the number isnt correct
#help-4
earnest dew
call it 22840
wooden dune
wrong
earnest dew
190(1+16^2/190^2)38pi
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Available help channel!
north scarab
I need help with 5
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I need help with 5
stark wedge
have you been able to do (1)-(4)?
north scarab
yep
shut mirage
close
stark wedge
you should not guess
you should be able to explain what step can get you from 2^3 = 8 to 2 = 8^(1/3)
i mean it is correct that 2 = 8^(1/3) im just unhappy that you explicitly said you guessed it
twin phoenix
do you think you can write 8^x using 2 as base?
north scarab
because 2^3=8 so if u want to reverse that, just take the cube root
north scarab
do you think you can write 8^x using 2 as base?
im not sure what u mean
shut mirage
because 2^3=8 so if u want to reverse that, just take the cube root
this is right
twin phoenix
im not sure what u mean
I meant you can write it as 2^3x right?
grizzled cape
anyone can do 77777777*= in 10 second
shut mirage
what about the other side?
stark wedge
anyone can do 7*77*7*77*77*= in 10 second
<@&268886789983436800> troll
stark wedge
how do i solve for x?
maybe stop walking in circles tbh
shut mirage
anyone can do 7*77*7*77*77*= in 10 second
!occupied
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anyone can do 7*77*7*77*77*= in 10 second
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).
idle void
anyone can do 7*77*7*77*77*= in 10 second
Please don't troll in the help channels
woeful trench
anyone can do 7*77*7*77*77*= in 10 second
Don't troll in the help channels please 
idle void
sniped :P
grizzled cape
@shut mirage ok
idle void
#discussion and #chill exist
woeful trench
What soda would you like 
proud sequoia
its not:(
proud sequoia
mb his all yourws
shut mirage
if you know 8^(1/3) = 2, you can reason from there what the power of 8 is for **1/**2
shut mirage
we write any fraction for instance
1/n as n^-1
recall this.
twin phoenix
if you know 8^(1/3) = 2, you can reason from there what the power of 8 is for **...
@north scarab what do you think you can do with the power here?
you already have the clue
weary pilot
no
shut mirage
8^(-3) blows up the number
weary pilot
theres negatives other than the integers
shut mirage
just so you're interested, 8^(-3) is 1/512
north scarab
oh I just did 8^(-1/3)=1/2 but I dont understand how it equates to 1/2, basically I guessed it
shut mirage
if you know 8^(1/3) = 2, you can reason from there what the power of 8 is for **...
you already reasoned this out. from here it is just one more step
north scarab
(s^(1/2))^2=s
shut mirage
what... has that to do with 2 and 1/2 being related as powers of 8?
north scarab
Im recalling my laws
shut mirage
did you guess again
yes.
shut mirage
oh I just did 8^(-1/3)=1/2 but I dont understand how it equates to 1/2, basicall...
here.
stark wedge
uhhhhhhhhhhhh maybe don't, dude.
north scarab
you already reasoned this out. from here it is just one more step
wdym
shut mirage
you know that the cube root of 8 (or 8 to the (1/3)th power) is 2.
you need **1/**2.
it's one step away.
shut mirage
to get 1/2?
north scarab
any hints
shut mirage
we write any fraction for instance
1/n as n^-1
^
stark wedge
yes
north scarab
why was that so hard to spot
shut mirage
that's why the reciprocal button on your calculator is labelled x^-1
stark wedge
why was that so hard to spot
bc you're out of practice with exponent laws tbh
north scarab
do u guys have any websites I could practice
stark wedge
these things need some amount of grinding in order to learn this particular pattern recognition
north scarab
log laws and stuff with exponent laws
stark wedge
ig khanacademy honestly
shut mirage
however and whichever you choose
stark wedge
not one
one sheet isn't enough practice imo
- you specifically have a lot of shit to grind
shut mirage
if that sheet is insufficient, here's another
https://madasmaths.com/archive/maths_booklets/basic_topics/various/logarithms_exam_questions.pdf
but after this, learn to search for practice problems and do as many of them as it takes for you to be super confident
craggy hemlock
Hello, I am trying to figure out a certain game mechanic but its too complicated...
The game says that cold in a mineshaft increases every 100 ticks (with 20 ticks equal to 1 second), and that each point of cold resistance slows cold gain by 1 tick. For example, with 90 cold resistance, cold should increase every 190 ticks. When cold reaches 100, the player dies.
The mineshaft is supposed to make cold gain 25% faster, so survival time should be shorter. However, in practice, the game doesn’t behave this way. The information given by the game about cold gain is incorrect, it doesn’t actually calculate cold the way it describes.
The game sends BRRR! Cold! In chat between 15 to 16, between 35 to 36, 57 to 58 and 86 to 87.
This is the time where cold gained starts to increase faster, its either some of my information is insufficient or im missing out on something.
craggy hemlock
Cold 0: 0s
Cold 1: 9.5s
Cold 2: 19.05s
Cold 3: 28.55s
Cold 4: 38.05s
Cold 5: 47.55s
Cold 6: 57.05s
Cold 7: 66.55s
Cold 8: 73.05s
Cold 9: 85.55s
Cold 10: 95.1s
Cold 11: 104.65s
Cold 12: 114.15s
Cold 13: 123.7s
Cold 14: 133.2s
Cold 15: 142.75s
Cold 16: 150.35s
Cold 17: 157.95s
Cold 18: 165.65s
Cold 19: 173.2s
Cold 20: 180.75s
Cold 21: 188.4s
Cold 22: 196s
Cold 23: 202.6s
Cold 24: 211.2s
Cold 25: 218.8s
Cold 26: 226.4s
Cold 27: 234.35s
Cold 28: 241.7s
Cold 29: 249.35s
Cold 30: 256.95s
Cold 31: 264.6s
Cold 32: 272.2s
Cold 33: 279.8s
Cold 34: 287.4s
Cold 35: 295.1s
Cold 36: 301.35s
Cold 37: 307.75s
Cold 38: 314.2s
Cold 39: 320.4s
Cold 40: 326.9s
Cold 41: 333.2s
Cold 42: 339.5s
Cold 43: 345.9s
Cold 44: 352.35s
Cold 45: 358.65s
Cold 46: 365.15s
Cold 47: 371.35s
Cold 48: 377.8s
Cold 49: 384.05s
Cold 50: 390.45s
Cold 51: 396.8s
Cold 52: 403.15s
Cold 53: 409.55s
Cold 54: 415.85s
Cold 55: 423.25s
Cold 56: 428.55s
Cold 57: 434.55s
Cold 58: 441.4s
Cold 59: 447.7s
Cold 60: 453.1s
Cold 61: 458.65s
Cold 62: 464.15s
Cold 63: 469.55s
Cold 64: 475s
Cold 65: 480.5s
Cold 66: 486.15s
Cold 67: 491.5s
Cold 68: 496.9s
Cold 69: 502.35s
Cold 70: 507.9s
Cold 71: 513.3s
Cold 72: 518.85s
Cold 73: 524.3s
Cold 74: 529.85s
Cold 75: 535.2s
Cold 76: 540.6s
Cold 77: 546.1s
Cold 78: 551.5s
Cold 79: 557.05s
Cold 80: 562.6s
Cold 81: 568s
Cold 82: 573.4s
Cold 83: 579s
Cold 84: 584.4s
Cold 85: 589.75s
Cold 86: 595.3s
Cold 87: 600.7s
Cold 88: 605.4s
Cold 89: 610.25s
Cold 90: 614.9s
Cold 91: 619.7s
Cold 92: 624.5s
Cold 93: 629.25s
Cold 94: 634.05s
Cold 95: 638.75s
Cold 96: 643.6s
Cold 97: 648.3s
Cold 98: 653.1s
Cold 99: 657.8s
Cold 100: 662.45s
Timestamps ^^^
Average 9.5s (0 -> 16 Cold)
Average 7.6s (16 -> 36 Cold)
Average 6.25s or 6.3s (36 -> 58 Cold)
Average 5.4s (58 -> 87 Cold)
Average 4.7s (87 -> 100 Cold)
Please 🙏, if someone here can help me figure this out im struggling for hours upon hours.
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@craggy hemlock Has your question been resolved?
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@craggy hemlock Has your question been resolved?
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@craggy hemlock Has your question been resolved?
craggy hemlock
aw
sonic halo
Huh
Sounds interesting
Lemme read it again real quick
Should change time to ticks
Easier to process imo
Ok 30 mins break
Have to go
craggy hemlock
Ok 30 mins break
jovial edge
Average 9.5s (0 -> 16 Cold)
Average 7.6s (16 -> 36 Cold)
Average 6.25s or 6.3s (...
using this information cold gain rate increasing as the time passes and not just 25% constantly
Tbh it looks weird as hell
cloud coral
Hello, I am trying to figure out a certain game mechanic but its too complicated...
yeah it seems like there are places where the rate is pretty much constant (but somehow there are spikes)
how much did you repeat this experiment?
do you always get spikes in the same places?
it seems the spikes are fixed rather quickly, so my best guess is
"the cold level is calculated based on time spent, but only updated periodically"
and if so, this predicts that the location of the spikes should be "random"
it could be determined by lag, but idk
idle sundial
cloud coral
seeing the size of the lag, my best guess is network lag
@craggy hemlock
actionables:
- is it played over a network?
- collect a second set of data
huh... but network lag would go late then back on time
not early then back on time
that's weird
and the difference is seconds, so this can't be inaccuracy of a timer function
but the values are to 1/20th of a second, and chances are you recorded it with a screen capture?
could it be inaccuracies in the screen capture software?
craggy hemlock
yeah it seems like there are places where the rate is pretty much constant (but ...
the spikes should always be in the same place
i assume changing my cold resistance has effect on it
i could test that in the evening today
craggy hemlock
<@910933497810653264>
actionables:
1. is it played over a network?
2. collect a...
- yes
- will do this evening
craggy hemlock
but the values are to 1/20th of a second, and chances are you recorded it with a...
i didnt record it, a person i know did
(he was cheating inside the gameplay with xray)
craggy hemlock
could it be inaccuracies in the screen capture software?
it could be SLIGHT inaccuracies
but its not likely
from what i know this game is very buggy and the wiki of the game usually have wrong information in it
it is a network with servers only in north america
therefore average player ping is 200ms
it isnt random for all i know, due to the times with certain cold resistances is always the same
and the game usually follows a certain formula
built on 1.8.9 minecraft (hypixel)
it most likely has to do something with the "brrr message"
since its the point where the cold gain gets faster.
there is also something weird with the small adjustments in speed between cold gain
of like a tick or two
but that could just be lag
cloud coral
ping is 200ms, probably explains most of the bumpyness
craggy hemlock
ping is 200ms, probably explains most of the bumpyness
yup, but its consistent bumps at certain points as u might see
actually nevermind its 1 test
it could still be lag
cloud coral
let's get a second test to establish if the bumps are consistent
craggy hemlock
would be 5 hours 30min until i can get accurate tests of the same person
or should i run recording of my own
i could do it now i guess
cloud coral
from a different player can be a good idea
it would basically separate the stochastic explanations from the consistent explanations
craggy hemlock
okay
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@craggy hemlock Has your question been resolved?
zealous pendant
Hello, I am trying to figure out a certain game mechanic but its too complicated...
“The information given by the game about cold gain is incorrect” Average skyblock wiki moment
I think that the lag spike near 58 cold is making it unclear when it drops down, i think it drops after 59 starting the new level at 60
Nvm I just noticed you said brr happens at 58
I think the wiki is just wrong then? Maybe do the test a few more times ideally in an empty server
Gl with that btw
frank salmon
this a serie of mathematic
This is easy
distant galleon
this a serie of mathematic
!occupied
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this a serie of mathematic
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).
craggy hemlock
Should I unclaim this help channel until I get more tests, or am I allowed to get the tests tomorrow without needing to reopen it
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@craggy hemlock Has your question been resolved?
ionic sluice
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steady charm
what about the second question?
quasi valve
vale dockBOT
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.
solar depot
the question is designedd that way
send the entire page ss so we can have some context
solar depot
im guessing it's from the book by Gilbert Strang
ionic sluice
yes
steady charm
i don't think this is a question you're meant to actually answer yourself, it seems to me more like "these are the questions this section will answer"
solar depot
i don't think this is a question you're meant to actually answer yourself, it se...
yes +1 on this
ionic sluice
i solved the first one
I tried putting different dimensions of v, first I put just 1 , then I got 1 dimension vector with 1 row , and 1 column thus 1 value, but when I tried putting the 2nd dimension, I got 2 dimesnion having 2 rows but one column,
but the second one was not proved
ionic sluice
I tried putting different dimensions of v, first I put just 1 , then I got 1 dim...
does this make sense tho...?
steady charm
i don't think this is a question you're meant to actually answer yourself, it se...
in fact it seems to me that the very next page is dedicated to providing the answers to the questions that were just raised?
ionic sluice
Oh okay...
thanks, i thought those were questions for me to be solved, i will be more careful when asking questions, thanks
!close
!end
steady charm
!done
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If you are done with this channel, please mark your problem as solved by typing .close
ionic sluice
.close
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Available help channel!
lusty blaze
please help
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please help
void plume
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@lusty blaze Has your question been resolved?
restive urchin
hippoty hopppoty this channel is mnow my property
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hippoty hopppoty this channel is mnow my property
umbral isle
Lol
restive urchin
$|x+3k|<4|x-k|$
rocky lotusBOT
V_ice
restive urchin
the questin says
solve thge inequality
and that k is a positive constant
idk what to do so i just whole squared both sides
hoping the x would cancewl out
but
i ended up with
$15x^2+26kx+7k^2>0$
rocky lotusBOT
V_ice
restive urchin
da hecc do i do now
wait
wait a minut
nope no idea
i was thinking of middle terming it
but
u cant get 26 from 105
oh wait
you can
oh wAIT
i cook
$(3x+k)(5x+7k)>0$
rocky lotusBOT
V_ice
stark wedge
$|x+3k|<4|x-k|$
restive urchin
sigma
LOL
nice pin
now wat
hmmmmmmmmmm
ive never solved a quadtraic inequality with two variable
s
stark wedge
you're solving for x
stark wedge
and the sign-change points are gonna be in terms of k
you dont "bring" anything anywhere
restive urchin
wdym sign change points
stark wedge
gonna be a hell no from me
restive urchin
oooooooooooooooooop
stark wedge
ab > 0 does NOT transform into any combination of a > 0 and/or b > 0.
restive urchin
o
stark wedge
restive urchin
thanks
we never learnt this at schooooooool
ok we did nvm but not specifically two variables
ig ill watch that
tysm boss
.close
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restive urchin
looks like i made a mistake middle terming it
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Available help channel!
mild rapids
Which of the following cannot be Fourier Series expansion of a periodic signal
Professor marked option b as correct.
I don't understand why are other options incorrect
stark wedge
all of the others are in fact periodic
while (ii) isn't because 1 and pi are incommensurate
safe fulcrum
(there is no rational common factor of 1 and pi)
mild rapids
Do you mean this ?? For context this is googles AI generated answer
How would I know if there were 4-5 terms ??
If I just have to take the ratios of coefficient of "t" and decide on the basis of rationality, how would I solve if there are more then 2 terms ??
safe fulcrum
Do you mean this ?? For context this is googles AI generated answer
How would I...
you shouldn't trust AI, but yes that's what I mean
mild rapids
you shouldn't trust AI, but yes that's what I mean
I don't thats the reason I posted it here
safe fulcrum
so cos t + cos(t/2) + cos(t/3) would work, cause a common divisor of 1, 1/2, 1/3 is 1/6
the issue comes from mixing rational and irraitonal numbers
cos(sqrt2 t) + cos(t) is not periodic
cos(sqrt2 t) + cos(e * t) is also not periodic
but cos(sqrt2 t) + cos(2 sqrt2 * t) is periodic
safe fulcrum
if you have even one irrational number, and the rest are whatever, except for something like $\cos(\sqrt2 t) + \cos(3 \sqrt2 t) + \cos(6 \sqrt2 t) + \cdots$, it won't be periodic
rocky lotusBOT
south
safe fulcrum
no worries!
safe fulcrum
Thanks man
if you're done type .close
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Available help channel!
subtle sparrow
I totally forgot how you calculate with roots 😭
Can anyone remind me and tell me how it works again on this equation? Sure, you multiply them by the number outside, but how?
stark wedge
,rccw
rocky lotusBOT
stark wedge
this isnt an equation
this is an expression
what... exactly do you wanna do with this? @subtle sparrow
subtle sparrow
😔
Basically I need to multiply the number outside the () with the inside ones
But I totally forgot how that works with √
stark wedge
Basically I need to multiply the number outside the () with the inside ones
i would've liked to see some written instructions.
subtle sparrow
It's in an equation, I just took out the part idk and wrote it down
stark wedge
but ok, if you dont yet see how to multiply these, then simply write the multiplication as a multiplication.
$\frac{\sqrt{2}}{2} \paren{2x - \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{2} \cdot 2x - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}$
rocky lotusBOT
Ann
stark wedge
like, if you dont see immediately how this stuff simplifies then just dont.
write down the step where you apply the distributive property, and only THEN worry about what comes next.
stark wedge
well how would you simplify $\frac{\sqrt{2}}{2} \cdot 2$ if you could let go of your rhizophobia for a bit?
subtle sparrow
I take math im french so I might not understand everything you're telling me
rocky lotusBOT
Ann
stark wedge
(||rhizophobia == fear of roots, but with greek words just for lulz||)
subtle sparrow
I have the answers, I just wanna know how to do it myself
stark wedge
i mean
again
for a start, you could just write down the multiplications
like yknow
lets set the roots aside
$3(4x+17) = 3 \cdot 4x + 3 \cdot 17$
rocky lotusBOT
Ann
stark wedge
do you agree this is legal to do @subtle sparrow
subtle sparrow
Yeah
stark wedge
right
so im saying for step 1 you should just write down $$\frac{\sqrt{2}}{2} \paren{2x - \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{2} \cdot 2x - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}$$ and from this point you see two clear avenues for simplification: $\frac{\sqrt{2}}{2} \cdot 2$ and also $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}$. do you agree?
rocky lotusBOT
Ann
subtle sparrow
I do
rocky lotusBOT
Ann
subtle sparrow
2√2/2?
subtle sparrow
I don't know
stark wedge
ok lets try to remove the scary element
how would you simplify $\frac{7}{9} \cdot 9$?
rocky lotusBOT
Ann
subtle sparrow
9×7/9
rocky lotusBOT
Ann
stark wedge
$\frac{9 \cdot 7}{9}$ simplifies to just 7 by cancelling out the 9's...
rocky lotusBOT
Ann
stark wedge
you would agree $\frac{ab}{ac} = \frac{b}{c}$ always (so long as $a, c \neq 0$) yes?
rocky lotusBOT
Ann
stark wedge
likewise $\frac{\sqrt{2}}{2} \cdot 2$ is \textbf{just} $\sqrt{2}$.
rocky lotusBOT
Ann
subtle sparrow
Right
stark wedge
and for simplifying $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}$ you will need to recall the definition of square root (racine carrée) to know how to simplify the $\sqrt{2}\cdot\sqrt{2}$ at the top.
rocky lotusBOT
Ann
stark wedge
But it had an x
yeah so?
subtle sparrow
√2x?
stark wedge
sqrt(2)x
you should write it like this so there's not even a chance to confuse the x as being under the root too
when it isnt
stark wedge
you dont need to put the bottom as 4
@stiff lily where was that example of yours with the 123456789 cancellation
subtle sparrow
It becomes 2,the top, below is 4
Then it becomes 1/2 no?
stark wedge
yes in the end it comes out as 1/2.
topaz tree
I am having issues understanding the following statement about linear independence for some vector sapce V, More specifically the last sentence, are they showing that a set containing a nonzero vector can't be linear dependent by contradiction?
abstract pollen
I am having issues understanding the following statement about linear independen...
yeah wait
lemme see
topaz tree
Also, why is it that $a^{-1}(au) = a^{-1}0$?
rocky lotusBOT
Pen
coral urchin
because $au=0$ by definition of linear independence
rocky lotusBOT
Cycadellic
topaz tree
because $au=0$ by definition of linear independence
not linear dependence?
coral urchin
ohp
topaz tree
thought linear independence was if there does not exist a nontrivial solution
coral urchin
sorry
let me be precise
a set of vectors ${v_1,...,v_n}$ are defined to be linearly independent iff for all scalars $a_1,...,a_n$
$$a_1v_1+...+a_nv_n\implies 0=a_1=...=a_n$$
rocky lotusBOT
Cycadellic
topaz tree
Yes
abstract pollen
I am having issues understanding the following statement about linear independen...
Yeah the statement is using a proof by contradiction to show that a set containing a single nonzero vector in a vector space V is always linearly independent
coral urchin
so, the setup for a proof by contradiction is to assume the negation of what you wish to show, and show a contradiction (or a false statement)
topaz tree
Yeah the statement is using a proof by contradiction to show that a set containi...
Ah, okay!
topaz tree
so, the setup for a proof by contradiction is to assume the negation of what you...
Okay, I was not being fuzzy then haha
abstract pollen
Ah, okay!
Suppose you have a set
{u} where u ≠ 0
To say {u} is linearly dependent means there exists a nonzero scalar
a such that au = 0
topaz tree
yup
abstract pollen
The last sentence manipulates this equation: If
au = 0 and a ≠ 0
then multiplying both sides by a^−1 (the multiplicative inverse of a), you get:
u = a−^1 (au) = a^−1⋅0 = 0
However, this implies u=0, contradicting the assumption that u is nonzero.
Therefore, such a nonzero scalar
a cannot exist, so the set is linearly independent
topaz tree
Missed that they multiplied both sides by a^-1, oopsie
Yup, all good now
Thank you Sonic and Cycadellic ❤️
abstract pollen
wlc
topaz tree
.close
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signal chasm
is so insane
stark wedge
did you just wanna complain or did you have a specific thing to ask
signal chasm
i literally
dont get how he found the other solutions
i got the k=-8 and 8.25
but this circle thing i dont get one bit
ruby sleet
but this circle thing i dont get one bit
If ABC is a right triangle, with right angle in A, then the circle that passes through points A,B and C has for center = the middle of segment BC
signal chasm
If ABC is a right triangle, with right angle in A, then the circle that passes t...
could u draw that please, i dont get what the middle of segment bc means
ruby sleet
Let's use the drawing you already have
A = (5,k), B = (2,3), C = (9,-1)
and assume the right angle is at A
if we name I = (5.5,1) the middle of the segment BC
then the circle that passes through points A,B,C
is the one drawn in your image
and has center I
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@signal chasm Has your question been resolved?
midnight pier
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midnight pier
can this be just be done by saying that u = w and then u + 3x = w, x needs to be 0 vector
for 3x to be 0 vector
and by that x is unique
viscid spade
Is there a def for V
midnight pier
V is a vector space
odd jackal
Suppose v+3x_1=w, v+3x_2=w then get the zero vector from here and your result
ebon glade
who said u=w
midnight pier
Suppose v+3x_1=w, v+3x_2=w then get the zero vector from here and your result
is this to prove uniqueness x_1 = x_2?
okay wth lmao so i know there is two steps first is to show existence of a x such that the equation is satisfied
and then show that the x is unique
first step is done
midnight pier
is this to prove uniqueness x_1 = x_2?
and this can prove
the uniquness
both together proves it?
midnight pier
who said u=w
also i just realized that was bad because u and w can be different as well
i need to prove for all u, w
midnight pier
Suppose v+3x_1=w, v+3x_2=w then get the zero vector from here and your result
i didn't thought about proving it this way ugh
is this used often to prove the uniqueness
viscid spade
Supposing two things having the same definition and ending with them being equal is the most common method
vale dockBOT
Available help channel!
frozen matrix
How the theta is projected from the inclined to the arrowed triangle? I'm lil confused. The topic is Statics of Rigid bodies but its just trigo smthn.
glass kelp
autumn carbon
Also a solution
vale dockBOT
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outer rock
vale dockBOT
outer rock
Anyone know to solve this
drifting hornet
Try finding the limit from different directions and see if they match
cobalt crow
did you tried evaluating the limit of the first case when (x,y) is approaching (0,0)?
we can maybe use polar coordinates, just throwing the idea out there, though I havent tried it myself but might work?
drifting hornet
Perhaps try bounding the numerator by sth which will cancel with the denominator
are you even here @outer rock ? We need to know ur progress
cobalt crow
@outer rock
junior ore
equate f(0+) = f(0) i think that should do it
cobalt crow
for example for the path x = y, we get 2|y|^2 / 2|y|, no?
for the function to be continuous at (0,0) the lim (x,y)->(0,0) f(x,y) needs to be equal to f(0,0)
and for the limit of f(x,y) as (x,y) is approaching (0,0) to exist all paths need to verify the same value
@outer rock u here?
vale dockBOT
@outer rock Has your question been resolved?
outer rock
See bro limit won't exist
Like anything we put it cant be equal to 10^2024
So basically what shud we put and check
outer rock
<@1330184550709461044> u here?
Ye
cobalt crow
So basically what shud we put and check
think of a couple paths
cobalt crow
what happens when m = 1 for y = mx^2?
outer rock
Lemme check
cobalt crow
what does the limit evaluates to?
outer rock
How to take out mod 😭
cobalt crow
How to take out mod 😭
try to draw in paper everything you are working on, in the exam you will need to write in paper, not in discord messages
outer rock
How to proceed
cobalt crow
,rccw
rocky lotusBOT
cobalt crow
How to proceed
take m = 1
cobalt crow
outer rock
Ye 👍
outer rock
Denominator can we take mod x common
?
outer rock
the path was y = mx^2?
Yep
cobalt crow
you missed a x^2 in the denominator
outer rock
you missed a x^2 in the denominator
Ye understood my bad 😭
cobalt crow
well, first we can factorize by |x|
cobalt crow
Denominator can we take mod x common
we can factorize by |x| in the num and in the denom
outer rock
Ok
cobalt crow
what do you get?
outer rock
why're u doing all this?
For showing is the function continuos
cobalt crow
now, since this is a single variable limit, evaluate the lateral limits
outer rock
now, since this is a single variable limit, evaluate the lateral limits
Wym bud?
cobalt crow
lim x -> 0^+ and lim x-> 0^-
outer rock
So shud i keep value of x as 0
cobalt crow
this is a limit as x is approaching 0, y is wrt to x, because y = mx^2 right? so I say take the lateral limits of x -> 0 because the absolute value is a piecewise
cobalt crow
why're u doing all this?
like, we are trying to see if the limit of f(x,y) as (x,y)->(0,0) approaches always 0 or if there is a path in which the limit does not evaluate 0, if we try a couple paths and see all of them go to 0, then we will need to proof the lim (x,y)->(0,0) of f(x,y) ALWAYS goes to 0 in order to show the f(x,y) function is not continuous on (x,y)=(0,0)
outer rock
this is a limit as x is approaching 0, y is wrt to x, because y = mx^2 right? so...
So what next
If I put x=0 then I get numerator 0
Denominator 1
cobalt crow
take the lateral limits of this, see if the limit exists and if so verify it goes to zero from both sides
outer rock
Bro wym lateral limits
cobalt crow
If I put x=0 then I get numerator 0
this is a limit as x -> 0, so for example x = -0.0001 is a possibility, this is why I say, check the left limit and the right limit as x approaches 0
outer rock
Ok
outer rock
Now we check y=mx^2 which we get 0
cobalt crow
this is good enough, no?
is still possible that, that limit doesnt exist if the lateral limits dont coincide
outer rock
Now we check y=mx^2 which we get 0
So after this what we have to check
Like which value
cobalt crow
Now we check y=mx^2 which we get 0
how do you know that limit exists? |x| = -x when x -> 0^-,
drifting hornet
how do you know that limit exists? |x| = -x when x -> 0^-,
are you claiming that we dont know whether the full (x, y) -> 0 limit exists or that we dont know that the x -> 0 limit exists (the one along y = mx^2)
cobalt crow
latter
outer rock
drifting hornet
latter
well, |x| is continuous, aint it? So we can plug in x=0, we get 0/1 and so the limit along y=mx^2 is 0
you dont need to check both sides just because of |x|
outer rock
well, |x| is continuous, aint it? So we can plug in x=0, we get 0/1 and so the l...
We don't know if modx is continuous
outer rock
So is this wrong @drifting hornet
cobalt crow
|x|^2 from the left is (-x)^2, but yeah the lateral limits seem to coincide to 0, so now that we checked that two paths coincide at 0, we can try to prove the f(x,y) limit approaches 0 always
drifting hornet
its right, it's just that ur original approach was fine as well. (The one where you computed the two-sided limit straight)
outer rock
Squeeze for what
cobalt crow
we want to prove f(x,y) goes to 0 as (x,y) approaches (0,0) ALWAYS
for that we will need to find an upper bound and try to use this using either epsidelta or squeeze theorem
cobalt crow
Squeeze for what
to prove the f(x,y) goes to 0 as (x,y)->(0,0)
zealous pendant
Polar coordinates?
outer rock
Can't we apply polar coordinates instead of saueeze
drifting hornet
yeah, i think my approach is a bit simpler. Polar coordinates would be fine, if the denominator was nicer
cobalt crow
Polar coordinates?
there are a couple of approaches
cobalt crow
but even if you end up using polar coordinates we still need to find an upper bound, no? in order to prove it using squeeze theorem
drifting hornet
its actually not that bad
outer rock
So we can't use polar?
cobalt crow
if we use polar coordinates we write the limit wrt one variable r, if the limit is simple enough that we can directly show that this is approaching 0 always, then we might not need squeeze theorem I think
zealous pendant
but even if you end up using polar coordinates we still need to find an upper bo...
It’s proportional to r and the theta bit is bounded between 1 and sqrt(2)
cobalt crow
anyways, after using polar coordinates what do you get?
cobalt crow
It’s proportional to r and the theta bit is bounded between 1 and sqrt(2)
i didnt know, thanks
zealous pendant
Wait sorry 1/sqrt(2) I forgot to inverse
cobalt crow
anyways, after using polar coordinates what do you get?
@outer rock
outer rock
I got like this
outer rock
<@1330184550709461044>
cobalt crow
I think we can say |cos^2(θ)| + |sin^2(θ)| is 1 no? is a special case of the cos^2 + sin^2 = 1 trig identity
outer rock
Ye
cobalt crow
also |r^2| is just r^2
outer rock
also |r^2| is just r^2
How
cobalt crow
doesnt matter, I see you already simplified with the denominator factorization
cobalt crow
so like, the denominator does not depend of r, so we can take that part out of the limit
cobalt crow
Costheta sin theta is bounded isnt
yeah so the whole denominator is bounded, and 0 * bounded = 0
outer rock
Ye when putting r=0 whole thing becomes zero
drifting hornet
importantly, the denominator is never 0
if it was 0 or approaching it, then we wouldnt be able to conclude that the limit 0
outer rock
So now what
drifting hornet
now you're done
you have shown that the limit is 0
Here is the pure squeeze approach btw $0\le\frac{\left|x\right|^{2}+\left|y\right|^{2}}{\left|x\right|+\left|y\right|}\le\frac{\left(\left|x\right|+\left|y\right|\right)^{2}}{\left|x\right|+\left|y\right|}=\left|x\right|+\left|y\right|\to0$
outer rock
So now we can say 0 not equal to 10^2024
rocky lotusBOT
MathIsAlwaysRight
outer rock
So limit not exist
drifting hornet
So limit not exist
Limit does exist
cobalt crow
So limit not exist
no
drifting hornet
but it doesnt equal f(0, 0)
outer rock
Ok limit exist but function are not equal
drifting hornet
the limit exists and equals 0. But it doesnt equal f(0,0), so the func isnt continuous at 0,0
outer rock
I see
drifting hornet
this is the 2D variant of this scenario. The limit exists, but its not continuous anyway
drifting hornet
Both y=mx^2 and polar coordinates
You usually just need to try some paths to convince yourself that the limit might exist and find its value
I usually do x = 0, y = 0, x = y and sometimes y = x^2
someone does y = mx^2
someone does y = mx
outer rock
I usually do x = 0, y = 0, x = y and sometimes y = x^2
What if when u put all u get 0
What does that mean
drifting hornet
then you have a reasonable evidence that the limit exists and is 0 and you should attempt to prove it
cobalt crow
the path you try depends of what the limit is approaching to, because if (x,y)->(1,0) then y = x^2 is not a valid path
drifting hornet
if you arent able to prove it, then you might go back and try several other paths to see if you are able to find one whose limit is different (that would mean the overall limit doesnt exist)
drifting hornet
the path you try depends of what the limit is approaching to, because if (x,y)->...
yeah, ofc you always need to adjust ur paths to the point
outer rock
Ok imagine I put y=x,y=x^2 but when I put y=mx^3 I get some other anser
drifting hornet
Ok imagine I put y=x,y=x^2 but when I put y=mx^3 I get some other anser
then the limit doesnt exist
outer rock
What does that conclude
drifting hornet
the limit exists if and only if the limit is same for all possible paths to the point
if you can find just one path for which the limit is different, the whole limit cant exist
the problem is that in practice you arent able to try out all the paths (because there are infinitely many) so u usually have to use polar coords and/or squeeze to prove that the limit exists
outer rock
@drifting hornet so during the question above when I put y=mx^2 I get 0 so its always better to try polar coordinates or just conclude limit exist
cobalt crow
Ok imagine I put y=x,y=x^2 but when I put y=mx^3 I get some other anser
usually, the paths you try are trying to make your life simpler, like you dont just try paths blindly but first like, try to find a path such that if you set y = something the single variable limit becomes simple to evaluate, you dont just try every single path, also, if two paths coincide but still the limit doesnt exist, then you will see that when you try to prove it using squeeze or something you cant?
outer rock
usually, the paths you try are trying to make your life simpler, like you dont j...
So if like these questions pops in my exam what shud be my intuition what path shud i go for first
cobalt crow
i said y = x because then you can factorize things from the numerator and the denominator
drifting hornet
<@951190447185874994> so during the question above when I put y=mx^2 I get 0 so ...
The strategy is generally this:
-
If you wanna prove that the limit doesnt exist, try out multiple paths. If one path has a different limit, you can conclude that the whole limit doesnt exist. If you cant find such path, go to step 2
-
If you wanna prove that the limit does exist, try polar coordinates and squeeze theorem. If it gets too complicated, it might be the case that the limit doesnt actually exist. Try returning to step 1 and try several other paths
cobalt crow
and it becomes a product of things instead of a sum
drifting hornet
So if like these questions pops in my exam what shud be my intuition what path s...
generally go for simple paths and for paths which will simplify the expression
drifting hornet
and always make sure the paths actually lead to your point
like you cant do x = y when the point is (0, 1) or you cant do y = x + 1 when the point is (0, 0)
outer rock
I see
During class test i actually tried y=x,y=x^2,y=x^3 all gained 0 so I wrote limits exist which is 0 but my friend tried y=mx^3 which gave another anser which is not 0
So I got cooked after test
My anser was wrong 😭
cobalt crow
you never tried to prove it
cobalt crow
During class test i actually tried y=x,y=x^2,y=x^3 all gained 0 so I wrote limit...
using y = mx is always better than y = x
outer rock
using y = mx is always better than y = x
I see
cobalt crow
The strategy is generally this:
1) If you wanna prove that the limit doesnt exi...
this are good guidelines to follow btw cobra
outer rock
So I just know y=something and polar coordinates methods
I haven't learned epsilon delta
Can I solve questions with both of them is that sufficient
cobalt crow
you cant just assume that because some paths work then the limit always evaluates to same thing
cobalt crow
So I just know y=something and polar coordinates methods
I haven't learned epsil...
squeeze theorem
no?
outer rock
No i haven't learned squeeze
cobalt crow
outer rock
I tried learning it but cudnt understand
cobalt crow
So I just know y=something and polar coordinates methods
I haven't learned epsil...
if after using polar coordinates and simplifying you get that the limit evaluates to sin(θ) + 2cos(θ) or something then you cant say anything because depending on thetha the limit will evaluate differently
outer rock
if after using polar coordinates and simplifying you get that the limit evaluate...
I see
outer rock
if after using polar coordinates and simplifying you get that the limit evaluate...
If like these situations come then squeeze is the only way
cobalt crow
no, what I am trying to say is, if you try to prove the limit f(x,y) always gets to same value but in the end you get something depending on thetha with no r, then you can see that depending on theta we get different results, aka we can't prove it always evaluate to the same value, then we go step 1 and trying to find a counterexample
outer rock
no, what I am trying to say is, if you try to prove the limit f(x,y) always gets...
Ok if that happens we have to try out y=something method
cobalt crow
try to find a few paths that will make your life simple
see what those single variable limits evaluate to
outer rock
Okey
I have an upcoming maths midsem exam so by solving pyqs will it be ok for me to pass
cobalt crow
So in these questions better to start with polar coordinates
for example, do you happen to know that sin(x)/x when x-> 0 is 1?
cobalt crow
yeah so, try to find a path that aligns with (x,y)->(0,0), such that we can use that to our advantage
zealous pendant
.solved
vale dockBOT
zealous pendant
(If you’re okay with that)
cobalt crow
for example, do you happen to know that sin(x)/x when x-> 0 is 1?
similarly sin(x^2)/x^2
dont want to give it away though
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tropic juniper
not sure how to approach this
gleaming kindle
not sure how to approach this
use appropriate paths
tropic juniper
how do i find what the appropriate paths are?
and how would i prove it discontinuous at every point?
ive only done it for a single point
wraith heart
how do i find what the appropriate paths are?
try simple ones. say ones parallel to the x and y axes
zealous pendant
Wouldn’t a better one be a circle centred at the origin of radius r=|z|? (Clockwise and anticlockwise)
Suppose for contradiction it is not discontinuous at every point, then pick any z which is continuous and then show it is actually discontinuous
vale dockBOT
@tropic juniper Has your question been resolved?
vocal panther
I figured how how to convert it to the kmap but i cant seem to figure out how they simplify it back into form.
rocky lotusBOT
Richard Mullin
hard nymph
Wait but what is you needed?
vocal panther
how it went fom the table there to m = xy + yz + xz
vocal panther
Take the vertical grouping
Like this
hard nymph
Yes
When it was 1 did you take the denied or not?
Anyway
What varies in the table in the vertical grouping?
vocal panther
What varies in the table in the vertical grouping?
Can you explain what you mean a bit more?
hard nymph
Which variable changes value
vocal panther
Which variable changes value
Which row?
hard nymph
Do you find that in the two boxes X for once it is 0 and once is 1?
vocal panther
No
hard nymph
Mmm
vocal panther
Its 1, 3 time in the second row
hard nymph
If you start from 011 and arrive at 111 what variable has it changed?
vocal panther
X
hidden valley
vocal panther
No, what am I canceling x with again?
vocal panther
Yes
hard nymph
What is not clear?
vocal panther
What was the table for if I just subtract the ending?
hard nymph
?
vocal panther
I think im still misunderstanding something
hard nymph
What
vocal panther
Whats this for?
hard nymph
What
vocal panther
.
I made it into a table but Im still confused on how to convert the table back into a simplified form of the equation, i understand that it has something to do with matching the 1s in the 0 and 1 part of the x part of the table but after that you jumped to solving it
hard nymph
Those are the expressions where the function is worth 1
You use kmap to get to the simplified form
xy+yz+zx
hard nymph
I wrote you above
vocal panther
Those are the expressions where the function is worth 1
Would that just make the same equation?
What are you looking at to make it different?
hard nymph
Let's do another method
Do you have the vertical grouping?
Write the two expressions
x'yz+xyz
= (x'+x)yz
=yz
This happens because precisely x changes from 0 to 1
The two horizontals
Starting from the one left left
The Y varies
So it will be xz
In the other change Z so it will be xy
So xy+yz+xz
vocal panther
so for the problem i have i do that twice?
vocal panther
is that what the circlings for?
vocal panther
Oh
vocal panther
So kinda like this?
hard nymph
Only the 1 you have to collect
Or collect all 0 or all 1
But the expression changes if you decide to collect 0
There will be a product of sums in that case
And if the variable will be 0 it will be not denied, while 1 denied
vocal panther
So how did u get xy and xz?
hard nymph
YZ you've already found it
vocal panther
I mean xz
hard nymph
You have to continue collecting until you have covered all 1
vocal panther
Collecting means im looking at everytime both rows are one right?
hard nymph
Yes
hard nymph
But why do you still have the 3 group in drawing?
hard nymph
I figured how how to convert it to the kmap but i cant seem to figure out how th...
In the figure as you can see, it groups 2
hard nymph
More precisely to powers of 2
.
For example, I had also had a 1 in box 001, you grouped them to 4, and the expression would become:
Vary y and x, then f (x, y, z) = z
It is always better to make the biggest groupings
vocal panther
Got it, thx
hard nymph
However less groupings you do better is (you always have to cover all 1 though)
honest stone
umbral isle
You just dont i guess?
vocal panther
Trapped here forever
vale dockBOT
@vocal panther Has your question been resolved?
umbral isle
Oh phew its alive again
river shale
.solved
open junco
Trying to solve this question any advice I cant get an answer
vocal tusk
sub y for 5-2x in 2nd eq
open junco
which gets me to 5x^2 - 20x+22=0 right
open junco
nah i cant tbh
vocal tusk
use quadratic formula
short solar
does it even work as a quadratic equation
ax^2 + bx + c = 0
if its in this form
its called a quadratic equation
open junco
ah ok I will try this, sorry really struggling with it
even tried google and it says theres no answer for it lol
open junco
is it because its negative?
umbral isle
Oh yeah b²-4ac would be negative in this case
short solar
yeah i mean the curves dont touch each other
open junco
brilliant
open junco
yeah maybe a typo tbh
short solar
do you have answer for this
open junco
I
short solar
maybe we can figure out where the question went wrong
open junco
ive wrote my working out but no answer is provided
through my googling theres similar questions floating around with x^2 - y^2 = 3 rather than +
short solar
yeah then you can conclude by stating due to discriminant being negative its kind of clear their are no real solutions
hyperbole
yeah possible
open junco
i thought I was having somekind of attack lol
well thank you very much guys settled my head
vale dockBOT
If you are done with this channel, please mark your problem as solved by typing .close
open junco
.close
vale dockBOT
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slate folio
Prove these inequalities are true with x>0
vale dockBOT
Prove these inequalities are true with x>0
frank salmon
3 arctan x < 3 arctan x+1
candid bolt
3 arctan x < 3 arctan x+1
its not so simple
candid bolt
arctan is strictly increasing.
slate folio
wait I might have an idea
frank salmon
arctan is strictly increasing.
So is it true?
frank salmon
ru allowed to differentiate
Nop
frank salmon
how do u know
In an innequality you cant differentiate the two sides but you can integrate
frank salmon
you can differentiate rhs - lhs
Yes if that what you meant
slate folio
and show it's >0
graphed it and it's false
vocal tusk
pretty sure there's an arctan sum formula as well
slate folio
as n tends to infinity the 2 sides both tend to 2π
frank salmon
pretty sure there's an arctan sum formula as well
Yes
vocal tusk
graphed it and it's false
so the question is wrong?
candid bolt
so the question is wrong?
no the derivative >0 hypothesis is false
vale dockBOT
vale dockBOT
Available help channel!
ripe saddle
thought it was A, answer sheet said it was B. I have no idea why
runic cargo
what did you write out
ripe saddle
I made two triangles given the dimensions in the question, subtracted their hypotenuses and it gave me A
stiff lily
can you show your work
cloud coral
hmm, do we assume flat earth
spark sluice
I made two triangles given the dimensions in the question, subtracted their hypo...
This should probably not have happened, you really just need to calculate the hypotenuse of one triangle
neon quartz
I made two triangles given the dimensions in the question, subtracted their hypo...
their hypotenuses doesn't necessarily form a straight line
runic cargo
try drawing a diagram it will seem more intuitive
cloud coral
okay, if round earth, then there would be a problem on how to add the distances, which wouldn't commute
neon quartz
thought it was A, answer sheet said it was B. I have no idea why
you can treat the location of the boat as coordinates
ripe saddle
hmm, do we assume flat earth
idk prolly
spark sluice
hmm, do we assume flat earth
I mean it doesn't matter because on the scale provided the earth is locally flat enough. If you want to be pedantic like this your should probably make your own help channel and ask your question there
neon quartz
then use Euclidean Distance
ripe saddle
then use Euclidean Distance
whats that?
runic cargo
just try and draw a diagram
cloud coral
maybe try overlaying the triangles to match at the radar station?
the radar station doesn't move, so that's a fixed landmark
spark sluice
maybe try overlaying the triangles to match at the radar station?
Do this and you will see why
You assumed that the hypotenuses are on the same line
neon quartz
whats wrong with what I did tho?
refer to my earlier message
neon quartz
their hypotenuses doesn't necessarily form a straight line
^
cloud coral
whats wrong with what I did tho?
probably not answering the question asked
yeah you can see that drawing it more accurately
sturdy tendon
Hello guys I need help with this question. I know it's in German but it translates to: "Match the graphical representations to the appropriate inequality (from A to F)!"
We did this in class but I still didn't understand it (the crosses on the letters are the answers cuz we already did it but I just don't get it). Please angel help me!!
sturdy tendon
It's 165 btw
neon quartz
Hello guys I need help with this question. I know it's in German but it translat...
solid lines means either greater/less than or equal to
dotted lines without equal to
unborn sable
If i have say y>=ax+b
i would have a solid line along y=ax+b, and the region would be everything above it (inclusive of the line hence the solid)
if it was > i would have a dotted line, and the region would be everything above it
marble arch
Replace the inequality with '=' sign, then check for appropriate values of x and y, that should give you the line
vale dockBOT
@sturdy tendon Has your question been resolved?
vale dockBOT
modest kettle
getting 0 now what
its not 0
verbal badger
You're getting 0/0, not 0. Big difference
modest kettle
yeah so what got you 0
marble arch
Factor out an x from the remaining ones and cancel it with the denominator
Then just put the limit
marble arch
Try what I mentioned
Open the ^5, the 1 cancels out, after which the x from the denominator would cancel out too
pliant palm
wait
pliant palm
Open the ^5, the 1 cancels out, after which the x from the denominator would can...
yea i did this
next
stark wedge
aight wait
stark wedge
yea i did this
show what you got from doing it.
marble arch
yea i did this
Just substitute x as 0
pliant palm
wait
stark wedge
before plugging any x in
marble arch
Not this, the step I had asked you to do
marble arch
Open (x+1)^5
This
neon quartz
notice that it's similar to the definition of derivative f'(a)
with f(x) = (x+1)⁵ and a = 0
marble arch
You still here @pliant palm?
vale dockBOT
@pliant palm Has your question been resolved?
vale dockBOT
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plucky pewter
hi, i need help with the task below.
vale dockBOT
hi, i need help with the task below.
plucky pewter
thanks in advance
N_1=(6/1)*f(-3)
N_1=0
but i am not sure how to solve for N_3
stark wedge
we're gonna need a reminder of what N_n and Ø_n mean
i suppose these stand for something in norwegian but i dont speak the language so i dont know
plucky pewter
we're gonna need a reminder of what N_n and Ø_n mean
i believe it means lower and upper darboux sum
N_n means lower darboux sum
while Ø_n means upper darboux sum
left would be N_n while the one to the right would be Ø_n
and n stands for the number of rectangles
does it make sense?
stark wedge
yes
you might want to sketch the graph of y=9-x^2
or at least note that it is a decreasing function so the upper and lower bounds on each rectangle can be found easily
plucky pewter
you might want to sketch the graph of y=9-x^2
plucky pewter
but why is this wrong:
N_3=(6/3)* f(-3)+(6/3) * f(-1) + (6/3)*f(1)
plucky pewter
augh ok this is going to be quite ass to write down
oh