forget it
#help-13
azure violet
but what's the answer?
and how do you get to it?
<@&286206848099549185>
bruh
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cedar kilnBOT
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valid bolt
How can this be?
cedar kilnBOT
How can this be?
valid bolt
$f''(x) = -\frac{1}{\left(1+x\right)^2}$
$|f''(x)| \le \frac{9}{16}$\
$|f(x) - T_1(x) | \le \frac{M}{2!}\epsilon^2$\
$\rightarrow |f(x) - T_1(x) | \le \frac{1}{2}\frac{9}{16}\frac{1}{9} = \frac{1}{32}$
wraith daggerBOT
Zim
valid bolt
Interval is [-1/3, 1/3]
valid bolt
How can this be?
if |f - T| <= 1/32, then it also has to be <= 1/8
valid bolt
So how is the epsilon defined already?
I set epsilon = |0 - 1/3| = 1/3
silent bobcat
So what is 0 and what is 1/3 here?
valid bolt
How can this be?
@silent bobcat The details are mentioned here
silent bobcat
Ow yeah so 0 is b where it is based, and 1/3 is related to the interval
Well I think you got it right yeah
Or else the epsilon is not set right
Because the rest is all correct
So it feels very weird
cedar kilnBOT
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valid bolt
How can this be?
Original question
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tardy pelican
Say I have a value ranging from 0-1 that can change at any time. How can I use that to calculate an angle of -89 to 89 degrees? For example, I'd need 0 to represent -89 degrees, 0.5 to represent 0 degrees, and 1 to represent 89 degrees
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prime oriole
Hello I need help
cedar kilnBOT
Hello I need help
prime oriole
SO Im learning transformations
what I need help with is finding what 15x-27 degrees is
<@&286206848099549185>
modern compass
Wait 15 minutes before pinging @Helpers
prime oriole
Oh sorry
modern compass
and the degree of a polynomial is the largest exponent on the varaible (in this case x)
cosmic steppe
Man you love breaking rules
cosmic steppe
what I need help with is finding what 15x-27 degrees is
What does this even mean
15x - 17 degrees as in?
prime oriole
Me what I'm sorry I didnt meen it whatever I did IDK I humbly apologise Im leaving sorry for whatever I did
prime oriole
No Im sorry GOODBYE
cosmic steppe
Bruh
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midnight zealot
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small lance
Abd
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azure violet
The Pantheon is a preserved ancient temple in Rome from approx. 120 .Kr. The inner part of the building is constructed as a cylinder with a hemisphere on top. The hemisphere forms the largest dome built in Roman times. The total height of the building is equal to the inner diameter of the base: 43.3 m. The cylinder and the hemisphere have the same height.
a) Show that the inner wall area of the cylindrical part of the temple is equal to the area of the dome.
b) 1 liter of paint covers 10m ^ 2. Approximately how many liters of paint can we assume agreed to paint one coat in the interior of the temple?
azure violet
I need help as quickly as possible
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floral otter
isn't the square root of 1 = 1, -1 ?
cedar kilnBOT
isn't the `square root of 1` = `1`, `-1` ?
crimson sedge
isn't the `square root of 1` = `1`, `-1` ?
I think because it gets connected to iota
Unsure tho
floral otter
what does that mean?
crimson sedge
But you are right
crimson sedge
Iota is complex numbers
It's like u can divide number in to two main parts
Real nos
And complex
floral otter
that sounds quite complex
crimson sedge
And iota is the way u represent complex numbers
crimson sedge
that sounds quite complex
Lol
crimson sedge
Yes
crimson delta
that said, the sqrt of 1 is just 1
there are two solutions to x^2=1
but the sqrt of 1 is defined to be just 1
crimson sedge
there are two solutions to x^2=1
Correct
crimson delta
not -1
floral otter
well I was calculating the zero point of f(x) = x^4 - 1 and when looking at the graph it's x1=1 and x2=-1
soo that the 4th root has 2 solutions right?
floral otter
yeah didnt have that topic yet
crimson sedge
soo that the 4th root has 2 solutions right?
Yes because -2²= 4
crimson delta
no
crimson sedge
And 2² is 4
crimson delta
if anything, (-2)^2=4
crimson delta
yes but you wrote -2 x 2
floral otter
okay u guys are confusing me haha
crimson delta
which is -4
crimson delta
doesn't matter. you first do the exponent and then the minus
floral otter
which is -4
why is that so tho
crimson sedge
I got confused
crimson delta
because $-2^2=-(2^2)=-(2\cdot 2)=-4$
wraith daggerBOT
Denascite
crimson sedge
**Denascite**
No no
crimson delta
but $(-2)^2=(-2)\cdot (-2) = 4$
floral otter
makes sense
crimson sedge
(-2)²
wraith daggerBOT
Denascite
floral otter
yeah
crimson delta
$x^4=1$ leads to $x^2=\pm \sqrt{1} = \pm 1$
wraith daggerBOT
Denascite
crimson delta
which then leads to either $x^2=1$ which implies $x=\pm 1$. or it leads to $x^2=-1$ which implies $x=\pm i$
wraith daggerBOT
Denascite
floral otter
f(x) = x^4 -1
f(x) = 0
x^4 -1 = 0 | +1
x^4 = 1 | 4thRoot
x1 = 1, x2 = -1
Yes correct
floral otter
which then leads to either $x^2=1$ which implies $x=\pm 1$. or it leads to $x^2=...
yup
.close
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granite stag
Hey @modern compass if ur still here , so basically getting 2 heads in a row, is( less likely than getting 3 heads or more) so can one conclude that the next outcome is more likely to be heads?
cedar kilnBOT
Hey <@334482942506041345> if ur still here , so basically getting 2 heads in a r...
@granite stag Has your question been resolved?
true oar
lament sonnet
what's your question?
true oar
whats the solution to everything 
pearl junco
whats the solution to everything <:blobcry:971797574916907028>
u here ?
true oar
yes
true oar
no and im not sure that what we're lookin for 
pearl junco
no and im not sure that what we're lookin for <:NervousSweat:789206248419491900>
dx and dy lol
pearl junco
i just know dy/dx is derivatives
lol so what have u learn on derivatives
pearl junco
i just know dy/dx is derivatives
lol where are u learning math from
pearl junco
uhh after answering that lemme got one final question : do u know limit
if yes, u're around grade 11, so i will find a simplest way to help u
if no, i think u should learn some more math
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pearl junco
i really hate language barrier
rain parrot
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rain parrot
can someone help with this question?
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cedar kilnBOT
any help pls
crimson sedge
Ok so figure 6 looks like a straight line equation, which is usually modeled by $y = mx+b$ . When you replace each variable with its respective representation, you get that $\log(d) = n\log(V) -1.77$. To get the actual formula, undo the log by raising 10 to the power of the equation.
$$10^{\log(d)} = 10^{n\log(V)-1.77}$$
wraith daggerBOT
Ausaramun
crimson sedge
Now use the properties of log and you get:
$$d=10^{\log(V^{n})-1.77}=V^{n} \cdot 10^{-1.77}$$
wraith daggerBOT
Ausaramun
crimson sedge
Clean up the equation and you get $d = kV^{n}$ where k = 0.017.
wraith daggerBOT
Ausaramun
crimson sedge
Does that clear your question?
crimson sedge
**Ausaramun**
What happened here
After raising it to the power of 10
Slightly confused on how you got to the next part after that
@crimson sedge ?
Hmm...
Like what happens when you raise it to the power of ten?
What cancels?
Well you just change the equation.
The log basically is the inverse of $10^x$ and $10^x$ is the inverse of log.
The log basically is the inverse of $10^{x}$ and $10^{x}$ is the inverse of log.
Hmm...
When you raise 10 to the log(x) power, you get x.
Ahhhh
Ok that's clears things up
Also
It says explain
But why are we doing calculations instead?
That's the explanation part
You explain how did he conclude that the final equation is true
Ahhh
I got it now, thanks
❤️❤️
arctic marsh
Hi everyone, I'm not sure whether this is the right channel to ask this, so in case it's the wrong one, I'm sorry.
I'm currently studying progressions, and I was asking myself "Is there a way to write the sum of a certain number of terms of a progression as a summation?"
For example, if I wanted to calculate the sum of the first 8 numbers of a(n)=6+30n/7, could I do it using a summation?
somber belfry
Hi everyone, I'm not sure whether this is the right channel to ask this, so in c...
so you want to plug in the values 1 to 8 for n correct?
arctic marsh
Yep, the sum should be 120
somber belfry
I mean normal sum sign should do
let me try with the bot here
$\sum{6+\frac{30n}{7}}$
,help
wraith daggerBOT
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arctic marsh
And i should write like 8 on the top of the summation and n=1 on the bottom?
somber belfry
yea
wraith daggerBOT
Eichhorst
arctic marsh
Oh, got it
Thanks much!
cedar kilnBOT
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sleek obsidian
how do I write it and how do I justify it?
cedar kilnBOT
how do I write it and how do I justify it?
cosmic steppe
how do I write it and how do I justify it?
Where is $\sec{(x)}$ undefined?
wraith daggerBOT
Umbraleviathan
small lance
because sec(x) is 1/cos(x), sec(x) is undefined wherever cos(x) is zero.
sleek obsidian
Where is $\sec{(x)}$ undefined?
π/2 and 3π/2 and more
small lance
so we say just tthose two. pi/2 and 3pi/2
cosmic steppe
Yes
small lance
yep
restive bloom
I've been reading up on kinematic motion, and I'm trying to calculate the point at which a particle's path intersects with a line when it's being affected by both vertical and horizontal forces. Say, gravity and wind. I've found several videos that explain how to find the point of intersection, as well as the particle's angle and velocity at the moment of intersection, but nothing that accounts for two-dimensional acceleration.
Does anyone know what the process for this would be, or where I could read more about it?
restive bloom
I'm not actually sure what flavor of math this falls under, so if it's more appropriate somewhere else, please let me know.
runic glade
I've been reading up on kinematic motion, and I'm trying to calculate the point ...
Just find the resultant vector of the forces and the acceleration becomes unidirectional
tacit jewel
hi i dont understand how i'd go about starting this
cedar kilnBOT
hi i dont understand how i'd go about starting this
tacit jewel
all 3 parts pretty much connect so im not sure how to do any of them
<@&286206848099549185>
<@&286206848099549185>
dire geode
do you know the definition of feasible
cedar kilnBOT
@tacit jewel Has your question been resolved?
tacit jewel
yes
dire geode
yes
what are you stuck on then?
Show that if x is feasible for (2), then tx is feasible for (3)
runic glade
There is no attachment
dire geode
the finite sum has an exact formula
crimson sedge
1/(n^2) converges to 0
And (1+2+3+...) diverges to positive infinity
But hold a sec
small lance
And (1+2+3+...) diverges to positive infinity
||actually that's -(1/12) 🤓 ☝️ ||
crimson sedge
||actually that's -(1/12) 🤓 ☝️ ||
e
Anyway, what is S_n
Because it's quite hard to understand you writing
I mean, what is the expression
Anyway, you can substitute 1+2+3 ... as $\frac{1}{2}n(n+1)$
wraith daggerBOT
Yustina
wraith daggerBOT
Yustina
crimson sedge
Equals $\frac{n+1}{2n}$
wraith daggerBOT
Yustina
crimson sedge
Now calculate the limit of this
It should be 1/2
you might know how to calculate this type of limit
stuck sigil
Help please
cedar kilnBOT
Help please
stuck sigil
The corresponding heights of two similar cylinders is 2:5 what is the ratio of their volumes?
@indigo steeple
indigo steeple
can you not ping random helpers who have the misfortune of having the first alphabet of their name be A
crystal raptor
read the rules, don't ping people
stuck sigil
Rules say after 15 minutes feel free to ping <@&286206848099549185>
crystal raptor
not paid to be here 🙂
stuck sigil
Interesting you want to be a helper but won’t help
stuck sigil
not paid to be here 🙂
Not my problem
amber ridge
they don’t have to help 24/7 and they don’t need to help with stuff if they don’t want to
stuck sigil
Then why be a helper
indigo steeple
yeah its not like i have a programming problem to solve or anything
stuck sigil
Lol
amber ridge
Then why be a helper
it’s fun to help people with interesting questions, not just homework problems
indigo steeple
volume is pi r^2 h
meaning height scales linearly to volume
so volume ratio is also 2:5
indigo steeple
why
indigo steeple
i mean just do $V_1 = \pi r_1^2 (2)$ and $V_2 = \pi r_2^2 (5)$
radius is same cause cylinders are similar
no nvm
wraith daggerBOT
AbdullahTrees
dire geode
Then why be a helper
No one's obligated to help you. Stop being entitled
stuck sigil
No one's obligated to help you. Stop being entitled
✌🏿
dire geode
.close
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dire geode
The fewer assholes we have in the server, the better
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robust hearth
how to i integrate xe^x dx?
cedar kilnBOT
how to i integrate xe^x dx?
astral crown
Use integration by parts
cosmic steppe
how to i integrate xe^x dx?
Integration by parts
robust hearth
k
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golden sundial
trying to do the d
cedar kilnBOT
trying to do the d
golden sundial
I know that x1+x2 = -b/a and that x1•x2=c/a
I have to find the value of k so that x_1+2x_2 = 1
true nebula
cedar kilnBOT
true nebula
How do I use the top 3 equations to solve for the bottom two?
Like I know $\frac{\partial w}{\partial x} = \frac{1}{x^2 + y^2 + z^2}(2x)$
wraith daggerBOT
beanbeanjuice
true nebula
actually that's not right but my point still stands
actually wait it is right
but how do i use that to solve for $\frac{\partial w}{\partial u}$
wraith daggerBOT
beanbeanjuice
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steady cloud
Hello, guys. I need some help. I have the following set and I need to know sup(A) and inf(A).
I think sup(A) = -1 (if I increase n, it gets closer to -1) and inf(A) = 0 (if I use n = 2, it turns to be 0).
I'm not sure of my answers so if you guys could explain or correct me, i'd be grateful.
celest ledge
sup(A)=max(A)=0
(2-n)/(n-1)=-1+1/(n-1) decreases, when n is 2 this is maximal
inf(A)=-1
steady cloud
So I basically flipped the answers?
celest ledge
Yeah
lapis flare
hey I have a mathematics question, Let's say if I created a list that contains a number of different values
taken from 1,000 people and at least from each person 10,000 was taken,
and it may range from 10,000 to 100,000, but at least its 10,000 from
each person. If the value is divided by 1,000 people, the value will be
reduced? Like dividing a million, 1000000 by 10000 it will be reduced,
but the question is will it still be reduced if the condition is applied that
it is not less than 10000 per person
lapis flare
how come?
fair geyser
Let's say if I created a list that contains a number of different values
taken from 1,000 people and at least from each person 10,000 was taken,
and it may range from 10,000 to 100,000, but at least its 10,000 from
each person.
If the value is divided by 1,000 people, the value will be
reduced? Like dividing a million, 1000000 by 10000 it will be reduced,
but the question is will it still be reduced if the condition is applied that
it is not less than 10000 per person
see it's exactly middle
it makes sense and then it goes incoherent
just explain again please, it was bad luck or something
lapis flare
can we talk in vc I think I can explain more clearly?
fair geyser
i don't like vc :(
lapis flare
you can mute
upper swallow
I'm not sure where I'm going wrong
cedar kilnBOT
I'm not sure where I'm going wrong
wraith daggerBOT
cedar kilnBOT
@upper swallow Has your question been resolved?
upper swallow
I did that
scenic mural
ill post pic
1 sec
make it into two integrals
one for root of x
and one for the other one
lilac crane
Is it this?
scenic mural
yes
upper swallow
why are both squared rather than just the whole thing?
lilac crane
Because you take the volume
Using the big radius
Minus the volume taken from the small radius
upper swallow
yeah and the formula I have for that is pi r^2 which becomes pi f(x)^2 and f(x) is sqrt(x) - x^2
so id get pi(sqrt(x) - x^2)^2
lilac crane
That's a case your lower function is y=0
See you take the whole volume using the top function, then you subtract out the volume obtained by taking the lower volume as the radius
upper swallow
yeah, i did that
scenic mural
its not the same
taking the difference of the two functions before calculating the volume is not the same as difference in volume between the two functions
upper swallow
huh
each bound should have its own volume calculation then minus one from the other?
scenic mural
yeah
scenic mural
can u show what u did?
upper swallow
i got that from 2x^7/2 / 7/2
scenic mural
if you calculate volume of top function and bottom function seperately
and then take the difference
you will get right answer
upper swallow
the last one gives 0
scenic mural
upper swallow
ah okay
cedar kilnBOT
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upper swallow
Like this?
upper swallow
(pi (1/2(1)^4)?
scenic mural
upper swallow
the formula is pi f(x)^2 though?
scenic mural
you already did it
scenic mural
show
scenic mural
no worries 🙂 hopefully it made sense
upper swallow
yeah it does, my tutor was just saying to do it as f(x) = sqrt(x) - x^2 then integrate from there
upper swallow
for area if its y=f(x) and y = g(x) should i always do them seperately?
scenic mural
i would think so, but recommend testing it to be sure
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heady spindle
Hi, I have to proof the following proposition, but I don't have any idea of how to do it :c
Prove that 3a^2 - 1 never is a perfect square
tropic oxide
@heady spindle are you familiar at all with modular arithmetic?
heady spindle
so so
tropic oxide
you can show that perfect squares are never congruent to 2 mod 3.
or equivalently that a perfect square is always either 0 or 1 mod 3.
heady spindle
wait, n = 3a^2 - 1 is equal to n = -1 mod 3?
cedar kilnBOT
@heady spindle Has your question been resolved?
steel stratus
can someone help to check if my work is correct specially the critical value
civic stag
help me pls
mortal plaza
help me pls
#❓how-to-get-help this channel is already claimed
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dire geode
.close
midnight zealot
.close
south dawn
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tidal ridge
hi! I am studying for an entry exam for a university and in the practice tests I am stuck with this integral. I've never seen an integral like this where you are treating the integer part of a number differently than the float part
tidal ridge
So [1,2] = 1 & {1,2} = 0,2
tropic oxide
fractional part
tidal ridge
But i do not understand how to integrate this stuff
tropic oxide
$\int_3^6 [x] \dd{x} = \int_3^4 [x] \dd{x} + \int_4^5 [x] \dd{x} + \int_5^6 [x] \dd{x}$
wraith daggerBOT
Ann
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carmine adder
Hi, I got a logarithm question, can anyone please help, thanks (I’m up to a certain part, idk if it’s correct tho)
delicate ember
why the 2?
delicate ember
so you now have
$2log_x(5) = log_5(x-1)$
wraith daggerBOT
ホタル
carmine adder
So answer is x = 26?
Not 26
sacred stone
I think it's $\log_5(x) -1$
wraith daggerBOT
秋水
carmine adder
Wait what does that mean
delicate ember
without the brackets thats how it should be interpreted
logx-1
would mean log(x) - 1
carmine adder
For the right side part?
delicate ember
yeah
carmine adder
O
sacred stone
then you know $\log_x (5) = \frac{1}{\log_5 (x)}$, you can solve the equation
wraith daggerBOT
秋水
carmine adder
$log_5(x) - log_5(5) = log_5(x-1)$
wraith daggerBOT
Techno
carmine adder
This?
sacred stone
$log_5(x) - log_5(5) = log_5(x-1)$
this is wrong
carmine adder
Wait what
dire elm
$log_5(x) - log_5(5) = log_5(x-1)$
But $\log_5 a - \log_5 b = \log_5 \left(\dfrac ab\right)$
wraith daggerBOT
pi over four
sacred stone
I mean in the original question is $\log_5(x) -1$, not $\log_5(x-1)$
wraith daggerBOT
秋水
sacred stone
no, I mean $\log_a b = \frac{1}{\log_b a}$
wraith daggerBOT
秋水
carmine adder
Oh yeah true
sacred stone
the original question should be $\sqrt{x}^{\log_5 (x)-1}=5$
wraith daggerBOT
秋水
carmine adder
No the original question should be $\sqrt{x}^\log_5 (x-1)=5$
wraith daggerBOT
Techno
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
carmine adder
Ok
dire elm
Wait do U mean they have to put brackets?
If they meant $\log_5 (x - 1)$, they would have put brackets around $x - 1$.
carmine adder
How about the other one?
wraith daggerBOT
pi over four
dire elm
I guess it’s the other one then
yeah
carmine adder
What is the next step?
carmine adder
**秋水**
Does it reach that step or is that just an example?
I’m confused on that to do next :( me bad at maths
sacred stone
$\sqrt{x}^{\log_5 (x)-1}=5$
wraith daggerBOT
秋水
sacred stone
take $\log_{\sqrt{x}} ()$ on both sides
wraith daggerBOT
秋水
sacred stone
then you get $\log_5 (x) -1 = \log_{\sqrt{x}}(5)$
wraith daggerBOT
秋水
sacred stone
notice $\log_{\sqrt{x}}(5) = 2 \log_{x}(5)$
wraith daggerBOT
秋水
sacred stone
then you get $\log_5 (x) -1 = 2 \log_{x}(5)$
wraith daggerBOT
秋水
sacred stone
then $\log_5 (x) -1 = \frac{2}{\log_5(x)}$
sacred stone
now you can solve $\log_5(x)$
wraith daggerBOT
秋水
dire elm
Shouldn’t this be 1/2 instead of 2
$$\log_{\sqrt x} (5) = \dfrac{\log_x 5}{\log_x \sqrt x} = \dfrac{\log_x 5}{\dfrac12} = 2\log_ x 5$$
wraith daggerBOT
pi over four
sacred stone
more general $\log_{a^b} (c) = \frac{1}{b} \log_a (c)$
wraith daggerBOT
秋水
carmine adder
Yeah I get that one
wraith daggerBOT
秋水
sacred stone
or x = 1/5
carmine adder
Ok
Ohh yeah I understand now
Tysm !!!
Thank you for you guys’ help! It’s is very useful and I appreciate it a lot
Damn discord is the future google
Uhhm I got another question
sacred stone
just take $3^{()}$ on the both sides
wraith daggerBOT
秋水
sacred stone
then $1-2 \times 3^x = 3^{2x+1}$
wraith daggerBOT
秋水
sacred stone
notice $3^{2x+1} = 3^{2x} \times 3^{1} = (3^{x})^2 \times 3$
wraith daggerBOT
秋水
carmine adder
**秋水**
Wait so u add like the 3 there on both sides so left side rid of log and right adds the 3 there?
sacred stone
actually you can get this directly using the definition of logarithm
you can start from this
carmine adder
Wait how?
sacred stone
$\log_a(b) = c$ then $b=a^c$, right?
wraith daggerBOT
秋水
carmine adder
Yes
sacred stone
so you get $1-2 \times 3^x = 3^{2x+1}$
wraith daggerBOT
秋水
carmine adder
Wait let me see
Then isn’t it supposed to be log something on the right
Since the log_3 is a
Oh wait
sacred stone
what? you have $log_3 (p) =q$, right? so $p = 3^q$
wraith daggerBOT
秋水
sacred stone
so you have $1-2 \times 3^x = 3 (3^x)^2$, right?
wraith daggerBOT
秋水
carmine adder
Yes
wraith daggerBOT
秋水
wraith daggerBOT
秋水
carmine adder
Then x = -1
sacred stone
yes
eternal basin
Okay so this will be a bit hard to say on mobile but essentially, .3 repeating = 1/3, .9 repeating = .3*3, 1/3*3 = 1
So there is no difference in infinitesimally small values and the number they approach
So why does limit work then for infinitesimally small values
If you have a removable discontinuity at 1, .9 repeating = 1, but would .9 repeating be defined in a function with 1 as a removable discontinuity?
if it’s not defined then how is .3 repeating * 3 = 1
If it is defined, same as above
cosmic steppe
Okay so this will be a bit hard to say on mobile but essentially, ```.3 repeatin...
Well, not exactly
The thing is that calculators have a limit
And they will round answers up or down
When you type 0.33333333333333 • 3 into a calculator, it'll get a number so close to 1, it'll say it's 1
eternal basin
1/3 * 3 = 1, yes?
cosmic steppe
Yeah
eternal basin
1/3 = .3 repeating?
cosmic steppe
Infinitely, yes
delicate ember
.9999... is equal to 1
cosmic steppe
We did
The thing is that a finite number of decimal places doesn't equal to the infinite number of decimal places
delicate ember
when we say lim x -> 1, x is never equal to 1
delicate ember
.9999... is not infinitesimally smaller than 1 it is 1
delicate ember
you can look up the proof online ig
cosmic steppe
Same shit
delicate ember
there should be a difference but there isn't (acc to me)
eternal basin
it’s not
cosmic steppe
Terminology my ass lol
eternal basin
infinite refers to large numbers
cosmic steppe
There's a reason I put it in quotation marks
eternal basin
wow that really feels like it shouldn’t work
like logically i understand it but
it feels so off
cosmic steppe
I mean the logic is kinda the proof ngl lol
delicate ember
Another way of representing it (without subtracting ig)
eternal basin
yeah that feels so weird lmao
cosmic steppe
But once again, you can think about having infinitesimal difference that's just negligible
What a goofy ahh word
eternal basin
I think u 2 are saying completely different things
eternal basin
No it’s not
It’s completely different
what @delicate ember seems to be saying
Is that there’s no infinitesimal difference
It just is
What you’re saying would break limits, I think
Unless I misunderstand
cosmic steppe
They're representing 0.9999... as a variable; regardless, it is still 0.999....
eternal basin
wtf are u typing
delicate ember
me?
cosmic steppe
10x - x would be 9x, but there would still be some negligible difference
delicate ember
10x - x would be 9x, but there would still be some negligible difference
I dont think so
cosmic steppe
Because x js 0.99999....
delicate ember
10 x - x = 9x is a fact
eternal basin
^
cosmic steppe
So wouldn't there be some discrepancy
delicate ember
no matter x
cosmic steppe
I mean algebraically I get it
cosmic steppe
Oh wait you know what
eternal basin
@delicate ember is saying it is literally =
cosmic steppe
Yeah lol
eternal basin
which is literally my question
eternal basin
Yes, your explanation made sense
@cosmic steppe this guy just kept saying completely different things which threw me off
eternal basin
Yeah my brain still hurts from that explanation, it feels so counterintuitive but I can’t tell where the specific part is that is so confusing
Again, logically it makes sense but wtf
delicate ember
https://youtu.be/jMTD1Y3LHcE
Watch this
cosmic steppe
At this point Mathematics is collapsing
The world around us is breaking apart and everything we knew, loved, touch and fathomed is dissipating into dust and ashes
eternal basin
Yo I get you’re trying to be funny
But if you don’t have anything to contribute pls leave
cosmic steppe
Perhaps we can show it as a summation?
To prove it
Because it's just 0.9 + 0.09 + 0.009 and so on
So $\sum_{n=1}^{\infty}9\left(\frac{1}{10}\right)^n$
Bruh
wraith daggerBOT
Umbraleviathan
cosmic steppe
If we take the sum of that:
$$\frac{0.9}{1-0.1}$$
$$\frac{0.9}{0.9}$$
$$1$$
wraith daggerBOT
Umbraleviathan
cosmic steppe
@eternal basin maybe this proves it
To me this makes more sense because you're adding an infinite number of decimals places
So if we take that sum
molten vine
**Umbraleviathan**
Could you explain how you got this fraction?
delicate ember
Infinite Geometric series
cosmic steppe
I just turned the fractions into decimals because typing a fraction into a fraction is ass
cosmic steppe
Infinite Geometric series
But that should prove it right
Because 0.9999... is quite literally the sum of 0.9 + 0.09 and so on
cosmic steppe
The only gripe I have is that the sum of an infinite geo series is defined by a limit itself
At least
delicate ember
yeah
cosmic steppe
By proof
molten vine
I just turned the fractions into decimals because typing a fraction into a fract...
I get how you got 0.9 as the numerator but how did you get 0.9 as the denominator?
cosmic steppe
I get how you got 0.9 as the numerator but how did you get 0.9 as the denominato...
1-r
1-(1/10)
9/10
0.9
eternal basin
Idk I still watched the video and I don’t get why it’s equal to and not just a limit
cosmic steppe
**Umbraleviathan**
Perhaps this?
cosmic steppe
now you're getting me to think about it lol
delicate ember
I guess one way to think of it intuitively would be
if 0.999... weren't equal to 1
that would mean there would be a number in between them
can you think of such a number?
eternal basin
but limits don’t involve numbers between the values either
by nature there is no number between it
it’s just a fancy way of getting around the fact that = doesn’t work anymore sometimes
so you say well technically it isn’t really equal to when for all intents and purposes it totally is
if there was a number in between you wouldn’t be able to calculate anything because for example
let’s say we are trying to find the limit of x->3 for (x^2-9)/(x-3)
the answer is obviously x+3, or 6
But we wouldn’t be able to get that if we didn’t pretend that for all intents and purposes limits are equal to
we would have to do 2.9 repeating
cosmic steppe
Which is where the summation comes in
And evaluates what 0.99999... could be
But I get what you're saying
Perhaps you can everything in https://en.m.wikipedia.org/wiki/0.999...
Because it's been a debate among mathematicians on how to prove it I guess
The sum that I showed is perhaps the easiest way to show it, but it's innately backed up by a limit proof, and I think you wanna stray away from limits, correct?
eternal basin
it’s not equal to
just asked my professor
it’s a limit, but not equal to
which makes 50000x more sense
eternal basin
because people are lazy according to professor
cosmic steppe
I see where you're coming from now
cosmic steppe
Lol based professor
woeful ore
Hi! So I think i made a mistake in my working (near the green star), I have the final answer from inputting a big number in the calculator but i’m not sure how to use this way to do it (we are given A will stabilise at a fixed size in the long term)
dire elm
So you have already gotten the eigenvalues and eigenvectors of $A$. You can use this information to write $A^n$ explicitly. Afterwards, you can take $\displaystyle \lim_{n\to\infty}$ of each entry.
wraith daggerBOT
pi over four
woeful ore
since (-1/2)^inf would be 0, wouldn't it just be (75 60) but that would be incorrect?
I'm not sure how to go about this sorry
dire elm
since (-1/2)^inf would be 0, wouldn't it just be (75 60) but that would be incor...
What did you get for $A^n$?
wraith daggerBOT
pi over four
woeful ore
Do you mean A^n when n->inf?
dire elm
No, just $A^n$ for any arbitrary positive integer $n$.
wraith daggerBOT
pi over four
woeful ore
0 1.25
0.4. 0.5?
dire elm
This is $A^1$. What about for general $n$?
wraith daggerBOT
pi over four
woeful ore
Ah im not sure then, is this something we should find or should be given?
dire elm
Since you have the eigenvectors $v_1$ and $v_2$, the matrix $P = \begin{pmatrix}v_1&v_2\end{pmatrix}$ would give a diagonal matrix $D = P^{-1}AP$. We say that it diagonalizes $A$. If you want to find $A^n$, you can find $D^n$ instead. $D^n = P^{-1}A^nP$, so $A^n$ is really just $PD^nP^{-1}$. Can you calculate that?
wraith daggerBOT
pi over four
dire elm
$P = \begin{pmatrix}v_1&v_2\end{pmatrix}$ means the first column is the column vector $v_1$ and the second column is the column vector $v_2$, so it's a 2 x 2 matrix.
wraith daggerBOT
pi over four
dire elm
No, $A^n$ should be in terms of $n$.
wraith daggerBOT
pi over four
dire elm
1 0
0 -0.5
Can you show me how you found this?
dire elm
You calculated $D$. Can you calculate $D^n$ from here?
wraith daggerBOT
pi over four
dire elm
No, when you raise a diagonal matrix to a power, you just raise every number to that power.
woeful ore
1^n 0
0 -0.5^n?
dire elm
For example,
$$\begin{pmatrix}1&0\0&-\frac12\end{pmatrix}^2 = \begin{pmatrix}1^2&0\0&\left(-\frac12\right)^2\end{pmatrix}^$$
wraith daggerBOT
pi over four
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
dire elm
1^n 0
0 -0.5^n?
Yes
woeful ore
^
wraith daggerBOT
pi over four
woeful ore
is A^n the same as D^n?
dire elm
No, it is $PD^n P^{-1}$.
wraith daggerBOT
pi over four
dire elm
Note that you can't simplify $PP^{-1}$ to $I$ because matrix multiplication is not commutative.
wraith daggerBOT
pi over four
woeful ore
:,)
1 0
0 1/2^n+1
wait
yeah nah im not sure sorry
I thought we could use this to figure the question out
dire elm
Is that what $c_1$ and $c_2$ means?
wraith daggerBOT
pi over four
woeful ore
from the other examples we did we used x0, yeah
dire elm
from the other examples we did we used x0, yeah
But how is it justified?
A method I thought of, is that you can use the fact that $x_0 = c_1 v_1 + c_2 v_2$. This is an equation involving vectors, which is actually just a system of linear equations. You can solve for $c_1$ and $c_2$.
wraith daggerBOT
pi over four
dire elm
This could be simpler
woeful ore
are we using
1.25 and -2.5
1 1
as v1 and v2?
dire elm
Yeah
woeful ore
c1= 36
c2=-6
dire elm
Yea
dire elm
So $x_n = 36 \lambda_1^n v_1 - 6 \lambda_2^n v_2$
wraith daggerBOT
pi over four
dire elm
I think it's .close
woeful ore
.close
vagrant leaf
i need help forming dis question into an equation ;-;
cedar kilnBOT
i need help forming dis question into an equation ;-;
vagrant leaf
i have no clue where to start
molten vine
Think of it as mowing fractions of the field
If it takes you 5h to mow the field that means you mow 1/5 of the field per hour
If you mow 1/5 of the field before school and he mows 1/3 of the field before school how much is left to mow
vagrant leaf
If you mow 1/5 of the field before school and he mows 1/3 of the field before sc...
ohhhhhh
i never thought of it as that way
wait so
its 1 - 1/5
which is 4/5
and teh i goooo 4/5 - 1/3
which is likeeeeeeeeeee
12/15 - 5/15
whichh is ssssss
7/15 😄
soooo
if his dad
plows the rest
thennnnn
i goo
(7/15)/1/3
which issss
D:
1.4 hoursss
.close
tall berry
cedar kilnBOT
tribal pier
ooh, limits are fun. What do you notice about trying to substitute in 0 for x in part a?
tall berry
ummm
tribal pier
actually, first off: are you familiar with what a limit is?
tribal pier
ok, so a limit is basically what happens as a particular variable gets super duper close to the value described
tribal pier
infinitely close
tall berry
okok
tribal pier
so it technically isn't, but it basically becomes that value
for most limits, you can start by trying to simply say "what if we just plug in the value for x?"
tall berry
does it work for a?
tribal pier
x, a, b, theta, whatever variable the limit asks for
oh, part a
try it and see what happens
tribal pier
this is true, which means we need to do more work
tall berry
guide me pls im lost 😞
tribal pier
it's undefined because x is in the denominator, right?
so let's use our algebra skills to try to shuffle things around until there isn't an x in the denominator anymore
tall berry
ok
tribal pier
so what do you notice about the numerator?
tall berry
that is x+3 to the power of 3
tribal pier
here's a hint if you need it:
||try a difference of cubes, 27=3^3||
tall berry
so try to open up braket?
reef venture
no
tribal pier
you could, but it might be easier to use a difference of cubes
reef venture
a^3 - b^3 has a formula
tribal pier
(a-b)(a^2+ab+b^2)=a^3-b^3
so if we apply that to our numerator with a=(x+3) and b=3 we get something like this:
((x+3)-3)((x+3)^2+3(x+3)+3^2)
from here you can simplify far enough. We're trying to find an x that we can cancel with the denominator
tall berry
srry still trying to solve
tribal pier
$\lim_{x \to 0} \frac{(3+x)^3-3^3}{x}$
cube, not square
wraith daggerBOT
cos(x)=1-x^2/2!+x^4/4! -....
crimson sedge
cube, not square
mb
tribal pier
all good
crimson sedge
**cos(x)=1-x^2/2!+x^4/4! -....**
anyways from here use limit defition of derivative @tall berry
tall berry
ok ill try
tribal pier
anyways from here use limit defition of derivative <@722186879570673833>
this isn't a derivative question, it's just a limit question tho?
tall berry
calculus is so hard
crimson sedge
this isn't a derivative question, it's just a limit question tho?
$f'(x)=\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
wraith daggerBOT
cos(x)=1-x^2/2!+x^4/4! -....
