes
thorny flameBOT
Frosst
jade thunder
oh it is
opal vault
|x| is the norm of x, so it's a non-negative real number
jade thunder
so you're saying that x^vec d = |x|^d
opal vault
**rafilou2003**
read again
jade thunder
**rafilou2003**
that looks pretty true, just from the definition of p and using only 1 dimension of the possible ones for your d
like if i just chose vec d in Z+^n to be (d, 0, 0, ...)
jade thunder
then sure x^vec d = x^d * 1 * 1 * 1 ...
opal vault
$x^{\vec d} = x_1^d \cdot ... \cdot x_n^0 = x_1^d$
opal vault
Oops sorry
thorny flameBOT
rafilou2003
jade thunder
yes
and if i take sup of this for all x
then i get whichever x has the biggest whatever component
to the power of d
with abs in front
but the sup is later it's outside
opal vault
So for $k\in {1,...,n}, \sup_{x\in \bR^n}(|x_k|^d|\partial^\alpha\varphi(x)|)<+\infty$
thorny flameBOT
rafilou2003
jade thunder
**rafilou2003**
so (\vec x = (x_1, x_2, ..., x_n), |\vec x|^d = \left(\sum_{i=1}^n x_i^2\right)^{d/2})
thorny flameBOT
Frosst
opal vault
so \(\vec x = (x_1, x_2, ..., x_n), |\vec x|^d = \left(\sum_{i=1}^n x_i^2\right)...
yes
But we will do something similar
jade thunder
ok sure
opal vault
We just proved that $\max_{k\in {1,...,n}}\sup_{x\in \bR^n}(|x_k|^d|\partial^\alpha\varphi(x)|)<+\infty$
thorny flameBOT
rafilou2003
opal vault
so we have $\sup_{x\in \bR^n}\max_{k\in {1,...,n}}(|x_k|^d|\partial^\alpha\varphi(x)|)<+\infty$
thorny flameBOT
rafilou2003
opal vault
So this means that, given the infinite norm of a vector :
opal vault
And vice versa, it's the vice versa we will use
opal vault
**rafilou2003**
From this, $\sup_{x\in \bR^n}(|x|_\infty^d|\partial^\alpha\varphi(x)|)<+\infty$
thorny flameBOT
rafilou2003
jade thunder
sure thing
opal vault
so, since infinite norm and 2 norm are equivalent
jade thunder
equivalent?
opal vault
$|x|_\infty \leq C\cdot |x|$
thorny flameBOT
rafilou2003
opal vault
with C a constant that doesn't depend on x
opal vault
**rafilou2003**
And so from this
$\sup_{x\in \bR^n}(|x|^d|\partial^\alpha\varphi(x)|)<+\infty$
thorny flameBOT
rafilou2003
jade thunder
the constant doesn't depend on x?
ah i see what you mean
the inf norm is less than 2 norm always
or equal
at best
opal vault
yes
jade thunder
where'd inf go
thorny flameBOT
rafilou2003
jade thunder
wouldn't that depend on x
jade thunder
i think i see it
the biggest |x| can get relative to ||x||_inf is if x was evenly spread
in which case you'd get (||x||_2 = \sqrt{nx_1^2})
thorny flameBOT
Frosst
opal vault
yes
opal vault
**rafilou2003**
so with this
jade thunder
so we first note that $\sup_{x\in \bR^n}(||x||_{\infty}^d|\partial^\alpha\varphi(x)|)<+\infty$
thorny flameBOT
Frosst
jade thunder
**rafilou2003**
then we use this
to get $\sup_{x\in \bR^n}(|x|^d|\partial^\alpha\varphi(x)|)<+\infty$
thorny flameBOT
Frosst
opal vault
yes
jade thunder
**Frosst**
and we know this is true because the inf norm of x is < inf
because x is a finitely large vector
or rather the biggest component of x still is only finitely big
sure ok
opal vault
**Frosst**
this is true because of the previous steps
we use that phi is in the schwartz space
opal vault
**rafilou2003**
which leads to this by choosing the correct vec d
opal vault
**rafilou2003**
so this*
opal vault
**rafilou2003**
so then we can swap sup and max easily
jade thunder
sure
opal vault
and then back to where we were
I'll repost this important intermediate result here:
If $\varphi \in \mathcal{S}$, then $\forall d\in \bZ_+, \forall \alpha\in \bZ_+^n$, $p_{d,\alpha}^*(\varphi) = \sup_{x\in \bR^n}(|x|^d|\partial^\alpha\varphi(x)|)$ is finite
thorny flameBOT
rafilou2003
jade thunder
ok this is true
opal vault
So in reality
jade thunder
like this?
thorny flameBOT
Frosst
jade thunder
ok let's pretend it was ^
opal vault
of course you need to use it according to d and |x|^d instead
opal vault
$p_{d,\alpha}^*(\varphi) = \max(\sup_{x\in \bR^n, |x|<1}(|x|^d|\partial^\alpha\varphi(x)|),\sup_{x\in\bR^n,|x|\geq 1}(|x|^d|\partial^\alpha\varphi(x)|))$
jade thunder
sure
thorny flameBOT
rafilou2003
opal vault
So
Since $\partial^\alpha\varphi$ is continuous on the compact set $\overline{B(0,1)}$, then we can find a constant $C$ such that $\forall x\in \bR^n, |x| < 1$, $|\partial^\alpha\varphi(x)| < C$
thorny flameBOT
rafilou2003
opal vault
But $(1+|x|^2)^{d/2} \leq 2^{d/2}$
opal vault
yes
jade thunder
ok
opal vault
this is still true (in the 1 radius ball at 0)
jade thunder
indeed
1 is indeed less than 2^d
we have at most 2 inside
so √2^d
which is less than 2^d
great
jade thunder
yes
opal vault
oh right 2^d/2 mb
thorny flameBOT
rafilou2003
jade thunder
**rafilou2003**
wait we had |x| < 1 not |x| <= 1
opal vault
So $1\leq \frac{2^{d/2}}{(1+|x|^2)^{d/2}}$
jade thunder
oh no, |x| can reach 1
how did this happen
thorny flameBOT
rafilou2003
opal vault
how did this happen
I mean can be as close to 1 as we want
opal vault
yes
opal vault
Let's keep the inequalities large so we don't get mixed up
if |x| < 1, then |x| <=1 and we apply the same thing
opal vault
So $C\leq \frac{2^{d/2}C}{(1+|x|^2)^{d/2}}$
thorny flameBOT
rafilou2003
opal vault
Let's do one better
jade thunder
that's also true
opal vault
$\forall x\in \bR^n, |x| < 1$, $|\partial^\alpha\varphi(x)| \leq \frac{2^{d/2}C}{(1+|x|^2)^{d/2}}$
thorny flameBOT
rafilou2003
jade thunder
**rafilou2003**
do we now choose this cool C
opal vault
yes soon
opal vault
**rafilou2003**
so this is our first part of the "cool" C
jade thunder
**rafilou2003**
here you just used
opal vault
Yes
jade thunder
ok good
opal vault
So now we can define our cool C, we'll "C" if it works
jade thunder
ha ha!
opal vault
$C_{d,\alpha} = \max(p^*{d,\alpha}(\varphi),2^{d/2}\sup{|x|<1}|\partial^\alpha\varphi(x)|)$
thorny flameBOT
rafilou2003
jade thunder
woah
that looks wrong
the p* is already using |x| < 1
why is the other one also < 1
opal vault
?
jade thunder
shouldn't the other one take care of the case when |x| > 1
since the p* is the one we choose for when |x| < 1?
opal vault
the one on the right takes care of the case |x| < 1 as we just saw
opal vault
**rafilou2003**
We took C = sup for |x| < 1 of...
jade thunder
no wait