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#help-17
thick niche
vocal sleetBOT
vocal sleetBOT
Available help channel!
honest leaf
How do i make a formula that includes a compounding component. For instance say i have the equation y=12x-5
And i want to know the sum of all value ys between x= 5 and 10.
vocal sleetBOT
How do i make a formula that includes a compounding component. For instance say ...
delicate bison
Please clarify your concern.
dark kiln
find the value manually so we know what you;re looking for
make an easier example if this is too hard
honest leaf
Im not even sure who you are talking to frog
twin meteorBOT
jesko
tardy lava
Why no latex?
twin meteorBOT
jesko
delicate bison
Please use latex. @honest leaf
honest leaf
So, i just recently learned about the summation symbol as i was looking for an answer, but it seems like its no different than brute forcing the solution. Is there a way to do it that does not require finding each value by hand? Or am i missing something about the summations use
tardy lava
So, i just recently learned about the summation symbol as i was looking for an a...
You don't have to brute force it, rather learn about the Sigma operator
dark kiln
it's 12(1+2+3+4...+10) − 12(1+2+3+4) − 5×6
honest leaf
You don't have to brute force it, rather learn about the Sigma operator
Right but learn what about the operator. As far as i can see from my investigation, it simply is notationally better. I cannot seem to find anything about how to use it to reach the final answer without still having to manually find the values between the index and stopping point
dark kiln
,calc 12(5.5)10 - 120 - 30
twin meteorBOT
Result:
510
Give me a moment
twin meteorBOT
jesko
twin meteorBOT
jesko
honest leaf
Barely but ill stare at it until it makes sense, thank you
tardy lava
Just so I know, can you give me an idea about how much experience you have in math
honest leaf
Up to discrete mathmatics, but wether or not i remember it is a different story
honest leaf
how do you quantify that
By giving a course. Same as if i said up to trig 1. -.-
tardy lava
I'm in high school so I know basic math and calculus. I have no knowledge of Discrete Math. But I can help you in Sequences and Series.
honest leaf
You can think of discrete math as calculus adjacent id say. Our courses just never got into things like the summation symbol
I also didnt follow it particularly well so im a bad example of it XD
tardy lava
If you can calculate very fast the brute force can help. But knowing some standard summation comes in handy.
You can learn all about it in Sequences and Series
I suggest you to invest some time in watching video lectures or books or a teacher if possible
zenith dagger
Discrete math, is that like learning probability, unions, how to calculate interest, doing matrices, things like that?
vocal sleetBOT
@honest leaf Has your question been resolved?
honest leaf
Discrete math, is that like learning probability, unions, how to calculate inter...
the best description i can come up with is its like using math to calculate an optimal path along a series of points.
vocal sleetBOT
the best description i can come up with is its like using math to calculate an o...
@honest leaf Has your question been resolved?
fleet flume
can someone confirm the answer for part d?
vocal sleetBOT
can someone confirm the answer for part d?
fleet flume
here is a correct diagram (the line that goes above is Cyclist Q)
here is information about cyclist P:
i think i can do it by setting their displacement equal to eachother maybe? and then use u + v /t a few times?
i know that p will have a displacement of 3.2* 12 + 120 I believe
so p displacement when it is passed by Q is 158.4?
so then i have 3.6 * T /2 + 3.6 * (48 -t) = 158.4 I believe
i cant tell if I should use 48 (from 54-6) or 54...
but when rearranging i get 14.4 = 1.8t
vast shale
0.5(45+54)*3.2=0.5(48+48-T)*3.6
T=8
vast shale
Well done, np
fleet flume
where does 0.5(45 + 54)3.2 come from?
this isnt my exam sheet and i calculated part a in a different way
they did this for part a, and i dont see how
vast shale
Area of first trapezoid = Atea of second trapezoid
Area here being the displacement since it's a velocity against time graph
fleet flume
the way i calculated part a was v*9/2 + v * 33 = 120
vast shale
Didn't check the previous parts just used the graph you drew
fleet flume
but i dont see how (42 + 33) * v /2 = 120
yeah... id like to stay away from using the graph if possible
fleet flume
i think i understand part d
vast shale
Or do you have other questions before that
fleet flume
if you could understand just part a?
vast shale
Alr lemme read that
fleet flume
the written solution for it? where it comes from?
fleet flume
they did this for part a, and i dont see how
.
fleet flume
the way i calculated part a was v\*9/2 + v * 33 = 120
my calculation for part a was this:
which is obviously equivalent but i dont know why this 42 + 33 /2 thing works..
vast shale
Do you not understand how to solve it in the first place or is it a certain form you don't understand?
vast shale
but i dont see how (42 + 33) * v /2 = 120
This form is the area of the trapezoid
fleet flume
oh, is that all it is? they didnt use some suvat equations to get it?
i can ask them next week..
vast shale
Suvat only works for objects changing velocity
What they did was just get the area under the graph of the journey AB of cyclist p
Not sure why you try avoiding using graphs tho, it saves alot of time in many cases
fleet flume
i see, thanks
vast shale
My pleasure
vocal sleetBOT
@fleet flume Has your question been resolved?
scenic ravine
Using the definition of the convergence of a sequence, prove that $a_n= \sqrt{1}{\sqrt{n}}$ converges to 0
\
We first define what it means for a sequence to converge. A sequence $(a_n)$ is said to converge to a real number $a$ , if for every positive number $\varepsilon$, there exists a $N \in \N$ such that whenever $n \geq N$, $\abs{a_n -a} < \varepsilon$
\
We this need to find a $n$, such that $ \frac{1}{\sqrt{n}} < \varepsilon$. This gives is $n = \frac{1}{\varepsilon^2}$.
\
We thus chose $n \geq \frac{1}{\varepsilon^2}$.
\
We chose $N = \frac{1}{\varepsilon^2}$
\
We thus get $\abs{a_n-a} < \varepsilon$
vocal sleetBOT
Using the definition of the convergence of a sequence, prove that $a_n= \sqrt{1}...
twin meteorBOT
ƒ( wai ina teacup)= I don't know
scenic ravine
I feel this is missing something
sly sierra
the reasoning is fine, you might want to state it a bit more clearly
like mention that a=0 for example
scenic ravine
Hmm, okay. Thanks
sly sierra
also maybe say what you're choosing for N
scenic ravine
I've probably asked this before , but in general hhow to chose n 😭 , it feels like black magic
sly sierra
(it's implicit that it's 1/eps^2, but it's good to say it)
in general the goal is to choose N that makes the desired condition |a_N - a| < epsilon true
and then verify that it remains true for all n >= N
scenic ravine
so reverse engineer a value of $\varepsilon$
twin meteorBOT
ƒ( wai ina teacup)= I don't know
sly sierra
no you have to take epsilon as given
some positive constant
and you find an N (which is a function of epsilon) that works
twin meteorBOT
ƒ( wai ina teacup)= I don't know
scenic ravine
so we have to find a $N$ , such that if $n \geq N$, $a_n-a|< \varepsilon$$
twin meteorBOT
ƒ( wai ina teacup)= I don't know
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
scenic ravine
$\abs{ \frac{n+1-n}{n}} \leq \varepsilon$
twin meteorBOT
ƒ( wai ina teacup)= I don't know
twin meteorBOT
ƒ( wai ina teacup)= I don't know
viral copper
right
scenic ravine
So this is my proof :
\
Chose $N > \frac{1}{\varepsilon}$
\
We then have $\abs{a_n-a} < \varepsilon$
twin meteorBOT
ƒ( wai ina teacup)= I don't know
viral copper
Right I assume you can flesh it out and be precise
viral copper
sure if you want to
scenic ravine
I mean I just looked at abbott's proof, what am I missing
viral copper
**ƒ( wai ina teacup)= I don't know**
You're saying this is your final proof?
scenic ravine
yeah
viral copper
You haven't defined anything
viral copper
I mean I just looked at abbott's proof, what am I missing
Abbott defined ε and N here
viral copper
Pretty much any proof will start with defining ε and N
viral copper
**ƒ( wai ina teacup)= I don't know**
additionally you need to say what a_n and a is here
scenic ravine
Got it
viral copper
but yeah you did the brunt of the work fine which was finding a suitable N
vocal sleetBOT
vocal sleetBOT
Available help channel!
tired matrix
what is the inverse of a factorial?
vocal sleetBOT
what is the inverse of a factorial?
tired matrix
like
n!=720
n=f(720)
what would be f?
and how would you get 6 from it?
i get that you divide by (5)(4)(3)(2)(1)
limber falcon
I don't know if there's a nice way to describe the inverse but you could think of it in terms of the inverse of the https://en.wikipedia.org/wiki/Gamma_function
if you define that in the corresponding interval in the positive reals
tired matrix
I don't know if there's a nice way to describe the inverse but you could think o...
how would you get the inverse of that?
would you just
vast shale
What I would do is the following, 720/2 = 360
360/3 = 120
120/4 = 30
30/5 = 6
6/6 = 1
tired matrix
x=int(t^(F(x)-1)e^-t dt)
vast shale
Idk if there's a dedicated formula tho
tired matrix
What I would do is the following, 720/2 = 360
360/3 = 120
120/4 = 30
30/5 = 6
6/...
that... works, actually
but this only defines it for integers
scenic ravine
,w derrangement
limber falcon
how would you get the inverse of that?
again I don't think there's an easy expression for the inverse, but since $\Gamma(x)$ is continuous and strictly increasing for $x\geq 1$ we can say there's a continuous, increasing inverse $\Gamma^{-1}: [1,+\infty)\to [1,+\infty)$
twin meteorBOT
derivada.schwarziana
tired matrix
yeah
limber falcon
there's some expressions in https://en.wikipedia.org/wiki/Inverse_gamma_function but honestly this is new to me 
tired matrix
let me have a little fun with this in my notebook and i'll get back
smoky pebble
ugly aah
limber falcon
i get that you divide by (5)(4)(3)(2)(1)
you can probably generalize something like this and say that (inverse factorial)(n!) is n! divided by (n-1)!, but that's almost just the definition of inverse function so it doesn't add much imo
smoky pebble
not sure why u would need that tbh
smoky pebble
math guys dont have a gf gang
limber falcon
either way since $n!=\Gamma(n+1)$, a way to define your ``inverse factorial of $n$'' function would be as $\Gamma^{-1}(n)-1$
twin meteorBOT
derivada.schwarziana
limber falcon
I think this works and can be defined for reals x \geq 1
tired matrix
so let's try to find what the formula for $\Gamma^{-1}(n)$ is
twin meteorBOT
Death on Two Legs
tired matrix
also because you're here
nevermind i can just do that on a graphing calculator
nevermind i need help
when does log_b(x)=b^x have only one solution
tired matrix
when does log_b(x)=b^x have only one solution
(b is a constant)
scenic ravine
when does log_b(x)=b^x have only one solution
For all b>0, \neq 1
no?
they are inverse functions
limber falcon
For all b>0, \neq 1
graphic calculator shows that this is not the case
interesting
the way I'd think about it is like,[\log_b(x)=b^x \iff x=b^{(b^x)}]
sharp rain
when does log_b(x)=b^x have only one solution
when b belong (0,1)
solar cape
Imma guess that it has something to do with e
twin meteorBOT
derivada.schwarziana
sharp rain
Yeah (0,1) works
solar cape
Imma guess that it has something to do with e
Nvm
tired matrix
Imma guess that it has something to do with e
probably
sharp rain
There must be another solution
tired matrix
then solve for F(x)?
i got $log_{t}(\frac{n'(t)}{e^t})+1=\Gamma^{-1}(n)$
sharp rain
When both b^x and log_b(x) become tangential to x=y
twin meteorBOT
Death on Two Legs
sharp rain
That value will also be a solution
solar cape
Technically is this also the same as saying find b such that b^x=x has only one solution
tired matrix
**Death on Two Legs**
for all numbers >=0
tired matrix
Technically is this also the same as saying find b such that b^x=x has only one ...
$b^x=x \
x=log_{b}(x) \
0=log_{b}(x)-x \
b^0=b^{log_{b}(x)-x}$
can you go somewhere from
oh dear lord
twin meteorBOT
Death on Two Legs
twin meteorBOT
lex.in.a.teacup
sharp rain
$(0,1) \bigcup e^{1/e}$
sharp rain
Ye
twin meteorBOT
lex.in.a.teacup
sharp rain
Better
twin meteorBOT
Death on Two Legs
sharp rain
We needed log to give us integers
tired matrix
$e^{\frac{x}{e}}$ equals n at $e^{\frac{x}{e}}$th power
twin meteorBOT
Death on Two Legs
sharp rain
We didn't want some terms in log
So I just assume b to be of format e^n
$log_{e^{n}}x = e^nx$
twin meteorBOT
lex.in.a.teacup
sharp rain
$\frac{log_{e}x}{n} = e^{nx}$
twin meteorBOT
lex.in.a.teacup
hexed umbra
this is very simple but im kinda confused, how do you find x = y in this situation
vocal sleetBOT
this is very simple but im kinda confused, how do you find x = y in this situati...
vocal sleetBOT
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.
hexed umbra
lone linden
The equation (x=2) only involves x
so y is irrelevant here
(note that the graph is a vertical line)
,w graph x=2
twin meteorBOT
limber oak
is there any math that can help me do this
a / (b + c)
Where I can't divide a directly with (b + c)
but I can divide
a with b
or
a with c
limber oak
something similar to this is also welcome
viral copper
I don't think there is
limber oak
I will wait, maybe someone know anything about this
jagged cargo
do you like, mean $\frac{a}{b+c} = \frac a b + \frac a c$
twin meteorBOT
limber oak
I mean like I can use that, if that exist
jagged cargo
but it's not a formula
limber oak
hmm... b and c should not be 0
jagged cargo
it just so happen that the equation is true for some value a b c
either way, can you show the full question
!xy
vocal sleetBOT
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.
limber oak
either way, can you show the full question
assume I have a problem
a / (b+c)
but I can't divide them directly
I can only do
a / b
and
a / c
because (b + c) is too big as number
jagged cargo
send a screenshot
jagged cargo
jagged cargo
assume I have a problem
a / (b+c)
but I can't divide them directly
I can only ...
wdym "too big as number"
is this from some programming language
limber oak
i need it so I can make performance algorithm for division
limber oak
is this from some programming language
yes
jagged cargo
i need it so I can make performance algorithm for division
can you elaborate
limber oak
the problem is simple
assume there is
a / i
where i is a very big number that can't be stored in any variable, and need at least 2 integer variable to store that value
as such i which is stored in 2 variable become (b + c)
and I need a way to do
a / i
jagged cargo
probably not the most efficient way, but i guess you can instead calculate the reciprocal of it
you want to output a/(b+c), correct?
limber oak
yes
jagged cargo
well instead, you can calculate $\frac1{\frac{b+c}a}$
twin meteorBOT
jagged cargo
from here, you can distribute them out and calculate b/a and c/a individually
if the sum is too big of a number, it might as well be infinity
and since 1/infty = 0, you know what to do
limber oak
i see... thx
vocal sleetBOT
Available help channel!
smoky pebble
if a circle is passing through the center the center of another circle it is impossible for them to touch only externally right?
vocal sleetBOT
if a circle is passing through the center the center of another circle it is imp...
sharp rain
Yeah
They may touch internally
Hmm
What if one of those circles have radius 0. It would lie on the circumference of the circle, that technically counts as external touching
Maybe
I could be wrong
@smoky pebble
smoky pebble
hmm
hushed pewter
if a circle is passing through the center the center of another circle it is imp...
what does "touch externally" mean?
brittle cipher
I think they mean like )( vs )) if that makes sense
internally being a circle inside the other circle
externally is outside of each other
limber falcon
yeah I'm assuming they mean that the circles are tangent to each other, i.e. )(
limber falcon
actually here they're also tangent 
didn't consider that
limber falcon
What if one of those circles have radius 0. It would lie on the circumference of...
but anyway I think this is correct
sharp rain
This is like playing on a technicality
It feels wrong
But kinda isn't
@smoky pebble what do you think?
vocal sleetBOT
@smoky pebble Has your question been resolved?
sharp rain
@smoky pebble 1
dapper sequoia
vast shale
Is the answer 3?
I am getting that
there could be many approaches to solve it
Here's what i did
- find AC side of right triangle ANC = root7
So AC=QP=BD= root 7
So now you can try taking equating areas ....of different figures using Side AB or CD to find it
dapper sequoia
i think the question is wrong
if you see
triangle anc and bod are congruent
an = bo
but if we use bpt
am/an = mb/bo
which gives us BO = 10/3
vast shale
I think i messed up somewhere
vast shale
i think the question is wrong
maybe
vocal sleetBOT
@dapper sequoia Has your question been resolved?
peak yacht
Hello! How do I get these determinates? 
vast shale
Hi there. what have you done uptill this point?
vocal sleetBOT
@peak yacht Has your question been resolved?
next meteor
vocal sleetBOT
next meteor
can someone explain part a
mint pewter
Substitute e^t for x and apply chain rule.
marsh folio
vocal sleetBOT
marsh folio
cant advance here
vocal sleetBOT
@marsh folio Has your question been resolved?
marsh folio
.close
vocal sleetBOT
dull quest
i dont understand how to solve this question
my first thought was to sub in the coordinates into the equation above and get k as subject of formula and then solve simultaneosly
but it didnt work so i tried to make a subject and solve simutaneosly also didnt work
vocal sleetBOT
dull quest
pale perch
vocal sleetBOT
Available help channel!
lone rapids
Hi I'm stuck on this question and not sure where I've gone wrong. Please ask if any of my working is unclear
vocal sleetBOT
Hi I'm stuck on this question and not sure where I've gone wrong. Please ask if ...
wintry vine
I would first give variables to center of each circle and their radii
apply constraints and solve
I will solve it my own and come back
wintry vine
still doing it
next step is do determinant =0
I don't have a calculator and I'm walking so I will be very slow
do you understand my method tho
wintry vine
^
wintry vine
by finding the centers x1 x2 y1 y2 and radii r1 r2
their eq of circle would then be (x - xi)^2 + (y -yi)^2 = ri^2
lone rapids
dont really understand but im more of an observer than understanding type of person. could you reach the conclusion i need to see how its done
vocal sleetBOT
@lone rapids Has your question been resolved?
wintry vine
from there you can get y1 with its formula
and then by extension r1
two circles since two solutions for each x1
just ask if any of the working isn't clear
also check the arithmetic I don't have a calculator as i said
the equation as mentioned would henceforth be
(x - x1)^2 + (y - y1)^2 = r1^2
lone rapids
why?
wintry vine
why?
4x1^2 - 4x1^2 - y1^2 + r1^2 after expanding the brackets
first two terms cancel
y1^2 = r1^2
lone rapids
ohh
wintry vine
actually y1 = +-r1 but I don't know where that leads us
lone rapids
ok i understand how u got to ur answer but the answer given in textbook is an integer for the centre of circle 1
is it the arithmetic thats wrong?
wintry vine
could be
at least I don't see a problem with my method, but let me think it over again
lone rapids
ill try to solve the question myself to see what i got wrong initially
wintry vine
vocal sleetBOT
@lone rapids Has your question been resolved?
vast shale
vocal sleetBOT
peak axle
,rccw
twin meteorBOT
vocal sleetBOT
@vast shale Has your question been resolved?
vast shale
<@&286206848099549185>
vocal sleetBOT
@vast shale Has your question been resolved?
vocal sleetBOT
@vast shale Has your question been resolved?
cloud pine
vast shale
can anyone help?
A·(BxC) = B·(CxA) = C·(AxB). So (axb)·(cxd) = c·(dx(axb)) = d·((axb)xc). You wrote d·(cx(axb)) instead
Look ur answer is the same but multiplied by -1
you can invert the subtraction because d·((axb)xc) = d·(-cx(axb)) = - [d·(cx(axb))], giving u the same answer
vocal sleetBOT
@vast shale Has your question been resolved?
vocal sleetBOT
@vast shale Has your question been resolved?
twin citrus
vocal sleetBOT
limber vortex
I think the third element on the top row of your matrix in the last picture should be -11
twin citrus
I think the third element on the top row of your matrix in the last picture shou...
you are right tnx
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
frail flame
Can someone help explain the solution, i dont get why we include the condition that a, b, cannot be equal because the product is a perfect cube? And i dont get why we divide by 6 at the end.
vocal sleetBOT
Can someone help explain the solution, i dont get why we include the condition t...
kind light
Can someone help explain the solution, i dont get why we include the condition t...
its a,b,c thats not equal
frail flame
what about it
kind light
well they cant be equal cause if they are all equal it would be a cube
frail flame
wont that be if a,b,c are all equal, i thought the question was only counting pairs not to be equal
like b cant be = a and a cant = c
kind light
?? yeah is that not what you said
wintry vine
it says at the beginning x < y < z
kind light
it says at the beginning x < y < z
and this for the /6
frail flame
but like cubes are made from 3 equal numbers
frail flame
and this for the /6
wait why 6
kind light
wait why 6
3! ?
wintry vine
yea
kind light
but like cubes are made from 3 equal numbers
the solution counted all, including equal, then subtracted them all equal and 2 being equal
wintry vine
xyz is not a cube because each prime had a component that isn't a multiple of 3
kind light
the all equal isnt subtracted cause its just not possible
frail flame
i think ill come back to this q, thanks for helping, i understood a little
gtg rn
open the channel again
kind light
if you are done, write ".close" to close the channel 
vocal sleetBOT
@frail flame Has your question been resolved?
final summit
M= 300 + 180sin(3t)
we have to integrate it and find area. so from 3t, period will be 2π/3. so we integrate it with limits 0 to 2π/3
now, since we are integrating it within full period, sine function value will be zero. so integral will come equal to 300.
then we have to draw a graph too. so after calculating integral, we get 300, its actually a physics question and we have to calculate Mean Torque from above Turning Moment equation. and it comes out to be 300, as I said. now instead of doing all this effort, can I say (and solve) while drawing its graph, that since it's "300 + 'a sine function' ", then I can just simply 'shift' the sine curve on Y axis by 300 units?
final summit
<@&286206848099549185>
vocal sleetBOT
@final summit Has your question been resolved?
sudden compass
hi
sudden compass
(sorry victor)
got a banger integral
$$\int_0^1 \frac{u^{\alpha}}{1-ku} \sqrt{\frac{u}{1-u}} du$$
twin meteorBOT
rak³en
sudden compass
**rak³en**
WA can evaluate for $\alpha = 0,1$ ($\alpha$ is a non-negative integer)
twin meteorBOT
rak³en
vocal sleetBOT
@sudden compass Has your question been resolved?
fast nexus
Hello👋🏾
fast nexus
Welcome to mathcord
Thanks
eager plume
been asking for help here a lot but
any ideas for this question?
vocal sleetBOT
been asking for help here a lot but
any ideas for this question?
@eager plume Has your question been resolved?
eager plume
<@&286206848099549185>
vocal sleetBOT
@eager plume Has your question been resolved?
still relic
can i have helpo
vocal sleetBOT
can i have helpo
still relic
bronze osprey
integrate both sides
smoky iron
yes you can
vocal sleetBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
bronze osprey
$\frac{d}{dx} (\ln y(x)) = \frac{1}{y(x)} \cdot y'(x)$ by the chain rule
so if you integrate $\frac{y'(x)}{y(x)}$ you get back $\ln y(x) + c$
still relic
so for my example what will be the result
bronze osprey
smoky iron
just ask chatgpt
bronze osprey
I was busy with other people sorry for the obvious typo
still relic
i know the answer i just dont know how to get there
bronze osprey
basically you get $\ln R = -kt + c \implies R = e^{-kt} \cdot e^c$
twin meteorBOT
south
still relic
just ask chatgpt
i cant put images in there
smoky iron
have you not updated?
smoky iron
No problem
vocal sleetBOT
@still relic Has your question been resolved?
dapper sequoia
vocal sleetBOT
twin meteorBOT
dapper sequoia
my answer of 4th question is coming to be different
boreal remnant
step 1: numerically approximate 3pi/2
dapper sequoia
around 4.7
vocal sleetBOT
@dapper sequoia Has your question been resolved?
dry ledge
can someone please help
vocal sleetBOT
can someone please help
vocal sleetBOT
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
scenic ravine
ask your question
dry ledge
isn't it
ok so
hold on one second
so the definition of arg(z)
is the angle from a point to the positive real axis
or just tan-1(y/x) of the complex number x +iy
scenic ravine
here, in chat
scenic ravine
No what I wanted was $\arctan(\frac{z-2}{z}) = \frac{\pi}{2}$
twin meteorBOT
ƒ( wai ina teacup)= I don't know
scenic ravine
You'll learn with time, it's fine
dry ledge
**ƒ( wai ina teacup)= I don't know**
also it's not arctan it's arg in that
scenic ravine
arctan = tan^{-1}
scenic ravine
yes, but there's a reason I've done this
dry ledge
alright
dry ledge
all good
twin meteorBOT
ƒ( wai ina teacup)= I don't know
scenic ravine
sure
dry ledge
alright
also how do u use the texti command
I can only seem to use this math bot for just roles
harsh canopy
$ before and after message
twin meteorBOT
Ram
dry ledge
multiply top and bottom by the conjugate of the denominator?
alright
imma just do it in my book rq
alr i got
$(x^2 - 2x + y^2 + 2iy)/(x^2 + y^2)$
twin meteorBOT
Ram
dry ledge
is that correct?
scenic ravine
yes
scenic ravine
so you have $\frac{x^2-2x+y^2}{x^2+y^2} + \frac{2y}{x^2+y^2} i$
twin meteorBOT
ƒ( wai ina teacup)= I don't know
scenic ravine
so what's arg(z) now
dry ledge
wait
ok
isn't arg(z) just that inside z?
OH
is it
arctan(y/x)
with the i part as y
scenic ravine
which is what here
dry ledge
and the first term as x?
scenic ravine
yes
scenic ravine
I mean cancel out the common terms
dry ledge
right
the x and y squared denoms get cancelled
and ur left with the first term numerator on top of 2y
in arctan
twin meteorBOT
ƒ( wai ina teacup)= I don't know
Compile Error! Click the reaction for more information.
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twin meteorBOT
ƒ( wai ina teacup)= I don't know
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smoky iron
$\arctan(\frac{2y}{(x-y)^2})$
twin meteorBOT
anshul6368
scenic ravine
$\arctan(\frac{2y}{(x-y)^2})$
thanks
dry ledge
wait hold on
scenic ravine
**anshul6368**
so when is this pi/2
dry ledge
**ƒ( wai ina teacup)= I don't know**
this was our expression
scenic ravine
yes, and what's the coefficent of i?
dry ledge
oh wait
it thought
that when you do arctan(y/x)
you don't include the i
like for e.g
if you had the complex number 1 +2i
and you wanna find the arg of that
you'd just do
arctan(2/1)
scenic ravine
yup!
scenic ravine
yes
twin meteorBOT
Ram
scenic ravine
so when is the arctan of this π/2
scenic ravine
when x=y
dry ledge
otherwise it can't be 90 degrees
wait waht
no but
when x=y isn't that for pi/4
pi/2 is 90 degrees
dry ledge
and you can only have y values for that
not x
yeah ig
arctan infinity
keep in mind we have to sketch this
scenic ravine
yeah
dry ledge
so what next
scenic ravine
you have y=x
dry ledge
but why
scenic ravine
so the locus is a line
dry ledge
and what's a locus
scenic ravine
x^2-2x+y^2 must be zero
dry ledge
x^2-2x+y^2 must be zero
so it's infinity alright
scenic ravine
yes
twin meteorBOT
Ram
dry ledge
so we get a circle
but this gives us a full circle
in the answers it's only a semi-circle above
scenic ravine
$(x-1)^2 +y^2 = 1$
how did you get this
dry ledge
complete the square for x
scenic ravine
that gives you (x-y)^2=0
dry ledge
$x^2 -2x + y^2 = 0$
twin meteorBOT
Ram
dry ledge
$(x-1)^2-1+y^2=0$
twin meteorBOT
Ram
dry ledge
and then yeah
scenic ravine
hmm
harsh canopy
Yes
cyan talon
Well you need to be sure that 2y/(x^2-2x+y^2) actually goes to + infty yea
dry ledge
so that's the reason?
dry ledge
Well you need to be sure that 2y/(x^2-2x+y^2) actually goes to **+** infty yea
yeah we did
harsh canopy
An alternative method would just be to use the argument properties
dry ledge
soooo
cyan talon
yeah we did
you didn't
cyan talon
Well you need to be sure that 2y/(x^2-2x+y^2) actually goes to **+** infty yea
just saying x^2 -2x +y^2 = 0 says nothing about the sign of this
cyan talon
If 2y is negative, the point ( x^2-2x+y^2, 2y) with x^2-2x+y^2=0 is on the negative y axis
So its argument is -pi/2 not pi/2
dry ledge
y can't be negative
cyan talon
Yea
left portal
If $n=2^3 \cdot 3^2 \cdot 5$, how many even positive factors does $n$ have
twin meteorBOT
938c2cc0dcc05f2b68c4287040cfcf71
queen root
**938c2cc0dcc05f2b68c4287040cfcf71**
u can find the total number of positive divisors of a number by adding one to each of its prime factors and multiplying them
to determine only the even ones, u can honestly see the odd ones by urself, theres 3, 3², 5, 5×3, 5×3², 1
so do subtraction
vapid solar
A positive even divisor of n is of the form $2^i3^j5^k$ where i is between 1 and 3, j is between 0 and 2, and k is between 0 and 1. So choosing an even divisor of n is like choosing such i, j and k. Si there is 3×3×2 even divisors, so 18
twin meteorBOT
TimourX
left portal
wdym?
left portal
u can find the total number of positive divisors of a number by adding one to ea...
why add one
scenic ravine
why add one
the number itself is a factor
left portal
A positive even divisor of n is of the form $2^i3^j5^k$ where i is between 1 and...
why i starts from 1 but j from 0
queen root
why add one
u can demonstrate it by pythagorean tables
its for determining all of the divisors of a number
vapid solar
why i starts from 1 but j from 0
Because I want the divisor to be even, so if i=0, then the divisor is not even. But j and k can be equal to 0, for example, 2² divides n
left portal
A positive even divisor of n is of the form $2^i3^j5^k$ where i is between 1 and...
3x3x2?
vapid solar
3x3x2?
There are 3 values that i can take (1,2 and 3), 3 values that j can take (0,1 and 2) and 2 values that k can take (0 and 1). So we obtain 3×3×2
vocal sleetBOT
vocal sleetBOT
Available help channel!
bitter pilot
It is given [ 12x +10 \overset{61}{\equiv} 6x + 23. ]
I concluded the following.$\$
Consider
[ 12x +10 \overset{61}{\equiv} 6x + 23 \Leftrightarrow 6x-13 \overset{61}{\equiv} 0. ]
Then that is equivalent to $6x = 61k + 13$ for some $k \in \mathbb{Z}.\$
In case of $x = 13\tilde x$ and $k = 13\tilde k$ are multiples of 13 with $\tilde x, \tilde k \in \mathbb{Z}$, it is
[ 6\tilde x = 61\tilde k + 1 ] and the right side is never divisible by 6.$\$
The other case of $x$ and $k$ not being multiples of 13 leads to
[ 6x = 61k+13 = 60k + 12 + k + 1 = 6(10k+2)+(k+1) ] which implies $0 = k+1$ and $x = 10k+2$ therefore for $k=-1$ it is $x = -8$ the only solution.
vocal sleetBOT
It is given \[ 12x +10 \overset{61}{\equiv} 6x + 23. \]
I concluded the followin...
twin meteorBOT
𝔸dωn𝓲²s
hushed pewter
Hi @bitter pilot. I do not understand what your question is
bitter pilot
Hi <@587715738484211754>. I do not understand what your question is
Oh my bad, I wanted to ask if it is correct
queen root
Hi <@587715738484211754>. I do not understand what your question is
values of x probably
o
bitter pilot
pls dont ping mods
hushed pewter
pls dont ping mods
dang I was just about to...
normal heath
It is given \[ 12x +10 \overset{61}{\equiv} 6x + 23. \]
I concluded the followin...
Why is 61k+1 not divisible by 6?
hard atlas
you are really hurting yourself by not using inverses
unkempt fiber
you are really hurting yourself by not using inverses
this
bitter pilot
you are really hurting yourself by not using inverses
i am sorry for being suicidal
hushed pewter
I do not undetstand why you are looking at a specific case where x and k are divisible by 13
hard atlas
in the real numbers you would immediately solve ax=b as x=a^-1 b
unkempt fiber
61·5+1 is divisible by 6
did not say it is
hard atlas
do the same in modulo
hushed pewter
i am sorry for being suicidal
you shouldn't jok about that online
wraith venture
it's not hard to manually find 6^-1
bitter pilot
I do not undetstand why you are looking at a specific case where x and k are div...
because then the equation could work out i thought
hushed pewter
because then the equation could work out i thought
Not seeing it myself, but I always sucked at number theory 
bitter pilot
Not seeing it myself, but I always sucked at number theory <:sadcatthumbsup:9323...
you would have basically on every term a factor of 13
but i think it's irrelevant now should restart
hushed pewter
Hello. Welcome to mathcord. For your own questions, you should open your own help channel. See #❓how-to-get-help (or even just #calculus or #precalculus for general (pre)calculus discussions)
hushed pewter
but i think it's irrelevant now should restart
With inverse, most likely. But that's the value of peer review (to invalidate all of your work and effort )
bitter pilot
it's fine lol
vocal sleetBOT
@bitter pilot Has your question been resolved?
bitter pilot
in the real numbers you would immediately solve ax=b as x=a^-1 b
[ 6x \overset{61}{\equiv} 13 \Leftrightarrow x \overset{61}{\equiv} 6^{-1} \cdot 13 ]
Since $\gcd(61,6) = 1$ it means the inverse exists. $\$
Applying Euclid I get
[ 61 = 6 \cdot 10 + 1 \rightarrow 10 = 1 \cdot 10 + 0. ]
Therefore $1 = 61 + 6(-10)$ which implies $6^{-1} \overset{61}{\equiv} -10 \overset{61}{\equiv} 51.\$
So I get [ x \overset{61}{\equiv} 6^{-1} \cdot 13 \overset{61}{\equiv} (-10) \cdot 13 = -130 \overset{61}{\equiv} -8 ]
The solutions are therefore [ \mathbb{L} = { x \in \mathbb{Z} \mid x = 8k, : k \in \mathbb{Z} }. ]
twin meteorBOT
𝔸dωn𝓲²s
bitter pilot
hmm funny
i got with the previous but faulty approach to the same I think, except i said it's the only solution, when it's not
I am kinda confused
x ≡ -8 ≡ 53
The solution set is wrong then?
hard atlas
x=-8+61k
bitter pilot
yes
ok thanks denascite for opening my eyes and to everyone else thanks too
.solved
vocal sleetBOT
vocal sleetBOT
Available help channel!
opaque geyser
Hello, can someone help me evaluate this limit?
$$[
\lim_{x \to +\infty} \left( \sqrt{x^2 + 3x} - \sqrt{x^2 + x} \right)^x
]$$
vocal sleetBOT
Hello, can someone help me evaluate this limit?
$$\[
\lim_{x \to +\infty} \left(...
twin meteorBOT
maxy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
opaque geyser
[
\lim_{x \to +\infty} \left( \sqrt{x^2 + 3x} - \sqrt{x^2 + x} \right)^x
]
twin meteorBOT
maxy
flat whale
try finding the log of the limit
opaque geyser
x * ln (...)??
flat whale
then uhhh l'hopital maybe
loud walrus
You only need to find the limit inside parenthesis
lilac pebble
loud walrus
I dont wanna spoil
vocal sleetBOT
@opaque geyser Has your question been resolved?
dim pumice
maybe let's just go through all the number systems there are and their differences
are you familiar with the natural numbers? 1, 2, 3, 4, ...
vocal sleetBOT
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Available help channel!
low haven
vocal sleetBOT
low haven
does exists a number n divisible by 103 such that $2^{2n+1} \equiv 2 \pmod{n}$?
twin meteorBOT
isomorphic to god
low haven
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
violet flare
Hi i need help here
Let Δ𝐴𝐵𝐶 be a triangle, Δ the midpoint of the segment AB, and E the midpoint of 𝐴𝐶. Let Z be the midpoint of CΔ, H the point of intersection of the lines AΖ and BC, Θ the midpoint of the segment ΔΗ, and Ι the midpoint of the segment BΖ. Prove that the points Ε,Ι, and Θ are collinear.
violet flare
in the unit of the exercise the most relative concept is Menelaus Theorem ,the converse of Menelaus and Thales Theorem , i was thinking of the converse but i cant find a triangle to apply it any ideas ?
queen root
ngl
queen root
Hi i need help here
Let Δ𝐴𝐵𝐶 be a triangle, Δ the midpoint of the segment AB...
i drew accordingly to the paragraph u provided
and visually
E, I and Θ arent collinear
violet flare
E, I and Θ arent collinear
at first i drew it too and got graph so they arent collinear , then i did it with geogebra and i got the image above,they look colinear
queen root
no what i meant its like
somewhere, theres a mistake in what u wrote
is there?
like theta isnt midpoint of ZH on ur figure, contradictory to what u said in the paragraph
oh so thats where the mistake is
theta is midpoint of ΔH not ZH
violet flare
no what i meant its like
yes you are right matte i change it
<@&286206848099549185>
vocal sleetBOT
@violet flare Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
hard hound
vocal sleetBOT
vocal sleetBOT
@hard hound Has your question been resolved?
heady ibex
You want to find for which angle the vectors (x, y) that satisfy the equation y = x do not change if you apply S-1RS to it
so you have the 2d plane
of (x,y) vectors
and you have this matrix
M = S-1RS
and so if you do Mv
you want it to be Mv = (k)v
not entirely sure if multiples of the vectors are allowed or if it is just Mv = v
vocal sleetBOT
@hard hound Has your question been resolved?
hard hound
You want to find for which angle the vectors (x, y) that satisfy the equation y ...
i don’t understandddd
do u mind showing ur working?
heady ibex
ok so y = x being an invariant line of some matrix means all vectors that satisfy y = x are not changed by it
so what you do is calculate what S-1RS is
apply it to some generalized vector v which satisfies y = x
and see for which theta it gives back v
vocal sleetBOT
@hard hound Has your question been resolved?
warm horizon
Prove by mathematical induction, for all $n\geq12$, that $5^n < n!$
warm horizon
I got to $5^{k+1} < 5*k!$ using assumption true for $n=k$ and idk how to proceed after that
twin meteorBOT
BOHO
warm horizon
to prove $5^{k+1} < (k+1)!$
twin meteorBOT
BOHO
warm horizon
sorry yeah i made a typo
tawny nacelle
Prove by mathematical induction, for all $n\geq12$, that $5^n < n!$
are you sure this is the exercise?
ah, I see
twin meteorBOT
BOHO
warm horizon
thats the question
warm horizon
well i think so, i just computed 5^12 and 12! and said it was less
tawny nacelle
mhm
finite owl
how can you rewrite (k+1)!
tawny nacelle
so now assume that $5^n < n!$, and try to prove that $5^{n + 1} < (n + 1)!$
twin meteorBOT
higher!
finite owl
so now assume that $5^n < n!$, and try to prove that $5^{n + 1} < (n + 1)!$
wat no
tawny nacelle
hm? 
finite owl
oh i mean yea sorry
loud walrus
No = yea
tawny nacelle
**higher!**
@warm horizon going off what Ryse said: try to rewrite (n + 1)! and 5^(n + 1)
loud walrus
finite owl
my head is cooked its late
loud walrus
I never ate cooked brain
limber falcon
I got to $5^{k+1} < 5*k!$ using assumption true for $n=k$ and idk how to proceed...
you know $5^{k+1}<5\cdot k!$, so it's enough to show that for $k>12$ you have $5\cdot k!<(k+1)!$. Try rewriting the factorial using the definition[(k+1)!=1\cdot 2\cdot 3\cdot \hdots \cdot k \cdot (k+1)]
twin meteorBOT
derivada.schwarziana
limber falcon
and likewise for the k! term
warm horizon
do i rewrite (k+1)! as (k+1) * k! so that i can cancel k! from both sides
limber falcon
yes that's one way to look at it
warm horizon
and then get k > 4
limber falcon
true
and if k\geq 12 (since we're proving this statement for n\geq 12), then surely it's true that k>4 right
warm horizon
yeah
limber falcon
so you're done with the induction step once you do that
warm horizon
If it is true for n=k, then it must be true for n=k+1. But it is true for n = 12, therefore inequality is true for n=12+1=13, 14, 15 and so on for all n ≥ 12 by mathematical induction
is this a suitable conclusion for this
vocal sleetBOT
vocal sleetBOT
Available help channel!
tawny nacelle
so I'm being asked to prove these three
vocal sleetBOT
so I'm being asked to prove these three
tawny nacelle
how does this look?
I feel okay about a) and b) (unless I've missed smth), but c) feels a little iffy for me
versed bane
how does this look?
all you need in c is the existence of a chart in which iota is smooth
so it’s a bit verbose already because you can just pick any chart in the atlas entirely in U and that same chart in M
and under those charts iota is the identity
tawny nacelle
hold on
zz
okay yeah that makes sense
I forgot existence didn't require me to find difference charts for some reason
oops
versed bane
wait what is the definition of a map being smooth in Lee
because you did this for all three parts
actually just (b) and (c)
tawny nacelle
wait what is the definition of a map being smooth in Lee
tawny nacelle
actually just (b) and (c)
yes, I could've chosen the same chart for b) too haha
versed bane
oh okay but that’s fine
like b is fine anyways
because you just proved it for every pair of charts about p
so actually it’s just fine
tawny nacelle
hmm, is this better? 
I still feel kinda weird about this
the last line in particular makes me feel uneasy
versed bane
yeah because it’s not right, it’s the identity on psi^-1(V)
tawny nacelle
ah.
grim lotus
V is an open subset of U so open in M
vocal sleetBOT