#help-33
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are you not in the course?
no
whoa
whoa is right im gonna be spending hours on the next couple
but hey last day of this course is always exciting
also for the answer to question 2 was it like 7.2 ish
@runic temple
with so root 26 times root 8 times 1/2 but that was wrong
since root 8 is the base of a to B
you can find AC using pythagorean
I thought BC was the height though
cool
and I guess you fit the bill by just simplifying the problems
so you would be perfect for this
i did participate in AMC in high school
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can someone help me with this? i dont understand the 1 equation 3 unknown
try to get a common denominator on the right side
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Is this a valid application of the central limit theorem
I'm just a bit worried because of the sqrt(n) in the denominator in the probability of the last line
How are you sure, sir?
clt just says that the distribution looks like a normal after sending n to infinity, so that means the cdf will look like a normal's cdf
just ignore the inequality with the infinity as it is true with probability 1 anyway
No I meant sure about the other thing
I understand ignoring the infinity side
like
CLT would say this is integral from 0.01/(sigma sqrt(n)) to infinity of the standard normal pdf
but I just replaced the lower bound with 0 because that felt right
although I'm not sure if it is
As n to infinity it approaches 0
why not
you have to get
(X_nbar - mu)/(sigma/sqrt(n)) in the middle term
if you do this then you get the left bound as 0.01/(sigma sqrt(n)) not 0.01
wait i think i've done this problem, is this 395
yes
can you send the original
oh ok then yeah
becomes integral infinity to infinity which is0
should be i -> 0, ii-> 1/2, iii-> 1
So is this valid fix
the lower bound is then sqrt(n) * 0.01/sigma which tends to infinity
Do you see my doubt about why this is valid or not?
0.01/sigma sqrt(n) is not a fixed value a
so I'm just wondering about if pulling the limit inside is fine to make it inf
the statement is like
the distribution tends to a normal as n grows large, the actual interval doesn't matter it can still depend on n
for all a,b with -inf <= a <= b <= inf
lim n->inf P( a <= ... <= b) = integrate standard normal pdf from a to b
but a depends on n
in the above
yeah, a and b can depend on n
_ _
it's better to think of the clt as saying that the distribution converges to something, not a specific probability statement
yea
unless a,b oscilate or something
Okay sir
does that seem reasonable
Well it always seemed reasonable to me I just don't want to lose points for applying the theorem in a way that wasn't explicitly written
yeah i think it is ok
clt is usually phrased in a stronger way
or maybe not usually, but often idk
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ty sir
oh i see
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are u familiar with compound/double angle results?
if yes, consider rewriting cos10 = cos(2x5)
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Let the function ( f(x) ) have the second derivative on ( (0, +\infty) ) and ( f''(x) < 0 ). Let ( u_n = f(n) ), then the sequence ( {u_n} ) diverges is the ( ) of ( u_1 > u_2 ) \
(A) Sufficient and necessary condition\
(B) Sufficient but not necessary condition\
(C) Necessary but not sufficient condition\
(D) Neither sufficient nor necessary condition
I'm cooked. No where to start
What exactly is the question
You sure it’s ui>uj and not the other way around?
some modifications still need to be made, "diverges is the () of" check this part
riyobi
it makes sense. the correct option is supposed to go in the ( )
okay that makes sense
This is entirely different now though
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
I used gpt to translate, it changed some meaning now I've modified
you should still post the original too
So this is correct now?
Yes
weird question then
I recommend you graph the question and try to look for clues
it doesn't look too hard, so I think you will be able to answer it
always do that by the way, it's a global hint, always try to visualize questions if possible
But f(x) is not a f with explicit expression, how to graph it?
make something up which satisfies given conditions
Well what does the condition f“<0 mean geometrically
Let's say f=-x^4
second derivative is positive on that one
don't even need to calculate you can imagine it xD
also the function you wanna come up with doesn't have to be elementary or something you know
you can just put axes and draw anything random with your pen but make sure it satisfies the conditions
try drawing multiple ones and ask yourself questions
can I draw one which diverges satisfying the condition ? can I draw one which diverges without satisfying the condition ? and so on
,w plot -e^x
Is this one's un diverging?
clearly diverges
as you go to infinity
it blows up to -infinity
is u1 bigger than u2 here ?
Yes
from this graph. generally if u1 > u2, does your function have to diverge ?
Yes
that's great
but does it have to be true ?
is it necessary
aka can you come up with another example in which it diverges but u1 < u2
Can't think of other examples
unleash your imagination xD
doesn't have to be a sine or a cosine or an e
just draw something with a pen
start with a function that has any value
make it rise
then keep it's second derivative negative
anything
again it doesn't have to have a symbol. or be something you have seen before
as long as you know what it means for the second derivative to be negative geometrically
you can just graph something without knowing what it is
okay how about this, what does it mean for the second derivative to be negative geometrically ?
like how would you describe it on a graph
,w plot -sinx-2cosx
,w plot sinx + 2cosx
okay so for this function for example you just drew
when is the second derivative positive
and when is it negative
and don't use any calculations
just look at it and tell me
if you don't know what it means geometrically you can ask by the way this is not an exam
For this, let u1=0.1 and u2=0.2, we get u2>u1
.
can you answer this question
it's important
you won't answer properly if you can't answer it
Yes ofc i know
then tell me
sinx + 2cosx
when does it have a positive second derivative and when doesn't it, you can name points, you don't have to be too precise
and make zero calculations
just tell me from looks
if f(x) = sinx + 2cosx
what is the sign of f''(0) and f''(pi)
that's not true xD
okay
Anyway, i knew, if idk I'll just ask
it's true then
okay so you know that it's related to concavity
all i'm asking is that you draw a concave down function that has u1 < u2
and see if you can make it divergent
non of these work by the way and i'm sure you know why
It's not divergent
no they don't have negative second derivative over (0,inf)
that's the condition in the original question
they go concave up in some regions
you wanna draw something that is always concave down
but starts rising
you can draw it using paint or anything , again it doesn't have to be something you would graph on wolframalpha
,w plot -(x-1)^2
Yes
does it diverge ?
Yes
D
are you sure
is it sufficient ?
like if that condition is true does it make the function diverge ?
Yes?
you tell me xD
we have established it's not necessary correct ?
now tell me whether or not it's sufficient and the question is solved
Necessary is : u1>u2 implies divergent?
yes
wait
no that's sufficient
sufficient means the condition is sufficient to make the function diverge
😭 it's so hard to remember it correctIy
necessary means that without it, it's impossible for the function to diverge
it can be annoying at first, the wording of logic
but just translate it literally
something that is sufficient means its mere existence is enough to make a statement true but it doesn't have to be there for the statement to be true, if it has to be there then it's necessary
the existence of rain is sufficient to deduce the sky has clouds, but it's not necessary to have rain in order to have clouds
But then we'll need to come up with another example to see if u1>u2 implies non divergent or not. If we can come up with one example, then it's not sufficient
that's a great note !
my question is can you come up with an example ?
if you can then you are right it's not sufficient
I can't, but that doesn't mean it's not true because maybe others can🤣
well in that scenario we prove that we can't, so no one else can 😂
Wait how did we prove
we didn't ... yet xD
I asked you if it was true or not and you said yes, you didn't provide evidence
can you prove that it's sufficient
lemme give you an example with a problem associated with it
you have a function f(x), f(0) = 0 and f'(0) = -1 , its second derivative is always negative, would that function be less or more or intersecting the line y = -x
Isn't the f(x) itself y=-x
this has zero second derivative
the other one has negative
so it should be falling faster
Yes
if you take the sequence y(n) = -n, that will diverge
well then what of the sequence you get from a function lower than that ?
Due to squeeze thm or something, it'll divergent too
not really squeeze theorem but it's a similar idea
proving the actual thing in your question however is slightly harder
because we don't have f'(x) but rather we have two points far apart

now check this out xD
f(1) > f(2)
a line passing through the points (1,f(1)),(2,f(2)) would have negative slope
which is f(2)-f(1) / (2-1)
the mean value theorem tells us the function f(x) has a point c in (1,2) which has exactly that slope
and since the function has negative second derivative
then f'(c) > f'(2) since c < 2
from here, for x > 2 the line passing through the two points must be larger than the function value.
and since that line is decending at a constant rate, again the sequence taken from it diverges to minus infinity
hence the sequence taken from the function below it also diverges to - inf
Does this idea have a name
I want to look it up as a reference
i'm not really sure if it does, it goes straight back to the definition of limits
the limit of a sequence an = -inf if for every B you can always find an N such that if n >= N then an < B. if bn < an after some threshould. then if you can find an N to make an < B then you can use the same N to make bn < B since it's lower.
this would prove by the definition of limits that if bn < an and an goes to -inf, bn must go to -inf as well
note than N is an integer and B is any real number
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can someone help me check if this is the correct answer for this question?
30%
!noans
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Total density= 0.50
Welp, wrong one used.
0.45 is correct
That is not true. Total density must be 1.
What is this density ?
It is probability density
From axiom of probability, the integral for the entire domain must be 1.
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Why do palindromic polynomials with odd degrees necessarily have -1 as a zero?
plug in 0 and see what happens, look at the first and last term, second and second last, etc
0?
oops -1 my bad
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Hello everyone, I need help with this one. Is the correct answer for s=1 and t=0 or s=5 and t=-1? And could u explain me how did u get the answer
you can always check yourself
does (s,t) = (1,0) or (s,t) = (5,-1) yield 1?
what does yield 1 mean?
could u explain me what does yield mean in this case?
yield as in
give you
return
make the function/expression in question evaluate to
i do not understand, could u explain it
i tried checking with the calculator, and both gives 793
great, so they yielded 793
it also means that your choice of integers are not unique
so if 793 is their gcd, then both pairs are fine
also i wanted to know if the question is + t3172 and for example i found t=-1 so is it wrong or correct since t is negative, which is different from the question
or can i write + (-1)3172 ? but t is still -1 right?
your task was to find a pair of integers (s,t) satisfying the above. you've found two pairs. -1 is an integer -- there is no problem
alright thank you
can i write my answer like this
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help
!status
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
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
sorry wrong qs but yes idk where to begin
i tried 360-(21x-1 ) to find RSTP
wait
i think
ok nvm
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can someone help? need to prove
@flat meteor Has your question been resolved?
just asumme a<b<c WLOG
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how do I solve this
any progress or ideas?
I think I need to find the slope but I don't know where to start
maybe first find the point of intersection???
ok I'll give it a try
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somone help
Don't ping random people/mods.
Last warning to stop pinging random users
We have #❓how-to-get-help for info on this.
ii dont got time
just say it
I don't care. Find time.
typical mod
you just have to wait for helpers to come to your channel
alr
ok fine
heres the sum again
someone welp
if they don't come in 15 mins
you can ping the Helpers
but only after 15 mins
ok fine
It says all this in the first paragraph of the channel I linked.
Hint 1 - || open the square what do you get ||
@wheat moth Has your question been resolved?
Hint 2 - || in the numerator u get the expression x⁴ +x³ +2x² +1 which can be written as x⁴ + x² +1 + x(x² + x)
Now divide it by the denominator to get two telescopic series ||
ok alr
im gettin 3/2
What the answer?
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.reopen
What level of math do I need to be able to do computer science?
Like pre calc? Or just algebra 2
Either game engineering or software engineering
Well, u can enter the university with only the high-school math u know
It would be nice if u know pre calc
a bit of linear algebra too
ig thats the answer
Oh so just regular linear and quadratic formula algebra works for computer science also?
what did u gry
For games you're gonna need calculus and trig
Show ur work 
In dm
oh no linear algebra is much more than just linear polynomials
For general software eng you just need good mathematical reasoning skills, but being good at precalc should give you a leg up
I tried what I told you in the hint 
Linear algebra is vectors, matrices, linear transformations
you should know about matrices and how to represent transformations with them and stuff
Oh so I guess I haven't reached it yet since im still in 9th
My school learning system is a bit slow
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Yes
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I need to solve this using trig identities to prove the equation is true but I'm rlly stuck on the step at the bottom. What do I do next?
Just split the fraction here
1/cos - cos^2/cos..?
There you go
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The 3rd question
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Alright
First thing, 1.024 is 1000+24 or 1 +24x10^-3?
(just to make sure I get the notation)
1.024 × 10^-7
Lemme rephrase my question
Does the dot signal thousands or decimals?
Decimals
May I ask why writing it in terms of power of 4
because if you write it in terms of the power of 4, its easier to solve it than multiplying all the decimal points of 0.04
The expression with x you've got is 4 times a multiple of 10
or smth like that ig
thats also a reason too
4 is a natural number, which is usually the most convenient number set to work with in these kinds of questions
No, rewrite 0.04 as (10^-k times 4)^x
Once you've got everything written in terms of 4 and 10, try to figure out the value of x
Should be pretty straightforward from there
Especially if they asked you to deduce it
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Thank you
No problem
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How in the world do I solve this lol
$\left( \frac{D^6k^6-2k^{12}}{D^{13}}\right)$
everg
Right, though I have no clue how to integrate
the t is still stuck to this function of D
@night mica oops
maybe Separation of variables could work
I can't
the v is preventing it
im wondering if I could try to substitute D for a trig function of some sort
@eternal sequoia Has your question been resolved?
is it even possible?
@eternal sequoia Has your question been resolved?
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can someone check my work, i have no idea how to do differential equations
Could u explain more deeply? I didn't get what to do after assumption
In the 1st step, you should be dividing by x on the right hand side
omg bruh
wow the whole thing is wrong classic
ill be back
des are literally giving me brain worms
is this ok
seems ok to me idk
Same here
Non separable?
Is that something specifically intructed?
like is it ia linear first order de
oh wait am i supposed to use substitution for this one where u=x+y
i think is ok
It should be fine
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im back with more questions
this would be a first order de right? does that mean i need to find the integrating constant
isn't the integrating constant like, your initial value thing? that y(0) = 9?
right i did
so is the integrating factor just e^(-cost)
sounds real
isn't g(t) piecewise? that'd only be one piece
yeah, this is the first one
anywayas i figured out the error i didnt divide c by the if
ah, I see it
wouls the answer be y=1+8/e
lol i typed in my answers and they were both wrong
what am i doing wrong i dont get it
did you substitute 8/e back into the original expression for y in the first case?
your y in this message is just a constant
oh wait
no i did not
so should the answer for the first one be y=1+ (8/e)/(e^-cost)
I believe it
let’s try
which you can simplify to 1+8e^{cos(t) - 1}
oh the first one was right the second one wasnt lol
there's another point for the second piece
the problem is your initial value thing
mainly, 0 is not in the interval [pi, 2pi]
ohhh
2pi is equal to 0 though right
so should i use that instead
wait thats literally the same answer
2pi isn't equal to 0 but I know what you mean
I still don't think so
so you can find two equations y1 and y2 corresponding to the two pieces of g
defined on their own respective domains
y is just the piecewise definition of y1 and y2, you know
right
since y is asked to be continuous, you want these two pieces y1 and y2 to be attached
for all points in the interior of the domain of y1 and y2, the corresponding piece, hence the function, is continuous at that point
it gets dodgy at the endpoints, in particular the place where y1 ends and y2 begins
that's what gives you your initial value condition for the second piece
so like, the intial value information given can only be used to solve the first de for the interval from 0 to pi
and i shuold make my answer for y2 equal my ansewr for y1 when t=pi cause thats when the piecewise connects them?
exactly
technically it's more like, the limit as x approaches pi from the right (coming from the domain of y2) should equal y1(pi)
but it's all the same really
so for my answer for y2 i should keep the part with (10/e)/(e^-cost) right
but instead of -1 at the front i should change it to reflect continuity?
I think the constant is the thing that ought to change right?
wdym
all of the DE shenanigans should still work up to this point, with arbitrary constant C
using your new initial value premise that y2(pi) = y1(pi) should give you a value for c that's probably different from 10/e
nice
I literally did the same thing btw
just now
and yeah, I agree with your choice of c
very cool
this is good news lol
you as well
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Hi, I was wondering how you would turn this problem into a function:
I am struggling with how to translate what is being explained into math
I imagine 27 is the y-value.
you mean how do you generalize its solution?
yeah if that is the correct terminology
I guess that sounds right because really I am wondering about any problem like this
particurarly if you had n amount of workers
but honestly just a function for this specific problem would probably help me understand how to think about this better
right, it's like the difficulty of modeling and generalizing something, sure
i have trouble deciding what the speed of the workers should be and what difference it makes that the workers are not working on the same dress
yeah i suppose that is what you'd call it
it's probably a lot uglier to assume that there isn't any kind of linear collaboration going on, if you will
like, all but one person could be done and have nothing to do
Ig the speed of the workers can be in terms of fractions of a dress per day
we can always describe the i'th worker's productivity as being a rate of r_i products per time unit
which turns out to be the actually useful thing
$D(t) = \frac{r_i}{t}$ ?? idk how to turn it into math at all lol
Neo
I genuinely can't remember how to model even basic things
so assume there is $s$ total product to be made and there are $n$ workers making the product, for which the $i$'th worker makes product at a rate of $r_i$ product per second, say. assuming productivity scales linearly, then the amount of time it takes to make $s$ product, $t$, satisfies $$r_1t+\ldots+r_nt=(r_1+\ldots+r_n)t=s,$$ yielding a time of $t=\frac s{r_1+\ldots+r_n}$
Flip
sorry I'm having trouble understand the i'th worker, is the i'th worked the sum of all the n workers?
the rate of all the n workers?
I'm not immediately sure how you'd do this if s were a positive integer and each worker can only work on at most one item at a time
ah that's okay to be honest just understanding this in of itself would be very helpful
nah, if I name my employees w_1, w_2, ..., w_n, then I'm saying that w_i works at a rate of r_i
but i am interesting in learning how that works too, that only one worker can work on an item at a time
oh okay so r is the rate
and r_n is the rate for that n worker?
I think it'd have to do with determining the first person to have nothing to do on their own, then hijacking the slowest worker's project if they're faster
then keep going like that
idk if this helps, but: I didn't define r. I defined n productivity rates, denoted r_1, r_2, ..., r_n
n is a reserved character denoting the number of workers, so we can't use that as an index
man im lost, but have i at least understood things right that the r in r_n is the rate for the n'th worker?
instead of exhaustively saying the first worker works at a rate of r_1, the second works at a rate of r_2, and so on, I just say that the i'th worker has a rate of r_i
since there are many different workers?
and you chose i arbitrarily or is that convention if n is already being used for something else?
ah, sorry, that's conventional. if I declare n items and say that the i'th item has attribute r_i, it's implied that i is like, a dummy index, that I use to quickly declare the n attributes. by context, the dummy variable i will take on integer values between 1 and n as I define each of the n attributes
if you like programming, it's like saying R = [r_i for i in range(1, n)]
i think maybe im being unclear
basically think of me as a high school student who is bad at math
i barely understand any of the terminology used on here but i am slowly learning
i somehow got into uni despite this
I am also Swedish so I don't know what range means
I think it has something to do with the number line or sets
ah word, my bad lol
this "range" is different from the "range" of a function
R are the real numbers?
I also wrote that to just be a different R, I'm sorry lol
no that's alright, i'm probably just out of my depth even with this question
I think we should ignore that example lol
maybe this would be better
actually no it wouldnt be
damn I suck at math
Problems arise because a man is bad at math.
https://static.macmillan.com/static/holt/famous-people/
My book, Famous People, is out now!
i will get back to you when i can at least formulate a question in some manner
thanks for trying to help
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hey, I'm trying to understand the relationship of this trig identity:
C_1cos(theta*t) + C_2sin(theta*t) = C_3cos(theta*t+ϕ)
but I don't really see how Acos(theta*t+ϕ) is a sum of these two vectors? I know that A = sqrt(A^2 + B^2) and that phi = arctan(B/A)
but when I try doing measurements it doesn't add up am I doing something wrong?
for context, it's for differential equations for Dampened harmonic motion
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can anyone tell me a very logic reason why am i bad at math and why the others is really simple to solve math equations?. Give me tips too to improve more
I think it's all about practice, mentality (attitude), and intuition personally
go study
what are examples of things you're struggling with?
it may be an issue of how you're being taught things
like the mindset tho, like math is just stressful subject and i dont wanna learn about it
and u get humiliated sometimes
that's gonna be an issue
sounds like a bad learning environment u are in
yups
math is a skill and as with all skills it comes with practice and experience
you're only bad at math because you haven't been exposed to it enough
study with your friends and classmates
can u like give me a good mindset how should i picture math
can you ride a unicycle
so abstract lol
i think u just learn it slowly using the method of your choice
dont think too much about this
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thankss
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Hello i'm having trouble in trying to visualize this box fractal and i can't determine what the number of self-similar pieces are
I assume the scaling factor is 1/3 so all i got for now is
D= log(n)/log(s)
D= log(???)/log 3
When i try drawing the box, it ends up looking like a cross and i'm not sure if that's correct
wut did u get
i tried this
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Let the function ( f(x) ) be once differentiable on the interval [0,1], and ( f(0) = 0 ), ( f(1) = 3 ), and ( \int_{0}^{1} f(x) e^x , dx = e ). Prove that: (1) There exists a ( c \in (0,1) ) such that ( f(c) = c^{2} + 2c ). (2) There exist distinct ( c_1, c_2 \in (0,1) ) such that ( f'(c_1) - f'(c_2) = 1 ).\
Any hints
riyobi
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@small stream Insane question by the way xD
I would give you a hint but, this style of hints is a bit too advanced compared to questions we have done before.
consider the function $g(x)=f(x) - x^2 - 2x$
Mohamed Mohsen
this would be my hint
explore this function a bit and see if you can relate it to your question
I'm really fed up with this kind of mean value theorem problem😭
g(0) = f(0) - 0^2 - 2(0) = f(0) = 0
g(1) = f(1) - 1^2 - 2(1) = 3 - 1 - 2 = 0
this problem is very neat though
beautiful observation
keep going
what more can you deduce ?
what does this function look like ?
hmnn
The exact shape of g(x) depends on the properties of f(x) so we dont know
[\int_0^1 f(x) e^x , dx = \int_0^1 \left( g(x) + x^2 + 2x \right) e^x , dx ] \Expanding this: [ \int_0^1 f(x) e^x , dx = \int_0^1 g(x) e^x , dx + \int_0^1 x^2 e^x , dx + \int_0^1 2x e^x , dx ]
[\int_0^1 x^2 e^x , dx = \left( 2 + e \right)]
[\int_0^1 2x e^x , dx = \left( e + 1 \right)] [e = \int_0^1 g(x) e^x , dx + (2 + e) + (e + 1)]
[\int_0^1 g(x) e^x , dx = -2e - 3]
riyobi
how did you calculate the x^2e^x integral
you are on the right track, I just think the calculation is not correct
you forgot the slash
at the end
[\int x^2 e^x , dx = x^2 e^x - \left( 2x e^x - 2 e^x \right) ]
[\int x^2 e^x , dx = x^2 e^x - \left( 2x e^x - 2 e^x \right) ]
oh yea it doesnt xD
riyobi
substitute the bounds
when you put 1
you have e
when you put 0, you get 2
but then you need to subtract not add
so it's e - 2
wait how
Nvm, mb🥲
Mohamed Mohsen
0(x=1)
-2
yea it's -2 at zero
don't forget that
so when you use the fundamental theorem of calc
it's 2
because you subtract the second bound
adding the two integrals is adding e - 2 + 2
what is that ?
[e = \int_0^1 g(x) e^x , dx + e ]
riyobi
Uhh wait what's the antiderivative of g(x)ex
bro xD
use this picture
Or we can get g(x)ex is odd function
what is the integral of g * e^x
i'm confused, you already solved it
why did we go back
0
But what can we derive things from it
oh boy you can derive alot from that
we have a function that has g(0) = g(1) = 0
and its integration times a positive function ( e^x in that case) is also zero
this must have one of two meanings
either the function g(x) = 0 or it has a zero in the interval (0,1)
tell me why
Symmetry
the function inside can't be fully positive or fully negative
yes
there must have been some ... not really symmetry but you need to have the function be positive in places and negative in places
or you need the entire function to be zero itself
but you know our function is differentiable and that means it's continuous
ofc so is g because you just added a polynomial to f
this means by the intermediate value theorem, there must be c inside (0,1) such that g(c) = 0
As I said, it's some thm, it's IVT 😭
forget about the theorem xD you can't get there without your understanding
you understand what you are doing then you write it using proper theorems
i wasn't thinking ivt when i got this idea
i was thinking the integral can't be zero if g had a singular polarity
I understand now
great and what does it mean for g(c) to be equal zero ?
(1) is proven
exactly
now think about part 2 and tell me what you think
I think it's easy so give it a shot now after you got part 1
part 1 was the hard work
Since g(0)=g(1)=0, we know there's a c s.t. g'(c1)=0
okay keep going
And since (1) we know that there's g'(c2)=0 ?
g"(m)=0 , umm wrong direction ig
i think it's a good practice to avoid naming the second derivative at all since the question mentioned a differentiable function only so it doesn't have to have a second derivative
[ f'(c) = \frac{f(b) - f(a)}{b - a} ] [ f'(c) = \frac{f(1) - f(0)}{1 - 0} = \frac{3 - 0}{1} = 3 ]
riyobi
Seems like I need a little hint
we are getting there xD
you are on the write track just using wrong points
you have 0,c,1 and we got c from part one, you can use those to make two points
[ f'(c_1) = \frac{f(c) - f(0)}{c - 0} = \frac{c^2 + 2c - 0}{c} = c + 2, ] [ f'(c_2) = \frac{f(1) - f(c)}{1 - c} = \frac{3 - (c^2 + 2c)}{1 - c}. ]
[ f'(c_1) - f'(c_2) = \left( c + 2 \right) - \frac{3 - (c^2 + 2c)}{1 - c}. ]
riyobi
I guess this will result in 1
Yes it is
Wow, this question is not that difficult as I imagine

Thank you so much🙏💕
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you are most welcome ^ ^
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Can anyone explain me what is mod 6
remainder after division by 6
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Hello all 😊
This is more of a theoretical question, but - We can obtain the amount of possible combinations of k elements from a set of n total elements with repetition allowed using the binomial coefficient (n + k - 1, k).
However, do we have a method / algorithm for obtaining the actual combinations?
For permutations, one could "itemwise" iterate over all possible options until all are obtained, but how could this be done for combinations?
Is there a more efficient / straightforward approach than generating all permutations and sorting + filtering out already existing combinations?
i think because order does not matter, one could just make up some order for these n elements and list up only the sorted combinations, which would then look like regular counting, just that whenever a digit wraps around, it has to start from the value of its left neightbor (after it has been incremented), for example
1 ... 1 1 1
1 ... 1 1 2
...
1 ... 1 1 n
1 ... 1 2 2
1 ... 1 2 3
...
1 ... 1 2 n
1 ... 1 3 3
...
1 ... 1 n n
1 ... 2 2 2
yeah you "iterate" the binomial coefficient
@dull mesa Has your question been resolved?
i don't know exactly how, it's not that easy, here's python's "combinations" for example
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I throw defective die 10 times, probabilty of even face occuring 5 times is twice of even face occuring four times , probabilty of even face occuring?
so what i did is
let p of even throw is x
So
(x)⁵(1-x)⁵ = 2(x)⁴(1-x)⁶
But its wrong
Idgt
Hello, can you please provide a picture of the question you're working on
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how do i start
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can you go another channel im already here
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bruh
?
hi
@dawn pendant Has your question been resolved?
Sketchy yes but mathematically it seems to check out, your conclusion also seems valid
But sketchy can be said about the entire exercise because never have I ever seen a simple mass spring damper system ma+cv+kx=0 being written in such a cursed way with momentum p🤮🤮
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Money money money
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Hello does anyone know how this equation is formed?
2a + 2pi*(b/2) = 400
...
🤔
We need some excuse to express b in terms of a
Oh, maximized area
No 💩
A = ab + pi*(b/2)^2
A(a) = a[2(200 - a)/pi] + pi(200-a)^2
hm
why can the 2/pi be seperated from the rest of the equation?
<@&286206848099549185>
I am not a calculus expert, but try to do 2 divided by 3.14 which is pi in number
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How is this hint useful?
|p(k+1)| = |p(k) U p(k+1)| ?
acc nah idk
ngl I was thinking of something like this
Inductive step: WTS p(k+1) = T(k+1)
T(k+1) = T(k-2) + T(k-4) - T(k-7)
and by inductive hypothesis we know:
p(k-2) = T(k-2), p(k-4) = T(k-4), p(k-7) = T(k-7)
or maybe for p(k+1)
let A = p(k-2) and B = p(k-4) and then A n B = p(k-7) somehow?
ye shit idk
<@&286206848099549185>
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I'm not quite sure how to formally do this but if you look at the two sets:
\begin{align*}
A={(a,b)\in\mathbb{N}^2|3a+5b=k-2} = {(a,b)\in\mathbb{N}^2|3(a+1)+5b=k+1}\
B={(a,b)\in\mathbb{N}^2|3a+5b=k-4} = {(a,b)\in\mathbb{N}^2|3a+5(b+1)=k+1}
\end{align*}
waler
Notice that if you take any solution (a,b) from set A, and 'transform' it to (a+1,b), you gain a solution to the (k+1)-cent problem where you must use a 3-cent stamp. That is a solution that has at least one 3-cent stamp.
So essentially, for every element in A, you would be able to build a solution for the problem "create a postage of (k+1) cents using at least one 3-cent stamp and some arbitrary amount of 5-cent stamp".
Similarly for the set B, taking any elements from set B, you would be able to build a solution for the problem "create a postage of (k+1) cents using at least one 5-cent stamp and some arbitrary amount of 3-cent stamp".
and if you now consider the union of A and B. Using any element from this set you can build a solution to the problem "create a postage of (k+1) cents using at least one 3-cent stamp or at least one 5-cent stamp". Which is equivalent to using either 3-cent stamp or 5-cent stamp to make a postage of (k+1) cents.
Which means that card(AUB) = p(k+1).
waler
So you can now prove that p(k+1) = p(k-2)+p(k-4)-p(k-7) using this and the hint given.
Though a problem is this is still very crude and I'm not sure what would be a more fomal way to represent it
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yea wait mb
yea whatever
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😭
it won't work here?
done

