gm
#help-36
lusty meadow
pliant shore
Just ask your question
final saddleBOT
@deft palm Has your question been resolved?
patent dome
How to properly use if then and if and only if. I understand the meaning but while writing proofs or definition I sometimes feel like if and only if should be used instead. For example a relation R on set A is transitive if for all x,y,z in A , xRy and yRz implies xRz. I think why double implication cannot be used. While writing this, I just got my answer. But is there any tip where I can speed up my thinking about which implication symbol is to be used so that my logic is more correct.
final saddleBOT
How to properly use `if then` and `if and only if`. I understand the meaning but...
desert mantle
using if instead of if and only if is never wrong
as long as you actually only need that direction
in your example you could also use if and only if. its just standard that for definitions we only use if
patent dome
So that's why some proofs explicitly write proofs from both direction I guess
desert mantle
well often the proofs are different depending on direction
patent dome
in your example you could also use if and only if. its just standard that for de...
The reason I find usage of if and only if wrong in that example, is the way for all x,y,z is used. suppose for all x,y,z , xRz implies xRy and xRz. Here y can actually if anything like stated in before (for all y)
patent dome
By if and only if we mean (if P then Q) and (if Q then P).
In original statement if xRy and yRz implies xRz. Here the first statement is true if we find such y and z which are in relation with x and then xRz is true.
But if we try to write it in backward, xRz implies xRy and yRz. We really don't know what y is. And by default it can be anything so it can be true or false and hence the implication will fail.
desert mantle
a relation is transitive if forall x,y,z in A, xRy and yRz implies xRz
patent dome
Ok I kinda don't know if I'm correct . I'm just confused.
desert mantle
I thought you were talking about that if
you could also put an if and only if instead of the if I wrote above
patent dome
Oh let me clear, I'm taking about the "if " which in my statement is used as "implies"
For x,y,z in A such that if xRy and yRz then xRz.
desert mantle
that one you cant just replace
patent dome
Replace what?
idle blaze
When you’re measuring the angle for a singular vector eg 6i + 3j do you measure the angle from the horizontal or vertical? As it was taught as the horizontal but I got it wrong today in a test and the working out was via the vertical ?!?
final saddleBOT
When you’re measuring the angle for a singular vector eg 6i + 3j do you measure ...
candid pulsar
When you’re measuring the angle for a singular vector eg 6i + 3j do you measure ...
hmm
can you show how you did it?
idle blaze
hmm
Well i’m walking right now so that’s a tad awkward but normally would you measure form the x or y axis
candid pulsar
Well i’m walking right now so that’s a tad awkward but normally would you measur...
from the x axis
in the counter-clockwise direction
near star
$\sqrt{a}-b$ is a root of $x^2+ax-b=0$ and $\sqrt{a}+b$ is a root of $x^2-ax-b=0$. find a+b
final saddleBOT
$\sqrt{a}-b$ is a root of $x^2+ax-b=0$ and $\sqrt{a}+b$ is a root of $x^2-ax-b=0...
near star
i've tried using vieta's formula with the quadratic formula but it didn't really work-
it's a question from a math competition that I remember LMFAO
soft zealotBOT
cheryy
brave shoal
i've tried using vieta's formula with the quadratic formula but it didn't really...
can i see your work
and are a and b integers or reals or etc
near star
the question didn't really specify, I believe this was all there was
near star
can i see your work
yes
so I found that if you use the quadratic formula the equation for the roots of the two equations just differ by the sign of the first term in the numerator
$\frac{-a\pm\sqrt{a^2+4b}}{2}$ and $\frac{a\pm\sqrt{a^2+4b}}{2}$
soft zealotBOT
cheryy
near star
and looking at the roots given in the question i assumed i use the plus
but just assumptions- i have no real evidence to pick that over the minus
so i got $\frac{-a+\sqrt{a^2+4b}}{2}$ and $\frac{a+\sqrt{a^2+4b}}{2}$
soft zealotBOT
cheryy
near star
oh yeah nevermind I didn't use the vieta's formula
i just tried equating these results to the roots given and doing some sort of simultaneous equation thing honestly i dont know what im doing
LMFAO i have no clue
maybe i should use vieta's formula in some way
near star
f(x)-2ax?
candid mulch
You can see it that way
But there’s a simpler one
near star
hmmm
candid mulch
Do you know odd and even functions?
candid mulch
Correct
candid mulch
$\sqrt{a}-b$ is a root of $x^2+ax-b=0$ and $\sqrt{a}+b$ is a root of $x^2-ax-b=0$. find $a+b$
soft zealotBOT
Afi
lunar skiff
$\sqrt{a}-b$ is a root of $x^2+ax-b=0$ and $\sqrt{a}+b$ is a root of $x^2-ax-b=0...
if its a root, then u can plug it into the equation x^2 +ax -b = 0
candid mulch
$f(x)$ has root $\sqrt{a}-b$ \
$f(-x)$ has a root $\sqrt{a} + b$ \
What about $-\sqrt{a}-b$?
soft zealotBOT
Afi
soft zealotBOT
cheryy
near star
**Afi**
since they need to have roots with the same absolute values as eachother im assuming $-\sqrt{a}-b$ would be a root of f(x)
soft zealotBOT
cheryy
candid mulch
That’s fair
near star
HMMM
lunar skiff
**cheryy**
but, i dont think u can 💯 assume that, since there might be two different roots for each quadratic
so, u dont know which one is equal to the other
near star
but ideally it would be $|q-k|=|k-q|$ and $|k+q|=|-k-q|$ right?
soft zealotBOT
cheryy
near star
im kinda suspicious because I didn't really use all the information i had
i just used one of the equations
oh wit
nevermind i made a mistake
uhh i'm not getting stuff out of what im doing LMFAO
i used vieta and subbed in the two roots i got for each of the equations
lunar skiff
so, this is the situation we are left at
candid mulch
For $f(x)=x^2+ax-b$ with roots $\sqrt{a}-b, -\sqrt{a}-b$
soft zealotBOT
Afi
near star
$\sqrt{a}-b-\sqrt{a}-b=-a$ thus $b=\frac{a}{2}$,
$(\sqrt{a}-b)(-\sqrt{a}-b)=-b$ thus $b^2+b=a$
and similarly for the second equation
we get $b^2-b=a$
but then solving these simultaneously we don't get anything useful
candid mulch
Yes you do
soft zealotBOT
cheryy
near star
oh what-
candid mulch
Actually
near star
😭
candid mulch
Gosh I’m getting confused
lemme look at the og question
near star
and we were expected to solve this in like <5 mins
actually <2 would be more correct
im gonna go kms
is there like a shortcut scam way of doing this
😭
well i should worry about that after i actually can get a legit answer LMFAO
lunar skiff
actually, we can see that the roots should be on the same distance from sqrt{a}, like this, and since sqrt{a} has to be positive
near star
actually, we can see that the roots should be on the same distance from sqrt{a},...
yeahh
but what can we do with that information
lunar skiff
so, we are left with two possible solutions
lunar skiff
,rcw
soft zealotBOT
near star
ah yes
candid mulch
Ok lemme write sqrta as c for brevity
x^2+ax-b=0 has root c-b
x^2-ax-b=0 has root c+b
Hence we know
x^2+ax-b=0 has roots c-b and -c-b
Apply Vieta’s formula
-a=c-b-c-b=-2b → a=2b
-b=-(c-b)(c+b) → b=a-b^2 →a=b^2+b
(a,b)=(0,0),(2,1)
near star
isn't it also possible that the root of the function below $\sqrt{a}$ is the first root of the function not the second
soft zealotBOT
cheryy
near star
Ok lemme write sqrta as c for brevity
x^2+ax-b=0 has root c-b
x^2-ax-b=0 has r...
yeah thats what i did above too
lunar skiff
this is if b is positive, if b is negative, the roots would change the signs of b
near star
this is if b is positive, if b is negative, the roots would change the signs of ...
what do you mean by that
lunar skiff
what do you mean by that
near star
oh yeah yeah
lunar skiff
however, a better way and a long way of solving this, is to just plug x as sqrt(a) - b into the equation x^2 +ax -b = 0
near star
however, a better way and a long way of solving this, is to just plug x as sqrt(...
yeah
lunar skiff
since its a root of the equation, then it achieves the equation
lunar skiff
and do the same with the other equation, then u will end up with two equations and two variables a and b
but i thought it might be overkill,
yoo, @candid mulch , soory but could u continue with him i need to go.
candid mulch
Hmm
near star
and do the same with the other equation, then u will end up with two equations a...
well i did get something but uh
$-4b\sqrt{a}+2a\sqrt{a}=0$
so it's
soft zealotBOT
cheryy
near star
$-2b+a=0$
soft zealotBOT
cheryy
near star
it's the same thing we got before
candid mulch
Yep
near star
back to ground zero baby
candid mulch
You have already solve it actually
near star
You have already solve it actually
as in?
candid mulch
**cheryy**
Here
candid mulch
b^2+b=a
2b=a
soft zealotBOT
cheryy
near star
well if we just get what b^2 is then we get a+b
candid mulch
You can straight up solve for (a,b) from there
near star
i tried that already :(
if you mean use them as simultaneous equations
i ended up getting like
b=3b or sm bs
near star
😭
candid mulch
b^2+b=2b
candid mulch
?
near star
and b^2-b=2b
and this as b^2-3b=0
you equate the two
and then b=3b yay kms
uk what maybe b=0 and a=0 and the answer is 0
LMFAO
candid mulch
Let c be sqrta for brevity
x^2+ax-b=0 has root c-b
x^2-ax-b=0 has root c+b
Hence we know
x^2+ax-b=0 has roots c-b and -c-b
Apply Vieta’s formula
-a=c-b-c-b=-2b → a=2b
-b=-(c-b)(c+b) → b=a-b^2 →a=b^2+b
We only have these two
a=2b
a=b^2+b
near star
and b^2-b=2b
i got this from the second formula
not formula what am i saying
uh
function
candid mulch
Hmm show me your work?
near star
$(b-\sqrt{a})(\sqrt{a}+b)=b$
soft zealotBOT
cheryy
near star
$b\sqrt{a}-a+b^2-b\sqrt{a}=b$
soft zealotBOT
cheryy
candid mulch
**cheryy**
You are missing a negative sign here
candid mulch
It’s just b^2+b=a at the end
candid mulch
…
a=2b
a=b^2+b
candid mulch
Solve for (a,b)
candid mulch
That’s it
near star
(╯°□°)╯︵ ┻━┻
(╯°□°)╯︵ ┻━┻
it was that
easy
im gonna
bye
(╯°□°)╯︵ ┻━┻
so a+b is 3
what the
bye
thank you bro
candid mulch
**cheryy**
You almost solve this at this moment
candid mulch
Aight I should get some sleep..
final saddleBOT
final saddleBOT
Available help channel!
junior mortar
I have been at this for 2 days now I dont get far enough with the info given here, its not for school or anything, I have been trying to relearn math since I dropped out of school when I was 13, and this is a year 7 book, it has this question, I would like a hit maybe or help understanding what I'm missing
final saddleBOT
I have been at this for 2 days now I dont get far enough with the info given her...
dire solar
are all the stars the same number or different numbers?
junior mortar
that is all the question, there is no more info about it
I can show you how far I get but maybe I'm wrong
junior mortar
the small 1s are the additional digits that will be added after the number is found
dire solar
the lower number is def 121
junior mortar
yeah I figured 100%
and that gives up the 8 uptop
since 1 x8= 8 at the third row
dire solar
right
junior mortar
and by having the 8 up there we need to make a 12 with the addition of the first number
and that has to be 2x2=4 then we get 12
dire solar
right
junior mortar
so that is as far as I can go any number I put in the last place makes no sense
I have tried every number
junior mortar
the problem I have is that we need to add 8 with a number to get 9 or 19, but you have to remember that we have a 1 left over form the 6+2+1+? which in any case gives a number bigger than 10 and less that 20
yes I tried 5 it gives 10+1 becomes 11. then 1+8+ the 1 leftover is not 9 ot 19
I feel like im missing something but still can figure it out
dire solar
interesting
hazy hemlock
Does star mean they are same?
junior mortar
I have tired 1-9 for the last place it doesnt work then I try to follow my steps back to see if something I did was wrong I cant find it
junior mortar
Does star mean they are same?
no
dire solar
Does star mean they are same?
no then it wont be solvable
junior mortar
stars are digits you need to find
hazy hemlock
Ok aight
dire solar
I have tired 1-9 for the last place it doesnt work then I try to follow my steps...
the question might be wrong?
junior mortar
the question might be wrong?
well why is it there then? its surprising
let me see if there is an answer book for this book
dire solar
sure
junior mortar
there is an answer but I dont want to look at it LOL
hazy hemlock
🙂
junior mortar
do you want to know
hazy hemlock
🙂 okay nice
junior mortar
thanks guys
hazy hemlock
🙂
remote crane
Can someone please help me find the I made a mistake? 
final saddleBOT
Can someone please help me find the I made a mistake? <:nozoomi:6518169868372541...
remote crane
(Assume it's a Riemann integral, and d(x-1) just means (x-1)' dx)
ah, lol. It's correct
nvm
.close
final saddleBOT
final saddleBOT
Available help channel!
late vault
i think the answer would be 108, because the angle is part or like attached to the center of the circle, but i dont want to submit it without being sure.
the fact that they added that 38 is RT feels like a red herring but i couldbe wrong.
final saddleBOT
i think the answer would be 108, because the angle is part or like attached to t...
late vault
does anyone know?
brave shoal
Can I see your work?
vital surge
looks like they just used the arc length = angle equality but with multiple misconceptions
brave shoal
Probably, because I got a different answer from OP's
late vault
Can I see your work?
i have no work
i thought that it would just be 108 but i feel like im misundersanding the prompt
brave shoal
Well do you know the formula for arc length?
late vault
not well no
this looks nothing like what i have been learning
oh wait i know the first one
oh wait
s = 108/360 * 76π
s = 27/90 * 76π
90s = 27 * 76π
90s = 2052π
s = 71.592
late vault
Well do you know the formula for arc length?
71.592?
brave shoal
Yeah
final saddleBOT
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Available help channel!
mortal berry
why can't I replace the numerator with $e^x$ in this limit? $$\lim_{n\to\infty} \frac{(1+\frac{1}{x})^{x^2}}{e^x}$
final saddleBOT
why can't I replace the numerator with $e^x$ in this limit? $$\lim_{n\to\infty} ...
soft zealotBOT
pocoyo
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
mortal berry
Then it would give me 1 but it's wrong
final saddleBOT
@mortal berry Has your question been resolved?
mortal berry
<@&286206848099549185>
mortal berry
<@&286206848099549185>
final saddleBOT
@mortal berry Has your question been resolved?
slim barn
Wait so what do you mean?
Where is e^x being put?
it is not equal to the numerator, why would you substitute it?
final saddleBOT
@mortal berry Has your question been resolved?
vital surge
why can't I replace the numerator with $e^x$ in this limit? $$\lim_{n\to\infty} ...
same reason you can't replace 1/x with 0
limits don't work that way; they're indeterminate for a reason
steep lion
final saddleBOT
steep lion
how do i do part (iii)
I have the partial derivatives
but how do I show it's continous
distant ridge
I don't know about this topic but maybe w polar coordinates when (x,y) --> (0,0)
steep lion
I don't know about this topic but maybe w polar coordinates when (x,y) --> (0,0)
but how
lyric summit
how do i do part (iii)
lyric summit
hmmm
have you ever computed partial derivative at the given point using the defintion ?
lyric summit
hmmm
you must use such a defintion:
if you have not used it before, then your exercise is not for you
lyric summit
how would you do that when it divides by 0
I may also give you a hint for the right side of the equality ive written for you, use there a squeeze theorem
steep lion
have you ever computed partial derivative at the given point using the defintion...
I have
I computed the partial derivative using that method in part (i)
and got that the partial derivative is 0
steep lion
Right side?
lyric summit
yes, you said, you did left side
correct?
so now, right side
if both sides will be equal
then you write, the partial derivistive with respect to x is continuous
at (0,0)
steep lion
then you write, the partial derivistive with respect to x is continuous
what is w?
lyric summit
so what ?
soft zealotBOT
Daniel
lyric summit
use your mind
steep lion
use your mind
yeah
lyric summit
i cant solve all for your i am not your tutor, i can only give you a hint
the rest you should compute in an analogous way
steep lion
I'm abit lost because of the divide by 0
lyric summit
I'm abit lost because of the divide by 0
🙂 by the way, inside your sinus function orignally in your paper, there is fourth power of y
steep lion
oh damn, I didn't see that fourth paper
lyric summit
what do you mean by this?
look below:
that is your first derivative with respect to x
lyric summit
the expression written before minus, goes to 0, as ishowed you above
yes i can
and you said that you did in i)
so now you shud do the rest
steep lion
and you said that you did in i)
I just applied the limit definition for part (i)
steep lion
okay I see
steep lion
and please continue
you can't conclude that the one to the right of the minus can go to 0 though
its divided by 0
part (i) we know that at (0,0) its 0
for the partial derivative
lyric summit
you can't conclude that the one to the right of the minus can go to 0 though
lol
i showed you :
and now , cope with the rest
lyric summit
the one onthe right is your task
steep lion
Could I please get a hint
lyric summit
either you show it goes to zero, or it does not reach any limit due to non-existence
steep lion
hmm, but the question ask you to show it's continous then it must be continous
lyric summit
it is written there : Is ...
steep lion
it is written there : Is ...
lyric summit
🙂
steep lion
wait so you already gave the solution?
lyric summit
i gave you a half , not all
steep lion
from here?
lyric summit
yes
the first component goes to zero, but the second one is also important
if seocnd ones goes to zero, then you wil be happy
then you say: it is continuous
but if the second component in aspect of limes does not exist
then tyoru partial derivative with respect to x is not continuous at (0,0)
so this part is a s you feel, very important now
steep lion
ahh, so it's not continous
lyric summit
but it needs proof
steep lion
because it cannot tend to 0
lyric summit
it is not enough
steep lion
hmm, I think in this case you choose a sequence
lyric summit
yes, two sequences
steep lion
i mean a counter example
lyric summit
are needed
steep lion
yeah
steep lion
(1/k^2, 1/k)
k approaches to 0
wait no
okay so one squence is (0, k)
as k approaches to 0
this would be equal to 0
lyric summit
yeah
look below:
I have offered the sequence, that when we apply it into our second component, goes to positive infinity
in an analogous way, you can offer the second sequence, and if this second one, appropriately selected, goes to not positive infinity, then the limit of the second component does not exist
for example, try this one:
final saddleBOT
@steep lion Has your question been resolved?
steep lion
look below:
was finding a counter example this entire time
steep lion
aren't these quite complicated sequences
lyric summit
aren't these quite complicated sequences
the presence of cosinus requires such an approach
final saddleBOT
@steep lion Has your question been resolved?
final saddleBOT
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Available help channel!
zenith flame
#help36
zenith flame
I need help with the graph
I don’t know if I did it right
<@&286206848099549185>
.close
final saddleBOT
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indigo plover
final saddleBOT
indigo plover
i dont wannt to sound like idiot
but like its impossible to do this without calculator?
Like they don't even allow us to use calculator in tests but then in hw they put this stuff ; (
yes
like decimals number
o
they said round 3 decimals
yeah
if i did that
i can do 1-1/2?
then like do 1/3^1/2?
prfct thnxs
.close
final saddleBOT
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slim saddle
final saddleBOT
silver maple
pigenhole principle
royal gust
Let's start small and work our way up. Is it possible for k to be less than 2?
That is, every single person has a unique number of children, a unique first and last initial, and a unique birthday?
slim saddle
Let's start small and work our way up. Is it possible for k to be less than 2?
...
im not sure..
final saddleBOT
@slim saddle Has your question been resolved?
final saddleBOT
@slim saddle Has your question been resolved?
final saddleBOT
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lime glacier
Is there a paper that shows me the proof of the maclaurin series expansion of tan(x) ?
I found this stackexchange question, but I'd rather use a peer reviewed paper since I want to cite it as a reference for an assignment
https://math.stackexchange.com/questions/2098941/bernoulli-numbers-taylor-series-expansion-of-tan-x
final saddleBOT
Is there a paper that shows me the proof of the maclaurin series expansion of ta...
@lime glacier Has your question been resolved?
pliant shore
Is there a paper that shows me the proof of the maclaurin series expansion of ta...
You should do the steps yourself, no need for a peer-reviewed paper
(And of course cite Maths SE since you're using it)
Also this is a really bad source if you just want the Maclaurin series
Wait no actually, I see what they mean
lime glacier
Yeah I just want the maclaurin series, I don't really need to derive it myself, I just need to show that its "correct"
pliant shore
But https://math.stackexchange.com/questions/2314942/finding-the-nth-derivative-of-functions-in-particular-y-tanx would be better imo
pliant shore
Yeah I just want the maclaurin series, I don't really need to derive it myself, ...
Oh ok that's interesting
Yeah just show that the 3rd derivative at x = 0 divided by 6 matches with the coefficient of x^3
And so on
lime glacier
hmm alright, I wanted to avoid doing anything myself lol, since my goal is just to analyse some properties (eg: it becomes more accurate with more terms etc).. and the assignment doesn't ask us for a derivation XD
pliant shore
Ah thank god
lime glacier
do you reckon stackechange is an appropriate source for a uni assignment?
lime glacier
But https://math.stackexchange.com/questions/2314942/finding-the-nth-derivative-...
thanks for this tho
pliant shore
No worries
lime glacier
alrighty thanks
pliant shore
No worries
tepid ivy
Hello, I have difficulty with integrals with exponentials, particularly the one that I find at the end of the differential equation, so I apply the Variation of parameters method, for example I end up with this
https://www.wolframalpha.com/input?i2d=true&i=Integrate[\(40)-2x-Power[x%2C2]\(41)*Power[e%2C-\(40)x%2BDivide[Power[x%2C2]%2C2]\(41)]%2Cx]
*
final saddleBOT
Hello, I have difficulty with integrals with exponentials, particularly the one ...
tepid ivy
could you tell me how to do this for example and in a more general way?
final saddleBOT
@tepid ivy Has your question been resolved?
final saddleBOT
@tepid ivy Has your question been resolved?
valid marsh
Let us call a whole number "lucky" if its digits can be divided into two
groups so that the sum of the digits in each group is the same. For
example, 34175 is lucky because 3 + 7 = 4 + 1 + 5. Find the smallest 4-
digit lucky number, whose neighbour is also a lucky number (i.e. the
whole number next to it is a lucky number as well).
final saddleBOT
Let us call a whole number "lucky" if its digits can be divided into two
groups ...
valid marsh
.close
final saddleBOT
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open heath
final saddleBOT
open heath
I've calculated the support reactions and the point force for W
for the triangle suppport, its 4Kn, and for the circular, 5kN
the point force for W is 6kN
I just dont understand how to find the shear force
final saddleBOT
@open heath Has your question been resolved?
ionic charm
Hello, recently I am learning Fubini's theorem. Would someone like to shed light on how to prove the integrability part?
final saddleBOT
Hello, recently I am learning Fubini's theorem. Would someone like to shed light...
ionic charm
final saddleBOT
@ionic charm Has your question been resolved?
final saddleBOT
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bold charm
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rain star
can someone explain exercise b?
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can someone explain exercise b?
rain star
this is jpw far I get
but the second term of the 3rd curve has to be multiplied by 2 for some reason
I want to solve this question systematically but I dont understand what im missing
tranquil pine
1+1=2
royal gust
So the path is:
x = t
y = 2t
And so,
dx = dt
dy = 2dt
Between 0 < t < 1
Make sure to be writing your parametrization out.
rain star
I could write a position vector in terms of t for each case right?
royal gust
That's the third path
rain star
yeah but I cant I write it also for path 1 and 2?
royal gust
ye
rain star
wait for the third path
x=t doesnt make sense
because when t=0, x should be 1
and when t=1 x should be 0
so it should be x=1-t
isnt this the complete parametrization then for each length?
@royal gust can you check my parametrizations if you have time?
royal gust
True! Good call, I did go the wrong way
Note my answer would have been the negative of the correct answer.
I like your parametrization
@rain star
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rain star
but upon evaluating i found for the third length -8/3 even though it should be -10/3
any clue?
the first and second integral being 0 and 4 respectively are correct
@royal gust
royal gust
Just from this work, it looks like the ∫ xy dx went missing
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dense ocean
Let ( f ), ( g ), and ( h ) be three arithmetic functions, and ( f^{\sim} ), ( g^{\sim} ) be the inverses of ( f ) and ( g ) respectively. Which of the following statements are correct? Check each correct proposition.
\begin{itemize}
\item If ( f ) is totally multiplicative, then ( f^{\sim} ) is also totally multiplicative.
\item If ( f ) takes values in ( \mathbb{Z} ), then ( f^{\sim} ) also takes values in ( \mathbb{Z} ).
\item If ( f ) takes values in ( \mathbb{Q} ), then ( f^{\sim} ) also takes values in ( \mathbb{Q} ).
\item The inverse of ( fg ) is ( f^{\sim}g^{\sim} ).
\item If ( f ) and ( g ) are multiplicative, then ( f \cdot g ) is multiplicative (where ( f \cdot g ) is the function associating to each ( n \in \mathbb{N}_0 ) the number ( f(n) \cdot g(n) )).
\item The inverse of ( f \cdot g ) is ( f^{\sim} \cdot g^{\sim} ).
\item If ( f ) is totally multiplicative, then ( f \cdot (gh) = (f \cdot g)(f \cdot h) ).
\end{itemize}
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Let ( f ), ( g ), and ( h ) be three arithmetic functions, and ( f^{\sim} ), ( g...
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lapis knot
how do i show that these limits DNE? for 1b for example
royal gust
Take two paths through the limit point, show they don't come to the same thing
lapis knot
is this what i'd do?
tiny gorge
how do i show that these limits DNE? for 1b for example
usual strategy is to find a path along which the limit doesn't exist, or two paths along which the limits exist but are different
lapis knot
usual strategy is to find a path along which the limit doesn't exist, or two pat...
how would i do that?
royal gust
I don't know how you turned 2x²/x⁴ into 2x⁴/2x⁴
tiny gorge
yea line 3 makes no sense
line 2 you have 2x^2 / x^4, which is 2/x^2
which has no limit as x->0, boom, done
there are actually two lines that are wrong
line 2 isn't the same as line 3, and line 3 isn't the same as line 4
lapis knot
line 2 you have 2x^2 / x^4, which is 2/x^2
and then do i cancel the x^2?
lapis knot
there are actually two lines that are wrong
line 2 isn't the same as line 3, and...
yeah pls ignore that idk what i was doing
tiny gorge
you mean cancel the x^2's in 2x^2 / x^4?
yes, that gives you 2/x^2
which blows up as x->0
so the limit DNE
lapis knot
which blows up as x->0
okay and what about when x=0 and y approaches 0?
tiny gorge
okay and what about when x=0 and y approaches 0?
your work there is fine
you got as far as $$\lim_{y \to 0} \frac{0}{y^2}$$
soft zealotBOT
Bungo
tiny gorge
which equals what?
lapis knot
which equals what?
undefined
tiny gorge
no
lapis knot
but isn't it 0/0
tiny gorge
if y is nonzero, what is 0/y^2 ?
tiny gorge
but isn't it 0/0
you don't just plug in y=0, that's not how limits work in general
lapis knot
if y is nonzero, what is 0/y^2 ?
so if its 0/any nonzero number then its 0
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shell panther
How do u solve question 3?
final saddleBOT
How do u solve question 3?
wintry kindle
but the solution is there
you know that the ride lasts 3.25 minutes
= 3.25 * 60 seconds = 195 secods
and after 10 seconds, a rotation is complete
meaning during that ride there are 19.5 rotations
shell panther
Yea
wintry kindle
after the 19 rotations, Elena just ends up where she was
shell panther
But at what degree will the ride end
wintry kindle
and 0.5 rotations later, she will have travelled another half circle
shell panther
But she starts at 210
shell panther
Ok
wintry kindle
which part confused you
shell panther
For the how far she traveled, do I just write 19.5 rotations
wintry kindle
likely in meters instead of rotations
how long do you have to travel for one rotation?
shell panther
10 seconds
wintry kindle
Do I write 2pi+30degrees
either do degrees or radians
in degrees: 30°
in radians: pi/6
wintry kindle
well feet
I refer to distance
the radius is 20 feet, meaning the a full circle is 2 * pi * (20 feet) long, so after one rotation you travel 2 * pi * (20 feet)
meaning after 19.5 rotations
you travel 19.5 * 2 * pi * (20 feet)
shell panther
So it’s 19.5 times 2pi times 20ft
wintry kindle
ys
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cyan night
I assume it's not as easy as just saying: Using Lagranges thm, I know 3 divides 33 as well as 2 divides 12?
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I assume it's not as easy as just saying: Using Lagranges thm, I know 3 divides ...
versed cave
That is exactly what you do. The order of G is 12, so the order of any element g in G will divide 12
cyan night
But does that mean that G must have an element of order 2. It's not just a possibility its a definite?
versed cave
I have a suspicion that G is cyclic and thus g^6 is in <g> = G for a generator g
So (g^6)^2 = g^12 = g^0 = 1
So g^6 would be of order 2
If that makes sense
tiny gorge
But does that mean that G must have an element of order 2. It's not just a possi...
this is cauchy's theorem
if p is a prime that divides |G| then G has an element of order p
a group of order 12 need not be cyclic or even abelian, for example A_4 (the alternating group on a set of 4 elements) has order 12 and is not abelian
cyan night
Doesn't Cauchy's thm only work for Abelian groups. We don't know if G is Abelian for the first problem
For the second I found online that claims that a = inv(a) thus a^2 = e but I don't see how that turned out
tiny gorge
no, cauchy's theorem holds for all finite groups
btw abelian groups have subgroups of all possible orders (if n is any divisor of |G| then G has a subgroup of order n)
for nonabelian groups that's not necessarily true
cyan night
Wait so does that mean that I could just use that for the second problem as well since 2 divides 12 and 2 is prime?!
tiny gorge
sure
assuming you have cauchy's theorem available
did you already cover it in your class?
cyan night
For some reason this is in my text book but I'm not sure why they limit it to only Abelian groups when the general theorem is just better
In conclusion I don't think that we learned it yet unfortunately so I have to find some other way around
tiny gorge
oh
they haven't given you cauchy's theorem for nonabelian groups?
which book is this? i can take a look to see what the context is and maybe get some idea what tools they want you to use for this problem
cyan night
We've just gotten through chapter 7
tiny gorge
ohh
so you haven't seen any form of cauchy's theorem yet?
haha one moment i only have a pdf of the 7th edition
lemme see if i can find the 10th
cyan night
I have a pdf of the tenth I can link a drive maybe. I'm not sure if can do that
cyan night
Take your time. I have all night to finish these two problems so I'm in no rush
As well as one other but I think I have it figured out
tiny gorge
Take your time. I have all night to finish these two problems so I'm in no rush
ah, it's so nice to see someone in the help channels who didn't late until the last 5 minutes to do their homework haha
cyan night
Well I was working on it yestarday in here but didn't find much success. And I knew this homework was going to be a bitch so I figured I'd grind it out now
tiny gorge
smart
ok, downloaded
so chapter 7 is cosets and lagrange's theorem
haha holy shit, he doesn't do the nonabelian version of cauchy's theorem until chapter 23
this is such a different organization compared with how i learned group theory haha
cyan night
Damn. It would have been so nice to use Cauchy's here
tiny gorge
yea, it makes it pretty much trivial
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cyan night
I'm doing something like this
And it works for everything except for 3 but I don't know what this means
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@cyan night Has your question been resolved?
tepid anvil
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tepid anvil
why is the lowest bounds of dr 1/2?
if the cardioid has some area going past the y-axis?
pliant shore
if the cardioid has some area going past the y-axis?
The inner curve is the circle of radius r = 1/2
The outer curve is the cardioid
For theta between -2pi/3 and 2pi/3
tepid anvil
The inner curve is the circle of radius r = 1/2
right right
wait im trippin
nvm
.close
lofty comet
Idk how to do 1-5
final saddleBOT
Idk how to do 1-5
crude wagon
Which part?
lofty comet
1.8
crude wagon
1.8 degrees?
steep sequoia
How much degrees is 4pi/7?
I'll give you a hint.
$\pi$ radians is equivalent to 180 degrees. Can you work out your question now?
lofty comet
Yea
soft zealotBOT
Enemagneto
lofty comet
102.9
crude wagon
right
lofty comet
Nice
crude wagon
which quadrant is that going to be in?
lofty comet
Uhhh
lofty comet
180-102.9
crude wagon
lofty comet
77.1
crude wagon
yes, this should help you roughly plot your radial line with angle
lofty comet
crude wagon
The angle you drew is less than 90
lofty comet
crude wagon
but 4pi/7 ~102.9>90
crude wagon
yes, but mark the reference angle
lofty comet
What u mean
crude wagon
This is good
lofty comet
crude wagon
which one is your reference?
lofty comet
Idk 💀
crude wagon
A reference angle is an acute angle enclosed between the terminal arm and the x-axis. It is always positive and less than or equal to 90 degrees.
lofty comet
77.1
lofty comet
So 77.1?
crude wagon
yes
lofty comet
Ok so I don’t make it a negative
lofty comet
Yea
crude wagon
-90 is the negative y axis
-180 is negative x axis
-270 is positive y axis
This should help
lofty comet
crude wagon
-360 is positive x axis
so where do you think -310 lies?
as you move clockwise you become more negative
lofty comet
crude wagon
alright
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rustic token
Can anyone please assist me with this
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Can anyone please assist me with this
silver maple
consider 1/x^2
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tranquil charm
$(1+x)\sum^{\infty}{k=0}(-1)^k\frac{x^k}{k!}-(1-x)\sum^{\infty}{k=0}\frac{x^k}{k!}=\sum^{\infty}{k=0}((-1)^k-1)\frac{x^k}{k!}-\sum^{\infty}{k=0}((-1)^k+1)\frac{x^{k+1}}{k!}$\
how come these two are equal?
final saddleBOT
$(1+x)\sum^{\infty}_{k=0}(-1)^k\frac{x^k}{k!}-(1-x)\sum^{\infty}_{k=0}\frac{x^k}...
inland kettle
...?
i suggest you use desmos
desmos uses latex too, so you can just enter the expression and paste it here
soft zealotBOT
Slowaq
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tender mica
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tender mica
can someone check is this division is correct
signal vector
Sign mistake
signal vector
Yes
tender mica
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rustic wedge
Let $P_n = \prod_{i=0}^{n} 2i + 1$
final saddleBOT
Let $P_n = \prod_{i=0}^{n} 2i + 1$
soft zealotBOT
rak³en
rustic wedge
From here, its easy to find the recurrence relation $P_{n+k} = P_n \cdot P_k$
soft zealotBOT
rak³en
rustic wedge
Is there a way to obtain a closed form solution in terms of n for P_n from the recurrence?
desert mantle
are the missing brackets intentional?
I assume no. its just the double factorial, isnt it?
rustic wedge
no, they r not
rustic wedge
I assume no. its just the double factorial, isnt it?
yeah, any solution in terms of the regular factorial tho?
cobalt lodge
does it actually satisfy the recurrence relation? ._.
if it holds why isn’t Pn immediately (P1)^n
desert mantle
google it
rustic wedge
does it actually satisfy the recurrence relation? ._.
if it holds why isn’t Pn i...
why wouldnt it..?
OH
it wont start from 0 🤦♂️
.close
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main nexus
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main nexus
so what i did so far is
p(x) = (x^2-5) q(x) +x+4
p(-x) = ((-x)^2-5) q(-x) -x + 4
but when i add it together i get
actually idek what to do next really
barren hound
(-x)² can be simplified
barren hound
well what is your goal
main nexus
find what its equal to
eager drum
Can someone help me find the piecewise for this
main nexus
yikes
eager drum
Once yall are done
main nexus
😭
barren hound
!occupied
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barren hound
find what its equal to
find what what is equal to?
eager drum
Someone else is already using this help channel. If you need help with a questio...
OH MB GUYS
main nexus
.
barren hound
yeah I see that
main nexus
p(x) + p(-x)
barren hound
what do you get for that?
main nexus
um
barren hound
and you're trying to find a remainder, correct?
main nexus
(x^2-5)q(x)+x+4 + (x^2-5)q(-x) -x+4
main nexus
and you're trying to find a remainder, correct?
yeah
barren hound
(x^2-5)q(x)+x+4 + (x^2-5)q(-x) -x+4
can you simplify this at all
main nexus
yes
barren hound
what form are you hoping to get it in?
barren hound
that seems like a good goal yes
main nexus
not sure how
barren hound
what's quotient form?
barren hound
yeah
barren hound
sure
barren hound
what does the quotient need to be?
just a polynomial, right?
and is q(x) + q(-x) a polynomial?
barren hound
then yeah I'd say that works
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lament tulip
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inland kettle
!status
final saddleBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
c
inland kettle
hm
what'd you get for part a and b?
explain your reasoning
@lament tulip
bro
inland kettle
b is 29!/2^14 x 14!
...?
sorry didnt really get that
can u explain part b via latex or smth
autumn flare
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autumn flare
Can someone help me solve questions number 16?
robust fulcrum
Can someone help me solve questions number 16?
Take the a in the denominator above
Write 1-a² as (1-a)(1+a)
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autumn flare
Take the a in the denominator above
I don't understand
robust fulcrum
I don't understand
From, 1/a - a, take LCM in the numerator
autumn flare
From, 1/a - a, take LCM in the numerator
Can you show process?
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granite tapir
@tired knot Fostering algebraic thinking a guide for teachers, grades -- Driscoll, Mark J -- 1999 --
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<@246345114350387200> Fostering algebraic thinking a guide for teachers, grades...
granite tapir
also check the The Art of Problem Solving books, perhaps
granite tapir
you can also make up your own puzzles, use computer agent to make variations
tired knot
nice of you :3
granite tapir
np