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#help-41
amber waspBOT
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shy moss
The power dissipation P in a resistor with resistance R is given by the formula P=U^2/R where U is the voltage across the resistor. Assume that U = 220V and R = 9 ohms. Estimate using the differentiability how the power changes if U is increased by 5 volts and R is increased by 0.3 ohms.
proven vapor
this is known as the total differential of a multivariable function
think of dP as a small change in P. P depends on the variables U and R, and delP/delU roughly describes how much P changes as U changes, so delP/delU * dU is the change in P for a small change dU in U. The other term is an analogous change in P but for a small change dR in R
shy moss
The main problem i'm facing is that the exercises are fairly easy at the start where i just have to differentiate wrt x,y,z etc. But now when i come to exercises where i have to use knowledge about differentiability, i become completely lost..
What you said makes no sense 
I did manage to differentiate wrt U and R respectively
"the concept for differentiability"
Are we taking the total change when U is increased by 5V and the total change when R is increased by 0,3 ohms and summing them together?
warm idol
Could you like post what your question is asking so we get a better understanding of what you're dealing with
warm idol
Oh right my bad
shy moss
that's the idea
So the total change when we increase U by 5 V
so we take 2U/R = 2*300/9 = 600 / 9 = 66,6 W?
no that can't be right
..
The total volt from start is 300
I have to multiply with the change
?
So 300 * 2U/R = 300 * 2*5/0,3
warm idol
Just substitute the values?
shy moss
= 10 000 ?
9 * - U^2 / R^2 = 9 * -300/0,3 = -9000
Nah idk
How does it work?
what is the value for
dU?
Does anyone know how this works?
shy moss
Yes it's 5 but why
shy moss
dU = derivative of U? = change of rate?
shy moss
<@&286206848099549185> Do we have anyone who's knowledgeable about partial derivatives in mult var calc and understand why we sum this part?
warm idol
dU does represent the overall change in U, for a derivative you're checking the change wrt another variable
idk if that's a good way of explaining but oh well lol
amber waspBOT
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vagrant acorn
amber waspBOT
vagrant acorn
can someone show me the mathimatical way of breaking out the x^2 the way that is highligted in red square
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mystic lotus
can someone show me the mathimatical way of breaking out the x^2 the way that is...
First they convert the double integral dx dy into an iterated integral dy, then dx; this is valid by Fubini's theorem. Next, since the inside integral is an integral dy, the x^2 is "constant" from the perspective of that integral, so we can factor it out.
amber waspBOT
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rain herald
I have an oval thats 30" wide and 7' long. The units I need to pack into it are 4" circles kept in a set straight line in a set of 4, so they can possibly be reduced to 4x16" rectangles. How many sets can I fit in this area and what is the optimal way to organize it?
rain herald
the "oval" might be better described as a 30" x 5' rectangle with a half circle on each side
is there any applications that solve them?
amber waspBOT
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@rain herald Has your question been resolved?
desert sedge
I got advanced functions next semester how cooked am I? outta 10
amber waspBOT
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amber waspBOT
@rain herald Has your question been resolved?
sinful sapphire
@rain heraldit's either 2 or 3
the area is 3 times larger
am i allowed to space them out?
like it's still a straight line
rain herald
<@129046667772035073>it's either 2 or 3
the answer's at least 15 lol
rain herald
the answer's at least 15 lol
would that change it?
sinful sapphire
you said you can only pack it in fours
rain herald
you said you can only pack it in fours
4" circles packed into fours, yeah
sinful sapphire
so 15 circles doesn't make sense
rain herald
15 sets of circles
15 x 4 if you want
the problem is you dont have space for a 2 full sets across
because a set is 16" and the width is 30"
sinful sapphire
but am i allowed to make the line longer?
sinful sapphire
no i don't mean like that
i mean like in my picture
can a line of 4 have empty space between circles
rain herald
it has to be obvious what circles are in what line
so like if two lines crisscross that wouldn't do
sinful sapphire
my picture takes all the space i just realized
it's unlikely to fit 3 sets i don't know why you say 15
sinful sapphire
probably
rain herald
1 foot is 3 sets, so 5' means at least 15
sinful sapphire
,calc (30)(7)/((4)(16))
grizzled pagodaBOT
Result:
3.28125
yeah i see it now
7 foot
i still don;t get if this is allowed
it's obvious what the lines are
sinful sapphire
well you make it exactly 30" wide
and stack it vertically
yeah i guess then the line doesn;t stand out anymore
rain herald
that could work. I guess you'd have to calculate how much you have to space out the circles to lose 2"
and that'd get you at least 30 sets
would it also not make sense to have the lines be diagonal?
Like your image you could have the sets like this
sinful sapphire
yes i can see that
you can't tell which ones i mean by the sets
and that's not allowed you said
you could do this
8 circles, inclined so it's 30 instead of 32
leave some space to distinguish
rain herald
you could do this
yeah its gotta be something like that
am I right in seeing that the offset would have to be something like 11"?
a 32" line would have to bend 11" down to fit in a 30" space?
then you'd only be able to fit it in for like 4 of the 5', so 24 sets + whatever you can fill in the empty spaces
amber waspBOT
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acoustic flicker
amber waspBOT
amber waspBOT
@acoustic flicker Has your question been resolved?
acoustic flicker
and 4x^2+12x+9
then ?
keen pawn
well, what have you tried so far
amber waspBOT
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amber waspBOT
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lavish zenith
anyone know how i can solve this equation where the unknown is alpha, an angle ?
lavish zenith
@solar gust 🙂
formal harbor
this is illegible, can you write the equation out more clearly?
lavish zenith
this is illegible, can you write the equation out more clearly?
sin(alpha)xcos(alpha) = (1/2)* 9.81 *(150/56^2)
unborn plaza
sin(2a)=2sin(a)cos(a)
lavish zenith
woww
how is that tru
i dont know all my trigonometry formulas
you got a list where all the important ones are ?
lavish zenith
https://en.wikipedia.org/wiki/List_of_trigonometric_identities
thanks !
solar gust
C'est de la physique ça
solar gust
Et pour montrer sin(2x) tu peux l'écrire comme sin(x+x)
lavish zenith
hmm
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keen pawn
Give an example, if possible of a sequence that is bounded , but contains no subseqeunce that converges . Would $a_n = sin(n)$ work?
grizzled pagodaBOT
math_rocks
sterile nymph
I don't think so, because you can almost certainly find a monotonically decreasing subsequence of sin(n)
You'll just begin getting very sparse
If you have a bounded sequence, then you can just pick a subset that either monotonically increases or decreases to get a convergent subsequence, I believe
distant herald
Give an example, if possible of a sequence that is bounded , but contains no sub...
look up bolzano weierstrass
keen pawn
oh right
I just learnt that 
, thought it was for a convergent seqeunce
oops
thanks
happy new year to the both of you
.close
amber waspBOT
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normal dove
what does this mean
amber waspBOT
what does this mean
violet blaze
use that min(i,j)+max(i,j)=i+j
amber waspBOT
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tough mica
Triangle $ABC$ is a right triangle. If the measure of angle $PAB$ is $x^\circ$ and the measure of angle $ACB$ is expressed in the form $(Mx+N)^\circ$ with $M=1$, what is the value of $M+N$?
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
tough mica
amber waspBOT
@tough mica Has your question been resolved?
glacial silo
What would this angle be? The one i highlighted @tough mica
verbal jewel
wow, still not solved after that? #help-32 message
normal dove
What would this angle be? The one i highlighted <@1187916384341078070>
chat u need to revise basic geometry
glacial silo
chat u need to revise basic geometry
I'm just trying to not give the answer straight away hahaha
normal dove
wont learn like that
glacial silo
not*
tough mica
glacial silo
We don't really need any of that to be fair
verbal jewel
you had this.
verbal jewel
you had this.
M
and ists given that M = 1.
glacial silo
you had this.
Oh, after finding this out cant you just plug 1 instead of M? @tough mica
From there you get (1)x + N = x - 90, and it shouldnt be to hard to find out what N is
tough mica
x + N = x - 90
glacial silo
Try simplifying further
tough mica
N = -90
tough mica
-89
glacial silo
👍
tough mica
happy new year
Hopefully you guys still mathin this year
Hopefully is a year full of math
thank you guys
.solved
amber waspBOT
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glacial silo
Hi! I need help verifying that my solution works.
\begin{align*}
3\sin(x-\pi) &= 2\cos(x) \
-3\sin(x) &= 2\cos(x) \
\text{Let } u = \sin(x) &\Rightarrow \cos(x) = \pm \sqrt{1 - u^2} \
-3u &= \pm 2\sqrt{1-u^2} \
9u^2 &= 4-4u^2 \
u &= \pm \frac{2}{\sqrt{13}} \
\end{align*}
- When $u$ is positive $\sin(x) > 0, \cos(x) < 0$. \
$$\begin{cases}
\sin(x) &= \frac{2}{\sqrt{13}} \
\cos(x) &= -\sqrt{1-\frac{4}{13}}
\end{cases}$$
$\Rightarrow x = 3n\pi - \sin^{-1}(\frac{2}{\sqrt{13}}), \quad n \in \mathbb{Z}$ \ -
- When $u$ is negative $\sin(x) < 0, \cos(x) > 0$. \
$$\begin{cases}
\sin(x) &= -\frac{2}{\sqrt{13}} \
\cos(x) &= \sqrt{1-\frac{4}{13}}
\end{cases}$$
$\Rightarrow x = 2n\pi + \sin^{-1}(\frac{2}{\sqrt{13}}), \quad n \in \mathbb{Z}$ \
- When $u$ is negative $\sin(x) < 0, \cos(x) > 0$. \
grizzled pagodaBOT
woomy
native stag
You should complete the sine.
glacial silo
It's not how I'd do it though
The inteded solution divides by cosine but im not sure how we can know for sure cos(x) =/= 0
glacial silo
You should complete the sine.
What does that mean?
glacial silo
The inteded solution divides by cosine but im not sure how we can know for sure ...
nvm this is obvious in this specific case but not generally applicable
native stag
$-3 \sin x = 2 \cos x \ \implies 2 \cos x + 3 \sin x =0$, yes?
grizzled pagodaBOT
Percy
glacial silo
yep
glacial silo
**woomy**
also this gives me two sets of solutions meanwhile
,w 3sin(x-pi) = 2cos(x)
grizzled pagodaBOT
native stag
you can divide by $\sqrt{2^2+3^2}$ which makes both the coefficients terms you can assign a trig expression too
grizzled pagodaBOT
Percy
native stag
so if $z = \arcsin(\sqrt{13})$ the the equation $\sin(z) \cos(x) + \cos(z) \sin(x) = 0$ which is just $\sin(x+z)=0$
(please check the details i might've fucked up the sub)
grizzled pagodaBOT
Percy
native stag
this gets you the answer in terms (provided you're comfy leaving in an arcsin ofc)
native stag
wait why are we dividing by this?
it makes them both be sin and cos terms
glacial silo
What does sin and cos terms mean? Like expressed as a sine?
glacial silo
Like, i reached here:
$$\frac{2}{\sqrt{2^2+3^2}}\cos x + \frac{3}{\sqrt{2^2+3^2}}\sin x = 0$$
grizzled pagodaBOT
woomy
glacial silo
But im not sure how this helps
native stag
it's like if you have $\sin x + \cos x$ you can write that as $\sqrt{2} \left(\frac{\sin x}{\sqrt 2} + \frac{\cos x}{\sqrt 2} \right)$ which is just $\sqrt 2 \sin(x+ \frac{\pi}{4})$
great
glacial silo
all good
grizzled pagodaBOT
Percy
native stag
there we go
glacial silo
ohhhh
i get it now
so going back to the original equation
$$\frac{2}{\sqrt{2^2+3^2}}\cos x + \frac{3}{\sqrt{2^2+3^2}}\sin x = 0$$
grizzled pagodaBOT
woomy
glacial silo
$sin(x) = \frac{2}{\sqrt{2^2+3^2}}$
grizzled pagodaBOT
woomy
glacial silo
hey, I need help with algebra
native stag
whom should I ask
the server
glacial silo
<#488120190538743810>
ask in an open help channel, everythings writting in here
native stag
just make a channel i'll see if i can do it
glacial silo
wait so
split sail
ask in an open help channel, everythings writting in here
alr ty
glacial silo
Lets say i have a general expression of the form $$A\sin x + B\cos x + C = 0$$
grizzled pagodaBOT
woomy
glacial silo
I would first bring the C to the other side
native stag
I think I might've gotten something a bit wrong-
glacial silo
yea something seems off
native stag
uhh
native stag
yeah no it does i just might've gotten something wrong
im checking hold on
what the fuck lol
yeah no
it's perfectly fine
hello again
minor self doubt over
native stag
i put it into desmos and forgot to take the square root and was all like 'oH whAt tHe Fu- it doesn't WoRkOoOooh'
glacial silo
Yes but its the form the solutions take
glacial silo
Cause this implies every half period theres one solution
nevermind makes sense
every period 2 solutions
the other one aswell every period two solutions
native stag
Lets say i have a general expression of the form $$A\sin x + B\cos x + C = 0$$
for things like this you should just put $\sin x = \sin z$ and find the general solution
grizzled pagodaBOT
Percy
native stag
it's like by far the best way to do this that i know
hm? what?
oh, no i just like z as a stand in variable
glacial silo
yea hahah lol
native stag
usually in my notes i use like $\zeta$ lol
grizzled pagodaBOT
Percy
native stag
also algebra person disappeared lol
glacial silo
yea never asked a question either lol
native stag
usually in my notes i use like $\zeta$ lol
NO NOT THAT, $\Zeta$
grizzled pagodaBOT
Percy
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reaction for more information.
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native stag
what?
grizzled pagodaBOT
woomy
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native stag
...okaay
glacial silo
$\varzeta$
grizzled pagodaBOT
woomy
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glacial silo
huh
native stag
left the server
very weird
amber waspBOT
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night sundial
how do they get this?
amber waspBOT
how do they get this?
formal yacht
Partial decomposition
keen pawn
notice that $\frac{1}{n} - \frac{1}{n+5} = \frac{5}{n(n+5)}$
grizzled pagodaBOT
math_rocks
night sundial
uuh
night sundial
**math\_rocks**
this is true
can we always do partal decomposition in the form x/(n(n+x)) into 1/n - 1(n+x)
winged quest
Even better, you can always decompose any denominator that factors
night sundial
do you mean with factors, that you can split the denominator in more than 1 part?
like making brackets
night sundial
notice that $\frac{1}{n} - \frac{1}{n+5} = \frac{5}{n(n+5)}$
i get that if i am given 1/n - 1/(n+5) you can indeed bring it to that form on the right
but how do i begin if i get the right part 5/(n(n+5))
amber waspBOT
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spark rock
@night sundial Where are you from?
night sundial
tr
spark rock
Sunucuda Türkçe konuşmak yasak mı
night sundial
bilmem de avrupa vatanasi oldugum icin turkcem pek yok yani
ingilizce konusmayi tercih ediyom 😂
amber waspBOT
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spark rock
bilmem de avrupa vatanasi oldugum icin turkcem pek yok yani
Anladim kral
stone bloom
is there a cone-like shape where the volume has a factor 1/2 instead of 1/3
stone bloom
like a shape with this formula
thick shale
There can be although not a right circular cone
stone bloom
what would it look like
small pendant
is there a cone-like shape where the volume has a factor 1/2 instead of 1/3
describe what restrictions you mean when you say cone-like
stone bloom
has a pointy end and has a circular base
small pendant
has a pointy end and has a circular base
um
yeah surely we can come up with such a shape it just won't be regular
stone bloom
chatgpt said the area of a cone cross section decreases quadratically
so maybe we need something that decreases linearly instead?
small pendant
so maybe we need something that decreases linearly instead?
yeah if the area of the cross section decreases linearly this formula is correct: however, this is impossible with a pointy end.
if we are in 3 dimensions
stone bloom
pointy end is just area 0, so why wouldn't it be possible?
area can decrease linearly until it hits 0
small pendant
pointy end is just area 0, so why wouldn't it be possible?
i mean its hard to explain
if you don't know delta-epsilon theory
if you are like epsilon away from the tip
where epsilon is very small
the radius you need "roughly" at that cross section would be proportional to root epsilon under a linear format
but if we need to go "down" by epsilon, and "away" by sqrt(epsilon) where epsilon is very small, this basically makes it flat
so there will be a "topmost" point
but it wont be "pointy"
as in like how the top most point in the unit sphere is not "pointy"
stone bloom
hmm then ig it doesn't have to be pointy, as long as there is a topmost point that has 0 cross section
small pendant
hmm then ig it doesn't have to be pointy, as long as there is a topmost point th...
oh that is possible then
it still won't be "regular" but it will be smooth now
stone bloom
is there a way I can see an image of how it'd look like?
stone bloom
ohh so it's just a parabola rotated then?
small pendant
yeah
stone bloom
wait but is the area of the cross section increasing linearly?
hmm wait ig it is
so I guess you could use calculus to show the volume would be this
small pendant
yeah
stone bloom
that's so weird
so if the area increases linearly then it looks quadratic and if the area increases quadratically then it looks linear
amber waspBOT
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tough mica
What is the sum of all integer values of $x$ such that $\frac{67}{2x - 23}$ is an integer?
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
quick ridge
What is the sum of all integer values of $x$ such that $\frac{67}{2x - 23}$ is a...
what are the divisors of 67
grizzled pagodaBOT
Result:
33.5
,calc 67/3
grizzled pagodaBOT
Result:
22.333333333333
what is the prime factorization of 67
quick ridge
you should know this is prime
but
it’s clear that it’s not divisible by 2
not divisible by 3 because sum of digits isn’t
not divisible by 5 because doesn’t end in 0 or 5
not divisible by 7 because 7 * 9 =63 and this is 4 more than that
then we don’t even need to check anything else
lusty saffron
not divisible by 7 because 7 * 9 =63 and this is 4 more than that
(alternatively, 7*10 = 70 and this is 3 less than that, which might be easier than remembering what 7*9 is)
quick ridge
I still need help finding x
to be an integer the denominator must be one of its divisors
it’s divisors are 67, 1, -67, and -1 if we’re considering all integers
grizzled pagodaBOT
Result:
45
,calc (1+23)/2
grizzled pagodaBOT
Result:
12
,calc (-1+23)/2
grizzled pagodaBOT
Result:
11
,calc (-67+23)/2
grizzled pagodaBOT
Result:
-22
now just add them
tough mica
,calc 45+12+11-22
grizzled pagodaBOT
Result:
46
ok ty
quick ridge
you’re welcome
amber waspBOT
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amber waspBOT
indigo cloud
can you make a quick drawing of the situation?
and what you mean by that usual laws of motion thing
amber waspBOT
@remote raft Has your question been resolved?
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@remote raft Has your question been resolved?
split sail
t,h is not enough to determine distance
it should also depend of the initial angle
oh wait
t is the angle
goeeeog
aight brb
split sail
are you sure there isn't any other condition
the most straightforward formula is d = v0 cost * T
if initial velocity is a constant for any choose of t,h then T is also solvable in terms of constants
half garden
No, not really, angle, height, and distance are the only variables we have pieces of data for
(https://docs.google.com/spreadsheets/d/10IZ9I7nJ4BU7oV1zqobtd6AsyNp1LsufusCY0i8-fAk/edit?gid=0#gid=0)
split sail
i would assume because the same cannon is used then the initial speed is always the same
maybe you can find v0 with 2 points of that data
half garden
plugging in points (30,2) and (30,32) (as well as (45,2) for good measure) I get different initial velocities, so it doesnt seem like the initial velocity is constant
amber waspBOT
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amber waspBOT
@remote raft Has your question been resolved?
hasty yacht
if you want to maximize d, there's a great geometrical solution
i don't know about f(t, h)=d but this should be a great insight
amber waspBOT
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hearty jackal
if you want to maximize d, there's a great geometrical solution
House House.
amber waspBOT
House House.
hearty jackal
.close
amber waspBOT
thick shale
:O
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Available help channel!
split sail
Part d
split sail
Var(W)=a^2Var(X) i think
Never seen this in my entire life
paper dagger
the variance of aX+b = a^2Var(X)
split sail
Is this something that should be memorised or did you derive it from another formula
paper dagger
Is this something that should be memorised or did you derive it from another for...
i guess memorized
split sail
It's not in my formula booklet either, but thanks
paper dagger
or u can derive it
paper dagger
It's not in my formula booklet either, but thanks
what formula booklet is it
split sail
From?
split sail
what formula booklet is it
Edexcel ial maths booklet s1
paper dagger
i guess u can go off of this
paper dagger
replace X with aX+b and see what you get id assume
paper dagger
Uhh idk what that means
split sail
If Y =ax + b that's coded data then VarY = a^2(Var(x)
I completely forgot about that for some reason
Thanks
cyan geyser
Can you help me solve this equation?
amber waspBOT
Can you help me solve this equation?
alpine flower
I can think of squaring both sides, but it gets hard to solve because of many terms
amber waspBOT
@cyan geyser Has your question been resolved?
alpine flower
its actually gets worser
cyan geyser
ugh, bigg noo
alpine flower
willow wagon
A game is being played by two players. The rule is one player can take either 1,2 or 3 sticks at once from 15 sticks then the other player will take 1,2 or 3 sticks from the remaining sticks and it will go on repeatedly until one stick remains in the end and the one picking it will lose the game. To ensure the victory one player must pick how many sticks at the beginning?
fresh ocean
Retrace the steps, imagine you are putting sticks back in the stack, when initial stack size is 1 stick
willow wagon
Retrace the steps, imagine you are putting sticks back in the stack, when initia...
Wdym by initial stack size is 1 stick?
fresh ocean
there is one stick on the ground and you and your opponent put 1/2/3 sticks on your turns
willow wagon
there is one stick on the ground and you and your opponent put 1/2/3 sticks on y...
Then?
fresh ocean
Think of optimal strategy
How would the game go if there were 2 sticks instead of 15... How about 3 instead of 15 and so on
willow wagon
One strategy I can now currently think of is if the stick number is odd which in this case is and if they could pick only one stick at a time then the first one to pick would probably lose. We can reduce the twice pick and thrice pick by subtracting 1 or 2 from the original stick number. And then maybe use something like the stars and bars formula to solve
fresh ocean
And then maybe use something like the stars and bars formula to solve
seems pretty unnecessary... You dont wanna count the possible ways a game can go
Hint: If you know how to play when theres 2 sticks left, you can always win if you arrive at 2 sticks no matter how many you started with. Similarly with 3, 4,5 and so on...
If you do this, you might observe a pattern
willow wagon
How would the game go if there were 2 sticks instead of 15... How about 3 instea...
2 seems like a valid option
Because it's the only even number
And If counted from reverse like if I keep subtracting 4's from 15 then I will 2 in the end
fresh ocean
|| You are at 2 sticks, means you can pick 1, and opponent has to pick the remaining stick and they lose. So winning strategy is to leave 1 stick. Same when theres 3 sticks, you pick 2. If theres 4 left, you pick 3. If theres 5, what would you do? ||
willow wagon
|| You are at 2 sticks, means you can pick 1, and opponent has to pick the remai...
That's the reason I can't be left with 1 or 4 stick at the end
fresh ocean
mhm
so you dont wanna end up in a situation where theres 5 sticks left. Can you imagine how to tacklle that?
can you extend this for more number of sticks?
willow wagon
can you extend this for more number of sticks?
Do I need to keep subtracting 4's?
I would be left with 3 then
And I think that's the answer
.close
plain fjord
hello, im trying to caluclate the unification, the intersection and the difference with these: a=⟨−3;0) b=⟨−1;2⟩
now ive tried to do this already, and my answers look like this
a∪b= ⟨−3;2⟩
a∩b=⟨−1;0)
a−b=⟨−3;−1)
b−a=⟨0;2⟩
ive checked the answers with my classmate and he seems to have the answers same, but not the differences. he has them like this:
a−b=⟨−3;2)
b−a=⟨1;2⟩
were not sure which option is correct, so if someone could tell me which one is correct, and why, that would be great, ty
amber waspBOT
hello, im trying to caluclate the unification, the intersection and the differen...
@plain fjord Has your question been resolved?
crisp stratus
hello, im trying to caluclate the unification, the intersection and the differen...
what does ⟨ mean?
are these intervals
crisp stratus
[ ]
plain fjord
or whatever the english tranlsation is
crisp stratus
these?
plain fjord
yeah
crisp stratus
ok
plain fjord
for some reason we use the ones i wrote down
thick shale
a∪b= ⟨−3;2⟩
a∩b=⟨−1;0)
a−b=⟨−3;−1)
b−a=⟨0;2⟩
these are correct
crisp stratus
hello, im trying to caluclate the unification, the intersection and the differen...
yours are correct
plain fjord
do you know how it was possible that he came up with the answers he had? we talked about it but idk
crisp stratus
i don't see how he arrived at his
plain fjord
alr, ty either way
.
plain fjord
yea :d
but we would understand ts like irts jusrt the mf teacher is expplaining math like you would explain physics or some completely other subject man
night sundial
i know i can write things in this form
amber waspBOT
i know i can write things in this form
night sundial
but how does it work with this
round widget
$$\frac{1+\frac{1}{n!}}{1+\frac{1}{(n+1)!}}=\frac{\frac{n!+1}{n!}}{\frac{(n+1)!+1}{(n+1)!}}=\frac{n!+1}{n!}\cdot\frac{(n+1)!}{(n+1)!+1}=\frac{(n!+1)(n+1)}{(n+1)!+1}$$
grizzled pagodaBOT
VincentBH
night sundial
for my analysis class in a test
is this sufficient
to say this goes to 1 as n approaches inf
round widget
is this on your test right now?
night sundial
old exam i guess
round widget
ah
night sundial
normally things cancel out when doing ratio test
this one stays a bit untrivial even after doing it
round widget
well you can write $$\frac{(n!+1)(n+1)}{(n+1)!+1}=\frac{(n!+1)(n+1)}{n!\cdot(n+1)+1}=\frac{(1+\frac{1}{n!})(n+1)}{(n+1)+\frac{1}{n!}}=\frac{(1+\frac{1}{n!})}{1+\frac{1}{n!(n+1)}}$$
distant herald
your series is 1 + 1/n!, why use the ratio test when it already fails the nth term test?
grizzled pagodaBOT
VincentBH
night sundial
your series is 1 + 1/n!, why use the ratio test when it already fails the nth te...
what is the nth term test
round widget
this goes to 1 if n to infinity
distant herald
a series being convergent implies that the limit of the nth term goes to 0
night sundial
this goes to 1 if n to infinity
are those steps mandatory or can i just plug n here
distant herald
meaning that if the nth term tends to anything other than 0, a series must be divergent
round widget
are those steps mandatory or can i just plug n here
ah ye I already did everything before the person said they wanted to take a limit haha
night sundial
a series being convergent implies that the limit of the nth term goes to 0
does this serie not converge for x in (-1,1)?
daring tusk
Hello
daring tusk
Just not seeing where they are getting this like
at all
It makes sense when I see it on desmos but idk how one would reach this conclusion otherwise
proven vapor
do you know what King's rule is?
tough mica
What is the maximum possible value of the greatest common divisor of two consecutive terms of the sequence $a_n = n! + n$, where $n \ge 0$?
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
tough mica
how do I use euclidean algorithm here?
viscid olive
i think so
pseudo crescent
how do I use euclidean algorithm here?
I'd only perform one step of it
or technically 0
rather than doing whole modulo, just do one subtraction
use the theorem saying gcd(a, b) = gcd(a, b - a)
keen comet
I'd only perform one step of it
Omg shut up
tough mica
tough mica
gcd(an, an+1)
A=a_n
B=a_(n+1)
gcd(a_n, a_(n+1)-a_n)
what do I do now? I applied the property
tough mica
n! + n
tough mica
gcd(a_n, a_(n+1)-a_n)=gcd(n!+n, (n+1).n!+(n+1)-n!-n)
pseudo crescent
https://tenor.com/view/luigi-typo-meme-gif-26876585
Um, by chance, do you know some method that works? I'll get a piece of paper and try to figure sth out as well
sterile swan
mby smth like $a_{n+1}=(n+1)!+n+1=(n+1)(n!+1)=(n+1)(n!+n-n+1)=(n+1)(a_n-n+1)$?
grizzled pagodaBOT
Bonk
sterile swan
then $gcd(a_n,a_{n+1})=gcd(a_n,(n+1)a_n-(n+1)(n-1))=gcd(a_n,(n+1)(1-n))$
grizzled pagodaBOT
Bonk
sterile swan
I understood until here
that part is writing out a_n+1 in terms of a_n
note that $(n+1)!=(n+1)n!$
grizzled pagodaBOT
Bonk
tough mica
$a_{n+1}=(n+1)(a_n-n+1)$
grizzled pagodaBOT
938c2cc0dcc05f2b68c4287040cfcf71
sterile swan
if you immediately go to that step it doesnt make sense
you need to go to the inbetween steps
$a_n=(n+1)!+(n+1)$
grizzled pagodaBOT
Bonk
sterile swan
$(n+1)!+(n+1)=(n+1)n!+(n+1)=(n+1)(n!+1)$
tough mica
(n+1)(n! +1)
grizzled pagodaBOT
Bonk
sterile swan
still following up to here?
tough mica
ye
grizzled pagodaBOT
Bonk
sterile swan
yes?
tough mica
ye
sterile swan
then we replace the n!+n with a_n
tough mica
(n+1)(an -n+1)
grizzled pagodaBOT
Bonk
tough mica
gcd(a_n, a_(n+1)-a_n)
sterile swan
so, $gcd(a_n,a_{n+1})=\gcd(a_n,(n+1)a_n+(n+1)(-n+1))=\gcd(a_n,(n+1)a_n+(n+1)(-n+1)-(n+1)a_n)=\gcd(a_n,(n+1)(-n+1))$
grizzled pagodaBOT
Bonk
tough mica
second equality
sterile swan
we are removing a n+1 multiple of a_n
tough mica
ok fair
sterile swan
if you go to this, a=a_n and b=a_n+1, then q=n+1
tough mica
ye i figured
pseudo crescent
What can we do after that though?
tough mica
^
pseudo crescent
I got to if p divides the gcd, then
p | (n-1)! + 1
and
p | n + 1
robust isle
gcd((n + 1)(n - 1), n! + n) = gcd(n + 1, (n - 1)! + 1)
keen comet
Uhhh
pseudo crescent
ye
keen comet
oooooooo
robust isle
write as gcd(n, (n - 2)! + 1)
amber waspBOT
I’m bored
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
robust isle
if n is composite then gcd is 1, so assume n is prime
then you compute (p - 2)! + 1 mod p
which is -1/(p - 1) + 1 which is never 0 unless p = 2
keen comet
no
tough mica
gcd(-(n+1)(n-1), n!+n)=gcd((n+1)(n-1), n!+n)?
robust isle
yes
keen comet
moans cutely
pseudo crescent
*moans* cutely
go to #chill , don't post off-topic here
robust isle
gcd((n + 1)(n - 1), n! + n) = gcd((n + 1)(n - 1), n((n - 1)! + 1)) = gcd((n + 1)(n - 1), (n - 1)! + 1)
because n and (n - 1)(n + 1) are coprime
then gcd((n + 1)(n - 1), (n - 1)! + 1) = gcd(n + 1, (n - 1)! + 1) because (n - 1) and (n - 1)! + 1 are coprime
tough mica
what is that coprime property
of the gcd, it wasnt mentioned in the pic
how is n coprime to (n-1)(n+1)
gcd(n,(n-1)(n+1))=gcd(n,(n-1)).gcd(n,(n+1))=1.1=1
sterile nymph
@tough mica gcd(n, (n+1)(n-1)) use the euclidean algorithm:
gcd(n, (n+1)(n-1)) = gcd(n, n^2 - 1) = gcd(n^2 - 1 mod n, n) = gcd(n-1, n)
From here, it should be obvious that n and n-1 share no factors, because the only possible factor that can divide both is 1.
or yeah, you can use the multiplicative property as well
amber waspBOT
@tough mica Has your question been resolved?
tough mica
wtf is wilsons theorem? why does it matter that n is prime
I understood until (n-1) and (n-1)! +1 is coprime
if n is composite, aka Not prime
what happens?
verbal jewel
What is the maximum possible value of the greatest common divisor of two consecu...
regardless of what you have already done you always should look at the first few elements of the sequence. sometimes it gaves you an idea whats happening, sometimes you can check your results ...
calcluate a_0, a_1, a_2, a_3, a_4, ....
tough mica
a0=1+0=1
a1=1+1=2
a2=2+2=4
a3=6+3=9
a4=24+4=28
verbal jewel
a0=1.
tough mica
my baaaaad
verbal jewel
from this few examples what is the greatest common divisor of two consecutive elements of the sequence?
verbal jewel
so at the moment you know that your answer is (at least) 2 - what ever you conclude with other methods you know it has to be at least 2.
tough mica
ye
verbal jewel
if you do some few elements more, you can create an assumption - isnt a proof but to get an idea where you possible need to go.
tough mica
a conjecture
verbal jewel
yes
tough mica
a5=120+5=125
a6=720+6=726
a7=5040+7=5047
a8=40320+8=40328
a9=362880+9=362889
let me bring calculator
sterile swan
generally there is not much to be learned from going much further into the terms
looking at the first few should be good enough
verbal jewel
generally there is not much to be learned from going much further into the terms
how many calculations are necessary to formulate an conjecture depends on the person. some needs more, some less.
sterile swan
how many calculations are necessary to formulate an conjecture depends on the pe...
well, if you dont see it at 5 you prob wont see it at 10
generally
amber waspBOT
@tough mica Has your question been resolved?
amber waspBOT
@tough mica Has your question been resolved?
grizzled pagodaBOT
ibratoch
consider $gcd(a_n,a_n+1)$
```Compilation error:```! LaTeX Error: Unicode character (U+2061)
not set up for use with LaTeX.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.49 consider $gcd
(a_n,a_n+1)$
You may provide a definition with
\DeclareUnicodeCharacter```
inland stag
fuck
grizzled pagodaBOT
Bonk
patent raptor
\gcd
sterile swan
or did you mean $gcd(a_n,a_{n+1})$
grizzled pagodaBOT
Bonk
inland stag
or did you mean $gcd(a_n,a_{n+1})$
thanks mate
sterile swan
$a_{n+1}$
grizzled pagodaBOT
Bonk
grizzled pagodaBOT
Bonk
sterile swan
$a_{n+1}$
grizzled pagodaBOT
Bonk
inland stag
so let's consider $gcd(a_n,a_{n+1})$
grizzled pagodaBOT
ibratoch
inland stag
$gcd(a_n,a_n+1)=gcd(n!+n,(n+1)⋅(n!+1))$
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
inland stag
$gcd(a_n,a_{n+1})=gcd(n!+n,(n+1)⋅(n!+1))$
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
inland stag
**ibratoch**
Compile Error! Click the <:errors:504266556054962196> reaction for ...
that's good btw
sterile swan
**ibratoch**
Compile Error! Click the <:errors:504266556054962196> reaction for ...
is this your answer?
inland stag
yep
inland stag
you still can use the property of GCD so
"The key property is gcd(a,b)=gcd(a,b−ka)gcd(a,b)=gcd(a,b−ka) for integers a,b,ka,b,k.
wtf
sterile swan
yes?
inland stag
$"The key property is gcd(a,b)=gcd(a,b−ka) for integers a,b,k"$
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
sterile swan
all that needs to be done is to simplify $gcd(n!+n,(n+1)(n-1))$
grizzled pagodaBOT
Bonk
inland stag
$a_{n+1}−(n+1)⋅a_n=(n+1).$
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
inland stag
$a_{n+1}−(n+1)*a_n=(n+1).$
grizzled pagodaBOT
Bonk
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
inland stag
there?
grizzled pagodaBOT
Bonk
inland stag
$-$
grizzled pagodaBOT
ibratoch
sterile swan
hmm
inland stag
$a_{n+1}-(n+1)*a_n=(n+1).$
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
sterile swan
whats the error?
sterile swan
are you on IOS?
inland stag
mac
sterile swan
ah, that might be the problem
inland stag
maybe
inland stag
There are three types of dash
long
moyen
and informatique one
i guess that was the issue
anyway back to our track
Thus, $gcd(a_n,a_{n+1})=gcd(a_n,n+1)gcd(a_n,a_{n+1})=gcd(a_n,n+1)$
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
inland stag
here we go
don't mind the last parenthese
now analyze gcd?
$a_n=n!+n$
-common divisors of n!+n and n+1
n! is divisible by 1,2,…,n thus n! is conforming to 0 (modn+1) if n+1 is a divisor of n!
$=> a_n≡n(mod n+1)$
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
inland stag
yeah fuck yourself for not take the bigger/equal to
patent raptor
\geq exists
inland stag
For n\geq 2, the term n+1 is in to coprime n, so $gcd(a_n,n+1)=1$
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
inland stag
didn't close it did i?
patent raptor
For $n \geq 2$, the term $n+1$ is coprime to $n$, so $\gcd(a_n, n+1) = 1$.
inland stag
love you
patent raptor
weird
inland stag
For $\n=0: a_0=1, a_1=2, and gcd(a_0,a_1)=1$
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
grizzled pagodaBOT
ibratoch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
grizzled pagodaBOT
𝔸dωn𝓲²s
inland stag
is that even a symbol lol
patent raptor
maybe you copied something from somewhere else anyway
inland stag
i've solved the thing i think
inland stag
maybe you copied something from somewhere else anyway
yeah I write on something else before copy pasta
so the maximum GCD is 2, occurring in n=1
I don't think it's good but whateverI'm tired
check if I've solved it or not
and mistaked maybe
bc the guy was stuck
lemme take the screens so
@split sail
forgot to add
patent raptor
lmao
dull pike
What a weird format
inland stag
outer hull
!nogpt
amber waspBOT
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).
patent raptor
😂
inland stag
remind me of that dude saying he solved rienmann hypothesis
with chat gpt glazes
i didn't know the proprity key until now
amber waspBOT
@tough mica Has your question been resolved?
amber waspBOT
amber waspBOT
Available help channel!
keen pawn
Trying to determine if this converges using the alternating series test
keen pawn
I think it doesn't converge as $\abs{a_n} \geq \abs{a_{n-1}}$
grizzled pagodaBOT
math rocks
crisp stratus
I think it doesn't converge as $\abs{a_n} \geq \abs{a_{n-1}}$
indeed
keen pawn
Thanks
prisma ember
you can also use nth term test i think
amber waspBOT
amber waspBOT
Available help channel!
misty cairn
can someone help me intergral of xdx/sqrt(3-2x-x^2)
amber waspBOT
can someone help me intergral of xdx/sqrt(3-2x-x^2)
tired girder
(x+1)/(√3-2x-x²) - 1/√(2² -(1+x)²)
For first integral consider (3-2x-x²) = z
tired girder
(x+1)/(√3-2x-x²) - 1/√(2² -(1+x)²)
The second one is a formula !
misty cairn
oki wait me
i cant solve the second
let me send u
A is done but B i used trig sub but its not ok
thank you i solved it now
cảm ơn
.close
amber waspBOT
tired girder
thank you i solved it now
Sry i didn't see your reply !
misty cairn
.close
amber waspBOT
Available help channel!
misty cairn
can someone help me question 6
amber waspBOT
can someone help me question 6
jaunty viper
can someone help me question 6
đề GT2 ạ?
jaunty viper
a bt làm k
em nghĩ là gần 0 thì anh dùng arctan(x) ~ x và mẫu số ~ 2x^(3/2)
jaunty viper
sao mẫu là 2x^(3/2)
là sqrt(4x^3) anh ạ
misty cairn
mà tại 0 là điểm kì dị có làm đc thế k
jaunty viper
mà tại 0 là điểm kì dị có làm đc thế k
đoạn này xét VCB ấy anh
misty cairn
í a là sao a vt ra đc k
dụ í bài này e nghĩ là xét arctan r đổi cận từ 0 đến pi/2 thôi
misty cairn
là sqrt(4x^3) anh ạ
sqrt(4x^3) cái này ở đâu vậy?
thôi cái này là divergent r
nếu dg VCB thì sẽ là x/3x^(3/2) thì còn 1/3x^(1/2) thì sẽ là div
jaunty viper
misty cairn
ok its diver 100%
jaunty viper
no dive mà 1/x^1/2
nguyên hàm của nó thì sqrt(x) bốc lên tử số mà anh ơi
nè
converge mà anh ơi
misty cairn
ok let me
misty cairn
vãi desmos sai r
đợi tí
thì mẫu còn x^1/2 đg k
đẩy lên là intergral x^-1/2
thì là 2sqrt(x) from 0 to infi mà
má k gửi đc ảnh
đây a ơi chuẩn ntn còn j
amber waspBOT
@misty cairn Has your question been resolved?
keen pawn
determine if this converges
amber waspBOT
determine if this converges
keen pawn
The seqeunce tends to 0 and $\abs{a_n}< \abs{a_{n-1}}$ . It thus satifies the alternating series test
grizzled pagodaBOT
math rocks(wai)
keen pawn
it convergees
amber waspBOT
@keen pawn Has your question been resolved?
indigo cloud
what about the cos
keen pawn
what about the cos
hmm, that alternates between -1 and 1
but when the alternating series test is used, we take the asbolute balue, right
indigo cloud
hmm, that alternates between -1 and 1
which needs to be mentioned
indigo cloud
yes
amber waspBOT
charred plinth
Real question only here: why is the second option a DNF of the original formula?, I get the rest of it
amber waspBOT
Available help channel!
eager shale
A half-full bottle of water has exactly as much as four identical empty bottles. We replace the bottle on the right tray with a full bottle of water.
How many empty bottles will he have to place on the left tray to balance the scales?
amber waspBOT
A half-full bottle of water has exactly as much as four identical empty bottles....
eager shale
.close
amber waspBOT
amber waspBOT
Available help channel!
hollow hound
amber waspBOT