#help-39
sharp vigil
so you see how e^x already appeared in the homogeneous solution
jolly kernel
yes
sharp vigil
so your initial guess for (polynomial) * e^x breaks because of that
jolly kernel
I kinda follow
My teacher said we should always find the y_c first just incase that it's similar (like here)
But I don't fully get it -- what to do now that it is the same
sharp vigil
if you have this sort of situation, where the solution you would guess is in the homogeneous solution, it won't work. but you can fix it by multiplying your entire guess by x
(just the e^x part to be clear, not the constant part)
jolly kernel
So I can do y_p = A + x(Bx+C)e^x ?
sharp vigil
yes
sharp vigil
you don't
jolly kernel
y_p = A + (Bx^2 + Cx+ D)e^x?
sharp vigil
you can, but it would just turn out to be 0
jolly kernel
gotcha
pearl pondBOT
@jolly kernel Has your question been resolved?
unborn swallow
How do I take the (hyper?)determinant of an n x n x n matrix? Can I just cofactor expand, ie for every element across 0,j,k I do el(0,j,k) * subtensor (1 to i_max, 0 to j_max \ j, 0 to k_max \ k) ? until the determinant of a 1x1x1 is just the element?
unborn swallow
my specific problem is I want to measure the nonlinearity of a nx1 vector of Rn -> R functions, so I take two partial derivatives to get the change in linearization, take the jacobian (determinant), then find the absolute max across Rn, which is simply taking J' = <0..0> and applying some numerical method from there
pearl pondBOT
@unborn swallow Has your question been resolved?
grand bough
here what is this N of the fft i don't understand aren't we supposed to do the fft for each sampled point ?
pearl pondBOT
here what is this N of the fft i don't understand aren't we supposed to do the f...
@grand bough Has your question been resolved?
unborn swallow
How do I take the (hyper?)determinant of an n x n x n matrix? Can I just cofactor expand, ie for every element across 0,j,k I do el(0,j,k) * subtensor (1 to i_max, 0 to j_max \ j, 0 to k_max \ k) ? until the determinant of a 1x1x1 is just the element?
my specific problem is I want to measure the nonlinearity of a nx1 vector of Rn -> R functions, so I take two partial derivatives to get the change in linearization, take the jacobian (determinant), then find the absolute max across Rn, which is simply taking J' = <0..0> and applying some numerical method from there
(which I just realised won't work, because if one of my n initial elements is 0, the whole determinant is zero, despite the nonlinearity definitely being nonzero)
(attempt 2 since I forgot I asked and no one answered)
pearl pondBOT
How do I take the (hyper?)determinant of an n x n x n matrix? Can I just cofacto...
@unborn swallow Has your question been resolved?
pearl pondBOT
@unborn swallow Has your question been resolved?
unborn swallow
better question; I have (a nx1 sized vector of functions fn (Rn -> R)) + Ax + Bx'. I wish to find the nonlinearity of each row.
Looking at one row at a time, I take two partial derivatives, I end up with a matrix that is (J(f) where Ji,j=d/xi d/xj fn(x)xi'xj') + (diagonal matrix where Ai,i=Aix''+Bix''')
however, if xi is not used in a given row (d/dxi fn = Ai = Bi = 0), then the determinant of this whole matrix at row n is zero
despite the function being very nonlinear
now, I could just ignore zero rows/columns when taking the determinant
but then, if I were to add in say x7 that was unused, by making Gn,7=1, then suddenly my whole determinant is multiplied by x7''
pearl pondBOT
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narrow crystal
how many of the first 10000 positive integer contain the digit pair 43 in that exact order
narrow crystal
i got 299 is it right?
i did it by like 4 3_ _
43xx
x43x
xx43
in the first one xx can be 00-99 so thats 100
second one x can be 0-9 so 10 x 10 = 100 also
then last is xx can be 0 - 99 also
so 3 x 100 = 300
then minus 1 cus the first and third one can appear as 4343 counting it as one
is this logic correct? or
tardy reef
is 4003 allowed? Like it also contains the pair of digits 4 and 3 and is in the desired order 
tardy reef
yea, then yours looks alright
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prime bramble
I want to show that if two elements x and y of a group are conjugates, then they have the same order
prime bramble
I've been able to do this for the case where x has finite order, but I'm entirely at a loss as to how I'd show this when x has infinite order 
or maybe it's that I'm confused about what it means for an element of a group to have infinite order
oh, hold on
infinite order just means that the order isn't finite
so if x = g^-1yg and |x| = infty, then |g^-1yg| = infty, i.e. (g^-1yg)^n neq e for any n
then by associativity, g^-1y^ng neq e for any n
so y^n neq e for any n, and we're done
.close
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west sapphire
short proof, just show that "conjugation by g" is an automorphism
prime bramble
that's basically what one does in the finite case without saying so anyways 
west sapphire
don't really even have to think about finite vs infinite if you do it that way
prime bramble
west sapphire
(well, assuming that you "know" that orders of elements are preserved under isomorphisms)
but maybe that's not a valid assumption here
"know" as in, have proved and can use in this context 😁
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latent quail
What are the prerequisites before solving this problem
pearl pondBOT
What are the prerequisites before solving this problem
latent quail
I did consult AI for relevent videos, but it doesn't seem to be the one I'm looking for
tardy reef
What are the prerequisites before solving this problem
there should be no prerequisites? Like you press a button, that completes a circuit and generates 4 signals C1 to C4 which help to identify the switch
if the current passes thru C, what happens at R?
latent quail
if the current passes thru C, what happens at R?
there are 4 ports of C
tardy reef
yea, so just explain how an activation at C1 affects R1, R2, R3 and R4
and so on all the way to C4
latent quail
Right
tardy reef
I mean opposite
R1 has a signal, and what pressing a switch does at C1
and so on
mb
latent quail
R1 has a signal, and what pressing a switch does at C1
When S1 is pressed, there would be current passing through C1 and R1 I assume
tardy reef
something like that yes
what happens is you send a pulse thru one of the R's or C's and listen at the other one
so if you hit a switch, it affects the output at one of those
latent quail
gotcha
I have a quick question
what is that resistor network doing?
Do they serve as any special purpose?
tardy reef
my guess would be to control the signal strength
so S1 and S5 register as distinct keys at C1
but im not too sure
latent quail
Right
One more question
is this the pulse's route when S1, as a button, is pressed?
tardy reef
yea
latent quail
So it has to pass through 2 resistos if it wants to reach R2
lemme clarify it, my wording's a big ambigous
tardy reef
my guess would be to control the signal strength
mb, I spent a minute actually thinking, and it would be just to keep the signal high. A device that reads keypresses, a microcontroller or smthn, would make signal at the R LOW (say its the input), and C would also be LOW. But pressing a switch would result in the C becoming HIGH.
latent quail
tardy reef
no this doesnt happen
latent quail
mb, I spent a minute actually thinking, and it would be just to keep the signal ...
it's alright
latent quail
no this doesnt happen
My idea is: when S1 is pressed it follows the red route
S5: blue route
tardy reef
no this doesnt happen
that would be impossible since the current would have to flow against the potential difference
latent quail
that would be impossible since the current would have to flow against the potent...
wait, so the current will not follow the blue route when only S5 is pressed?
tardy reef
blue is right, red is not
latent quail
How come?
tardy reef
ok, so the way it would work is like, you have R as signals that you set and C as the signals that you read ok?
now you set R1 as LOW and other Rs as HIGH
latent quail
Right, lemme summarize it
It's like we send 4 signals R1 to R4, and observe the behaviors of C1 to C4
to determine which button(s) S1 to S16 are turned on
tardy reef
yea
latent quail
Right, so the signals are either 1 or 0, is that correct?
tardy reef
yea
tardy reef
wdym 1 signal input?
latent quail
Lemme draw
S1 is closed and S5 S9 S13 are opened
the result would be 1 at C1
smth like this
tardy reef
yes
latent quail
mb, I spent a minute actually thinking, and it would be just to keep the signal ...
got it, I'll just ignore them since they're trivial to the question lol
tardy reef
||When a switch is pressed it should|| ||make the R and C it connects LOW while the rest remain HIGH||
latent quail
R1 to R4 0010
C1 to C4 XXX1
latent quail
||When a switch is pressed it should|| ||make the R and C it connects LOW while ...
I'm not sure what are LOW and HIGH mean, is it another way to represent 0 and 1?
tardy reef
yea
latent quail
Right
latent quail
My doubt is: I don't think the value of C1 to C3 affect how we determine the status of S12
tardy reef
well, you want C4 to be 1 to detect S12 right?
latent quail
yes
tardy reef
so what about switch S11?
what would you prefer it to show on C3 when S11 is pressed?
latent quail
so what about switch S11?
opened?
tardy reef
so 1?
tardy reef
R1 to R4 0010
C1 to C4 XXX1
now imagine here, its R: 0010 and C:1111. Now tell me is it S9, S10, S11 or S12?
which one did get pressed?
latent quail
now imagine here, its R: 0010 and C:1111. Now tell me is it S9, S10, S11 or S12?
I'd say S12
but can't they all be pressed simultaneously?
tardy reef
I'd say S12
why? doesnt S11 operate on XX1X? 1111 is also in this form!
tardy reef
but can't they all be pressed simultaneously?
thats the problem. you wanna see what happens on a single key press
latent quail
why? doesnt S11 operate on XX1X? 1111 is also in this form!
yeah, but if 12 is closed then 1111 will not be the case
tardy reef
mhm
latent quail
because there is no bridge to the last exit
tardy reef
R1 to R4 0010
C1 to C4 XXX1
I literally replied to what you said. Its you who said C4 has to be 1 when S12 is pressed
and since the other dont matter, why cant they be 1 as well?
latent quail
hmm
I have a dumb question
S9 needs to be 1 for R3 to deliever signal to C1, right?
tardy reef
by 1 you mean pressed state? if so, yes. it needs to be pressed so that c1 gets low and detected
latent quail
Right, so if S12 is pressed the current can flow from R3 to C4
latent quail
yeah, but I still don't understand why C1 C2 C3 matter in this case 
if S9 S10 and S11 are not pressed, can the current flow to C1 C2 and C3?
like only S12 is pressed
tardy reef
if S9 S10 and S11 are not pressed, can the current flow to C1 C2 and C3?
no.
that not flowing is by design
latent quail
now imagine here, its R: 0010 and C:1111. Now tell me is it S9, S10, S11 or S12?
Right, so for this question. S9 to S12 are all pressed
because C1 to C3 are 1s
tardy reef
yes
thats the point
so when S12 is pressed (as asked in your question), it wont lead to XXX on C1, to C3
it should not be left ambiguous
latent quail
it should not be left ambiguous
yes, but they can be either 0 or 1, can't they?
tardy reef
no
latent quail
Because we're just looking for the route directly to C4
tardy reef
also, 0 is LOW and 1 is HIGH. R=0 and C=0 is when keypress gets noticed. coz them being low allows Vcc to be detected thru them
so, it should be 0s at C4 and R3 when S12 is pressed
and 1s everywhere else
latent quail
yeah, I'm probably having problem with the noun definition. Pressed means "the passage is closed" and Not pressed means "the passage is available"
tardy reef
pressed switch allows signal to pass thru it, which is same as closed. Unpressed would be open switch and no current thru, inf resistance
also, i gtg, I'll be back later if its still unresolved
bye
latent quail
Alright, appreciate it
I'll come back later
@tardy reef Thanks again, appreicate that you're willing to teach a
NOOB
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pearl pondBOT
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little terrace
Hi, I'm doing an assignment that requires me to mathematically verify my results (it's an assignment about creating "roads" of different speed limits using functions and justifying using the rate of curvature). In what ways could I do this? All the guides online don't really have anything beyond saying to either use technology or calculate error off of real world data, but I don't really have any data to compare to (since all my functions were found using trial&error). If someone could help I'd really appreciate it!
crystal dew
verify what specifically though? the speed limits? the rate of curvature? something about the roads themselves?
(tl;dr, overbroad question)
(perhaps showing the parts you need verified would help, plus context)
little terrace
sorry, yeah that would help
honestly i think this might be something only my teacher can answer
super sorry for wasting your time but tysm 🙏
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pearl pondBOT
crystal dew
no worries about wasting time. that's what helpers are here for
little terrace
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smoky ivy
pearl pondBOT
smoky ivy
Help please
pearl pondBOT
@smoky ivy Has your question been resolved?
slender viper
consider simplifying the sum inside the square root?
normal magnet
x tends to 0 and so does sinx. It is easy to prove that x >= sinx for all positive x.
Hence, x - sin(x) is positive and tends to 0 as x -> 0.
Hence limit x->0 (x-sinx) = 0
Therefore,
For x->0, sinx ~ x
Or
limit x->0 sinx/x = 1
frank violet
It is easy to prove that x >= sinx for all positive x.
yeah I see that's not easy to prove
neat oasis
what is the question here?
frank violet
It's might be checking his proof
neat oasis
> It is easy to prove that x >= sinx for all positive x.
yeah I see that's not e...
the taylor series expansion for sinx
frank violet
the taylor series expansion for sinx
That's like using rocket for a snow ball fight
neat oasis
well if it gets the job done what's wrong
normal magnet
That's like using rocket for a snow ball fight
More generally, 'modern artillery'
normal magnet
what is the question here?
Yes.
neat oasis
another way to prove x>sinx doesn't come to mind that quick
frank violet
well if it gets the job done what's wrong
Maybe it doesn't get the job done
neat oasis
shit
frank violet
Cuz smtime they're not allowed to use that
normal magnet
> It is easy to prove that x >= sinx for all positive x.
yeah I see that's not e...
Is the rest of my proof correct, considering that I have proved this fact?
frank violet
Is the rest of my proof correct, considering that I have proved this fact?
Hmm, I don't think so
x >= sinx is wrong on the left hand limit iirc
normal magnet
Let the limit be from the positive side. I tried to prove limit x-> 0+ sin(x)/x = 1
frank violet
ehhh, could be, idk that's a weird way to prove lim sinx/x
normal magnet
Is the fact that x >= sin(x) for all positive x usd in the proof?
For example, I could say:
sin(x) tends to 0 and so does sin(2x). Clearly, sin(2x) - sin(x) for small positive values of x.
Hence, sin(2x) - sin(x) is positive and tends to 0 as x->0.
Hence limit x->0+ (sin(2x) - sinx) = 0
Therefore,
For x->0+, sin(x) ~ sin(2x)
Or
lim x->0+ sin(2x)/sin(x) = 1
frank violet
What i want to say is, 2x>=x but 2x/x=2
fraction is like dealing with the ratio
So I don't think you can do that argument
fallen lotus
Is the fact that x >= sin(x) for all positive x usd in the proof?
i think there was a number (dottie number??) before which sinx > x
normal magnet
Yes.
Thank you, I understand.
I have another attempt at the proof.
Let me send it.
Nvm.
It uses the same fallacy.
Thanks for the help.
.close
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brazen blade
hai
cursive wraith
but its cos = 0
yeah, so
toxic lichen
$2z = 0$ is the same thing as $z=0$
jolly parrotBOT
Ann
cursive wraith
now that you have cos(theta) = 0 as info
you can find r, and then you can find the points (x,y) that work
pearl pondBOT
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wary vault
hello, suppose i have a sequence a_n. if it is descending and bounded, then how can i explain that a_1 is a real value?
toxic lichen
mmmmm
why... would a_1 be anything but real
like i guess if you're really into exotic shit you could say "oh i guess a_n is a sequence of points in some linearly ordered set that may or may not be R"
but also like...
!xy
pearl pondBOT
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.
humble root
like i guess if you're really into exotic shit you could say "oh i guess a_n is ...
a_n is descending so would that imply some sort of order?
I'm not confident on spaces that have order, R is the only one I know 🤷
toxic lichen
that... is what i said, yes.
wary vault
like i guess if you're really into exotic shit you could say "oh i guess a_n is ...
lol i just wanna be sure that i can state a_1 is real so that i can continue my proof of something else
toxic lichen
can you show this something else
cedar sentinel
what was the original proof?
humble root
that... is what i said, yes.
Oh, I see sorry.
toxic lichen
i really dont think you should be having any trouble with declaring a_1 is real if it is indeed real
decreasing-ness and bounded-ness be damned
tropic saddle
from the question it should be obvious whether you can assume that the a_i are real or not
weary ledge
lol i just wanna be sure that i can state a_1 is real so that i can continue my ...
The boundedness is the only condition you require to prove this i think
tropic saddle
it should be made clear in some way in the question
wary vault
hmm okeoke maybe i didnt think simply enough
tropic saddle
or if your course only considers real sequences then its also clear
pearl pondBOT
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misty thunder
how do we solve stuff like this?
pearl pondBOT
how do we solve stuff like this?
versed remnant
how do we solve stuff like this?
ig u could find a normal vector from that equation of plane
and can find a vector of the same direction using the equation of line
and use dot product , a . b = |a||b| cos theta , and find p
misty thunder
im not sure i understand what you mean
versed remnant
im not sure i understand what you mean
which step
this is lil a fast drawing i can make
misty thunder
and can find a vector of the same direction using the equation of line
this one
versed remnant
this one
do u know how to find a vector equation of a line
using a point and a parallel vector?
its just that , but in reverse step
misty thunder
using a point and a parallel vector?
yes
misty thunder
and use dot product , a . b = |a||b| cos theta , and find p
so a here is (1,1,3) and b here is (-1, 1, p)?
versed remnant
ye
misty thunder
ohh okay
misty thunder
okay because the normal vector is perpendicular to the plane so we do 90-60
i got the correct answer!
tysm
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pearl pondBOT
versed remnant
np <3
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void grail
pearl pondBOT
void grail
I still struggle with ε-δ definition FOL
pine jay
What have u tried, or is there something specific you want to go over, tactics in general or?
void grail
What have u tried, or is there something specific you want to go over, tactics i...
Maybe what pattern to look for?
What are possible ways to approach this?
glass meadow
You already have the values x=3 and g(x) = 3/5, so all you need to do is consider |g(x) - 3/5| < ε and |x-3| < δ
Given ε, what δ would work?
pine jay
A good approach is to start backwards, i.e start from |g(x) - 3/5| < and work your way to something that is independent of x and contains delta, now this is supposed to be less than epsilon. So we solve for delta and we’re done
Sometimes we can make that job easier by making further assumptions that’s appropriate for delta
pine jay
aslan can u tell what epsilon delta definition is?
Formally or intuitively?
Also create a channel for that
void grail
lemma limit_eps (ε:ℝ) (εpos: ε>0) : ∃δ>0, ∀x, 0<|x-3|<δ → |g x - 3/5|<ε := by sorry
pine jay
This is occupied
void grail
just a gist of it
∀ ε, ∃ δ > 0, ∀ x, (0 < |x − a| < δ ⇒ |f x − L| < ε)
tame cave
I believe this is the definition
i didnt understand what u wrote
void grail
i didnt understand what u wrote
One sec, I'll try to write
pine jay
i didnt understand what u wrote
It formalises the notion of arbitrarily getting closer and guaranteeing we’re within a certain threshold set by epsilon, a lot easier when you think of it in terms of sequences (without the delta business)
tame cave
It formalises the notion of arbitrarily getting closer and guaranteeing we’re wi...
ok thanks!
pine jay
This in turn formalises what we’d consider a limit
pine jay
∀ ε, ∃ δ > 0, ∀ x, (0 < |x − a| < δ ⇒ |f x − L| < ε)
The reason we have to express it like this, is because math expresses statements using symbols from the language of first order logic
But we can sort of dumb it down and use natural language sometimes
We benefit however from being precise
void grail
i didnt understand what u wrote
Choose an ε whatsoever, however small, there is a δ for which the distance of f(x) and L is ε.
No matter how near f(x) we go on the y axis (choose any value, no matter how small: ∀ ε > 0)
there is a δ > 0 (∃ δ > 0)
tame cave
Choose an ε whatsoever, however small, there is a δ for which the distance of f(...
alright ty! respect for the efforts 🫡
void grail
alright ty! respect for the efforts 🫡
I struggle with this! I would like to write it accessibly
pine jay
A good approach is to start backwards, i.e start from |g(x) - 3/5| < and work yo...
@void grail are you making progress with this?
I can give some tips on how to think about this
void grail
<@747618696092057650> are you making progress with this?
Yes, in fact learning to write it accessibly seems like a good aim, because if I would understand it well enough to explain it then that would be great maybe
pine jay
Yes, in fact learning to write it accessibly seems like a good aim, because if I...
It really does, I think first understanding intuitively what happens with sequences makes this much more accessible
Want hints on how to proceed on the concrete problem?
void grail
alright ty! respect for the efforts 🫡
void grail
Want hints on how to proceed on the concrete problem?
Thank you @Aslan, this is plenty of help, I will look at sequences and the idea of working backwards and the other suggestion, thank you all!
🙌
pearl pondBOT
@void grail Has your question been resolved?
rare holly
hello whys this wrong
pearl pondBOT
hello whys this wrong
toxic lichen
can you show the exact thing you put in there and the exact feedback you got
rare holly
?
toxic lichen
show how you typed it into the answer box.
then show what exactly the system said after you did it
rare holly
oh wait
that ans is correct
but i did it the wrong way around
i was gna ask why is tht wrong
this
bc the red arrow goes down so we dont chnage sign
basically when do i change the signs
void grail
basically when do i change the signs
Have you tried drawing these sums?
toxic lichen
im not gonna give a coherent rule of "when to change the signs and when not"
but your idea was that $\ora{FD} = \frac12 \ora{BD}$, and $\ora{BD} = \ora{BA} + \ora{AD}$
jolly parrotBOT
Ann
toxic lichen
the minus sign comes in at the point where you work out $\ora{BA}$ as $-\bd{m}$
jolly parrotBOT
Ann
void grail
basically when do i change the signs
pearl pondBOT
rare holly
tysm
void grail
tysm
Addtion is commutative by axiom in any vector space over a field.
∀ u, v ∈ V, u + v = v + u
I think.
rare holly
Addtion is commutative by axiom in any vector space over a field.
`∀ u, v ∈ V, u...
idk wht this means but ok
void grail
idk wht this means but ok
It means that you can add one vector to the other, or the other to the one, order doesn't matter .
Draw m first, then n, or n first, then m, same path.
rare holly
i need help bru
void grail
When you wrote-m+n,
I think you wrote (-1)m + n
then with m-n:
m+((-1)n)
multiply by (-1) flips the vectors direction
@toxic lichen why warning sign?
toxic lichen
<@183668144404037632> why warning sign?
inappropriate level of formalism, imo.
void grail
i need help bru
Maybe
void grail
i need help bru
rare holly
yh
pearl pondBOT
@void grail Has your question been resolved?
lapis lynx
Im a little confused on why for the final jib, it only has a length of 1m
lapis lynx
I thought that it should be 2 because it states that a jib touching another tower doesnt result in a crash since the distance between the towers is 2
Output is the the length of jib in meters such that the covered area is maximized btw
somber adder
not sure what they mean with that last line... maybe just ignore it?
lapis lynx
tbh i interpreted it as the length of jib can equal the distance between the two towers
somber adder
Well clearly the output doesn't show that, so just assume that's not the case
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shut flicker
hi i was trying to do this equation "find all solutions to 4x^2 = y^3 + 1. ive found that x=0 y=-1 works, that y>= -1, and i feel like this may be the only solution, but im trying to prove it. i tried some mod arithmetic, may not be useful. i thought maybe if we can show that y^3+1 is never a perfect square unless y=-1 that this would show it is the only solution (dk if true tho), would this be the way forward? thanks
shut flicker
should be a y^3 in the 4th line acc
fallen sundial
should be a y^3 in the 4th line acc
hmm i would suggest you to take x^2 as t for some time
where t is always positive
and draw a graw of 4t=y^3+1
then just sqrt the value of t to find the two corresponding values of x
shadow spindle
finally a server that i can get help on
fallen sundial
finally a server that i can get help on
use one of the unoccupied channels though pls
bitter vortex
hi i was trying to do this equation "find all solutions to 4x^2 = y^3 + 1. ive f...
Are we trying to find integer solutions? Or reals or what
pulsar flax
Are we trying to find integer solutions? Or reals or what
integer prob
shut flicker
Are we trying to find integer solutions? Or reals or what
sorry its a diophantine equation so yes integer i think
shut flicker
and draw a graw of 4t=y^3+1
interesting didn't think it that way
fallen sundial
sorry its a diophantine equation so yes integer i think
thats sad
so all integer soln's hmm
bitter vortex
interesting didn't think it that way
Isolate y and factor the x’s for first step
You will notice the factors are coprime
You can basically create an equation of a^3 - b^3 = 2 using that fact
And show the only solution for that is 0,-1
shut flicker
so y^3 = (2x+1)(2x-1), how did u jump to a^3 - b^3 = 2
bitter vortex
2x+1 and 2x-1 are coprime meaning they can be represented as a perfect cube because their product is a perfect cube (this is a fact from number theory)
You can show they are coprime by showing that some gcd of both 2x+1 and 2x-1 must be 1
shut flicker
i acc never thought about how 2 odd integers were coprime damn
bitter vortex
So 2x-1 = some a^3 and 2x+1= b^3
shut flicker
consectuve
bitter vortex
Well they are coprime if their gcd is 1
So you just have to show that
You need something that must divide the difference of 2x-1 and 2x+1
Which would be 1 or 2
But since 2x-1 is always odd your only option is 1
shut flicker
So 2x-1 = some a^3 and 2x+1= b^3
sorry im still not too sure abt the jump to this
bitter vortex
sorry im still not too sure abt the jump to this
This comes from showing they are coprime
By definition two coprime integers that have a perfect cube product must themselves be perfect cubes
shut flicker
howd u prove that
bitter vortex
It’s a long proof using unique prime factorization
To my knowledge it’s just an accepted fact
shut flicker
okok
bitter vortex
In my course it was anyway
carmine juniper
Why is this wrong?
pearl pondBOT
Why is this wrong?
lucid bridge
nothing i think
rich spindle
hi i was trying to do this equation "find all solutions to 4x^2 = y^3 + 1. ive f...
i think its false if you want to put square root you have to do the absolute value
toxic lichen
Why is this wrong?
doesn't look wrong; who's saying it is?
rich spindle
Why is this wrong?
yea doesent look wrong
shut flicker
i think its false if you want to put square root you have to do the absolute val...
ok yh
carmine juniper
Oh!
toxic lichen
did you look in the wrong place in your answer key maybe?
carmine juniper
did you look in the wrong place in your answer key maybe?
There’s an error in the results from the text book
solemn shell
Guys would you like to help me in physics and chemistry ??
carmine juniper
Actually
toxic lichen
Guys would you like to help me in physics and chemistry ??
!occupied
carmine juniper
Guys would you like to help me in physics and chemistry ??
Open new channel bro
pearl pondBOT
Guys would you like to help me in physics and chemistry ??
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).
toxic lichen
Open new channel bro
she's literally got those she/her pronouns and you're calling her bro?
carmine juniper
There’s an error in the results from the text book
@lucid bridge @rich spindle
Idk that it was a girl
rich spindle
im a boy
solemn shell
I'm new here and i don't know what to do
buoyant tide
I'm new here and i don't know what to do
read #❓how-to-get-help
solemn shell
read <#488120190538743810>
Thx
lucid bridge
she's literally got those she/her pronouns and you're calling her bro?
its not that deep relax
carmine juniper
.close
pearl pondBOT
pearl pondBOT
Available help channel!
carmine juniper
Am I on the good path? Cause I can’t ^4square root 25…
pearl pondBOT
Am I on the good path? Cause I can’t ^4square root 25…
spare lark
Cube also go to 25
spare lark
No but when you write 4th root (...)³ you have to make that cube to 25 also, not just x and y
lucid bridge
what is (4t)^2?
buoyant tide
$4^2 \neq 4\cdot 2$
jolly parrotBOT
Yeatte
fresh stump
Someone help me pls
nocturne grail
Am I on the good path? Cause I can’t ^4square root 25…
you forgot about the 25 when you spread the ^3.
nocturne grail
Who could do sum trigonometric summitions?
do not advertise about your question in other help channels.
if anyone that can help you sees your question, they will.
fresh stump
Ok sorry i apologise
nocturne grail
its alright! patience then.
nocturne grail
you forgot about the 25 when you spread the ^3.
@carmine juniper
quick venture
pearl pondBOT
tulip ore
x(x+8) dy/dx = y
x(x+8) dy = y dx
move y to the left side, x(x+8) to the right side
x(x+8)/y dy = 1 dx
1/y dy = 1/(x(x+8)) dx
then ∫
∫ 1/y dy = ∫ 1/(x(x+8)) dx
quick venture
im asking why its 1/x(x+8)
tulip ore
they divided both sides by x(x+8)
quick venture
why
tulip ore
if you had x(x+8) dy = y dx, you cant really integrate both sides
you can integrate (something with x in it) dx and (something with y in it) dy though
so with x(x+8) dy = y dx,
they need to move the x(x+8) to the dx side and the y to the dy side
because ∫ 1/(x(x+8)) dx is more doable than ∫ x(x+8) dy
midnight haven
pearl pondBOT
pearl stag
yo
midnight haven
is my answer right 😭
pearl stag
one sec
midnight haven
okkk tysmm
little socket
looks right to me too 
dapper nymph
do u have a question
Yes
plush bramble
pearl pondBOT
@dapper nymph Has your question been resolved?
proper nova
pearl pondBOT
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.
pearl pondBOT
Available help channel!
shadow reef
What did i do wrong with these two? Im trying to find derivative
toxic lichen
the second one is definitely more of a planning error
as in you dont need the quot rule for that one
but you also have an execution error: $\dv{x} \ln(10) = 0 \neq \frac{1}{10}$
jolly parrotBOT
Ann
shadow reef
as in you dont need the quot rule for that one
How can i solve it without the rule?
toxic lichen
recognize that 1/log(10) is a constant multiplier
you would not use the quotient rule to differentiate $\frac{x^3}{123456789}$ now would you?
jolly parrotBOT
Ann
shadow reef
I see 🤔
shadow reef
recognize that 1/log(10) is a **constant multiplier**
Sorry it's been a long time since I've worked with log, could you explain this part to me?
toxic lichen
this doesn't actually have anything at all to do with log
10 is a constant
any function applied to the constant 10 will also produce another constant
sqrt(2) is also a constant
sin(0.17) is also a constant
e^7 is also a constant
capisce?
@shadow reef
shadow reef
Wait I'm not quite getting it, are you saying I can take out 1/log(10) from f(x)?
toxic lichen
idk what you mean when you say "take out"
you can and should view 10^x/log(10) as (1/log(10)) * 10^x, sure
what im trying to drive home is that 1/log(10) is just a number
misty thunder
how
misty thunder
part a and c
pearl pondBOT
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1. I don't know where to begin.
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3. I got an answer but I was told that it's wrong.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
serene sentinel
so take you base case n=2, take induction hypothesis for n=k. Then you gotta prove that n=k+1 hold
so it suffices to prove
and that's the same as proving part (a)
misty thunder
how did you get this
serene sentinel
when we assume that our inductive hypothesis holds for n=k, we get
so to take it to n=k+1, I added $\frac{1}{\sqrt{k+1}}$
jolly parrotBOT
Donkey
misty thunder
yes i understand how it shows up on the LHS but why does it show up on the RHS
shouldnt it be 2sqrt(k+1)-2
serene sentinel
when you add stuff to an inequality you gotta add it on both sides yeah
serene sentinel
shouldnt it be 2sqrt(k+1)-2
we're trying to prove that
serene sentinel
when we assume that our inductive hypothesis holds for n=k, we get
this is a known inequality from our inductive process
so we work a bit on this
and try using it to prove the whole theorem
misty thunder
so it suffices to prove
and here you just divide by 2 to get part a right
serene sentinel
yup exactly
void grail
pearl pondBOT
serene sentinel
l'hopital shd do the trick on both
void grail
l'hopital shd do the trick on both
Thank you very much, I will try it
fathom scaffold
I don't recommend using L'Hopital
void grail
I don't recommend using L'Hopital
ε-δ definition easier?
fathom scaffold
Why would you need that here
prisma kernel
Yeah L'hopital seems way more straightforward
tiny hound
ε-δ definition easier?
delta Eplison is pretty overkill for most calc bc limit analysis
void grail
also I rationalised top and
x⟶0: (0+1)/(√(1+0+0)+1)=1/2
void grail
delta Eplison is pretty overkill for most calc bc limit analysis
When do you use it?
tiny hound
you don't lmao it's not in the calc bc curriculum anymore
void grail
you don't lmao it's not in the calc bc curriculum anymore
Was it moved to analysis curriculum?
tiny hound
When do you use it?
but in actuality you use it when formally defining limits and a more rigorous way to do limits. In actuality tho I say don't use it unless the question says so, which in AP calc bc they don't
this is for AP calc bc, right?
void grail
I need to learn it for BSc calculus/analysis exams, and I am also curious to learn it
tiny hound
yeah that makes sense, it is more of a real analysis thing really
It's useally easier to do literally anything else lmao
void grail
yeah that makes sense, it is more of a real analysis thing really
Alright, I didn't kow that, it's good to know
void grail
It's useally easier to do literally anything else lmao
It teaches logic also
tiny hound
I meant when actually doing limits, I do think it is good to teach
void grail
I meant when actually doing limits, I do think it is good to teach
Sequences are easier more intuitive?
void grail
not sure what you mean by that
Ah I thought it was maybe easier to learnl imits from the perspective of sequences, but I guess what I was suggested was it was easier to learn the ε-δ definition from the perspective of sequences
You are saying there are other, simpler ways to find limits of functions
tiny hound
well you just did
L'hospital's rule
and also algebraic manipulation to cancel out holes
void grail
maybe x⟶0: (0+1)/(√(1+0+0)+1)=1/2 is based on the epsilon delta definition logic somehow
tiny hound
there's also natural logging both sides of the limit
void grail
is x⟶0: (0+1)/(√(1+0+0)+1)=1/2 canceling out holes?
tiny hound
and also algebraic manipulation to cancel out holes
what I mean by this is say we have ((x+4)(x-3))/(x+4)(x-8). If we want to take the limit as x approaches -4, we can just cancel out x+4 and plug in
void grail
algebraic manipulation, like factoring the denominator, and saying x≠ whatever makes the denominator 0, so that's a sort of cancelling the asymptote sending the function to another function which is preserving the limit maybe
tiny hound
in ii here you could try multiplying numerator and denominator by the conjugate of the top expression
tiny hound
algebraic manipulation, like factoring the denominator, and saying x≠ whatever m...
yeah in that sense
void grail
Is this topology?
tiny hound
algebraic manipulation, like factoring the denominator, and saying x≠ whatever m...
but generally not an asymtote
tiny hound
Is this topology?
no it's calculus lmao
tiny hound
but generally not an asymtote
generally these cases are useally holes
void grail
oh asymptote is maybe convergence to infinity
tiny hound
kind of
an asymtote can converge to a finite value
but yes generally it would involve infinity iirc
also another thing
in AP calculus bc the way we get around delta Eplison is by just by heuristically saying that the limit is describing what's happening around a neighborhood of the function
void grail
Alright, thank you very much for your help @tiny hound , awesome, and thank you everyone
Also negating the epsilon delta seemed like a way to understand something about it, in my experience
tiny hound
yeah you don't practically need it if you can just grasp an intuitive understanding of a limit
unless you are asked
void grail
like ∃ε∀δ¬(…)
tiny hound
tho it is still good to know
void grail
yeah you don't practically need it if you can just grasp an intuitive understand...
and that maybe helps understand the underlying formal defintion / logic
tiny hound
yep, why it's good to do
fathom scaffold
I didn't bother with it and neither did my teachers
tiny hound
way things useally go: if you have to do solve something, don't overcomplicate it
fathom scaffold
Like it exists - ok. But its usefulness is barely there
tiny hound
delta epsilon useally overcomplicates things
tiny hound
Like it exists - ok. But its usefulness is barely there
imo it is good to have a bit of an understanding but also save it for real analysis
fathom scaffold
Yeah, it's definitely not for my calculus brain
pearl pondBOT
@void grail Has your question been resolved?
misty thunder
how do we do these
pearl pondBOT
how do we do these
verbal whale
compact ridge
using the result from part a, raise both sides to the power of 4
then using the double-angle identity for cos(2x), it's a quadratic in u = cos x
misty thunder
what does it mean by quadratic in u=cosx
compact ridge
what does it mean by quadratic in u=cosx
e.g, 5 (cos x)^2 + 3 (cos x) + 1 = 0 is a quadratic in cos x
misty thunder
ohh so we can solve for cosx using quadratic formula
compact ridge
yep!
I think it's factorisable but you should check for yourself
if it's p1 it will be factorisable
misty thunder
yeah it is p1
but i often just use quadratic formula in any case unless i need the factored form
misty thunder
can we not factor when theres irrational coefficients?
compact ridge
can we not factor when theres irrational coefficients?
you can
depends if you can spot it or not
it's not really worth bothering over if you can't
misty thunder
ohh
compact ridge
I would do, it's either (4u + .....)(u + ....) or (2u + ....)(2u + .....)
and then the first case doesn't work, but it's not easy to see why unless you just try every case
misty thunder
i got cos(x) = 3sqrt(3)/2 or sqrt(3)/2
heavy onyx
compact ridge
cool and hopefully you know what to do now, for 0 < x < pi/2
there's something blatant about cos(x) = 3 sqrt(3)/2, for example
compact ridge
mhm, cause you can't have cos x = 2 or anything (with absolute value) more than 1
pearl pondBOT
pearl pondBOT
Available help channel!
shut flicker
yo ive been absolutely tweaking on this question, i rly dont want someone to tell me the solution as i want the satisfaaction of working it out myself, but instead can someone criticise my current approaches and why i shouldn't have approached it like this, question probably easy as well😔 prove that 14^n + 11 is never prime. thanks🙂
spare lark
What does n belong to
shut flicker
i think integers if thats what u mean
lilac ledge
toxic lichen
14^0 + 11 isn't prime either
i think there's gonna be some modulo bullshit going on
tulip ore
red, does the problem say that n is natural
tulip ore
strange, can you screenshot the problem then
shut flicker
shut flicker
what n is could be specified somewhere else
sorry what do you mean
as in is it meant to specify what n is somewhere
compact ridge
there's a clear pattern for when n is odd and when n is even
I just tried around on Wolfram Alpha
but it's not hard to show, nowhere near what you've done so far
shut flicker
ok thanks
tulip ore
yo ive been absolutely tweaking on this question, i rly dont want someone to tel...
yes your current approach is correct
compact ridge
as in, when n is odd, they're all 0 mod X
and when n is even, they're all 0 mod Y
shut flicker
there's a clear pattern for when n is odd and when n is even
but this would require computing like n = 4 and higher which would take a while, dont think im allowed to
tulip ore
this typo is problematic
14^n + 11 is equivalent to something else (mod 6) when n is even
when n is odd, consider the following
its possible to figure out an argument that knocks out all odd n the same way there is for even n
now what causes n = 1 to not be prime?
try that out and youre done
pearl pondBOT
@shut flicker Has your question been resolved?
shut flicker
thanks a lot, so my typo was that 14^n+ 11 = 2 (mod 6), it should be = 3 (mod 6) right? so that can never be prime, since it would be in the form 6n+3 = 3(2n+1), hence, have a factor of 3, so if we consider when n is odd, i noticed that when n is 1, we get 25, which is a multiple of 5, so i tested at 14^(2k+1), which is always congruent to 4 (mod 5), and when you plus 11, congruent to 1 (mod 5), we get smth congruent to 0 (mod 5), therefore, also cannot be prime, perhaps i could also done an induction proof to show that 14^2k+1 + 11 is always divisible by 5, thanks for all your help😄
fast canyon
pearl pondBOT
fast canyon
Im getting 280 ms^-1
but the answer is 285 ms^-1
first im confused whether the intial vertical velocity we'll take is 0 or 50sin30.
but trying both of them im still getting 280.
v² = u² + 2as
v = sqrt[0+2(9.8)(4000)] = 280
v = sqrt[(50sin30)² + 2(9.8)(4000)] = 281
of part a
pearl pondBOT
@fast canyon Has your question been resolved?
errant snow
Im getting 280 ms^-1
but the answer is 285 ms^-1
first im confused whether the ...
You're only considering the vertical component of its velocity. Remember it also has a horizontal component due to the plane's initial velocity.
(Also remember to use 9.81 and not 9.8 since it explicitly says to use 9.81)
fast canyon
You're only considering the vertical component of its velocity. Remember it also...
oh so we will use 50sin30 right? its asking us velocity when it hits the ground, the horizontal component remains the same throughout so we have to use the vertical
errant snow
oh so we will use 50sin30 right? its asking us velocity when it hits the ground,...
Yes, we use 50sin30 for the vertical component.
But remember that we also need to include the horizontal component once we've found the vertical component
fast canyon
But remember that we also need to include the horizontal component once we've fo...
okay so now we have both vertical and horizontal should i find the resultant velocity?
oh yeah now im getting 285
wowir
thanks a lot
errant snow
Yea
fast canyon
.close
torpid wolf
for (ii) if question A1 is chosen then one of the questions is guaranteed and doesn't matter so you have 3 options left and 2 questions to answer for part A so it's (3C2) and then again for B1 we exclude it so surely the answer should be (3C2)(6C4) and then for (iii) the first case is that A1 is chosen so it should be (3C2)(6C4) and then the second case is that A1 is not chosen so we have 3! options for part a. and then 7 options so 6(7C4) for the second part so both cases added together it's 3(15) + 6(35)
torpid wolf
except im not even close
to the correct answer
im just trying to explain my thought process
rare scaffold
for (ii) if question A1 is chosen then one of the questions is guaranteed and do...
are you sure there are 3! choices for section A questions in part (iii) case 2? that makes it sound like order suddenly matters
torpid wolf
are you sure there are 3! choices for section A questions in part (iii) case 2? ...
wait so if the order isnt the same but they are the same options they are not considered distinct?
rare scaffold
the solution to part (ii) is a probability, so don't forget to divide the number of A1 => no B1 by the total number of exam routes
rare scaffold
wait so if the order isnt the same but they are the same options they are not co...
how many ways can you choose 3 items from a set of 3?
torpid wolf
how many ways can you choose 3 items from a set of 3?
1
rare scaffold
cool
torpid wolf
so 3(15) + 35
not 3(15) + 6(35)
ah ok
that was my mistake
i dont fully understand when something should be distinct and when it isnt
ig
rare scaffold
it's not a question of distinctness, you're in a setting where order doesn't matter and suddenly changed your counting approach to one that counts permutations
in part (i), choosing 3 items from a set of 4 yields (4 choose 3) = 4 options, so removing one of the candidates should yield less
more importantly, the counting problem itself doesn't change between the two variants, like you're still choosing k things from a set of n items (without replacement & order doesn't matter), just when (n,k) = (4,3) and (n,k) = (3,3)
latent quail
--
latent quail
Original question ^
slender viper
this breaks into series and parallel pretty easily
vestal tapir
what's the deal with variable resistor?
slender viper
not sure, are we supposed to write the result in terms of the resistance of the variable resistor
vestal tapir
that's the whole mystery
latent quail
okay so, my doubt is that how do we transform it from LHS to RHS?
latent quail
what's the deal with variable resistor?
ig it can be considered a fixed resistor with 10K $\omega$
vestal tapir
step by step, convert the rightmost part
jolly parrotBOT
e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#
slender viper
okay so, my doubt is that how do we transform it from LHS to RHS?
repeat the following steps:
replace series resistors with an equivalent resistor
replace parallel resistors with an equivalent resistor
vestal tapir
fold r6 and vr1 and r5 into one resistor, repeat
jolly parrotBOT
Result:
3.5294117647059
-# It's quite funny you said that
latent quail
oh, I thought they belong to the same system
nvm, figured
Right, I drew a simpler diagram for demonstration
I just need some confirmation rq
Presumably BEF can be simplified as
$\frac{1}{B} + \frac{1}{E} + F$?
jolly parrotBOT
e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#
latent quail
**e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7\#**
latent quail
Could you elaborate plz, thanks in advance 
latent quail
yes, but what's the difference between the diagram I drew and this one?
latent quail
hmm, if that's the case can I add up R6 and VR1 in the first place
and consider R5 in the nxt process?
vestal tapir
yes
latent quail
if that's the case, I think this works perfectly
lemme calculate it rq
$VR_1 + R_6 = 12$\
$\frac{1}{12} + \frac{1}{3} = \frac{5}{12}$\
$\frac{5}{12} + 5 + 2 = \frac{89}{12}$\
$\frac{12}{89} + \frac{1}{5} = \frac{149}{445}$\
$\frac{149}{445} + 1 = \frac{594}{445}$\
jolly parrotBOT
e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#
latent quail
Right, looks legit.
@vestal tapir Could you take a look of it?
vestal tapir
not really
i mean, if the answer is right, you got it
i would search for a circuit simulator, draw it and see if it agrees
latent quail
Nah, I just need the core concept of how to do it 😂
Thanks for helping me
Have a good one
.close
latent quail
OPEN
latent quail
serene sentinel
open
latent quail
Right, for the first question I do not understand why that formula is true in the first place, let alone proving its authenticity
latent quail
with R = R2 for V_out
wait so Vout serves as the output of voltage, right?
how about DVcc? I thought it's a power source as well
tardy reef
yea
it is a power source, but you split it to get a lower V
Vcc gets split into V_out and Vcc-Vout due to the two resistances
latent quail
So basically there exists two route for the current to go
and both ways are being used at the same time.
I'm not sure if I'm on the correct path
tardy reef
yes
well tbf atp you are only measuring
at part 2 it would get split
coz technically voltmeters have inf resistance and as such no current would flow thru the other part in an ideal case
latent quail
I see, but how does that formula come from?
I have no idea how to derive that from the current information
tardy reef
calculate the current and then V=IR
this is how
latent quail
I tried applying V = IR but it doesn't seem working
latent quail
@tardy reef is it correct that I simplify the disgram on LHS to this?
I was writing my reasoning while I found out this crucial doubt in the midway
tardy reef
<@617550509477330944> is it correct that I simplify the disgram on LHS to this?
wait what? whats this? why is there two R1s there?
is the one on top supposed to be RL?
also note that R1 is in series with R2 as well as RL while R2 and RL are in parallel
so either way the diagram is incorrect
additionally, voltmeters are in parallel, so V should not be there
latent quail
additionally, voltmeters are in parallel, so V should not be there
ah yes
latent quail
wait what? whats this? why is there two R1s there?
no, I was trying to draw the diagram on RHS into a simpler, conciser manner
My idea is that the current is spilt into two routes, red and blue
tardy reef
yes the idea that it splits is correct
but that also means originally both red and blue pass thru R1. Is that why you have R1 on both?
latent quail
yes
tardy reef
then where is RL?
latent quail
because the current must pass through R1
latent quail
then where is RL?
makes sense 💀
tardy reef
so... can you now update the diagram?
latent quail
Certainly
tardy reef
yes
also, then in that case, you should draw the resistance outside the parallel configuration, and show it in series with the other resistances
coz the actual circuit has only one of R1, and all your circuits must reflect that
latent quail
latent quail
Right, so I assume we need to find the current flowing to the upper route
because that's what we are looking for
tardy reef
also uhh, there is a reason why we have symbols for various circuit elements, so it would make sense if they actually get used 💀
latent quail
also uhh, there is a reason why we have symbols for various circuit elements, so...
I suck 😭
I wonder why tf you know so much
this is physic field, thought you're majoring in math
tardy reef
I wonder why you dont? Is that not in your highschool syllabus?
tardy reef
this is physic field, thought you're majoring in math
I am engineer, but tbh this is highschool syllabus
latent quail
I wonder why you dont? Is that not in your highschool syllabus?
I managed to attend cs with my English and Math grade. My physics was like 44 out of 100 in my entrance exam
It's a pain in my ass since HighSchool since I never study when I was a high schooler
anyways, Imma finish this
tardy reef
anyways, Imma finish this
latent quail
The total current flowing into the upper route is: DVcc $\times\frac{R_L}{R_L + R_2}$
jolly parrotBOT
e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#
tardy reef
simple dimensional analysis shows that you are wrong
latent quail
Applying the formula V = IR\
$V = DVcc\times \frac{R_L}{R_L+R_2} \times $
yeah, I need help 💀
tardy reef
so what is the Req for the circuit?
latent quail
you mean the sum of all resistors?
tardy reef
this thing that you did before
latent quail
$\frac{1}{R_L} + \frac{1}{R_2} + R_1$
jolly parrotBOT
e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#
tardy reef
nope
latent quail
how come?
tardy reef
lets look at the R2 and RL, what is the single equivalent resistance of those two?
tardy reef
how come?
again, dimensional analysis
resistance is not same as 1/resistance
1/R1 would have unit 1/ohm
and 1/ohm cant add with ohm
latent quail
**e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7\#**
So this one is incorrect as well
because I was using the same approach
tardy reef
So this one is incorrect as well
no you have 1/Rt there
so actual resistance Rt would become correct
you are just one step away there
its incomplete not wrong
tardy reef
yea, then its wrong
tardy reef
well, luckily in this particular problem Rt is 1, so 1/Rt would also have the same value
latent quail
Right, so
$\frac{1}{R_T} = \frac{1}{R_L}+\frac{1}{R_2}$\
$R_T = \frac{R_LR_2}{R_1+R_2}$
jolly parrotBOT
e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#
tardy reef
yep
latent quail
is the diagram I draw equal to the diagram on LHS?
because I think the first subquestion is about the diagram on LHS
tardy reef
no...