maximum ≠ point of maximum
#help-10
hidden compass
hidden compass
What is a maximum then
The function evaluated at the point of maximum
karmic hedge
The function evaluated at the point of maximum
wym by that sry
junior granite
wym by that sry
bro just read the question thoroughly
hidden compass
wym by that sry
If the point of max occurs at x = 3, then f(3) is the maximum
karmic hedge
If the point of max occurs at x = 3, then f(3) is the maximum
Ok, makes sense
So the max occurs and t=4 and t=5, so why is there not 2 maximums?
hidden compass
Because f(4) and f(5) are the same
karmic hedge
Because f(4) and f(5) are the same
Yes, I agree
So why is 1 the answer and the other isnt?
hidden compass
I didn't follow your exercise, sorry
karmic hedge
t^2 -9t +20 = 0 so the 2 values of t I got were 4 and 5
Ok I will come back to this another time
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
grave lagoon
do I just plug the numbers into simpson rule after finding w ?
obtuse pebbleBOT
do I just plug the numbers into simpson rule after finding w ?
grave lagoon
I mean they already gave me w, which is 60/6 = 10
obtuse pebbleBOT
@grave lagoon Has your question been resolved?
obtuse pebbleBOT
@grave lagoon Has your question been resolved?
tacit zephyr
i have x^2 = y^2 mod n under the condition that x != +-y mod n how does it follow that n is not a prime number?
obtuse pebbleBOT
i have `x^2 = y^2 mod n` under the condition that `x != +-y mod n` how does it f...
tacit zephyr
what i have so far:
x^2 - y^2 = 0 mod n => n divides x^2-y^2=(x+y)(x-y)
short spire
(x-y)(x+y) is a multiple of n and we know that x ≠ +- y, so what is the only possibility for n here?
tacit zephyr
if x=+-y would hold then n would be 0 right
short spire
If that was the case then n could be anything, e.g. a prime number
tacit zephyr
(x-y)(x+y) is a multiple of n and we know that x ≠ +- y, so what is the only pos...
"so what is the only possibility for n here?" yeah i dont get that step
ahh i think i got something
short spire
We know now that the factors must be atleast non zero. For n to possibly be a prime number one of these factors has to be 1 in such case.
So suppose x+y = 1, then x-y = 1-2y
i.e 1-2y is a multiple of n
tacit zephyr
isnt n then a multiple of 1-2y?
because n is a multiple of (x-y)(x+y) and you set that to (1-2y)1
short spire
Yup as I mentioned above
tacit zephyr
you said "(x-y)(x+y) is a multiple of n" but isnt that wrong?
(x-y)(x+y) = 0 mod n means (x-y)(x+y) = kn for some natural number k
so n is the multiple of (x-y)(x+y) and not the other way
short spire
If (x-y)(x+y) = kn, then it’s a multiple of n…
tacit zephyr
but the idea that i got was:
let d be the gcd(x-y, n) = n then it follows that x-y = 0 mod n => x=y mod n (Error with our condition that x!=+-y mod n)
let d be the gcd(x-y, n) = 1 and we know n divides (x-y)(x+y), then it follows with some theorem i forgot the name (if a divides b*c and gcd(a,b)=1 then a divides c) that n must divide x+y => x=-y mod n (Error with our condition)
so we showed that the gcd(x-y, n) is not n and not 1 so it has to be something in between and it follows that n has a non trivial divider
tacit zephyr
If (x-y)(x+y) = kn, then it’s a multiple of n…
maybe i translate that wrong idk
yeah makes sense i translated that wrong in my head sry
but yeah anyway
.close
obtuse pebbleBOT
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Available help channel!
brazen gorge
i noticed an interesting pattern that i can't exactly explain
obtuse pebbleBOT
i noticed an interesting pattern that i can't exactly explain
brazen gorge
$\zeta(1+10^{-n}) \approx 10^{n}$
warm shaleBOT
artemetra
warm shaleBOT
safe ravine
Lol
brazen gorge
,w zeta(1.0001)
warm shaleBOT
brazen gorge
surprisingly accurate
safe ravine
I k that
brazen gorge
i wonder why this happens
safe ravine
As zeta(1) is undefined
brazen gorge
yes
warm shaleBOT
safe ravine
See
brazen gorge
i get that, but i don't get why it would be 10000 instead of idk 21654
safe ravine
Which?
brazen gorge
i conjecture $\lfloor\zeta(1+b^{-n})\rfloor = b^n$
warm shaleBOT
artemetra
brazen gorge
,w approximate log_2(floor(zeta(1+2^(-4))))
warm shaleBOT
Requested by arte_m_etra
brazen gorge
bruh
brazen gorge
,w zeta(1+2^(-11))
warm shaleBOT
brazen gorge
,calc 2^11
warm shaleBOT
Result:
2048
So u want what generalise something?
brazen gorge
ig
safe ravine
Or only discuss
stoic yacht
btw wolfram seems to indicate that the difference between those tends to the euler mascheroni constant or however you spell it
regardless of the base
safe ravine
See
dusk widget
$\zeta(1+10^{-n}) \approx 10^{n}$
while I don't have anything to contribute to this, I wanna ping @warm quartz about this, because it seems up his alley
stoic yacht
regardless of the base
(as long as a > 1)
warm shaleBOT
brazen gorge
$\zeta(1+b^{-n}) \approx b^n + \gamma$
safe ravine
Euler macheroni constant is 0.57721...
warm shaleBOT
artemetra
stoic yacht
Euler macheroni constant is 0.57721...
... subtract 128 from your answer
i'm saying the difference tends to the constant
warm shaleBOT
safe ravine
,w zeta(24)
warm shaleBOT
warm shaleBOT
safe ravine
Fr 🔥
stoic yacht
,w zeta(25)
warm shaleBOT
brazen gorge
**artemetra**
can be extended to subtraction too:\
$\zeta(1-b^{-n}) \approx -b^n + \gamma$
safe ravine
Zeta becomes 1
warm shaleBOT
artemetra
stoic yacht
Zeta becomes 1
... that's what i said
safe ravine
,w solve limit n tends to infinity zeta(n)
warm shaleBOT
safe ravine
Fr
brazen gorge
sorry oppailol but you are not really helpful rn lol
safe ravine
Bro what u want I'm ready to help
Your conjecture is
zeta(1+b^-n)=approx(b^n+gamma)
safe ravine
What
brazen gorge
$\zeta(1+i\cdot b^{-n}) \approx \gamma - i\cdot b^n$
warm shaleBOT
artemetra
stoic yacht
not really sure what happens in the second plot if the limit is apparently gamma
brazen gorge
just trying out different directions from which i can approach the pole
safe ravine
The pole
Fr
Lmao bro
Can we break down the Riemann hypothesis
And show that your conjecture has a link to it
Can we do it ?
brazen gorge
i think i found the answer
safe ravine
Can we?
stoic yacht
never seen that 😭
safe ravine
Check Bose integral and relation b/w gamma(s).Zeta(s)
brazen gorge
brazen gorge
wait hold up gamma here is a function not the constant
safe ravine
gamma(s).Zeta(s)=
brazen gorge
safe ravine
O to infinity.
x^s/(e^x-1)
This is
Was it s or s-1
?
Aah
So now what?
We have the image
@brazen gorge
Tell
brazen gorge
nah idk anything lol
brazen gorge
i honestly do not get this at all
stoic yacht
nah idk anything lol
i believe the error between zeta(1 + 10^(-n)) and 10^n satisfies gamma < zeta(1 + 10^(-n)) - 10^n < 0.645
as long as n > 0
brazen gorge
hm, why 0.645
stoic yacht
new conjecture:
gamma < zeta(1 + a^(-n)) - a^n < pi^2 / 6 - 1
for n > 0 and a > 1
brazen gorge
looks great!
brazen gorge
new conjecture:
gamma < zeta(1 + a^(-n)) - a^n < pi^2 / 6 - 1
for n > 0 and a ...
this explains why the euler mascheroni is there, at least
stoic yacht
ohh
and i think i've seen expansions around the pole that have gamma come up a bunch in the coefficients
i believe you can show that zeta(1 + a^(-x)) - a^x is decreasing on this interval
making the limit of gamma a lower bound
and making zeta(1 + a^(-0)) - a^0 = zeta(2) - 1 = pi^2 / 6 - 1 an upper bound
safe ravine
We need Riemann
Hypothesis to be proved here
Fr
@brazen gorge
@stoic yacht
Can we 3 prove it
stoic yacht
i suspect that three undergrads absolutely cannot prove the riemann hypothesis 💀
safe ravine
I suspect that in this world everything is truly possible
Though it may take time
@brazen gorge
,w zeta(1/2)
warm shaleBOT
safe ravine
,w zeta(1/4)
warm shaleBOT
brazen gorge
i suspect that three undergrads absolutely cannot prove the riemann hypothesis �...
lmao
safe ravine
, w zeta(1/8)
warm shaleBOT
safe ravine
,w solve limit n tends to sum from 1 to n zeta(1/2^n)
warm shaleBOT
warm shaleBOT
warm shaleBOT
stoic yacht
Can we find some pattern
wolfram says it diverges
brazen gorge
this is the reason why this happens
brazen gorge
so hard to discover something new in math lol
stoic yacht
still what are the odds someone's played around with bounding zeta(1 + 10^(-x)) - 10^x
ofc that generalizes here
but still maybe we had reason to suspect
brazen gorge
still what are the odds someone's played around with bounding zeta(1 + 10^(-x)) ...
non zero considering i did it lol
anyways thanks again neil and everyone else
.close
obtuse pebbleBOT
safe ravine
@brazen gorge . reopen
obtuse pebbleBOT
✅
safe ravine
The reason is that
1
Only 1 is the factor
Compute sum from 1 to n-1 [zeta(1/2^n)-1]
brazen gorge
$\lim_{x\to \infty} \zeta\left(\frac{1}{2^x}\right) = \zeta(0) = -\frac{1}{2} \neq 0 \implies$ the sum diverges
warm shaleBOT
artemetra
brazen gorge
safe ravine
Why
stoic yacht
zeta(2^n) tends to 1 as n tends to infinity
opaque dome
Zeta(2n) aswell
brazen gorge
brazen gorge
wait wtf
brazen gorge
Now check 2^n+1
+1 where
brazen gorge
stoic yacht
we're not disagreeing with you that it converges
safe ravine
If you compute it comes in terms of ln(2)
I had the result derive I can try again
See this is bcz
opaque dome
What if you add the same sum but taking the derivative of zeta ? 🗿
stoic yacht
What if you add the same sum but taking the derivative of zeta ? 🗿
this almost definitely converges since the derivative of zeta tends to 0 quickly
safe ravine
Derivative of Zeta will be
-ln(n)/n^s
Add it
No
I am wrong
It will be ln(1/n)/n^s
opaque dome
-ln(n)/n^s
Kinda harder
safe ravine
Fr
brazen gorge
i think i broke WA
brazen gorge
full bonkers mode
opaque dome
Zeta(odd) become constants
Indeed
safe ravine
Try 3n
Lol
3n
Yes if u see
Galois proved
Overall that na
Odd is tough
Fr broh
opaque dome
Fr
stoic yacht
hold up. 5n, polynomial of degree 5...
i wish it was more obvious where the quintic was coming from
safe ravine
,w summation from n=0 to n=inf (zeta(7n)-1)
warm shaleBOT
brazen gorge
???
hold up this factors
(x + 1) * (x^4 + 9*x^3 + 31*x^2 + 49*x + 31)
,w solve x^4 + 9x^3 + 31x^2 + 49*x + 31 = 0
warm shaleBOT
brazen gorge
wdf
brazen gorge
modern mathematicians are so lazy. all they know is use wolphramalpha, graph in desmos, algebra in sage and ask on forum
opaque dome
Im not a modern one then, crazy
But if there is a step by step solutions then it will be intersting to see how this computer does it
warm quartz
while I don't have anything to contribute to this, I wanna ping <@29001085825856...
@brazen gorge partial summation to extend the zeta function at a the domain you want near s=1 then find the asymptotics of it. That will answer your question
brazen gorge
<@474467399207223307> partial summation to extend the zeta function at a the dom...
i already figured it out, it's due to this
but thanks nonetheless!
warm quartz
i already figured it out, it's due to this
Do you see why this is the reason?
I just want to check
brazen gorge
Do you see why this is the reason?
i don't have much experience with complex analysis so not entirely but i assume what this expression means is that around 1, zeta behaves like the 1/x function
warm quartz
i don't have much experience with complex analysis so not entirely but i assume ...
More or less you are right. The way to think about it is. If you fix s = a+ib and set s=1 then the function is defined on all values of b except at b=0. That's exactly when s = 1 which is what you suggested
$\zeta(s) = \frac{s}{s-1}-s \int_{1}^{\infty} \left{ x \right} x^{-s-1}dx$
warm shaleBOT
Deltoid
obtuse pebbleBOT
@brazen gorge Has your question been resolved?
brazen gorge
More or less you are right. The way to think about it is. If you fix s = a+ib an...
i see, thank you!
heady vault
x^(x+1)=(x+1)^x i know that answer is between 2 and 3 but how do I solve it?
open folio
the domain where x in x^4-2x^2-8<0 is defined is R, correct ?
obtuse pebbleBOT
the domain where x in x^4-2x^2-8<0 is defined is R, correct ?
dark stirrup
I can think of a few numbers that do not satisfy this
normal cairn
well its a polynomial with the highest-power term having a positive sign so we know that its definitely positive somewhere
seems like youll have to solve the inequality
marsh geyser
You did not make the questions understandable then
wooden cipher
we are looking for x in R, yes
open folio
we are looking for x in R, yes
Ok thanks
I have another question
When we have a polynomial<0 or >0
And its solutions are complex, then how do we know if the solutions are all of R or that no solutions without experimenting ?
junior granite
the domain where x in x^4-2x^2-8<0 is defined is R, correct ?
Will it not be $x\in(-2,2)$ ?
warm shaleBOT
77²
open folio
yes it is
wooden cipher
if it doesnt intersect the x axis in R, then it must be on one side of the axis
e.g. 1+x^2 has roots +-i and is always above the x axis
obtuse pebbleBOT
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Available help channel!
honest narwhal
Can someone explain this to me please?
obtuse pebbleBOT
Can someone explain this to me please?
normal turret
Are you given the formula for the surface area of a cone?
honest narwhal
Are you given the formula for the surface area of a cone?
No unfortunately that's why I'm confused
normal turret
this is for algebra or geometry, yes?
honest narwhal
Geometry measurements
normal turret
because you could solve this problem with calculus if not given the formula, but if the answer is yes then you should be given the formula
warm shaleBOT
Savage_Cat
normal turret
thats your formula
and you're given the slant height and diameter
notice how the formula doesn't have input values for those measurements, it requires radius and height
are you able to determine those variables with the given information?
trail wraith
Can someone explain this to me please?
Surface Area=πr(r+l)
Substituting the given values:
Calculating step-by-step:
Calculate
13×69=897
Multiply by
π:
3.14×897≈2817.93
approx 2818
radius = 13ft
and slant = 56
honest narwhal
are you able to determine those variables with the given information?
I got 2347.89
trail wraith
Surface Area=π×13×(13+56)
cacluate by substituting
honest narwhal
Surface Area=π×13×(13+56)
cacluate by substituting
Okay I got 2878.82
I forgot to add the 13 earlier 😅
trail wraith
how? do 13(13+56) = 13 into 69 = 897
then pie*897 = 3.14 into 897 = 2818 approx
honest narwhal
I was removing the squares now I got
I got the whole thing confused because of it
My bad
trail wraith
(h^2 + r^2) = slant height^2 or L^2...just remove root dont include it
obtuse pebbleBOT
@honest narwhal Has your question been resolved?
rare cargo
why is D incorrect?
obtuse pebbleBOT
why is D incorrect?
opaque dome
why is D incorrect?
How do you do it?
rare cargo
find the time when the rate of change is maximized
opaque dome
So the max of the derivative ?
obtuse pebbleBOT
@rare cargo Has your question been resolved?
rare cargo
So the max of the derivative ?
yes
opaque dome
So the sign of second derivative, and deduce the variations of the derivative
rare cargo
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
outer spindle
Why is y’ = -x/y
obtuse pebbleBOT
Why is y’ = -x/y
outer spindle
Here
marsh geyser
Use chain rule
outer spindle
Chain rule
outer spindle
Because y=sqrt(4-x^2)
This
outer spindle
Can you walk me thru the steps if that (sorry this is a 6 week class)
Chain rule = f’ (g) times g’
F = sqrt x
G is 4-x^2
marsh geyser
Continue
stoic yacht
this approach works ofc by just differentiating y = sqrt(4-x^2), it might be a little bit faster to find y' for x^2 + y^2 = 4
outer spindle
How y’ was made, I am following you so far
stoic yacht
y' = 1/2 (4-x^2)^(-1/2) (-2x), and (4-x^2)^(1/2) was y
so you get y' = 1/2 (1/y) * (-2x) = -x/y
outer spindle
Something raised to the negative 1/2 is 1/sqrt (x)
I think this is the part that gets me
stoic yacht
$y' = \frac{1}{2} \cdot \frac{1}{\sqrt{4-x^2}} \cdot -2x$
warm shaleBOT
neil
outer spindle
Yes
marsh geyser
Now simplify the 2 and done
stoic yacht
$y = \sqrt{4-x^2}$
warm shaleBOT
neil
stoic yacht
so $y' = \frac{1}{2} \cdot \frac{1}{y} \cdot -2x$
warm shaleBOT
neil
stoic yacht
btw another way to get this is
y = sqrt(4-x^2)
y^2 = 4 - x^2
d/dx (y^2) = d/dx(4-x^2)
2y * y' = 0 - 2x
y' = (-2x) / (2y)
y' = -x/y
stoic yacht
a third pretty cool way to get the answer, is that we learned in geometry that the tangent line to a circle at a point is perpendicular to the radius of the circle at that point. the slope of the radius of the circle is y/x, so the perpendicular slope of this is -x/y
outer spindle
Why is y’ = -x/y
.
outer spindle
**neil**
.
Okay got it
Says x and y pen ran out of ink
Can I just keep reposting questions or do I just keep reopening @stoic yacht
Prob have 4 more im trying to go over
stoic yacht
you can probably keep posting
stoic yacht
product rule says d/dx( f(x) * g(x)) = f'(x) * g(x) + f(x) * g'(x)
technically g(x) = y but you just use chain rule
so you get d/dx(xy) = d/dx(x) * y + x * d/dx(y)
d/dx (x) is just 1
and d/dx(y) is just y'
so d/dx (xy) = 1 * y + x * y' = y + xy'
outer spindle
product rule says d/dx( f(x) * g(x)) = f'(x) * g(x) + f(x) * g'(x)
Yes
In lecture one sec
Ok so you took the actual derivatives of variables which is actually 1
outer spindle
and d/dx(y) is just y'
Because you’re not differentiating for y you just left that alone as y’
Gotcha
More questions coming soon I am taking notes on side for lecture but will have quiz relating to this at the end
.
Next question
Consider the curve y^2 = x^2 + sinxy
Find dy/dx
@marsh geyser if ur still available
<@&286206848099549185> 🙂 if I could just have someone walk with me through this
It’s alright
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
arctic glade
obtuse pebbleBOT
obtuse pebbleBOT
@arctic glade Has your question been resolved?
foggy quail
i would suggest combining all of the terms and factoring
,, 1 + \frac{\sqrt{n+1}-\sqrt{n}-1}{\sqrt{n}\sqrt{n+1}}
warm shaleBOT
Bair
foggy quail
should get something like this
and now you can bring the 1 in
,, \frac{\sqrt{n}\sqrt{n+1} + \sqrt{n+1}-\sqrt{n} - 1}{\sqrt{n}\sqrt{n+1}}
warm shaleBOT
Bair
foggy quail
now can you factor the numerator?
obtuse pebbleBOT
@arctic glade Has your question been resolved?
obtuse pebbleBOT
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Available help channel!
fathom veldt
how did they get 8/n?
obtuse pebbleBOT
how did they get 8/n?
stoic yacht
they expanded (1+4i/n)^2
(a+b)^2 = a^2 + 2ab + b^2
the 8 comes from the coefficient of 2 in that binomial expansion
vagrant nova
Hi! Sorry for reaching out at 1 AM EST, but I need help with answering Part A and Part B, if someone is able to show their work for both Part A and Part B, and then able to explain how they did that, it would be amazing. I am just trying to look for some kind of help that I could rely on for assistance.
obtuse pebbleBOT
Hi! Sorry for reaching out at 1 AM EST, but I need help with answering Part A an...
wicked pebble
Do u know rule of cosine?
wicked pebble
So u just need to apply it into this question
vagrant nova
i know, but i need someone to show me, so i can learn
wicked pebble
vagrant nova
yes, i know the cosine law, but i need help solving it, could u help me solve it?
wicked pebble
So think about Luff as c, Leech as b and a as foot
vagrant nova
okay
wicked pebble
yes, i know the cosine law, but i need help solving it, could u help me solve it...
I will, be patient man
vagrant nova
okay
part a:
wait one sec
c^2=10^2+13^2−2⋅10⋅13⋅cos(52∘)
c2=100+169−2⋅10⋅13⋅0.6157
c2=269−159.962
c2=109.038
wicked pebble
c2=100+169−2⋅10⋅13⋅0.6157
good try but u used wrong sides
vagrant nova
c= square root of 109.038
wicked pebble
wait a min
vagrant nova
c≈10.44feet
wicked pebble
vagrant nova
oh, is that it?
wicked pebble
yep
vagrant nova
and just follow simplifying each equation?
wicked pebble
and just follow simplifying each equation?
First one is for the Leec
vagrant nova
First one is for the Leec
part a
wicked pebble
u solve it and got the length of the Leec
vagrant nova
and then part b is the area
wicked pebble
Yep go on
vagrant nova
alr
vagrant nova
wait just to make sure that 5th character for the first problem is an x? or what character is that?
after the +x^2-2
wicked pebble
yep that is an x
vagrant nova
is that a multiplication sign before it?
wicked pebble
no need to solve that by hand
vagrant nova
or decimal?
wicked pebble
mulitplication
vagrant nova
okay okay just making sure, it looked like a decimal for a sec
wicked pebble
no problem at all
vagrant nova
do we solve with quadratic formula?
wicked pebble
do we solve with quadratic formula?
don't try to solve it on your own
plug it into a calculator
u don't know what cos52 is
wicked pebble
no
u got that value from previous try
which was wrong
now just look into the equations i sent
and plug them in
u should get 16,49605
wicked pebble
i used the quadratic formula
u can find it on your own ofcourse but no need
vagrant nova
iranian, but close i guess
wicked pebble
yeah
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
agile coyote
I need some help with logic/set theory
obtuse pebbleBOT
I need some help with logic/set theory
dusk widget
agile coyote
in a proof
could i use this equivalence
or equivalences?
just confirming
it seems logical and valid to me
like if we are given
that set A and B are disjoint
would it be correct to express it in terms of quantifiers
to apply proof techniques?
brazen gorge
yeah
brazen gorge
looks good
agile coyote
yeah
alright, just wanted to confirm
because i was doing an exercise
where the author used contradiction
but i just expanded the disjoint form
.close
obtuse pebbleBOT
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olive basalt
olive basalt
.reopen
obtuse pebbleBOT
✅
obtuse pebbleBOT
@olive basalt Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
past solar
Last for today, I don't understand what they mean as "expressed as powers of 3 or sum of distinct powers of3"
obtuse pebbleBOT
Last for today, I don't understand what they mean as "expressed as powers of 3 o...
@past solar Has your question been resolved?
past solar
<@&286206848099549185>
swift hull
Last for today, I don't understand what they mean as "expressed as powers of 3 o...
can be expressed as powers of 3 means it can be expressed as 3^0, 3^1 , 3^2 etc...
expressed as sums of distinct powers of three means it can be expressed as a sum of powers of 3 without repeating
ie:
3^1 + 3^2 + 3^4
so 10 for example can be expressed as the sum of distinct powers of three
since its
3^2 + 3^0
18 cant since its
3^2 + 3^2 and thats repeated
3 can since its just 3^1
etc...
past solar
Hmmmmmm
How does one go at this?
I get that 3^ odd power given 3 mod 4 and 3^even power gives 1 mod 4
Now what is next?
blazing zealot
Now what is next?
1024/4 = 256
the number of terms a1, a2, a3, a4...... a1024 divisible by 4 is 256
past solar
I'm pretty sure that's not how it works
blazing zealot
I'm pretty sure that's not how it works
1=3^0
3 = 3^1
4 = 3^0 + 3^1
9 = 3^2
10 = 3^0 + 3^2
You know
3^0 ≡ 1 (mod 4)
3^1 ≡ 3 (mod 4)
3^2 ≡ 1 (mod 4)
3^3 ≡ 3(mod 4)
past solar
Hmmm, I see, but does it all work through all n, because there might be cases where there are only 2 mod 4, or 3 mod 4(caused by adding all odd powered n)
blazing zealot
And let`s add 3^0!!!!
3^0 + 3^0 ≡ 2 (mod 4)
3^1 + 3^0 ≡ 0 (mod 4)
3^2 + 3^0 ≡ 2 (mod 4)
3^3 + 3^0 ≡ 0(mod 4)
past solar
3^0 + 3^0 is not possible because it has to be distinct btw
blazing zealot
Each term in the sequence can be represented uniquely as a sum of distinct powers of 3, analogous to binary numbers where each digit represents whether a certain power of 3 is included in the sum (1) or not (0). For example:
1=3^0
3 = 3^1
4 = 3^0 + 3^1
9 = 3^2
10 = 3^0 + 3^2
We need to find which of these sums are divisible by 4. Consider the binary representation of numbers. If the binary number 𝑏𝑛𝑏𝑛−1…b1b0 represents a number N where 𝑏𝑖∈{0,1}, the equivalent sum in terms of powers of 3 can be written as:
N=b0⋅3^0+b1⋅3^1+b3^2+⋯+bn⋅3^n
To be divisible by 4, the sum must yield 0 when taken modulo 4. Let's investigate the modulo 4 properties of powers of 3:
3^0 ≡ 1 (mod 4)
3^1 ≡ 3 (mod 4)
3^2 ≡ 1 (mod 4)
3^3 ≡ 3(mod 4)
obtuse pebbleBOT
@past solar Has your question been resolved?
granite dragon
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obtuse pebbleBOT
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2. I have begun but got stuck midway.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
represent alpha in r(cos(x)+isin(x)) form and see what happens when you consider 1/(conj(alpha))
hint: ||you get 1/r (cos(x)+isin(x))||
granite dragon
isnt it 1/r cosx - isin x
brazen gorge
no because you also take the conjugate first
$\frac{1}{r\overline{(\cos(x)+i\sin(x))}} = \frac{1}{r(\cos(x)-i\sin(x))}$\
$= \frac{1}{r} (\cos(-x)-i\sin(-x)) = \frac{1}{r} (\cos(x) + i\sin(x))$
warm shaleBOT
artemetra
granite dragon
Oh you meant it like that ok yep
obtuse pebbleBOT
@granite dragon Has your question been resolved?
meager hearth
obtuse pebbleBOT
brazen gorge
Let $y = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$. Then $y=\sqrt{6+y}$
warm shaleBOT
artemetra
brazen gorge
solve for y
obtuse pebbleBOT
@meager hearth Has your question been resolved?
plain nebula
obtuse pebbleBOT
plain nebula
why is the working out differnet
arent they both binomial expansions
is one just tech active or smth
obtuse pebbleBOT
@plain nebula Has your question been resolved?
plain nebula
@analog vault
<@&286206848099549185>
can someone remove these ppl named helper
its so annoying
slim cove
I'll check your work
obtuse pebbleBOT
slim cove
.reopen
obtuse pebbleBOT
✅
slim cove
(You forgot to react to the bot message asking if you were done)
slim cove
It's because the first one asks for the probability that X is equal to a specific number P(X=0)
And the second one asks for P(X=0 or X=1 or X=2 or X=3 or X=4 or X=5)
So if you wanted to work it out the same way
You'd have to do P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)
This is kinda tedious, so there's a built-in calculator function to do it for you
plain nebula
oh i meant question 2 and 3
i alr know part a and b
my bad
ohh wait u were talking about q3 already
so its the same thing just diff working
.close
drowsy olive
Hi I was wandering if you guys could make an annoucement and post a survey i need more results for a maths investigation 🙂
obtuse pebbleBOT
Hi I was wandering if you guys could make an annoucement and post a survey i nee...
molten stirrup
Hi I was wandering if you guys could make an annoucement and post a survey i nee...
does this count as advertisement ?
what do u think ;-;
wooden cipher
try messaging @lyric cosmos and see what they decide
long plinth
.close
drowsy olive
does this count as advertisement ?
not rlly, I mean its a survey
obtuse pebbleBOT
not rlly, I mean its a survey
thin pawn
.close
obtuse pebbleBOT
drowsy olive
I texted the mod mail bot so I'm sure ill find my answer there :))
ty for your help guys!
obtuse pebbleBOT
Available help channel!
pliant zodiac
hi guys
junior granite
The curves meet at the points where their x and y coordinates are same
hot dawn
find their intersection points by subbing one equation into the other
2x+5y=1 --> x=(1-5y)/2 --> subs into x^2+5xy...
etc
hot dawn
check your square term
junior granite
also in second step you calculated 4y^2 incorrectly
junior granite
guys, for this, do we square it first or times it by -4 first?
By order of operations
do squaring first
pliant zodiac
2x+5y=1 --> x=(1-5y)/2 --> subs into x^2+5xy...
thank you so much
i’m going to try it first
junior granite
gg
pliant zodiac
.end
junior granite
!done
obtuse pebbleBOT
If you are done with this channel, please mark your problem as solved by typing .close
pliant zodiac
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
timid silo
obtuse pebbleBOT
timid silo
what is happening here
molten stirrup
find cos x
timid silo
oh?
molten stirrup
yeyeyeey
timid silo
is the cross section the triangle's outline or the height?
molten stirrup
out liem
timid silo
alright so hypo and op
molten stirrup
u want 2 that
timid silo
must be sin then
timid silo
sin x = 24/36
molten stirrup
ye adj and hypo
molten stirrup
here look at angle x
timid silo
if its the adj
molten stirrup
24 is opp
timid silo
oh
molten stirrup
SRY
timid silo
so sin x = 24/36
molten stirrup
so sin x = 24/36
sin aply in right triangle
36 belongs not to a right one
it is actually half of 36
timid silo
wait im confused what is a cross section?
molten stirrup
like one area in the whole thing
1 sec
lets redraw the thing
look at triangle ACH
Cos x = HC / AC ye ?
molten stirrup
no AH is 24
timid silo
oh alright
molten stirrup
it says the depth (height) is 24
timid silo
how r u doing that lol
so the value of x when we use tan x = 24/18
x= 53.13
we do cos 53.13
which is 0.6
timid silo
the answer is 0.6 right?
timid silo
obtuse pebbleBOT
timid silo
@molten stirrup come back
queen cloak
timid silo
what is happening here
junior granite
!occupied
obtuse pebbleBOT
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).
molten stirrup
open a new one for ur own safwety ^^
timid silo
anyways
molten stirrup
Didn't know it was private
not rlly
it is just occupied
hmm tough one
1 sec lememe think
molten stirrup
yue
junior granite
@timid silo do you want to me to give a mindblowing method to solve this question?
timid silo
<@456226577798135808> do you want to me to give a mindblowing method to solve th...
please
molten stirrup
so what is <B then
it would be wrong to write that way
timid silo
it would be wrong to write that way
they did that tho
molten stirrup
they did that tho
they only did that for A and C
since angle B has been cut
u cant write it that way
junior granite
sinA = cosC
which means that sinA = sin(90-C)
timid silo
alright
junior granite
implies that A = 90-C
junior granite
Now you can apply that same logic for required thing
Angle DBC = 90 - Angle ABD
got it?
timid silo
wait let me cook
timid silo
its A
molten stirrup
the given sin A = cos C only lets us know that it is a right triangle bu gives no info about where point D is
wait how now im the one who needs explaining
seems like wer switching roles
timid silo
Angle DBC = 90 - Angle ABD
how do i get ABD
junior granite
how do i get ABD
bro you don't need to calculate that
junior granite
Angle DBC = 90 - Angle ABD
just plug that information in sine or cosine that you want
you'll se yourself
junior granite
Required answer = sin(ABD) - cos(DBC)
implies sin(ABD) - cos(90-ABD)
implies sin(ABD) - sin(ABD)
timid silo
abc = 90
the left angle is C-90
the right angle is C
and them all
2C = 180
C = 90
junior granite
bro no
junior granite
yeah those long geometry methods
I am not that good in geometry so I can't help much in that
timid silo
no worries thanks tho
junior granite
Required answer = sin(ABD) - cos(DBC)
implies sin(ABD) - cos(90-ABD)
implies sin...
I just solved the question here
timid silo
@molten stirrup are u majoring math?
molten stirrup
so sin abd = cos dbc?
junior granite
yeah
molten stirrup
<@1113321524909322270> are u majoring math?
no im just bored
junior granite
since Angle ABC is 90 degrees
junior granite
that's the method I was explaining
timid silo
if u guys understand this please explain
molten stirrup
oh dang
molten stirrup
sry guys i just got blown minded
junior granite
see sin(A) = cos(C)
sin(A) = sin(90-C)
A = 90-C
junior granite
from here you can verify that angle ABC is 90
timid silo
from here you can verify that angle ABC is 90
since 180 - ( 90-c +c)
great
so know ABC = 90
now*
junior granite
okay so now from the diagram we can see that angle ABD + angle DBC = 90degrees
junior granite
Required answer = sin(ABD) - cos(DBC)
implies sin(ABD) - cos(90-ABD)
implies sin(ABD) - sin(ABD)
junior granite
what?
timid silo
can the triangles being similar have something to do with this?
junior granite
can the triangles being similar have something to do with this?
bro you can just see that in diagram
ABD + DBC = ABC