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#help-27
woven radishBOT
wicked rover
its just *only* the number being the lower index
ah some books do it
iirc folland analysis
i do it to be lazy
astral lodge
ah ok
vocal mural
I got to $f(x) - x f(x) = a_0 + {1 \over ({1-x)^2}}$
woven radishBOT
eugene
vocal mural
I dont think i know how to rewrite -xf(x) as a generating function
wicked rover
I dont think i know how to rewrite -xf(x) as a generating function
LHS = f(x)(1-x)
vocal mural
how did you get 1-x
wicked rover
factor
vocal mural
nice okay
wicked rover
thats all 🙂
vocal mural
Lol i just realized
they wanted me to use the n+1 formula
since the hint gives the answer
$a_{n+1} = a_n + (n+1)$
woven radishBOT
eugene
wicked rover
theres no need to redo if all calculations are right
i just rewrote the recursion to make plugging it in easier
you shouldve done index shifts somewhere tho…
thats where the hint comes in
vocal mural
Yeah but also, the problem specifically states use the recursion found in Problem 1 part 3 which is the n+1 formula
Problem 1 part 4 is the a_n formula
vocal mural
you shouldve done index shifts somewhere tho…
yeah i did this, but ill just redo it just in case
wicked rover
if u didnt do the index shift before using the hint then some bad math happened…
id even say that the two index shifts, one where i rewrote the recursion and one before using the hint, are necessary
vocal mural
I started with $$ f(x) - a_0 = \sum_1a_nx^n= \sum_1a_{n-1}x^n + \sum_1nx^n $$
the last part became x / (1-x)^2
wicked rover
yeah u had to use my rewritten recursion here
woven radishBOT
eugene
woven radishBOT
eugene
vocal mural
I dont have to use a_0 btw since we knew a_0 = 0
It was way easier to do with the a_n+1 formula actually
devout snowBOT
@vocal mural Has your question been resolved?
dim lantern
devout snowBOT
dim lantern
Idk ig I'm dumbb
bitter quarry
0 A as the current would avoid the resistance and go to the wire with no resistance
bitter quarry
you can set the other wire to be some resistance x and the equivalent resistance would be 1/R = 1/5 + 1/x, as x goes to 0, 1/x goes to inf and 1/5 can be neglected so 1/R is just 1/x and R=x which is 0
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sleek hamlet
devout snowBOT
sleek hamlet
guys how do i do this?
pseudo basin
can you come up with a sequence of basic transformations, described in words, that gets you the right graph?
meaning like translations or vertical/horizontal stretches that sort of thing
sleek hamlet
move 1 unit left and 4 units down
dilated 3 from y axis and 2 from x
but here is the form they're asking for
pseudo basin
dilated 3 from y axis and 2 from x
hmm id like a bit of specifics on that one
actually you need to be careful with the ordering in all of these
here actually how about you try this: you write down the sequence of steps, but after each step, write the resulting equation of the curve
including the original
so like
Start: y = 1/x
TRANSLATE 1 LEFT
y = 1/(x+1)
STRETCH HORIZ. 1/2
y = 1/(2x+1)
...
ok
ok so far ive got 3/x from a dilation of factor 3 from x axis
how do i get to 3/2x?
pseudo basin
maybe stretch horizontally by 3/2
this will get you y = 1/(2x/3), which simplifies to 3/(2x)
sleek hamlet
oh i see dilate by 3/2 from x
sleek hamlet
how will that be slightly tricky?
pseudo basin
translating 1 left won't do it
sleek hamlet
ohh i see so translate 1/2 to the left
i see its either option A or E, but am confused on the dilation thing
devout snowBOT
@sleek hamlet Has your question been resolved?
fossil locust
i see its either option A or E, but am confused on the dilation thing
so if $x \mapsto x/2$, that's a compression factor of 2 because the x-coordinate is being divided by 2
woven radishBOT
south
fossil locust
now you apply the $x \mapsto x - 1/2$ and combining this with the previous transformation, you get $\frac{1}{2}x - \frac{1}{2}$
woven radishBOT
south
fossil locust
so you go from f(x) to f(2x) to f(2(x + 1/2)) = f(2x + 1)
and up to here you get y = 1/(2x + 1) as desired
also, you need the brackets in 2(x + 1/2) because the compression is being applied first and not the translation
you start out with 2(x) then you replace x with x + 1/2, and the brackets stay
devout snowBOT
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fickle slate
What is the range of the function
devout snowBOT
What is the range of the function
soft umbra
Example?
fickle slate
$y= 3sin\theta , \theta=\sqrt{\frac{\pi ^2}{16}}-x^2$
frozen aurora
use \pi for pi
woven radishBOT
Triaengle
frozen aurora
is x^2 inside the square root or no
fickle slate
it is inside
frozen aurora
it is inside
$y= 3\sin\theta , \theta=\sqrt{\frac{\pi ^2}{16}-x^2}$
woven radishBOT
artemetra
frozen aurora
it is inside
what is the largest theta can be?
fickle slate
The sqrt term cannot be less than 0
frozen aurora
yes
stone stump
(you need to get comfortable thinking only in radians)
frozen aurora
bro, someone in help-25 needs you
lmao 😭 i honestly would've helped you but i know quite literally nothing about vector calculus
stone stump
The sqrt term cannot be less than 0
you said yourself that negative angles are not possible
frozen aurora
so acc to me the domain is -45° and 45°
domain or range?
fickle slate
domain
frozen aurora
yes
but i am asking about the range of theta
is the book talking about the domain or range of theta?
fickle slate
is the book talking about the domain or range of theta?
both
frozen aurora
can you show a picture?
fickle slate
wait
frozen aurora
cuz the domain of theta is -45° to 45° (or, better said, -pi/4 to pi/4)
but the range of theta is 0 to pi/4
fickle slate
Oh my bad it was range
but how is the domain -π/4 to π/4 but the range is only positive
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devout snowBOT
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rustic isle
Is the relation {(0, 1), (2, 3)} transitive?
devout snowBOT
Is the relation {(0, 1), (2, 3)} transitive?
pseudo basin
on what domain?
winter torrent
i don't think the domain matters assuming 0, 1, 2, and 3 are all distinct objects
faint zinc
@rustic isle it's an interesting edge case, because transitive says if you have aRb and bRc then you must also have aRc. But you do not have any b that satisfies aRb and bRc.
So it is vacuously true
pure flower
devout snowBOT
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wicked fractal
devout snowBOT
wicked fractal
in this question i did it by using two seperate distributions then finding the "at least more than 2" probability for each of them, then multiplying them
nvm i got it
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viral lynx
devout snowBOT
viral lynx
cos(x)/(cos(x) + sin(2x))
cos(x)/(cos(x) + 2sin(x)cos(x))
1/(1 + 2sin(x))
So cos(x) = 0
x = pi/2, 3pi/2
Are those the 2 discontinuities? Was that all I had to do?
Oh wait I need coordinates right
limpid dove
sinx can also lead to a zero
viral lynx
sinx can also lead to a zero
That name 😂
limpid dove
1 + 2sinx = 0
limpid dove
That name 😂
i like math (a bit too much)
viral lynx
But to get the coordinates I just put pi/2 and 3pi/2 back into f(x) right?
limpid dove
RD is a hole, so cosx = 0 will be the points
viral lynx
Like f(pi/2) and f(3pi/2)
limpid dove
yes
limpid dove
well, you take the limit
viral lynx
❤️
viral lynx
well, you take the limit
wait what sry
limpid dove
but since youve already factored out the cosx, just plug in f(that)
viral lynx
Ahhh yeyeye
I get you
Alright, I can manage that hehe
Thank you! (again)
❤️
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devout snowBOT
limpid dove
np
woven radishBOT
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regal berry
Given x show that x meets the marked condition
devout snowBOT
Given x show that x meets the marked condition
regal berry
i am not sure if I did right what I did or what to do now
So if someone could pleasa help me
regal berry
If only I could read that 😅
Sorry
mystic scarab
!original
devout snowBOT
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.
regal berry
!original
This is the original problem-
I literally said above the picture. What we are asked to do
lime plaza
Like write it neatly
sullen island
what are a and b anyway
regal berry
what are a and b anyway
variables
regal berry
Idk? all they said is that given x = ... x, real number show that the marked condition is true
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devout snowBOT
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valid garden
X axis - skip count by 1, y axis- skip count by 5.
hey guys, i just need help with what number should i start the scatter plot on, (what u guys think would be ideal) thats all!
devout snowBOT
@valid garden Has your question been resolved?
valid garden
🥺
valid garden
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devout snowBOT
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wraith pumice
devout snowBOT
wraith pumice
what's the first step here?
wraith pumice
hmm
wraith pumice
top by x^5 and bot by x^10 inside the sqrt and so x^5 outside
sqrt(x^10) = x^5?
pseudo basin
x -> ∞ of course x will be positive
olive snow
depends if he write algebra with or without the limit
final osprey
if the degree of x in the numerator and denominator is the same
olive snow
thats why i precise
final osprey
the limit will simply be the ratio of their coefficients
wraith pumice
top by x^5 and bot by x^10 inside the sqrt and so x^5 outside
so u get -40 / sqrt(64x^10 + 7x^9) / sqrt(x^10) right?
wraith pumice
if the degree of x in the numerator and denominator is the same
so they're the same here bc the sqrt makes x^10 be x^5?
final osprey
so they're the same here bc the sqrt makes x^10 be x^5?
yas
pseudo basin
depends if he write algebra with or without the limit
assuming x ≥ 0 is literally fine lol
olive snow
so u get -40 / sqrt(64x^10 + 7x^9) / sqrt(x^10) right?
rather writing sqrt(64x^10 + 7x^9) = sqrt(x^10(64+7/x)) = x^5(sqrt(64+7/x))
final osprey
so technically I can rewrite as numerator / 8x^5 + sqrt(7x^9)?
not really
final osprey
just deal with their coefficients
final osprey
so we get -40 / x^5(sqrt(64+7/x)) ?
wait let me write something out for you
youre on the right path
but take 64 out as well
out of the sqrt
wraith pumice
but take 64 out as well
how? ur adding 64 and 7/x
so I'd think u can't go any furthber
,w lim x to infinity (-40x^5 + 5x^3 -8x^2)/(sqrt(64x^10 + (7x^9))
woven radishBOT
woven radishBOT
wraith pumice
final osprey
yes
final osprey
okk
final osprey
u divided numerator by x^5 right
i factored out the term with x^5 in both the numerator and denominator
<@&268886789983436800>
wraith pumice
i factored out the term with x^5 in both the numerator and denominator
I see
final osprey
i factored out the term with x^5 in both the numerator and denominator
cause that will give 1/x or something similar for all other terms
anything divided by infty is zero
so we can get rid of that
x^5 will cancel each other out and we'll be left with -40/8
wraith pumice
yeah
makes sense
I'm given another problem
is this just the same methodology
I should be able to solve it myself if so
final osprey
yess
wraith pumice
okay perfect, should be doable then
final osprey
it's the same methodology but pay attention to the degree of x in the numerator and denominator
wraith pumice
yeah fs
final osprey
tropic girder
Hi,
I just saw a video that explained how to factorise a second degree equation by using the sum and product method but without Guessing the numbers.
It seems very efficient to find the roots, and even with complex numbers (Hence, in this case i (and I don't know yet if there are others)
I was wondering, do you think this may be more efficient than using the Completed square form method in order to factorise ?
Perhaps you need more details, don't hesitate to ask me
soft umbra
Hi,
I just saw a video that explained how to factorise a second degree equation...
The completion of square is the most efficient way in terms of simple quadratic function
heady plinth
Other complex numbers are just multiples of i and multiples of i plus a real number
Can you send the vid
winter patrol
sum and product method isn't efficient if the roots aren't nice
soft umbra
There is an alternative, which applies formula to find the solution. It’s used when the roots are complicated
winter patrol
factorisation is most efficient if you can easily identify the pair of values
wraith horizon
I would always try for the sum and product method and if it doesn't look like it's going to give nice (integer) solutions then I'll reach for the quadratic formula
tropic girder
Other complex numbers are just multiples of i and multiples of i plus a real num...
Yeah we can find the i + other numbers thanks to his method
And what he says makes sense so that's why I would have liked to use it instead of the completed square form
tropic girder
The completion of square is the most efficient way in terms of simple quadratic ...
Do you know in which cases it would be more efficient to use this alternative?
winter patrol
if the goal is to factorise any quadratic,
i'd recommend completing the square
and difference of two squares
tropic girder
I would always try for the sum and product method and if it doesn't look like it...
I know this method perfectly, but it is long to use
I rather to use the Sum and product way (when the numbers are nice), but I often use the completed square form
However, I am still trying to look for efficient methods to work that I may like
And ∆ is far from being my favourite, even if it was when I started this year
winter patrol
their approach using symmetry of roots in a slighlty different way makes it look tedious
tropic girder
if the goal is to factorise any quadratic,
i'd recommend completing the square
a...
You think so ?
What about factoring with a complex number
Is this as simple as the method that I shared ? (I never worked with imaginary numbers yet)
tropic girder
their approach using symmetry of roots in a slighlty different way makes it look...
I kinda liked the symmetry because I tend to think the same way
tropic girder
complex number for a root?
Then I'll continue to use the little completed square form method since it seems more efficient
winter patrol
x^2 + 4x + 10
completing the square, also applies the principle of symmetry
= x^2 + 4x + 4 + 6
= (x+2)^2 - (-6)
then factor as a difference of two squares
= (x + 2 - i * sqrt(6))(x + 2 + i * sqrt(6))
for the last example,
cuts down so much
tropic girder
Indeed
I'll keep it then
I wonder why we learnt the quadratic formula first when there were so many more efficient methods that also work
My classmates are stuck on that formula while it can be so long to do 🥲
winter patrol
its designed to solve any quadratic by just blindly plugging in values
with the sole goal of getting the roots
devout snowBOT
@tropic girder Has your question been resolved?
tropic girder
its designed to solve any quadratic by just blindly plugging in values
Ah yeah
Simply by doing a job step by step
Uh
wraith pumice
devout snowBOT
wraith pumice
can someone tell me in what scenarios a horizontal asymptote exists, vs a vertical asymptote, vs a slant asymptote?
or could there be multiple?
etc
final osprey
a vertical asymptote exists when x-> a certain constant, the value of f(x) -> infinity
a horizontal asymptote exists when x-> infinity, the value of f(x)-> a certain constant
wraith pumice
a horizontal asymptote exists when x-> infinity, the value of f(x)-> a certain c...
okay, I'll try and make sense out of these
thanks
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devout snowBOT
final osprey
can someone tell me in what scenarios a horizontal asymptote exists, vs a vertic...
slant asymptote is usually the quotient when the numerator has a greater degree than the denominator
oop
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Available help channel!
wraith pumice
@final osprey sorry if you don't like being pinged but
devout snowBOT
<@852139670136553483> sorry if you don't like being pinged but
final osprey
yess
native galleon
does this seem about right?
Looks right.
final osprey
an equal sign is missing in the second point for va
wraith pumice
an equal sign is missing in the second point for va
oh yeah right
follow up question regarding slant asymptotes
I graphed (x^2+3x+2)/x-1 in desmos
and apparently there's a slant asymptote at y=x+4
makes sense but when you do long division you get this
does the 6/x-1 just despawn or something?
final osprey
makes sense but when you do long division you get this
yess
the quotient is x+4
which is the asymptote
6 is the remainder
so you write it as (x+4) + 6/x-1
wraith pumice
so you write it as (x+4) + 6/x-1
but this isn't the slant asymptote right?
wraith pumice
I see
thanks so much!
for the future if I have follow up questions is it fine if I ping u or no?
final osprey
suree
i'll help if i know the concepts
devout snowBOT
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Available help channel!
trail spoke
devout snowBOT
trail spoke
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devout snowBOT
worldly tangle
Je comprends pas la 4.a 💀(correction)
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velvet wren
could any one explain it to me how to do it
devout snowBOT
could any one explain it to me how to do it
velvet wren
like 7
polar chasm
Yeah, that too. But the fact that they are both multiples of 28 is stronger
velvet wren
how do u get the answer
polar chasm
The LCM is a multiple of 35 which is 7 * 5
note that the LCM is divisible by 5
that means that at least one of our numbers must also be divisible by 5
try using those facts above to get a pair of numbers that's as small as possible and satisfies the criterions
devout snowBOT
@velvet wren Has your question been resolved?
scenic gull
I'm looking for analysis of zero-sum-utility auctions, but auction theory texts I've looked at don't cover zero-sum anything and game theory texts don't explore auction-like zero-sum games.
devout snowBOT
I'm looking for analysis of zero-sum-utility auctions, but auction theory texts ...
scenic gull
Specifically, I'm curious about dominant and equilibrium bidding strategies for sealed bid 1st & 2nd price actions of single items.
Assuming n bidders with independent valuations (v) from 0..m uniform distribution, the forbidden tools told me v*n/(n+1) is the nash equilibrium bidding strategy for zero sum 1st price auction. It passes the sanity check of being less than v but more than the positive-sum bidding strategy of v*(n-1)/n. I also did a Monte Carlo Sim and n/(n+1) did do the best amongst simple multipliers of v.
devout snowBOT
@scenic gull Has your question been resolved?
raven pasture
This might be more suited to one of the topic channels rather than help-xx
devout snowBOT
@scenic gull Has your question been resolved?
crisp fog
Hello. I was working on the solutions to these. I have no idea how to solve d. Usually, I work backwards from the answer when I'm stuck. But here, I'm completely clueless
devout snowBOT
Hello. I was working on the solutions to these. I have no idea how to solve d. U...
astral lodge
Hello. I was working on the solutions to these. I have no idea how to solve d. U...
What have you tried?
misty crest
crisp fog
I'm sorry, I'm having a hard time answering. Hang on
astral lodge
why is there no t-value 
crisp fog
It's basically simplyifying the equation for a coujple of these.
I tried eliminating two because I got confused with r(2-t)-r(2)
crisp fog
As soon as I realized that didn't make sense, I got really confused
acoustic leaf
Δt is not the same as t. treat it as a different variable
astral lodge
I tried eliminating two because I got confused with r(2-t)-r(2)
just plug it in normally
you will get an i an j-component that is simplifiable
crisp fog
The 2 and r(2) are where I;m stuck. If it was r(2-delta(t)) I would understand a little, but the answer has a 4, and two separate instances of the number two, with a square
astral lodge
do not fret
calculate $\textbf{r}(2-\Delta t)$ and $\textbf{r}(2)$ seperately, and tell me what you get
woven radishBOT
vengeance
crisp fog
Okay, so like two separate functions, and merge them together?
astral lodge
Okay, so like two separate functions, and merge them together?
yes
astral lodge
you will get an i an j-component that is simplifiable
again, refer to this
crisp fog
Okay, that makes sense thank you. I don't know why I got wrapped up in that so much. I can work with this thank you!
.close
wheat folio
If t=tanh x/2 prove that sinhx= 2t/1-t² and coshx= 1+e²/1-e²
devout snowBOT
If t=tanh x/2 prove that sinhx= 2t/1-t² and coshx= 1+e²/1-e²
@wheat folio Has your question been resolved?
violet wind
Can you just use the formula for tanh?
wheat folio
I'll try
devout snowBOT
@wheat folio Has your question been resolved?
devout snowBOT
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Available help channel!
cyan axle
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cyan axle
I’m having trouble with the kvf rules
How do you decide the order of the loops
It makes sense that from a i have a,f,e,d but why no c or b? Isn’t that also in the loop?
Same for the other side b, c, d, e, f why doesn’t it cause an and if it basing if off loop 1 why does it include f?
midnight echo
Wdym the order of the loops? There are three closed meshes in that circuit, abcdefa, abcfa, efcde
midnight echo
you can go afedcba, thats fine too
cyan axle
Would it complicate anything?
midnight echo
but you cant go afedcfa, if thats what you mean?
midnight echo
Would it complicate anything?
no
You just need to be careful about your signs, that's all
technically d and c are part of the same node, so one of those are redundant
you can do your analysis completely disregarding one of the two, since the properties at d = properties at c
cyan axle
Yeah about those signs
The next step looks like this
How do I know which current can be added together/simplified?
Because it’s not like 5 and 10 are in parallels
I’m having problem with a lot of Kirchhoffs laws
This is a practice test problem
midnight echo
You should be a bit more organized. What is your ground (referece)? It looks like node b
cyan axle
I’m trying but this is the answer sheet I have
I don’t understand how you figure out which currents simplify like that
midnight echo
Okay, your prof is using mesh current analysis here. It's one of the two methods (along with nodal voltages) to systematically solve circuits. To do mesh current, first you need to find all the closed loops in the circuit.
cyan axle
I also don’t really understand why the prof starts where they did….
All the book and assignment examples started at the power source so I’m like completely lost here
midnight echo
You can incorporate the power sources at any point. Have you found the two meshes?
smoky nimbus
I also don’t really understand why the prof starts where they did….
You can start anywhere in the loop. All that matters the most is the signs
midnight echo
a mesh is simply a closed loop
cyan axle
No, no I haven’t
cyan axle
You can start anywhere in the loop. All that matters the most is the signs
How do I know where the loop starts and ends though
midnight echo
it must start where it ends
cyan axle
??
smoky nimbus
How do I know where the loop starts and ends though
... It's a loop
smoky nimbus
You can't go from abcd and end it there
cyan axle
Ah
smoky nimbus
It has to close
cyan axle
So afcb and fcde work?
smoky nimbus
Yes
midnight echo
yes, but its good practice to be consistent in your direction. either clockwise or anticlockwise
cyan axle
I just need to go back to where I started?
But how do I know which resistors values can be added together for simplification
Just whatever’s in series?
midnight echo
they cant be in this question since there are power sources
midnight echo
no
cyan axle
Man
smoky nimbus
But how do I know which resistors values can be added together for simplificatio...
You're not simplifying resistors. KCL is related to the current of the circuit
It's based using V = IR
smoky nimbus
The terms with the i3
cyan axle
Yes but where does that come from
midnight echo
Once you have both loops determined, you just apply kvl to each loop
smoky nimbus
-5 * i3 - 5 * i3 = -10 * i3
midnight echo
Draw out your loops first
midnight echo
draw your loops and we'll be able to explain it
smoky nimbus
Looking at this how do I determine which have like current values
Typically the resistors
Because of V = IR
cyan axle
draw your loops and we'll be able to explain it
Wdym draw a loop I thought you just determined a loop and base it off that 😭
smoky nimbus
You draw the loops, to help you identify the signs of each component so you can make the equation
cyan axle
??
midnight echo
Yeah that's all
cyan axle
So I have fabcf and fcdef
midnight echo
Yeah
basically those two rectangles
Now, lets think of Loop 1. From KVL, we know that the total voltage difference from the start of the loop to the end (the starting point) is 0
right?
cyan axle
Yes assuming it’s closed right?
midnight echo
yes, and it is. since its a closed shape (we end where we start)
so let's call the current flowing through b (in the clockwise direction) I1
What is the voltage at point b? If point a is the ground?
cyan axle
It should still be 20 right?
cyan axle
But it’s not said in the image that there is a negative side so I assumed no battery just charge
This was the question before it
I’m ok with this one tho
midnight echo
I mean the drawing of the battery implies direction
cyan axle
Okay so assuming that the a side is + and b is -?
midnight echo
sometimes the + and - is ommited but we still need to consider it
yeah
its not really an assumption, the long line means +, thats convention
cyan axle
Wouldn’t it be easier if I went ccw then?
midnight echo
no cause the other loop is gonna be poisitive lol
cyan axle
Yeah
midnight echo
youre gonna have to deal with one negative V regardless
cyan axle
Alright
orchid kettle
Ts pmo fr
midnight echo
so at b its -20 V, what is the voltage difference as it goes from b to c?
cyan axle
Ts pmo fr
You’re probably like atleast three years older then me get off your high horse
orchid kettle
You’re probably like atleast three years older then me get off your high horse
Ts real
cyan axle
so at b its -20 V, what is the voltage difference as it goes from b to c?
Would this be from V_b=Vc+IR?
midnight echo
Yeah, so that's -5I1 right?
smoky nimbus
https://youtu.be/8f-2yXiYmRI?si=lUw_c3HvP8MhCCKT&t=870
This channel is good in general and you should definitely watch the entire video if you get a chance, but I linked the start of the example he does. It's about 20 minutes long. It may clear things up for you
midnight echo
Consider voltage to be a charge that is dissipated or added, everything is relative. You should never divide a previously obtained voltage
The voltage at point b is -20, since the resistor is dissipative (causes energy loss), we lose -5*I1 voltage as we go from b to c
cyan axle
Why I_1 tho
midnight echo
well that's the current flowing from b to c, right?
it hasnt split yet, so the current at c is the same as the current at b, which we defined as I1
cyan axle
So it’s I1 because it’s the first one?
midnight echo
You can call it I_banana, its just some current
midnight echo
tes
cyan axle
The junction rule
smoky nimbus
It splits into I2 and I3
The method you are doing, does not split into 3 currents
smoky nimbus
Nothing to do with nodes
cyan axle
Not the nodes but the path
smoky nimbus
You have two loops
cyan axle
But having two loops doesn’t change it being there?
smoky nimbus
You're going to confuse yourself
midnight echo
You're skipping steps. Just follow the system
The reason this system exists is since its consistent
smoky nimbus
Keep it simple. You labeled two loops, one loop that has current I1 and the other has current I2
cyan axle
But in the profs example we have I3?
Okay I’ll wait
I’m not going anywhere with that
midnight echo
I1 and I2 are the currents at the origin of the loop. Ie. the current that enters and exits the respective power sources
at point C, we have -20V-5*I1, correct?
smoky nimbus
I’m not going anywhere with that
My suggestion, still is watch the video I linked. It may clear up a lot of your confusion
cyan axle
Yes that’s the equation so far and following the loop we go form c to f?
midnight echo
Yes, now, when we go from c to f, the current I1 gets influenced by the current coming from loop 2 in the opposite direction, I2.
You can visualize it like the flow of water, at the junction between c to f, I1 is going left, but I2 is going right
make sense?
midnight echo
Yeah, so logically the current between C and F will be... (we're still on Loop 1)
cyan axle
But the way I’m thinking about it why is loop 1 able to keep moving forward if the voltage in loop two is pushing back with nearly double the voltage?
midnight echo
It might not be able to!, we assume I1 is positive, but the result of our calculations might be that I1 is negative, which means that it is indeed in the opposite direction (ccw)
But for now, we assume that its able to, in order to determine the value of I1
Once we get the value (and the sign of) I1, we'll be able to conclude if its able to or not
cyan axle
So it’ll be the resistance it’ll normally have plus the resistance by the current from loop 2
midnight echo
I'm just asking for the current from c to f, nothing to do with resistance
cyan axle
Yup
midnight echo
You have I2 flowing right
cyan axle
Yes
midnight echo
what is the net flow left
just the current
dont think about the resistor for now
cyan axle
I1-I2
midnight echo
exactly
which is "I3", but its best to keep it as I1-I2
So the voltage loss from c to f will be..?
cyan axle
3_I1-I2
cyan axle
Don’t I have to go back to a?
midnight echo
yeah, can you finish it up?
recall that the current collects back as it is about to enter a, so the current entering a will be the same as the current leaving b
cyan axle
?
cyan axle
I’d be better off starting at d for loop 2 right?
cyan axle
Or maybe e?
midnight echo
I would do d but both is fine
cyan axle
If I start at d I’m not allowed to go just through that am i?
midnight echo
youre allowed to go through the battery yeah
you just need to add 35V
bc thats whatthe battery does
cyan axle
I’ll just start at e for my sanity then
midnight echo
sure, but you still have to go through the battery to get back to e fyi
cyan axle
Just adds 35?
midnight echo
yup, just like how the other battery subtracted 20
cyan axle
Okay I’ll start at d just to get it away to start then
But wouldn’t the net just be 0 then
From d to e
midnight echo
from d to d yeah
cyan axle
D to d or d to e?
midnight echo
you still need to do the +35 though, just like how you did it for the other loop
d tod
you end where you start, thats the definition of a loop
you're overthinking it, just start whereveer and try to do it
cyan axle
Like that?
Starting at d
Stars negative goes through the battery and all the resistors then back at d is gains it again?
midnight echo
starts at d, which is 0 (our starting point). gains 35 as it goes through the battery, loses 10I2 as it goes through the first resistor, loses (I2-I1)*3 as it goes from the second resistor, and then we're back to d
and all of that must sum to 0
I think you confused yourself with the battery, thats why you have a bunch of 35s
cyan axle
Why isn’t starting at d -35?
Yeah this is my first time seeing an example going through the battery somehow
midnight echo
it can be -69, doesn't matter. Voltage is a relative value, meaning we care about the voltage difference between two points. Between d and d the voltage difference is 0, regardless of what the "absolute" voltage is. That's all that matters,
we have "some amount" of voltage at d, we gain 35 as we go through the battery, lose some through the reisstors etc but we END UP with the same "some amount" at d, so it makes sense to define d as 0
cyan axle
Okay so since I start with -35 and I’m going through it to e I can just say 0
Then it’s 0 minus whatever it loses going through the resistors
midnight echo
you start with 0 at d, you gain 35 through the battery, and lose some amount through the resistors, you must end up with 0
not -35, this time you go through the + ve junction of the battery dont forget
midnight echo
Yeah this is my first time seeing an example going through the battery somehow
you did so in the previous loop too
midnight echo
Why do i start with 0 there but not at loop 1?
you did
cyan axle
So it’s the direction I’m going in relative to the battery?
midnight echo
lets try to verbalize what we did for loop 1. We started at b (0), we lose some amount throughg the 3 resistors, and finally lose 20 more through the battery
at the end of the process, we ended up at b, which we defined to be 0
cyan axle
But with loop 1 I never went through the batter I went from b to a
midnight echo
and then we went from a to b...
why do you think we have the -20?
because we went through the battery
when we went from a to b to close the loop
midnight echo
we started with -20 because we implicitly went through the battey as we went from a to b
yeah exactly
we started at a, went through the battery (losing -20), and then went through the three resistors
losing some more
cyan axle
So with d I go through the battery but wouldn’t that just be +35
midnight echo
going from d to e?
midnight echo
So with d I go through the battery but wouldn’t that just be +35
yeah exactly
cyan axle
So d is -35 going through it I get +35
cyan axle
From d to e is zero?
midnight echo
no, its +35 since we go through the battery
(and whatever amount lost through that resistor)
cyan axle
Man
But if I went from a to b why isn’t it +20
I also went through the battery and a is grounded
cyan axle
I hate how this is written
midnight echo
remember, + end is long, - end is short
cyan axle
I keep assuming going through it is +
midnight echo
nope
midnight echo
yeah
woven radishBOT
cyan axle
That
midnight echo
yeah
cyan axle
=0 ofc
midnight echo
and both are equal to zero,right?
cyan axle
Yuh
cyan axle
Lemme write the grounding stuff down Before I solve this
midnight echo
Don't forget that voltage is inherently a relative concept. No point on a circuit (or anything irl) has XX volts, it has XX more volts than some arbitrary datum/reference. The battery adding 35 volts just means that the exit of the battery has 35 more volts than the entrance, not that the entrance has "0" or the exit has "35" volts. That's the intuitive definition of KVL. If we go in a full loop and end up back in the same place, the amount of volts at the starting point should be the same as the volts in the ending point (since they are the same physical point)
cyan axle
I am so glad I’m not going for electrical engineering then
midnight echo
sometimes I think I should've went into that instead lol
cyan axle
I’m allowed to simplify like that right?
cyan axle
sometimes I think I should've went into that instead lol
What did you choose instead
midnight echo
mechanical
midnight echo
More pay/opportunities in electrical maybe? Microchip/gpu engineering make crazy money, can't really come close to that in mech, but maybe its a grass is greener illusion thing idk
cyan axle
There is a lot of excitement in that rn
But for me tbh I couldn’t be interested in it if you paid me
Probably why I’m struggling so bad in this when there isn’t even calc
midnight echo
Fair enough. I hated enm in high school but warmed up to it in the few electronics classes I took in uni
cyan axle
I actually went to trade school for manufacturing before I realized they didn’t care about college ed
I liked cad and that was kinda the whole reason
The profs example comes out kinda different?
Its kinda saying that the current from the two 5I and the 10Inare the same
And it’s just adding everything as like terms
midnight echo
what dud you get for I1 and I2
cyan axle
Didn’t solve for it yes I have to solve it as a system no?
The main issue I have is I don’t really have a lot of like terms in mine
midnight echo
yeah, but the 3 should be outside the parens
midnight echo
you can just solve for I2 in terms of I1 using the top equation and plug it into the bottom equation
algebras gonna get a bit messy eiter way
cyan axle
Yeah a bit
Idk how the prof did it but it’s in a way where she could just subtract 1 from 2
Yeah screw it im using a calculator
midnight echo
dont think so
midnight echo
What did you get?
cyan axle
I_2= 79/32 and I_1= -31/32
Off by a bit compared to prof
I have all the work so I’ll just ask the prof/TA
cyan axle
Huh….
wraith pumice
for each vertical asymptote, analyze x=a. analyze lim x-->a from the left of f(x) and from the right.
devout snowBOT
for each vertical asymptote, analyze x=a. analyze lim x-->a from the left of f(x...
midnight echo
what are your vertical asymptotes?
wraith pumice
what are your vertical asymptotes?
idk
I mean, I know to find it you need to set the deominator to 0 and then solve
if the demoniator is 0 when the numerator isnt then it's asymptote
and u gotta do extra stuff
this is the graphs I have based on a previous related problem x to +- infinity of f(x)
VA seems to be x=-4 but like idk
why is it saying asymptotes as if there's multiple
devout snowBOT
@wraith pumice Has your question been resolved?
upbeat fjord
where can i ask for help?
devout snowBOT
where can i ask for help?
upbeat fjord
i dont know how to do letter a
this is my brothers homework hes in 7th grade and hes asking for help but not even j know how to do it
acoustic leaf
,rccw
woven radishBOT
manic sedge
xy=0.32
x+y=1.2
acoustic leaf
so you are trying to find two numbers which multiply to 0.32 and add to 1.2
it may be easier to think about if you work in fractions instead of decimals
upbeat fjord
i did. i know how to do it, i just can’t seem to do the decimal one.
im really confusdd
upbeat fjord
it may be easier to think about if you work in fractions instead of decimals
thank you i will try this
manic sedge
Fractions are far better actually, like solving it algebraically would require quadratic
prime narwhal
devout snowBOT
prime narwhal
After I do the question 5a how do I identify restrictions?
rocky marlin
Do you know what restrictions are?
rocky marlin
Alr what is it cause i have no idea but if you tell me and refresh my memory i can prob do ot
prime narwhal
Like so for 5a would it be x can't equall -1
rocky marlin
Ohhhh ok hold up
Ok so first re write the equation so its a fraction
So the first part over the second part
devout snowBOT
@prime narwhal Has your question been resolved?
scenic osprey
I'm given this question.
During a certain epidemic, the number of people infected at any time increases at a rate proportional to the number of people that are infected at that time. In a city, where the population is 67,500, if 1200 people are known to be infected, how many days will it take until 85% of the residents of the city are infected?
if i'm not given an initial condition, how can i find this answer?
scenic osprey
i was not given a second point in time where it shows the number of infected people
crisp tapir
what equation did you get
i think the limits will be from 1200 to 0.85*67500 and 0 to t
scenic osprey
P(t)=1200e^kt=57375
crisp tapir
P(t)=1200e^kt=57375
yes
scenic osprey
so if i'm not given a second point in time, is this question not possible to do? because i need to find k in order to find the question
crisp tapir
youre right
scenic osprey
🤔
crisp tapir
maybe youre supposed to answer in terms of k
scenic osprey
"how many days will it take until 85% of the residents of the city are infected?"
crisp tapir
are you given the proportionality constant in the question
scenic osprey
are you given the proportionality constant in the question
that was the full question
okay, so i just asked someone else. i guess our teacher gave us y(1) 😓
wasn't written on the paper
crisp tapir
bruh
scenic osprey
mb lol i didn't know
normal lagoon
0.018% infected at one point in time. And increases at a rate proportional to the number of people that are infected at that time. So if we assume day 1 is the point where 0.018% of the population is infected, day 2 will see a 0.018% increase in the infected individuals in the population.
Okay nevermind
Seems like you were given more info.
crisp tapir
its instantaneous
scenic osprey
thanks guys, sorry for inconvenience 😂 😭
normal lagoon
I was thinking if it grows every day via the % of the population increases it would reach 85%+ in a lot of days.
devout snowBOT
scenic osprey
0.018% infected at one point in time. And increases at a rate proportional to th...
this is what i thought before my friend told me there was an initial value
crisp tapir
its not a linear function lmao
devout snowBOT
Available help channel!
normal lagoon
I know but it WAS the only information we were given before we found out there was an initial value.
devout snowBOT
I know but it WAS the only information we were given before we found out there w...
crisp tapir
fair enough
normal lagoon
I was told that I can use regular PMI to solve both of these problems.
Can we start with the first?
Base-kit - Any positive integer so the lowest value is n=1
Now what is the next step? Proving the hypothesis applies for some k element of positive integers?
pseudo basin
Now what is the next step? Proving the hypothesis applies for some k element of ...
nope
you assume that the thing holds for k and prove from that that it also holds for k+1
devout snowBOT
@normal lagoon Has your question been resolved?
devout snowBOT
@normal lagoon Has your question been resolved?
devout snowBOT
winged bluff
So basically prove that it applies for k? How so?
Not really no
you prove that if it holds for k, it holds for k + 1 (if it holds for a number it holds for the next)
youve already proved it holds for 1 so if you prove that if it holds for holds for the next, then youll be able to prove it holds for 1,2,3,4 and so on (all positive integers)
as for how to do this youll just plug n=k+1 into the expression and algebraically manipulate it until you can use your assumption that it holds for k to prove it
devout snowBOT
Available help channel!
normal lagoon
,
normal lagoon
Not really no
you prove that if it holds for k, it holds for k + 1 (if it holds ...
So what should I do next? Prove that a^(k) - 1 = (a-1)(a^(k-1)...+a+1)? If so how do I start.
Cause what I can't seem to figure out is how to eliminate the (a^(k-1).... down to what can only be (a^(k-1) +1)
Is there a way to keep this up? I'm also working on some job applications and currently writing cover letters.
winged bluff
So what should I do next? Prove that a^(k) - 1 = (a-1)(a^(k-1)...+a+1)? If so ho...
nono
prove that
a^(k-1) - 1 = (a-1) (a^(k-1)...+a+1)
cuz remember we're plugging in k+1 for n
winged bluff
Is there a way to keep this up? I'm also working on some job applications and cu...
yeah when the bot asks if your questions been resolved react with X(unless its solved ofc) pretty sure thatll make it not close
devout snowBOT
@normal lagoon Has your question been resolved?
normal lagoon
nono
prove that
a^(k-1) - 1 = (a-1) (a^(k-1)...+a+1)
cuz remember we're plugging...
I thought we needed to prove it for k first?
winged bluff
yeah
normal lagoon
Then prove it for k+1
winged bluff
youve already prived it for 1
thats our k in this case
proving it for k just means proving it in general
if we couldve dine that from the start without all of this hassle we woulda just done that
so we prove it for k+1 ASSUMING its true for k
(the assumption holds when k is 1 since youve already proved that)
winged bluff
so we prove it for k+1 ASSUMING its true for k
(the assumption holds when k is 1...
so after we do that it shows it holds for 1+1 which is 2, that shows it holds for 2+1 which is 3, so on and so on using the if its true for k its true for k+1, proving it for all positive numbers
normal lagoon
How would I word this if I was moving on to k+1? Let me picture what format I am looking at.
Are we supposing something here?
I am confused on how to format the next part of the proof.
winged bluff
look i dont know much about formatting proofs using the for all there exists and epsilon stuff tbh but I think you just say base case k = 1, then prove it holds for k+1(when you get to the part you need to use your assumption for say as has been proven for the base case k = 1)
then after that you just say so the statement holds by induction
im pretty sure you dont have to explain the concept of induction during the prove, just show its requirements are met
(holds for base case, if it holds for k it holds for k+1)
normal lagoon
I'm trying to think if there is any theory in Divisibility and Primality that can help explain k and k+1 sitaution without showing the proof by calculation.
Hang on.
Wait did I really prove when n=1 that the theory holds?
Would I need to go a^0 - 1 or just -1?
winged bluff
Wait did I really prove when n=1 that the theory holds?
i think so yeah
you showed that both expressions are equal when n = 1
normal lagoon
This is what I mean.
When n=1
Do I use the top or the bottom?
Cause the top will give me a result of (2a-2)
Also I forgot to change the sign of the 2nd equation on the left to +
winged bluff
the bottom yeah just (a-1)(1)
bc we're dealing with n = 1 so a^(n-1) is just a⁰ wed only stsrt including a¹ if n were 2 or more
normal lagoon
Okay so I proved base kit.
Now I can skip proving for K?
And move onto k+1
Just trying to format it correctly since that counts towards points in an actual exercise.
winged bluff
Now I can skip proving for K?
the base case IS k
proving for k in general is just proving the theoreom
do you jnow what k represents?
normal lagoon
Any positive interger k, which represents n in this case.
Wait would we need to prove it for a as well?
winged bluff
Any positive interger k, which represents n in this case.
not really no since theres no reason for it not to hold for a particular a
(a-1)=(a-1) is true for ALL a
winged bluff
Any positive interger k, which represents n in this case.
yeah so when in induction you LET k = 1, so 1 is called the base case in this case
then you prove if the statement holds for a hypothetical number k , it holds for k+1
not prove it holds for all k, just that if it holds for EVEN A SINGLE k it holds for the next one
so since youve proved it for 1 you just let our hypothetical k = 1
and show it holds for k+1 if it holds for k
normal lagoon
Okay wait. Gonna try and format all this properly then come back and see if it makes sense.
I mean my answer so far.
winged bluff
that means it holds for all positive integers
normal lagoon
Wait still trying to format it properly.
devout snowBOT
@normal lagoon Has your question been resolved?
normal lagoon
Okay I think I've rewritten it a few times but I got something. Almost done with the first example xD
Will post in a bit. Just need a bathroom break after an interview.
How do I finish this proof?
normal lagoon
Is it done? I'm not even sure its properly finished.
normal lagoon
<@&286206848099549185>
mighty knoll
Why are you pinging random users
mighty knoll
Doesn‘t mean you can ping random users. You can ping @ helpers once every 15 minutes but that‘s it.
remote karma
<@&286206848099549185>
he did
winged bluff
honestly the k+1 proof isnt very clear to me so itd be better if you cleared it up first then we'll think about finishing/polishing the proof
mighty knoll
<@&268886789983436800> most likely a troll
winter torrent
<@731617267959463977>
dont do this
remote karma
Ok
normal lagoon
honestly the k+1 proof isnt very clear to me so itd be better if you cleared it ...
How do you think we can clear this up. I'm trying to think of a better way to show that (a^(k+1)-1) = (a-1)(a^(k+1-1)....+a+1)
SLR. Was almost asleep from waiting earlier.
remote karma
m3nny, Btw I have sent you a DM.
normal lagoon
Don't do that.
remote karma
Dm?
winged bluff
How do you think we can clear this up. I'm trying to think of a better way to sh...
its not that the way you've shown is wrong its just its unclear, maybe you could type it instead of writing it for me to look at it better
remote karma
Its not related to this
normal lagoon
its not that the way you've shown is wrong its just its unclear, maybe you could...
Ah sorry. I'll fix my hand writing in the draft. I'll try typing it out in a bit.
winged bluff
tldr hes a person im 95% sure is a troll and i used to think he was genuinely asking for help but now he keeps making alts bc he keeps getting banned
winged bluff
I know him
yayo right?
normal lagoon
<@&268886789983436800> Please.
normal lagoon
tldr hes a person im 95% sure is a troll and i used to think he was genuinely as...
Alright blocked this one.
I'll work on statement 2 tomorrow. I haven't eaten dinner. Just need to finish doing statement 1.
Typing my proof for k+1 in PP now.
normal lagoon
its not that the way you've shown is wrong its just its unclear, maybe you could...
It was basically proving you can cancel out most of the parts of (a-1) product except for a^k+1 and -1.
That is probably the roughest part of my proof from earlier.
winged bluff
this seems right yeah but just 2 things
1.the jump from step 1 to step 2 is abit confusing so itd probably be better if you broke it up into multiple steps
2.unless im missing something the proof didnt rely on the statement holding for k at all? if so that means it didnt need need to be proven using induction in the first place
now that i think about it yeah this could be done just using some algebra (like you showed) why are we using induction
normal lagoon
It was recommended that if you don't have all the theorems and collilaries memorized to prove these kinds of statements, you can use PMI.
Plus proving that it can apply for all positive integers helps.
If it works for the smallest possible, it can work for k+1 which means all positive integers upto positive infinity can use this statement.
So given that it can explain statement 1, I think we have done statement 1. Now we can close this and I'll try and do statement 2 tomorrow.
.close
devout snowBOT
devout snowBOT
Available help channel!
trim shoal
Hello, I would like some help in order to understand the following,
Radioactivity is a random, spontaneous process, however it has a definite half-life. I understand that this is because of something that allows us to predict randomness in large numbers, for example in 10,000 coin flips, it is highly probable for the numbers of heads to be 5000, but why?
devout snowBOT
Hello, I would like some help in order to understand the following,
Radioactivit...
glossy dew
i guess god liked it that way
trim shoal
I have also tried to run a coin flip simulation, which also gives a ratio close to one for many trials, however again, I feel like I am lacking some sort of intuitive understanding of the problem.
Simulation:
[Flips]: 500000 [Heads]: 249992 [Tails]: 250008 [Ratio]: 0.9999360020479344
glossy dew
for some reason we can assume everything in the world is perfectly random
if we take a too large quantity of anything
gases are random
the universe is random
coin flipping 10000 times is random
trim shoal
Then why does order seemingly emerge from randomness in this case?
glossy dew
good question
im not sure either
only thing we can do is contradict it
if it is truly not random
and it favours a result
then it isn't random and has a bias
a truly random system has no bias
wise scarab
Hello, I would like some help in order to understand the following,
Radioactivit...
it's not actually that probable for the number of heads to be exactly 5000
this is a consequence of the law of large numbers
trim shoal
Yes, I'm failing to understand why the law of large numbers holds true
wise scarab
which says something about the convergence of the limit of iid random variables
wise scarab
There must be some intuitive way of understanding it
well, think about it. if you sum n iid rv X_i, the expectation is going to be nE(X), right?
adding iid random variables is basically like conducting multiple independent identical trials
as you add more and more, the partial sums divided by n is probably going to tend toward the expectation, because that's sort of what the expectation is supposed to tell us
of course, this is all heuristics to understand lln i guess
see https://en.wikipedia.org/wiki/Law_of_large_numbers#Proof_of_the_weak_law for a formal proof assuming finite variance, which is a pretty reasonable assumption
