@analog vault
#help-10
cyan sinew
what part in the second line was not clear?
do you know the basic expansion of a binomial expression?
unique shuttle
no no i jus got it thanks
cyan sinew
cool
unique shuttle
thank you for ur time
cyan sinew
run .close if done
obtuse pebbleBOT
knotty jacinth
.close
obtuse pebbleBOT
Available help channel!
crude marsh
Hello i need help on question 1 B and 1 C
obtuse pebbleBOT
Hello i need help on question 1 B and 1 C
bitter crater
Hello i need help on question 1 B and 1 C
do you know the general equation of a circle?
bitter crater
it is
wait sorry I thought you wanted hep with A part, could you send a better photo, in this the B and C part aren't visible
timid silo
firstly, yes it is an equation of circle
then
do you know the homogoenous 2nd degree eqn ?
crude marsh
nope
bitter crater
the special thing about a tangent is that the perpendicular distance from the centre of the circle is equal to the radius of the circle
do you know how to find the perpendicular distance from a point to a line?
crude marsh
no
bitter crater
well no problem, there is another thing we can do
since the line cuts the circle only at one point
I meant touches
then there is only one value of x and y which satisfy both right
crude marsh
yes
bitter crater
so if we subsitute y = k in the equation of the circle we will get a quadratic in x
and since there must be only one point common between circle and line that must mean the quadratic will only have one root
crude marsh
if we do that it will be (X-a)^2+(K-b)^2=r^2
bitter crater
if we do that it will be (X-a)^2+(K-b)^2=r^2
You can put it in the equation given to you
crude marsh
and since there must be only one point common between circle and line that must ...
oh okay
crude marsh
b^2-4ac =0 or something
bitter crater
you made the quadratic?
crude marsh
what quadratic
bitter crater
so if we subsitute y = k in the equation of the circle we will get a quadratic i...
.
bitter crater
yw
also i just realized due to the conditions of this question there was a much easier way to do this
rather than doing all this quadratic stuff
crude marsh
oh how?
bitter crater
since the line is of form y= k
it will be parallel to the x axis
that means the distance between it and centre of circle will be = |k-b|
where b is the y coord of centre of circle
and if you equate the above to the radius of the circle
you will directly get k
as perp. distance from centre to tangent is radius of the circle
crude marsh
and if you equate the above to the radius of the circle
what do you mean by this?
bitter crater
what do you mean by this?
I meant |k-b| = r
bitter crater
bitter crater
so that means that it will be |k-2| = 25
the radius is 25?
crude marsh
yes i worked it out
bitter crater
then sure
crude marsh
if we do that it will be (X-a)^2+(K-b)^2=r^2
using the equation
ok
what about C if possible..?
bitter crater
that would required some geometry
crude marsh
okay
crude marsh
not really.
bitter crater
what about now?
crude marsh
yes
bitter crater
so perpendicular from centre to a secant does what to the secant?
bitter crater
well thats the definition of a perpendicular
crude marsh
sorry im kind of in yr11 and doing y12 work..
bitter crater
major thing it does is that it bisects the secant in half
bitter crater
sorry im kind of in yr11 and doing y12 work..
its ok
crude marsh
major thing it does is that it bisects the secant in half
oh ok
bitter crater
sorry im kind of in yr11 and doing y12 work..
I'd suggest asking your teacher or perhaps use some videos online to learn about the circles and their geometry a lil bit
I think that would solve most of your problems
crude marsh
I'd suggest asking your teacher or perhaps use some videos online to learn about...
okay
its fine thank you though for B
bitter crater
major thing it does is that it bisects the secant in half
yw, you can use this fact and pythogorous theoram and then a similar thing I did before with the tanget to solve C part
you can have at it on your own
crude marsh
yw, you can use this fact and pythogorous theoram and then a similar thing I did...
oh yes
oh one more question
how would you work out the radius of a circle with 2 coordinates?
bitter crater
how would you work out the radius of a circle with 2 coordinates?
2 coordinates of what?
you mean two points on a circle?
crude marsh
yes
bitter crater
you have nothing else?
bitter crater
oh its the diamter
crude marsh
oh wait
oh didnt see that
wait so you work out the length of the line and divide by 2?
bitter crater
for the radius yes
crude marsh
whats the equation for length again?
bitter crater
root((x2-x1)^2 + (y2-y1)^2)
bitter crater
for any two arbitary points with no speciality like the diamter you can have infinite number of circles passing through them, you need atleast 3 pts to determine the circle
bitter crater
yw
obtuse pebbleBOT
@crude marsh Has your question been resolved?
timid silo
obtuse pebbleBOT
kind hollow
?
golden night
timid silo
idk
obtuse pebbleBOT
@timid silo Has your question been resolved?
tardy epoch
Did you try checking small cases of n
obtuse pebbleBOT
@timid silo Has your question been resolved?
fallow atlas
how should I approach 125?
obtuse pebbleBOT
how should I approach 125?
fallow atlas
fallow atlas
how would i get y=sin^2x in terms of x
jagged ocean
No need
wanton dune
i should determine wheter this set of matrices is linearly dependent
obtuse pebbleBOT
i should determine wheter this set of matrices is linearly dependent
wanton dune
I know that the third matrix is sum of 1first one + 2second one
so the set is lin dependet
gilded needle
I know that the third matrix is sum of 1*first one + 2*second one
ok, what can you conclude from that?
gilded needle
so the set is lin dependet
^ correct
wanton dune
but how do i determine it generally
is there some algorithm? or i just have to see it like in this example
gilded needle
probably the most straightforward way generally would be to convert the matrices to row vectors or column vectors, stack them into a matrix, and then do row reduction
wanton dune
like i would put those three columns from first matrix under each other and ill get first column and then the same for remaining 2? so ill get 9 by 3 matrix?
gilded needle
yes correct
or equivalently you could take the three rows of the first matrix and combine them into one long 1x9 row
do the same with the other two
then stack the three rows to get a 3x9 matrix
either way will work
wanton shoal
i have no idea how to attempt this probklem
obtuse pebbleBOT
i have no idea how to attempt this probklem
@wanton shoal Has your question been resolved?
steady umbra
Why isn’t it possible to solve it
obtuse pebbleBOT
Why isn’t it possible to solve it
jovial portal
Maybe because it is diverging or the root gets negative
probable because $x^3 >> x^2$ therefor the root gets negative
warm shaleBOT
Tobi
steady umbra
The root can get negative, can’t it?
jovial portal
Let x=2 then we have $x^2- x^3 = 4-8 = -4$
warm shaleBOT
Tobi
flat geyser
it does converge
steady umbra
But you can take a 3 root of a square number
past sand
Why isn’t it possible to solve it
It is, your calculator just isn't good enough
steady umbra
> Why isn’t it possible to solve it
It is, your calculator just isn't good enoug...
How do you solve it then?
flat geyser
How do you solve it then?
do you know l'hopital?
past sand
Personally I ask WolframAlpha
steady umbra
do you know l'hopital?
Yeah
flat geyser
Yeah
$\sqrt[3]{x^{2}-x^{3}}+x\to x\left(\sqrt[3]{\frac{1}{x}-1}+1\right)\to\frac{\left(\sqrt[3]{\frac{1}{x}-1}+1\right)}{\frac{1}{x}}$
warm shaleBOT
Combustion
steady umbra
So the answers 0 to the original question?
flat geyser
no
jovial portal
Are you familiar with complex numbers?
steady umbra
Haven’t got there yet
jovial portal
because for real values the upper border is x=1 then the function wont be defined because the root is negative
flat geyser
because for real values the upper border is x=1 then the function wont be defin...
it's a cubic root
flat geyser
**Combustion**
just apply l'hopital on this as it's 0/0
simplify, and you'll get the answer
steady umbra
Ight thanks
My original question was to find the asymptotes of the function (everything under the cubic root)
I figured that there was no horizontal and vertical ones
And I’m stuck on the obligate one
robust raven
this limit should be solved in a such way:
$\lim_{x \to \infty } \left( x-\sqrt[3]{x^{3}-x^{2}} \right)={{\cdots }}\\a-b=\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}$
warm shaleBOT
Joanna Angel
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
lone egret
Do i just plug in n, evaluate like normal and multiply my answer by 2?
obtuse pebbleBOT
Do i just plug in n, evaluate like normal and multiply my answer by 2?
gilded needle
yea, unless there's some more context that gives n a specific value?
iron mason
Hi
placid locust
hii
you can try to convert sqrt x to exponent form
that will make it easier to see what you can do
placid locust
Yes indeed
iron mason
Do I subtract 5/2 with 1/2?
placid locust
da
iron mason
On both sides
placid locust
You can simply 4/2 as well in the exponent
obtuse pebbleBOT
@iron mason Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
topaz vault
how i have to do
obtuse pebbleBOT
how i have to do
fossil crag
Might want to expand
topaz vault
u mean 3x/2 - 5/2?
fossil crag
The whole product
topaz vault
and the other its x^2+2x+1?
fossil crag
Nono it's x²-1
topaz vault
truue
fossil crag
I'm asking to write (3x/2 - 5/2)(x²-1) as a sum
fossil crag
can u show me?
Do you know what (a+b)(c+d) is?
Heard of FOIL technique? Or maybe a similar name
topaz vault
no
topaz vault
yes
fossil crag
Ok, so what does it mean?
topaz vault
a sum of products by using the fact that multiplication distributes over addition
fossil crag
Ok
fossil crag
how i have to do
So can you expand the product inside the integral?
obtuse pebbleBOT
@topaz vault Has your question been resolved?
torn jewel
obtuse pebbleBOT
torn jewel
confused with step 4
i expandedthe numeraotr and got
x^2 + 2x + x^2 -32x
but idk what is inserted in the brackets for both?
or what kinda process the step is undergoing
torn jewel
true'
timid silo
i get u..
timid silo
k
fossil crag
Don't worry about the 'size of the boxes'
They're just made big enough so you can put virtually anything in it
timid silo
They're just made big enough so you can put virtually anything in it
yh but sometimes in the actual exam, the size of box can help you to find answers
so i think thats why she got confused
fossil crag
yh but sometimes in the actual exam, the size of box can help you to find answer...
Ooof sometimes is like a coin toss, never rely on it
timid silo
true
obtuse pebbleBOT
@torn jewel Has your question been resolved?
midnight gorge
can someone help me
obtuse pebbleBOT
can someone help me
polar fossil
it may be helpful to imagine printing 100 business cards. How much would that cost?
obtuse pebbleBOT
@midnight gorge Has your question been resolved?
midnight gorge
oh myb i was afk
midnight gorge
it may be helpful to imagine printing 100 business cards. How much would that co...
100*0.15
polar fossil
<@726569994065805323> Has your question been resolved?
make sure you answer the bot or else she'll close your channel
polar fossil
yes, so then how much, in total, does it cost per card?
or in other words, how much would you have to sell each card for, to break even?
midnight gorge
more than the rate
carmine bear
I have to find the value of $C$ in $$\psi(x) = Ce^{-\alpha x}\left(1-e^{-\alpha x}\right)$$ using the fact that $$\int_0^\infty \psi^*\psi \dd x =1$$ but i don't know what the complex conjugate of something like this would be its probably something obvious but i cant figure it out
obtuse pebbleBOT
I have to find the value of $C$ in $$\psi(x) = Ce^{-\alpha x}\left(1-e^{-\alpha ...
warm shaleBOT
I can't believe you've done this
obtuse pebbleBOT
@carmine bear Has your question been resolved?
carmine bear
.close
wintry iris
What would be the right bounds for this one?
obtuse pebbleBOT
What would be the right bounds for this one?
@wintry iris Has your question been resolved?
tame narwhal
I wanna say the inner integral bounds should be from y to 1, but not 100% sure
if you don't get an answer, try asking in #probability-statistics
naive mountain
A 3 digit multiple of five is created using the digits 0,3,4,5 and 7. How many numbers are possible if all digits in the number must be different?
obtuse pebbleBOT
A 3 digit multiple of five is created using the digits 0,3,4,5 and 7. How many n...
hexed gull
A 3 digit multiple of five is created using the digits 0,3,4,5 and 7. How many n...
consider how many possibilities you have for the first digit
and then how many for the second
and then how many for the third
multiplying them yields the number of possible 3-digit numbers
zenith raft
i think you should start with the third
hexed gull
either exclude 0 for the first or start with third yeah
zenith raft
well i meant the number of numbers you can make starting (100s place), for example, with 3 and starting with 5 are different
zenith raft
multiplying them yields the number of possible 3-digit numbers
so this isn't exactly right
hexed gull
well i meant the number of numbers you can make starting (100s place), for examp...
oh I misread the task
multiple of five
naive mountain
So like, I put down just 3 empty _ to represent digits. And for the first _ I said there are 5 possibilities because we’re only usint the digits 0,3,4,5 and 7. Is that correct or am I supposed to consider like ALL 3 digit multiples of 5
hexed gull
yeah then ofc start with last digit, as it's 0 or 5
naive mountain
oops I jsed a _
I pit down 3 empty lines
is what I meant
just to visualize it
hexed gull
So like, I put down just 3 empty _ to represent digits. And for the first _ I sa...
if you start with the first then you need to exclude 0
since e.g. 035 wouldn't be a three-digit number in that sense
hexed gull
given the criterium that it's a multiple of five
it's easier if you instead start with the last digit
since it's limited to 0 or 5
otherwise you can't have a multiple of 5
hexed gull
ys
so consider the two cases
last digit = 0
then the second digit is either 3/4/5/7
4 options
and the first digit is one of the remaining others
3 options
thereby you have 3*4=12 possible numbers if it ends in 0
now I'll let you ponder the case for last digit = 5
naive mountain
makes sense thanks
outer moat
URGENT!
Briefly, how do I check if a function is self dual?
obtuse pebbleBOT
URGENT!
Briefly, how do I check if a function is self dual?
outer moat
Here the dual and the function arent equivalent are they
And also, is there a faster way to understand this with a truth table?
obtuse pebbleBOT
@outer moat Has your question been resolved?
outer moat
I got how this works. Anyone knows how to identify a self dual function by a truth table?
flat belfry
hello all i need help really quickly how do i find this inverse derivative, I actually have no clue where to start :/
If f(x) = x(sqrt(x-4)), then what is the value of (f^-1)'(2)
obtuse pebbleBOT
hello all i need help really quickly how do i find this inverse derivative, I ac...
flat belfry
this is not not difficult
cedar iris
I gochu
flat belfry
im assuming i find the inverse of f(x), get the derivative of that, then just plug in 2 but i cant seem to get the inverse for f(x)
flat belfry
fr?
cedar iris
You just need to find f'(f^-1(2)) which is f'(4.2242)
tardy epoch
Inverse function theorem gives you a formula
cedar iris
and then do 1/that
flat belfry
Inverse function theorem gives you a formula
i kinda get that formula
its basically chainruling g(f(x)) = x or something
to get 1/g(f(x))
tardy epoch
i kinda get that formula
Yes so use it
flat belfry
but that requires me to find the iverse of f(x) because it says g(x) equals the inverse of f(x)
and i cant figure out how to do that
sub x and y but then i get some annoying stuff
tardy epoch
but that requires me to find the iverse of f(x) because it says g(x) equals the ...
No it doesn't
flat belfry
is this website just straight up wrong
cedar iris
The question is wrong. thats not possible I think
because f'(f^-1(2)) is complex
the formula is 1/f'(f^-1(2)).
However f^-1(2) is appriximately 4 which will result in derivative of f being complex
additional proof
flat belfry
ok so i just gotta sub 2 for y in f(x) and plug that into f'(x) under 1
tardy epoch
additional proof
!nogpt
obtuse pebbleBOT
additional proof
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
cedar iris
!nogpt
Bruh. But still thats additional proof and it literally used code interpereter which means it used working python libs to do that
which checks out with wolfram so chill
tardy epoch
Bruh. But still thats additional proof and it literally used code interpereter w...
...
tardy epoch
Bruh. But still thats additional proof and it literally used code interpereter w...
!nogpt
obtuse pebbleBOT
Bruh. But still thats additional proof and it literally used code interpereter w...
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
chrome crypt
daring sorrel
It is wrong though
cedar iris
It is wrong though
explain
flat belfry
oh great
cedar iris
I doubt that
daring sorrel
“The result indiciates the function is not real valued at x=2” this is not relevant to the question
We r looking at the inverse
tardy epoch
snitch <@261933205387477002>
Don't break rules then. Not that hard
flat belfry
dude idk even what to do now
daring sorrel
dude idk even what to do now
Dont find the inverse
cedar iris
Dont find the inverse
He doesnt need to bru
daring sorrel
Yeah I said dont
cedar iris
read above @daring sorrel
flat belfry
he said dont
cedar iris
yea but we already know problem is wrong
or at least theres no real solution
so dont complicate things
daring sorrel
f^-1(x)=y is the same as y=f(x)
cedar iris
yes, but what's your point
flat belfry
i just gotta do 2=x(sqrt(x-4)) and solve for x
then ill have f^-1(2)?
cedar iris
yes
flat belfry
bad typo mb
daring sorrel
Anyway this will be near 4 like u said
cedar iris
excactly
daring sorrel
And infact its greater than 4 as wolfram showed
cedar iris
and then f'(~4) = no real sol
daring sorrel
So u can plug it in into the derivative and get a real solution
daring sorrel
Bro
daring sorrel
When a>b a-b>0
cedar iris
im hallucinating its too late
daring sorrel
Just like chatgpt when u ask math
cedar iris
sometimes lol
daring sorrel
Yeah in math we dont like sometimes
flat belfry
I would just like to know the process for finding (f^-1)'(c) when given f(x) and c
daring sorrel
Or in pretty much any subject tbh
flat belfry
because intructional stuff ive found is all wierd and they use different variables
cedar iris
yes but generally I can fact check ir cuz im not bein dunb
cedar iris
Or in pretty much any subject tbh
you are significant;y underestimating its capabilities
tardy epoch
yes but generally I can fact check ir cuz im not bein dunb
Can you just stop talking
cedar iris
Can you just stop talking
if u hv a problem leave
daring sorrel
Im not bro trust me I use it for math all the time and 90% of the time its wrong
tardy epoch
You've done nothing but confuse OP
cedar iris
Im not bro trust me I use it for math all the time and 90% of the time its wrong
u have gpt4?
cedar iris
You've done nothing but confuse OP
thats my bad, but now its over so chill
daring sorrel
I will stop talking about this to avoid spamming channel
tardy epoch
you are significant;y underestimating its capabilities
And now you're just flooding their channel with off topic bs
flat belfry
.close
obtuse pebbleBOT
daring sorrel
Rip
flat belfry
sorry yall i am not great at this stuff
cedar iris
sorry yall i am not great at this stuff
no worries sorry for messing up
obtuse pebbleBOT
Available help channel!
inner hemlock
let f be a function such that f"(x) = 6x + 8, Find f(x) if the graph of f is tangent to the line 3x-y=2 at the point (0,-2)
obtuse pebbleBOT
let f be a function such that f"(x) = 6x + 8, Find f(x) if the graph of f is tan...
royal basin
well
inner hemlock
Ik the equation tangent to f(x) has a slope of 3
royal basin
the fact that the graph of f is tangent to y=3x-2 at (0, -2) lets you find f(0) and f'(0)
f'(0) specifically at this stage in particular
inner hemlock
wait
why do I do that?
oh
nvm I get it
you set x to 0 to find the constant
right?
since the slope is a constant
f'(0) is also 3
royal basin
since the slope at 0 has to match that of the tangent line
inner hemlock
o
so at pont 0,-2
point* they sharethe same slope
when x is set to 0 for the equation with the constant
we are left
with f'(x) = C
but since we know another equation
we use that and find the slope of that at the specific point 0.-2
and that would be our missing c value
thanks
.close
inner hemlock
how do i find f(x) when given 3 hints
obtuse pebbleBOT
how do i find f(x) when given 3 hints
inner hemlock
f'(x) = ax^2+ bx
f'(1) = 6 and f"(1)=18
integral of fx from 1,2 is 18
if you could tell me hints
how to get it
unreal musk
put 1 into f’(x), that should be equal to 6
Then differentiate that f’(x), that gets you f’’(x), similar idea (put x=1, you get 18)
From that you should be able to find an and b, then integrate for f(x), then integrate again between 1 and 2…
inner hemlock
i get a+ b = 6
wait im slow
then its a sytem of equations and you solve
what about c
how do i find that
Or is there no C to find?
unreal musk
There will be 
when you integrate f’(x)
unreal musk
For which, you make use of that third piece of information, the integral $\int_1^2 f(x) \dd x = 18$
warm shaleBOT
@unreal musk
inner hemlock
how do I integrate (x^2+1)/(x^3+3x-5)^3
obtuse pebbleBOT
Available help channel!
solid jungle
How exactly do i do this problem? I dont understand what formula to use
obtuse pebbleBOT
How exactly do i do this problem? I dont understand what formula to use
smoky vigil
,tex .unit circle
warm shaleBOT
Moosey
smoky vigil
$\sec(\theta)=\frac{1}{\cos(\theta)}$
warm shaleBOT
Moosey
smoky vigil
$\cot(\theta)=\frac{1}{\tan(\theta)}=\frac{\cos(\theta)}{\sin(\theta)}$
warm shaleBOT
Moosey
solid jungle
How do i figure out the arc length tho? It says I need to find the sec and csc ratios for those arc lengths but i'm a bit lost on how to work my way up
floral kayak
How do i figure out the arc length tho? It says I need to find the sec and csc r...
Arc = θ × r
in this case r=1
it should be the same anyways :/
You don't really need the arc length formula
solid jungle
Arc = θ × r
Yeah thats for the first point, but for the 2nd point in question 77 you dont know r u just know the y value
solid jungle
Ohh
floral kayak
y^2+x^2=1
floral kayak
the sqrt wasn't really necessary but it is the orignal formula
solid jungle
yes \:)
Next, to find the value of theta in order to find the arc length, do i first find the angle from the horizontal line to the point and add pi/2 to find principal angle?
floral kayak
not too sure about w and t or what that is
but I assume that would be it from appereance
solid jungle
Im still talkinf about the first queation
floral kayak
oh
I mean assuming you have the cosinus
you can solve the secant
or you can solve using arc length 🙂
bright pumice
Or you can just derive the angle given you have a y increase of 0.6 and the hypotenuse of 1 you can use Pythagorean theorem and then sohcahtoa I guess if that makes you happier
solid jungle
or you can solve using arc length 🙂
Yeah and to get that i need the r which i have and need theta, so do i just use cosine law to find theta in relevance to quadrant 2 since it’s located there, then add 90 to compensate for the fact it’s not in quadrant 1, then convert it to radians from degrees then use the arc length formula to find arc length?
Idk if that sounds too complicated? Is there a shorter way maybe?
solid jungle
Yeah you can do something like that too haha
I mean if theres a shorter path im all in for it I want to save time
bright pumice
Your x is cos (theta) and your y is sin (theta)
You already have the y value of 0.6
So sin (theta)=0.6
It's doesn't get more complicated than than unless you want to over complicate it by going around
solid jungle
So sin (theta)=0.6
Yes that part is clear
Now
By theta do you mean
This angle a or angle b that i drew?
obtuse pebbleBOT
@solid jungle Has your question been resolved?
worldly bolt
For this I need to write an explicit formula for the sequence. How do I do that?
median dome
do you see any pattern in the numbers?
plucky rivet
do you see any pattern in the numbers?
Like increasement between each two terms
median dome
correct
worldly bolt
Ok
plucky rivet
Like 3,5,7,9
Do you see the increasement of ' the increasement'
worldly bolt
+2
plucky rivet
Yes, define a sequence based on this increasement
royal basin
not quite
worldly bolt
Oh
royal basin
will help to make our notation a little more precise here
worldly bolt
Oh yea because if I did n = 2 then it would be 4 not 5
royal basin
let's use a_n to denote your original sequence
worldly bolt
Ok
royal basin
then the differences are: \
$a_2 - a_1 = 3 \ a_3 - a_2 = 5 \ a_4 - a_3 = 7 \ a_5 - a_4 = 9$ \
and we are assuming that these differences will keep going as an arithmetic progression
and we must not forget the initial condition of $a_1 = 4$ as well
warm shaleBOT
Ann
royal basin
i am indexing the terms of your sequence starting at 1 following common practice
read through this, ensure the notation makes sense to you, and ping me once you have done so.
then we will continue.
worldly bolt
then we will continue.
Ok I got it
royal basin
ok right
now, give me a moment to think of how next to phrase this while minimizing the chance for off by one errors
worldly bolt
Ok
royal basin
are you familiar with sigma notation?
worldly bolt
Yes
warm shaleBOT
Ann
royal basin
and $a_{k+1} - a_k$, by assumption, is $2k+1$ (verify this. the sequence $b_k := a_{k+1} - a_k$ is assumed to be an arithmetic progression with common difference 2)
warm shaleBOT
Ann
royal basin
and $a_1$ is 4
warm shaleBOT
Ann
worldly bolt
Alright
royal basin
do you see how to continue from here
worldly bolt
Oh nvm I skipped the a_1 + part
royal basin
as in "shouldn't this be something else?" or as in "how did you come up with that?"
worldly bolt
How did you come up with that
royal basin
let b_k := a_{k+1} - a_k
then we know b_1 = 3, b_2 = 5, b_3 = 7 and b_4 = 9
we assume b_k continues as an arithmetic progression
does this make sense to you?
worldly bolt
Yea so b_5 would be 11 and so on
worldly bolt
No
royal basin
... right ok let's get that sorted out first then
first off do you know what an arithmetic progression is?
worldly bolt
Yes
worldly bolt
It’s when numbers in a sequence go up by the same amount. I think it can also go down too but I’m not sure. So it’s like the 3,5,7,9 are all going up by 2
royal basin
yes, arithmetic progressions can go down.
anyway
you might have seen this:
$u_n = u_1 + (n-1)d$
warm shaleBOT
Ann
worldly bolt
Oh ok yea
royal basin
right
here for b_k our first term is 3 and the common difference is 2
b_k = 3 + (k-1)*2
and simplify that
worldly bolt
Wouldn’t it be *4 if the common difference is 4?
worldly bolt
For this I need to write an explicit formula for the sequence. How do I do that?
So could I do that with this?
worldly bolt
Oh right
royal basin
its sequence of first differences, what i call b_k, is
worldly bolt
Ok
royal basin
do you have anything else left to ask here
worldly bolt
So I have to make my formula with b_k
worldly bolt
let b_k := a\_{k+1} - a_k
And my formula would look like this?
royal basin
no
you first construct b_k (reproducing what i did already) then from b_k construct a_n
you will have b_k = 2k+1
royal basin
$a_n = 4 + \sum_{k=1}^{n-1} (2k+1)$
warm shaleBOT
Ann
royal basin
work out this sum
worldly bolt
7?
royal basin
.... you should not be getting something that doesn't depend on n lol
you said you were familiar with sigma notation
worldly bolt
Yea but not when I have it mixed with arithmetic sequences
royal basin
pretend it's not mixed in, then
worldly bolt
That’s what I did. I completely ignored n and did 4 + 2(1) + 1 to get 7
I don’t know n-1
royal basin
no, you can't just ignore a crucial part of the notation...
worldly bolt
But I don’t know the value of n
royal basin
n is just n
ok let's seee
let's set this problem aside for the time being
do you know what $\sum_{k=1}^n k$ is
warm shaleBOT
Ann
royal basin
wrong on all counts
worldly bolt
I know what to do when there’s a number in place of n but I’m not sure what I have to do when it’s a variable
royal basin
ok sounds like you are not as familiar enough with sigma notation as i thought
ok
let's back up one
worldly bolt
I’ve only been introduced
royal basin
do you know how to find the sum of the first n terms of an AP
worldly bolt
Yea you add the first however many terms together
royal basin
this notation clashes with what we did earlier, and this formula will be inconvenient, but yes.
worldly bolt
Oh this is what my teacher taught in the lecture
I wrote this directly from her notes
royal basin
the problem for us is that (1) our sequence is called b_k rather than a_k and (2) we need the sum of the first n-1 terms and not n and (3) we do not immediately know the last (n-1'th) term of our sequence
though (3) is the least worrying of these
$b_{n-1} = 2(n-1) + 1 = 2n-1$
warm shaleBOT
Ann
royal basin
no
... sorry i do not know how to explain this further without just repeating myself a hundred times
worldly bolt
Why can’t I put it in a_n?
Why does it have to be b_k
Is it because a_n is arithmetic?
Nvm I’ll ask my professor
…thanks anyway
.close
timid silo
ANY JEE ASPIRANTS?
royal basin
WHY ARE YOU SHOUTING???
ALSO NO, NOT EVERYBODY HERE IS FROM INDIA, LET ALONE A JEE ASPIRANT.
BUT THERE PROBABLY ARE SOME
I'M GOING TO CLOSE YOUR PREVIOUS CHANNEL FOR YOU!
@timid silo
tardy epoch
....
obtuse pebbleBOT
timid silo
.reopen
obtuse pebbleBOT
✅
zenith raft
grim copper
??
buddy this is a help channel, not a discussion channel LMAO
you shouldn't be occupying two channels anyways
timid silo
ANYONE CLOSE ONE
versed pendant
hey all how would you differentiate both sides of the eqn. wrt t?
obtuse pebbleBOT
hey all how would you differentiate both sides of the eqn. wrt t?
obsidian isle
pseudo swift
obtuse pebbleBOT
@versed pendant Has your question been resolved?
lofty perch
obtuse pebbleBOT
lofty perch
Please can I get help with number 5
obtuse pebbleBOT
@lofty perch Has your question been resolved?
obtuse pebbleBOT
@lofty perch Has your question been resolved?
nova python
Hello, I need some food for thought on how to write this proof (I am a beginner). I already know why R<=D<=2R is the case, but I don't know how to write it down. (I translated the problem from another language into english, that's why the format is kinda bad).
If you dont want to read the mathematical definitions:
d(u,w) is the shortest distance from u to w.
D is just the longest distance without any detours in the entire graph.
R just takes the minimum of the longest possible d(u,w) for each individual node.
obtuse pebbleBOT
Hello, I need some food for thought on how to write this proof (I am a beginner)...
nova python
R<=D is trivial, the real problem is with 2R
it would be nice if someone could just give me a hint instead of the answer as i want to solve it myself
obtuse pebbleBOT
@nova python Has your question been resolved?
lusty goblet
Assume AB is the longest path in the graph. Q is a node with the smallest distance to the the furthest mode equals R. What is the length of AQB?
nova python
👍
nova python
Nope! Thanks for the help
obtuse pebbleBOT
@nova python Has your question been resolved?
cinder spindle
yo
pallid aurora
possibly
obtuse pebbleBOT
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/
@cinder spindle Has your question been resolved?
young laurel
can anyone tell me what the spectrum of a matrix is?
obtuse pebbleBOT
can anyone tell me what the spectrum of a matrix is?
robust raven
can anyone tell me what the spectrum of a matrix is?
$\text{The set of all eigenvalues of A is called the spectrum of A }\\\text{and is denoted by σ(A)}$
warm shaleBOT
Joanna Angel
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
young laurel
But is it like the diagonal matrix of eigenvalues or just a set eigenvalues
robust raven
set
young laurel
so for example the matrix [1 0; 02] its spectrum would be [1,2]?
robust raven
$\sigma\left( A \right)=\left{ \lambda:det\left( A-\lambda
I \right) =0\right}$
warm shaleBOT
Joanna Angel
runic holly
help 😭
worn coyote
Find a,b such that 3 curves connect
runic holly
Find a,b such that 3 curves connect
how do you do that 😋
worn coyote
For example, at x=-1 3x+a should match -x^2+2x+6
runic holly
so equal them??
worn coyote
Otherwise the function won't be continuous
worn coyote
so equal them??
Yeah equal them at x=-1 and solve for a
And then you can use a similar argument to find b
runic holly
solets say to find a i equal 1 and 2 and fill in x as 1?
then i do the same thing with 2 and 3 to find b?
im like hopeless at this 😭
robust raven
yes right but formally , it should be written in a such way:
$\lim_{x \to -1^{-}} f\left( x \right)=\lim_{x \to -1^{+}} f\left( x \right)\\\lim_{x \to 2^{-}} f\left( x \right)=\lim_{x \to 2^{+}} f\left( x \right)$
warm shaleBOT
Joanna Angel
robust raven
after writing those limits, yes, correct
runic holly
ok so lets say for finding B, id use equations 2 and 3, but unlike 1 and 2 there so definiet equals sign theres a range, so in that case what would i do
robust raven
i do not see the difference
runic holly
so like in the first one it was x is less than 1, the second 1 its x is greater or qual than -1 less than 2
robust raven
x less than -1 you wanted to write ?
runic holly
um well youre meant to input x=-1 when trying to find a right?
so to find b what value would you input in for x?
runic holly
ohhhhhhhhhh sorry i didnt read it properly, i kept focusing on the wrong one 💀
robust raven
ok)
robust raven
🙂
runic holly
cheers and thanks for the help
robust raven
yw )
runic holly
.close
chilly scroll
Is there a way to do this with graph?
obtuse pebbleBOT
Is there a way to do this with graph?
prime jungle
I feel like this could be solved with an algorithm. Generally, you know that the sum of x and y should equal a multiple of 5. In addition, the sum of x and y should be no greater than 50 and no less than 5 (lowest possible is 2 but anything lower than 5 would not work). Does this help?
chilly scroll
oh yea that method works
i was just thinking if there was a way with graphs
like x + y =5k a strgt line but i dont know what to do with it
prime jungle
There probably is. I think that a multivariable function f(x,y) would work best. Is that taught in your class?
chilly scroll
nah
I saw somewhere probabiltiy density function
So i thought of this approach
But i dont know much about it
oh can the intersection between the area of y 1 to 25 x 1 to 25 and y = 5k - x be the points?
fossil crag
try to answer the question for x = 1, then x=2,... up to x=5 and try to see a pattern
prime jungle
Using your x + y = 5k, letting k be a natural number 0 through 50, then I would look along the slope to find coordinate pairs that would match what you are looking for. For example:
x+y=5k, k = 2 (5k = 10)
Looking along the slope, you can see that (3,7), (5,5), and (2,8) are pairs that match what you are looking for.
chilly scroll
Can this work?
obtuse pebbleBOT
@chilly scroll Has your question been resolved?
livid linden
Can y' be independent of anything or has to be independent of x only ?
obtuse pebbleBOT
Can y' be independent of anything or has to be independent of x only ?
steel escarp
How do i solve this? I have attatched my current workings but I dont think its going somewhere
obtuse pebbleBOT
How do i solve this? I have attatched my current workings but I dont think its g...
steel escarp
damn you ms paint (new attachment below)
It wont work if I equate the 2 equations, will it?
steel escarp
you don't need derivatives at all here
I was having a problem solving it. How do I know that I can just equate them? Sometimes I just cancel my equations out to 0=0
polar fossil
well the question is "what values of c make this have two intersections" so we care about their intersections
steel escarp
well the question is "what values of c make this have two intersections" so we c...
But if we equate them, how do we garuntee 2 intersections, not only 1?
polar fossil
using c
steel escarp
what kind of equation do you get?
Am I supposed to be able to solve x^2 + 2cx - 4x + 4 - c = 0 when it has 2 variables?
should I use simultaneous eq?
polar fossil
you won't be able to find values for x
but you'll be able to say when that equation has 2 solutions
yes using the discriminant
steel escarp
ah
polar fossil
be mindful of the fact that the c in "b^2 - 4ac" means a different thing than the c from your equation
steel escarp
$(c\approx 5,541\ c\approx -0,541)$
warm shaleBOT
PiggyAwesome
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
steel escarp
huh seems kinda unsatisfying
polar fossil
you should get a range for c, not two values
and generally you'll want to leave it in exact form
steel escarp
you should get a range for c, not two values
c < (5−sqrt(37))/2 or c > (5 + sqrt(37))/2
polar fossil
i would not recommend that
polar fossil
yea
steel escarp
smoky vigil
$\left(2c-4\right)^{2}-4\left(4-c\right)>0$
warm shaleBOT
Moosey
steel escarp
polar fossil
is this not what you did?
smoky vigil
you will get two open intervals. This results from solving a quadratic with respect to c. The quadratic will be zero at two points and dip below the x-axis, and on the other sides of the zeros, it will be positive.
steel escarp
you will get two open intervals. This results from solving a quadratic with resp...
How do I know this without visual proof?
smoky vigil
**Moosey**
Well I gathered it from the visual proof. This can be turned into a quadratic that factors nicely as well
no need for any quadratic formula even
steel escarp
**Moosey**
Idk man ive never done that before im doing AS lvl atm
smoky vigil
well, when you set the curves equal, you move everything to one side, and you get a quadratic. All you care about is when this quadratic has b^2-4ac>0 (when you have two real solutions, this corresponds to the line and the curve intersecting twice)
smoky vigil
$\left(2c-4\right)^{2}-4\left(4-c\right)>0$
this is the resulting inequality you get
steel escarp
well, when you set the curves equal, you move everything to one side, and you ge...
but why did the quadratic formula only give 2 answers if there are multiple real ones?
smoky vigil
the quadratic formula tells you the zeros
if you do b^2-4ac=0 this corresponds to where the curves intersect exactly once
smoky vigil
**Moosey**
so if this equals zero, it will only be the c values where the line intersects exactly once with the parabola
it's why we need >0
finding where this is 0 is helpful anyways though, as it gives us the transition points we need (i.e. where it goes from 2 intersection points, to 1 , to 0)
steel escarp
i dont know enough theory to understand this yet unfortunately
smoky vigil
You're focused on where $x^{2}+(2c-4)x+(4-c)$ has two real zeros. This parabola comes from setting the two curves equal
warm shaleBOT
Moosey
smoky vigil
x^2+2cx+4= (4x+c) -> x^2+2cx+4-(4x+c)=0 -> x^2+(2c-4)x+(4-c)=0
a=1, b=(2c-4), c=(4-c)
steel escarp
when you say 2 real zeroes, what does that mean? Is it 2 real intersections at y=0?
smoky vigil
yes
steel escarp
ah
smoky vigil
now, since we're focused on purely the case where there's two real zeros, we want the discriminant of this quadratic (b^2-4ac), to be greater than 0
smoky vigil
$\left(2c-4\right)^{2}-4\left(4-c\right)>0$
which gets us this. We can solve this by simplifying and factoring, with no need for quadratic formula
$\left(2c-4\right)^{2}-4\left(4-c\right)>0$
warm shaleBOT
Moosey
steel escarp
$\left(2c-4\right)^{2}-4\left(4-c\right)>0$
I see. But when you expand, why does it make a sussy range?
smoky vigil
[(2(c-2))^{2}-4(4-c)>0]
Applying $(xy)^{2}=x^2 y^2$, and distributing $-1$ to $(4-c)$
[4(c-2)^{2}+4(c-4)>0]
Divide both sides by $4$
[(c-2)^{2}+(c-4)>0]
Expanding $(c-2)^{2}$
[c^2-4c+4+c-4>0]
Canceling like terms
[c^2-3c>0]
[c(c-3)>0]
warm shaleBOT
Moosey
smoky vigil
c(c-3)>0 only when c and c-3 are both negative, or c and c-3 are both positive
this comes strictly in the case where...
c<0 and c>3 respectively
it's not a continous interval
steel escarp
c(c-3)>0 only when c and c-3 are both negative, or c and c-3 are both positive
oh because c(c-3) / c > 0 / c = c-3>0 if c is positive else c-3<0 so c<3
steel escarp
doesnt the sign flip when you divide by negative?
smoky vigil
look what happens when you plug in for example -0.1 for c
(-.1)(.-1-3)=(-.1)(-3.1)=positive
but once we go past 0...
(.1)(.1-3)=(.1)(-2.9)=negative
and until we get to 3...it's negative, but once we get to a number slightly above 3...
(3.1)(3.1-3)=(3.1)(.1)=positive
so we get two intervals for where c will give us two intersections, with an interval between them that doesn't give valid c