ur right
#help-23
devout shale
think about it this way
if u just did dual-enrollment instead
you'd need like 12+ APs
valid rock
school doesnt offer dual enrollment
devout shale
to match the credits
valid rock
unfortunate
valid rock
you'd need like 12+ APs
mm
devout shale
exactly but what im saying is obviously the dog inside you
is capable
and the other dog is chasing his tail
yk what I mean?
So touch that one dog
but leave the other
valid rock
ok ill take ap calculus bc
severe pond
🗣️🗣️🗣️
valid rock
u motivated me
devout shale
Mashallah
valid rock
inshallah we get a 6 on the exam
valid rock
calculus sounds like an easy word so
devout shale
literally
devout shale
it's easy bro
trust
if when u were a toddler
I showed u 2+2=4
you'd cry like a baby
valid rock
myfunction = function()
print("Calculus")
end
devout shale
that's what this is
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shadow sparrow
how should i do this? do i write down a parametrization of a torus then write maps to that onto two circles?
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how should i do this? do i write down a parametrization of a torus then write ma...
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lost flare
Hello, I would like ask about this question that I have been stuck on.
torn ivy
use the binomial expansion of (a+b)^n where a = 1 and b = -x/2
torn ivy
the coefficient of x² in the expansion would be (n choose 2) / 2²
lean otter
dunno
lost flare
I got it wrong. The second coefficient should be +-1
I get it now
I only have to equal x/2^(n+1) =9
if n = 18
lost flare
Using combinatorics.
lean otter
flat frigateBOT
Krish
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lost flare
Equalling the equation with 9 gives me n = 0.
lean otter
n = 0 is 1
mossy lotus
Have you written the binomial expansion in the form similar to what lenny suggested?
lost flare
Yes
mossy lotus
so what is the term when k = 2?
lost flare
nC2 = (1^n-2) (-x/2^n+2)
lost flare
I also have to multiply with 2n, right?
mossy lotus
But you kinda made a mistake there
the term is nC2 * (-x/2)^2 * 1 ^ (n-2)
it shouldnt have equal sign
lost flare
I understand.
mossy lotus
So, what is the coefficient there? (Just remove the x from the expression)
lost flare
-1/2
mossy lotus
No no no
the whole term
it has many more constants other than simple -1/2
coefficiet is all of them together
lost flare
-1/2 ^ n+2
mossy lotus
coefficient is everything in a term except for the variable x
theres one more
also what you wrote is wrong
nC2 * (-x/2)^2 * 1 ^ (n-2)
so
The coefficient is:
nC2 * (-1/2)^2 * 1^(n-2)
and that whole is given as 9
lost flare
yes I understand now. Is nC2 equal to 2n?
mossy lotus
you should get a quadratic in n
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lost flare
<@&286206848099549185>
I have no idea how to solve equations with factorials.
<@&286206848099549185>
pliant void
I have no idea how to solve equations with factorials.
You mean factors? Like middle term break?
mossy lotus
you dont end up with factorials. Like I said, the equation reduces to quadratic in n
lost flare
But, the combinatorial expansion contains factorials.
mossy lotus
Yes and factorials like n! become n * (n-1) * (n-2) * ....
lost flare
I see.
mossy lotus
so the numerator and denominator both have factorials which cancel out most of the terms
lost flare
So, we can cross off (n-2)!.
mossy lotus
yes
lost flare
So, how much of the terms are cancelled out between (n!) and (N-2!)
mossy lotus
write out the products in both
mossy lotus
(n-2)! = (n-2) * (n-3) * (n-4) * (n-5) * ...
this is wrong
lost flare
So, all that would be left is n and n-1
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modern bloom
Why did they make f(x) as x^T x, and why ▿f=2x
safe radishBOT
Why did they make f(x) as x^T x, and why ▿f=2x
proper bramble
Why did they make f(x) as x^T x, and why ▿f=2x
its like the vector variant of x²
modern bloom
why?
proper bramble
x^T*x = x1² + ... + xn²
modern bloom
Wow ! I'm not familiar with this form
proper bramble
Wow ! I'm not familiar with this form
sry i messed up a bit xd
proper bramble
Wow ! I'm not familiar with this form
its just the inner product of x with itself
modern bloom
I understand the process of calculating this vector, but I don't understand why it's in an addition form
Because it's hard to relate this addition form to x^2 in 1D->1D case
proper bramble
because it behaves like a quadratic function
f(2x) = 4f(x)
in general f(ax) = a²f(x)
you can look up quadratic forms
modern bloom
I know quadratic forms are just like this :
proper bramble
I understand the process of calculating this vector, but I don't understand why ...
if you calculate the product normally like you would do for any matrix youd find this sum
proper bramble
I know quadratic forms are just like this :
third isnt a quadratic form
modern bloom
Wait. We can write ∂(x² + y² +... + z²)/∂x = ∂(x²)/∂x + ∂(y²)/∂x +... + ∂(z²)/∂x ?
proper bramble
Wait. We can write ∂(x² + y² +... + z²)/∂x = ∂(x²)/∂x + ∂(y²)/∂x +... + ∂(z²)/∂x...
well yeah
modern bloom
If it's ND to ND, then how can we write expression for f(x)
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lean otter
a + b + 2c = 15. Find number of positive integral solutions. How do i approach this using permutations and combinations, if not for the coeffecient of c being 2, i could've solved it by "distribution of alike objects" concept pretty easily but now with the tweak in the question im left clueless somehow, any help would be appreciated !
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a + b + 2c = 15. Find number of positive integral solutions. How do i approach t...
tardy mango
a + b + 2c = 15. Find number of positive integral solutions. How do i approach t...
Hint: How many solutions (a,b) are there for a given c?
Hint 2: ||Sum this over all possible c||
lean otter
right ! thanks, i didnt think of that at all. much appreciated
lean otter
Hint 2: ||Sum this over all possible c||
got the answer right, again, thanks for the help 🫡
.close
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balmy sky
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dry wolf
Sure, however depending on the level you are in you might be expected to guess a whole number exponent and then optimize the coefficient via least squares
You could use the python method polyfit for that for example
balmy sky
actually i dont know the least squares method
dry wolf
it is in numpy
balmy sky
so do i just pass the points in it as an argument?
dry wolf
and make sure to use np.array for the inputs
dry wolf
btw
maybe its nicer if you can fit a square root
But I am just guessing
Maybe they want you to vary the power
lol
Just try the results in desmos
balmy sky
maybe its nicer if you can fit a square root
so i should put 0.5 as the degree?
dry wolf
Yeah+
dry wolf
just try it
dry wolf
instead of (1,2),(2,3)
(1,2) and (2,3)
Ohh wait
shitty example
I meant
Instead of ((8,9),(6,7),(3,5)) make (8,6,3) and (9,7,5)
balmy sky
Instead of ((8,9),(6,7),(3,5)) make (8,6,3) and (9,7,5)
i did
those are x and y
i made that tuple separately
the x and y basically store that only
dry wolf
can you show me how it looks in desmos?
dry wolf
try m,d = np.polyfit(...)
balmy sky
try m,d = np.polyfit(...)
it just returns 1 coefficient
that np.polyfit outputs only 1 coefficient, that is 42.16666667
so like how could that m, d work since there is only one value in the array that the function returns
@dry wolf
balmy sky
can someone help me with this please?
lost vector
can someone help me with this please?
take log ig?
shell field
it may not work because x^0.4 is not a polynomial I would suggest using scipy.optimize.curve_fit
balmy sky
it may not work because x^0.4 is not a polynomial I would suggest using scipy.op...
they don't want polynomial as an answer tho
they just want a power function
shell field
for this u have to define a function f(x, a): return ax^0.4
u cannot use np.polyfit if you dont have a polynomial
curve_fit(f, x, y)
balmy sky
u cannot use np.polyfit if you dont have a polynomial
yea exactly thats what i was thinking. i dont know about polyfit but the doc says its for polynomials
dreamy veldt
they just want a power function
Wdym
balmy sky
curve_fit(f, x, y)
thats the scipy library you mentioned above right
lost vector
N=k*A^b
take log both sides
log(N)=log(k)+b.log(A)
take log N as y
log A as x
log k as c
so y=bx+c
then fit a linear model to the points (log(A),log(N)) for vals of b and c
dreamy veldt
c4^a=5
c40^a=9
10^a=9/5
a=log_10(9/5)
c=5/4^log_10(9/5)
dreamy veldt
c4^a=5
c40^a=9
10^a=9/5
a=log_10(9/5)
c=5/4^log_10(9/5)
Does that solve it?
lost vector
N=k.(291)^b
N=3.10*A^0.308
lost vector
its coming approx 18 species
balmy sky
ig this is good enough
balmy sky
its coming approx 18 species
yupp
dreamy veldt
3.51*A^0.255 I'm getting
balmy sky
N=k*A^b
take log both sides
log(N)=log(k)+b.log(A)
take log N as y
log A as x
lo...
so you basically put the log of the points in a linear model?
lost vector
so you basically put the log of the points in a linear model?
yep
balmy sky
3.51*A^0.255 I'm getting
balmy sky
yep
ahh so then i have to just put the values of m and c of the resulting linear model in that power function
dreamy veldt
Desmos has an approximation feature
Put you're X and Y coordinates in lists
Then type
Y~cX^a
And it'll find some values
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storm maple
hello, i found this passage on a book which i dont understand. can somebody please help?
safe radishBOT
hello, i found this passage on a book which i dont understand. can somebody plea...
storm maple
"Since real numbers are uncountable, then there must exist real numbers that are not algebraic. In fact, almost all real numbers are not algebraic (trascendental."
the fact that real numbers are uncountable implies that the real numbers are an extension of the rational numbers
empty gyro
Not necessarily, no
storm maple
though sqrt(2) is a real number but not trascendental
empty gyro
It just means that no surjective function from rational to real exists
storm maple
It just means that no surjective function from rational to real exists
yes, or more generally that there is no biijective function between R and N, right?
empty gyro
To say the reals are an extension of the rationals is another thing
empty gyro
yes, or more generally that there is no biijective function between R and N, rig...
Surjective is more precise, but sure
Anyway, what is your question exactly
storm maple
Anyway, what is your question exactly
i dont understand why if real numbers are uncountable then there must exist real numbers that are not algebraic
as you said, if real numbers are uncountable then there is no surjective function from the rationals to the reals
hence the irrational numbers like sqrt(2) and pi are the numbers that make the real numbers uncountable
but unlike pi which is trascendental, sqrt(2) is algebraic
that's what i don't understand
steep lily
the algebraic numbers are also countable, so you need to have 'more' reals than algebraic numbers
lapis shadow
Given any real u can't find a natural number that maps to it. But given a real that is algebraic, u can
Ish
steep lily
you might think of them as being in between rationals and reals, but the same size as rationals
lapis shadow
It makes intuitive sense doesn't it
storm maple
Given any real u can't find a natural number that maps to it. But given a real t...
so if we consider the set X = rationals + algebraic irrationals (sqrt(2), golden ratio, etc)
but trascendental numbers like pi and e are not in X, then X is countable?
steep lily
it turns out that yes
storm maple
the algebraic numbers are also countable, so you need to have 'more' reals than ...
now that makes sense
how do we prove this formally? the cantor diagonal argument shows that the reals are uncountable but it doesn't distinguish between the algebraic irrationals and the trascendental irrationals
steep lily
there's only countably many polynomials with integer coefficients, and each has finitely many roots, so the set of algebraic numbers is also countable
storm maple
okay so
just saying that algebraic numbers are countable and the real numbers are uncountable is enough to prove that the trascendentals are uncountable?
steep lily
yes
because otherwise they're countable, and the union of two countable sets is also countable which contradicts the reals (union of algebraic and transendental) being uncountable
storm maple
because otherwise they're countable, and the union of two countable sets is also...
oohh okay makes very much sense
thanks very much! 🙂
all clear now
.close
shrewd topaz
im not sure how to analyze this
safe radishBOT
im not sure how to analyze this
shrewd topaz
when you approach from the left the function is continuous
but when you approach from the right at x =3 the f(x) is 0
so the function isnt continuous at 3?
quiet plume
Well first off, to the left of x=2, you have just a line, which we know is continuous.
To the right, we have a parabola. Also continuous.
So the only place something bad could happen is at x=2.
shrewd topaz
huh
but when x =2 the top equation is -14
and the bottom is -11
OH wait so visually the graph looks like this?
as x approaches 2 from the left the f(x) approahces -14 but from the right it approaches -11
shrewd topaz
Well first off, to the left of x=2, you have just a line, which we know is conti...
is my understanding correct?
quiet plume
Yes. That means that the limit at x=2 does not exist.
Since a function is continuous at a point a if its limit about that point is equal to f(a), well the limit has to exist in the first place, so in this case f is not continuous at 2.
shrewd topaz
so if c = 2 and f(c) = 2 the point is continuous @ 2?
so the graph would look like this but there wouldnt be a hole at c
quiet plume
We want that $\lim_{x\to 2} f(x) = f(2)$. That implies three things :\
flat frigateBOT
Azyrashacorki
quiet plume
- the limit on the left exists
- f(2) is defined
- they are equal
If any of those fail, then f is not continuous at x=2.
So yes it would look like the graph.
Essentially,
- ensures that both sides of the function approach the same value. This prevents, like in this case, that both parts are completely disconnected.
- ensures that there is no hole there
- ensures that f doesn't just have a point sticking out
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thick turret
Hello I have a question
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Hello I have a question
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prime locust
Why are they integrating over the interval 0,x
i think it’s just to cancel the “d/dx” out on the left
basically whenever you have something like this where you want to solve for F(x):
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Ryan
prime locust
you can first rewrite the formula using t’s instead of x’s:
d/dt (F(t)) = f(t)
[sorry i accidentally deleted the typeset message]
then integrate both sides with respect to t, from t=0 to t=x
flat frigateBOT
Ryan
prime locust
which also can be written like this, just using F'(t) notation:
flat frigateBOT
Ryan
prime locust
and then you can apply the fundamental theorem of calculus to the left-hand side:
flat frigateBOT
Ryan
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prime locust
**Ryan**
all of those steps, the whole point is to just solve for F(x). and as you can see here we have F(x) on the left-hand side, just with a -F(0) attached to it
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thick turret
Okay, can I give an example and try to make sense from that
safe radishBOT
Okay, can I give an example and try to make sense from that
thick turret
So we just integrate from 0 to x?
prime locust
first rewrite the x's as t's
thick turret
Okay, as a dummy variable
prime locust
yes
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cunning pasture
smartest kid in the cosmos
lean otter
have to apply rolles theorem,dk how tho
safe radishBOT
have to apply rolles theorem,dk how tho
hearty egret
it sounds pretty strange as the degree of the polynomial is odd so it must have a least one real root (lim p(x)=+- infty as x goes to +-infinity )
hearty egret
it just seems falso to me (so you cannot solve it) and I would ask to a professor (or who gave it to you) some explanations
oh
maybe it says that you cannot have every root of the polynomial to be real
lean otter
oh yah
that would make sense
ps:this question is supposed to be from application of derivatives
so my obv guess when i see related to roots is rolles theorem
hearty egret
ok...
lets say you have 5 real root
then between these root you have 4 root for the first derivative
lean otter
yah
hearty egret
the between these roots you have 3 root for the second derivative
the between these roots you have 2 root for the thirth derivative
third
lean otter
but we know all of them cant be real
hearty egret
so maybe you just have to compute 3th derivative and find that cannot have 2 root becouse of the inequality you are assuming
lean otter
wait ill try
hearty egret
so you get a contraddiction if you assume that every root is real
3th derivative is $60x^2-24a_0x+18a$
flat frigateBOT
everg
lean otter
oh so D will be less than 0 since not all roots are real
lean otter
yes i got the answer
hearty egret
actually D<0 as the third derivative can t have double roots as (what I have written before)
hearty egret
gg
lean otter
tysm
hearty egret
np 😄
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balmy sky
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shrewd topaz
uhh how do i factor t^4 - 625
safe radishBOT
uhh how do i factor t^4 - 625
lean otter
uhh how do i factor t^4 - 625
hello
lean otter
u know 625 = ?^4
shrewd topaz
yes 5
shrewd topaz
t^4 - 5^4
lean otter
!original
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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.
balmy sky
i can try to figure it out myself i just forgot the algebra rules lol
do you remember any identity like a^2 - b^2
shrewd topaz
do you remember any identity like `a^2 - b^2`
so i do (t^2)^2 and 25^
balmy sky
25^2
shrewd topaz
then just t^2(t-5)(t+5)?
shrewd topaz
a = t b = 5
shrewd topaz
t^2
shrewd topaz
b=5^2
balmy sky
nice
shrewd topaz
(a-b)(a+b)
shrewd topaz
(t-5)(t+5)
balmy sky
isn't a = t^2 and b = 5^2
shrewd topaz
so i have t/t^2(t+5)
shrewd topaz
isn't a = t^2 and b = 5^2
well i just wanted to make sure i factored it right
balmy sky
well i just wanted to make sure i factored it right
what will you get if you put those values in (a-b)(a+b)
a = t^2 and b = 5^2
shrewd topaz
well thats why i took out ^2
so id have a = t and not t^2
oh right
(t^2+25)(t^2-25)
shrewd topaz
(t+5)(t-5)(t+5)(t-5)
balmy sky
(t+5)(t-5)(t+5)(t-5)
a^2 + b^2 ≠ (a-b)(a+b)
shrewd topaz
shit i thought it went both ways
balmy sky
no no
you can look up the identities and their proofs here
would help you from getting confused ig?
shrewd topaz
okay i was wondering what exactly i had to look up
balmy sky
np there are some good resources on khan academy too
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modern bloom
Let f(x,y) has second-order partial derivative, satisfing 3( \frac{\partial^2 f}{\partial x^2} - \frac{\partial^2}{\partial x \partial y} +k \frac{\partial^2 f}{\partial y^2} = 0 ). Find the values of ( k ) and ( \alpha ) such that the above equation can be transformed under the transformation ( u = x + 4y ) and ( v = 3x + \alpha y ) (where ( k ) and ( \alpha ) are constants) to an equation that contains only terms with ( \frac{\partial^2}{\partial u \partial v} )
safe radishBOT
Let f(x,y) has second-order partial derivative, satisfing 3\( \frac{\partial^2 f...
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riyobi
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@modern bloom Has your question been resolved?
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@modern bloom Has your question been resolved?
sick scarab
chain rule is the way to go
modern bloom
Could you elaborate a bit
sick scarab
ill show you how to transform the first term only
you do the two other terms
and then plug them in the diff eq
just the chain rule over and over
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severe pond
so if you moved the sec^2 term to the left then multiplied by cos^2 on both sides
what would you get
mhm
what’s tan*cos^2
mhm
yup
what is the double angle identity for sinx
do you remember
sin2x=2sinxcosx
you can show this from the angle sum identity of sinx
by 2
yes
you’re welcome
use ".close"
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crude star
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crude star
There is something confusing me
howcome L1 is regular but L3 isnt (ik its not finite but L1 isnt either?)
kind tinsel
pumping lemma
crude star
My profs comment on this was "For the first one it doest matter how many 0s and 1s there are as long as 0 appear ebfore 1 but for the third one we care about the amount"
crude star
pumping lemma
I see but I thought if its not finite thn it cant be regular
and L1 doesnt seem bounded
m and n
kind tinsel
I see but I thought if its not finite thn it cant be regular
the space of all words is regular for any alphabet
kind tinsel
well, a language is regular if there is a finite automaton that accepts the language
crude star
oh true but there was this comment tripping me out where my prof said that regular languages cant count so finite is okay and infinite isnt
quiet plume
howcome L1 is regular but L3 isnt (ik its not finite but L1 isnt either?)
A finite language is regular, but not all regular languages are finite.
crude star
Alr that makes more sense
so there are some infinite cases aswell
that are regular
bet I get it now
severe pond
why part 2?
kind tinsel
you should probably internalize the pumping lemma if you want to know more about how to show that some languages are not regular
severe pond
did you fail
severe pond
ahh
kind tinsel
you should probably internalize the pumping lemma if you want to know more about...
that actually gives a condition on when infinite languages are regular (read: tells you a case when they are definitely NOT regular)
crude star
you should probably internalize the pumping lemma if you want to know more about...
so if they asked me like "Is this regular or not" I should always start with the pumping lemma proof ?
and see if I end with a contradiction or not
instead of checker whether its possible to produce a finite automaton
checking
coz I feel like that would take more time
kind tinsel
so if they asked me like "Is this regular or not" I should always start with the...
well, you should internalize a general pattern for languages, you just have to see a bunch of languages and say "yeah i can prove this one is regular" "yeah this one is not regular by the pigeonhole principle" "yeah this one is not regular by the pumping lemma" etc
you just need a bunch of examples
safe radishBOT
quiet plume
For more intuition, the idea is that for 1, you just have to construct an automaton that accepts a bunch of 0s and then a bunch of 1s.
For 3. the automaton would need some "memory" to know how many 0s you put in the first place to write the exact same amount of 1s after. But finite automata can't do that, you need an extra structure for that called a pushdown automaton.
This is formalized by the pumping lemma as smay suggests.
As you get more comfortable with them through examples, you can usually make an educated guess at whether or not the language is regular.
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velvet lark
can anyone please help me with how this simplification works?
safe radishBOT
can anyone please help me with how this simplification works?
quick crater
can anyone please help me with how this simplification works?
1->2 is combining fraction
2->3 is factor
velvet lark
so how does 2 -> 3 work>
crude star
factor $x_{n-1} - 2x_{n}$
flat frigateBOT
Calc III Victim (Pt. 2)
quick crater
so how does 2 -> 3 work>
$$x_{n-1}-2x_n+\frac{-1(x_{n-1}-2x_n)}{x_{n-1}x_n}$$
$$x_{n-1}-2x_n-\frac{(x_{n-1}-2x_n)}{x_{n-1}x_n}$$
$$(x_{n-1}-2x_n)+(-\frac{(x_{n-1}-2x_n)}{x_{n-1}x_n})$$
flat frigateBOT
Skill_Issue
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@velvet lark Has your question been resolved?
velvet lark
then why did it become a multiplication on step 3?
when it is still a addition/subtraction problem at step 2
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velvet lark
.close
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strange plover
Can someone explain this? There's a difference with the placement of the green dot in each graph
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Can someone explain this? There's a difference with the placement of the green d...
strange plover
junior leaf
is there any other context to what the piecewise function is?
either way I think the answer should be A as B looks like it isnt even a function as there are two outputs at x=0
exotic cypress
isn't a function 1 to 1 tho? cuz inverse functions are and they need restricting the domain to get a function
junior leaf
functions dont have to be one to one but it does need to be one to one to have an inverse
strange plover
I'm confused
junior leaf
its kind of hard to see in the picture, but graph A is the function following the red curve, but then it jumps at x=2 to the blue line
the green dot is saying that the value at exactly x=2 is 6
unless theres a hole in the red curve on graph b at x=0 that i cant see from the picture
since the green dot is also at x=0, graph b is saying that the function is equal to 0 and also 6 at x=0
(which is not possible for a function)
strange plover
ooohh
does it still count as a function if it lies in a same value? like in the Graph A x=2
junior leaf
do you mean if the green dot in A was at y=4?
strange plover
no, I meant the dots blue, green and red lies in a same value of x axis
junior leaf
ohh
if they are on the same value, then it isnt a function, but usually what people do is indicate things with a filled in dot / open dot
oopsd i forgot to label it just imagine that vertical line is x=2 for instance
this is saying that the function is the red line all the way until x=2, but not including x=2
at exactly x=2 it swaps to the blue line, so its never overlapping ontop of each other
in graph A, it would have to be something like this to be a valid function
so exactly at x=2 its the green dot, and before and after its the red/blue
if the dots were filled in above, then it would mean at x=2 there would be 3 values
strange plover
ooohhhh
junior leaf
basically the dots filled in/not juist tells you if it includes that point/not
strange plover
oohh so the dots that are filled are included and the dots that aren't filled is excluded
junior leaf
yep
thats why i was thinking there could be a dot i cant see in the picture for graph B at x=0, since then it would be a valid function
usually in piecewise functions, you could have a function defined to be for instance x^2 for values of x < 0, and then 2x for values x >= 0
you have to pay attention to the inequalities used and whether or not they are strict or not to tell which dots should be closed/open
so in this example it should look like this:
actually this wasnt really a good example since they happen to be the same value at x=0 anyways lol
but you get the idea
junior leaf
yea im kind of confused too since the question doesnt seem to actually give you a piecewise function to match
it only shows graphs
pastel cloud
graph B right?
i think graph A is tryna show discontinuity
with a green dot in the middle
but i could be wrong
lol sorry i shouldn't pitch in
junior leaf
is there a hole in graph B at x=0?
junior leaf
then the answer is A then
the question is sort of weirdly worded then, they are more of asking which one is a valid piecewise
strange plover
the question is sort of weirdly worded then, they are more of asking which one i...
precisely
okay, so it's not a valid function if a dot lies in x=0?
junior leaf
yep
well because theres two dots at x=0
one from the green dot and one from the red line
strange plover
oohhh
last question bro
I'm really confused in piecewise function
how do you solve this
junior leaf
ok, so you can think of f(x) as being broken up into 3 functions
and it uses the different functions based on what values of x you are putting in
the first line says that for all inputs less than 2, its using the function x^2
so for f(-2), this would be using x^2, and it would become (-2)^2 = 4
junior leaf
np, glad it was helpful
stone pasture
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keen zinc
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@stone pasture Has your question been resolved?
chilly valve
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chilly valve
How do I check if this series converges or diverges
I think I need to use a comparison test but I can’t figure out the bounds
I tried using a divergence test but got a 0 so that doesn’t work
thick pasture
@chilly valve how do i calculate standard deviation
chilly valve
??
thick pasture
standard deviation non calc
chilly valve
theres a formula. please use a different channel
thick pasture
ok im new which one?
chilly valve
the ones that are free
median vigil
rough storm
How do I check if this series converges or diverges
you can choose a b sufficiently large enough s.t. bn>2n+log(19n)
rough storm
note that log(19n)=log(19)+log(n)
chilly valve
i think i can put down 3n
rough storm
good choice :3
chilly valve
actually make that 10n
rough storm
now note in the denominator the largest power
rough storm
mhm
junior smelt
(you can also remember that you have log(x) < x always 
)
chilly valve
thats true
rough storm
so if you have purely sum n/n^1.5 what series does this simplify to
chilly valve
1/sqrt(n)
rough storm
yes
rough storm
okay, so this is sort of what we want to compare it to.
so we actually want to find a series we can compare it to with a smaller value
chilly valve
yep
0 <an<bn
the one we got is bn?
and trying to get a good an that we can evaluate?
rough storm
ignore my reasoning from before. We want to find bn<2n+log(19n)
this bn is clear
you can just keep it as 2n :3
or n
don't matter
chilly valve
rough storm
our goal is to find a sequence b_n s.t. b_n < a_n for all n
chilly valve
So far like this?
chilly valve
All good
rough storm
yes
chilly valve
I think we need to get rid of the 4n and 15cos
So that we get either a geometric or harmonic p series
Once things cancel
rough storm
now, for the denominator, we want something larger in it.
rough storm
it does, but there's a nicer sort of way to do it
we want 2n/() to be less than our original fraction, yes?, so we want the denominator to have a power atleast equal to or GREATER than nsqrt(n)
whats a nice power that's greater than nsqrt(n)?
chilly valve
2
chilly valve
ah I see
Wouldn’t we have to be careful though
If we chose 3, that’ll give 2n/n^3 so p=2
And that converges
rough storm
well its obvious that a divergent series will eventually take a value greater than a convergent series
chilly valve
That’s true
rough storm
we want to find a divergent series that's sequence we are summing over always less than the original sequence, because this allows us to say, 'hey, since this sequence that's always smaller we're summing over goes to infinity, that MUST mean the bigger one does too!'
does that make sense?
rough storm
I should say that the +4x and such makes it a bit more complicated
chilly valve
Evaluate 2n/n^2 and say that it diverges since p=1. After that I’ll say that since this diverges and it’s smaller, then my original one has to also diverge
Yep
rough storm
so you may need to add a shift. or simply shift the sum over
chilly valve
It’s a very ugly series
Everything else in the lectures was very simple
I might use a graph online to help out with that
rough storm
what is the maximum value 15cos^2(x) will take?
chilly valve
15
chilly valve
Yep
chilly valve
5x
rough storm
8x would be better, then could have (x+4)^2=x^2+8x+16
flat frigateBOT
Result:
2.9444389791664
in all honesty, you simply need to show there's a cutoff point somewhere where your smaller series is always in lower value than your original series
chilly valve
I see
rough storm
in other words, there's only a finite number of values where original series < new series
chilly valve
Yep
Also one more question
I can use a divergence test for this one right
I take the limit of cos(…) and move the limit inside
Lim(fraction)=0 since that’s a standard limit and cos(0)=1=/=0
rough storm
yes
rough storm
oh if you can use limit comparison test rather than purely comparison it would also help tremendously
for the one we were doing
rough storm
woven axle
Am I correct?
chilly valve
@woven axle looks right
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pine wren
how did we figure out that its divergent out of whats in the red box?
hallow knot
If the limit of an doesent go to 0 the serie diverges
That what is saying in the red box
I don't know if understood your question
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@pine wren Has your question been resolved?
pine wren
If the limit of an doesent go to 0 the serie diverges
but cant a serie be convergent towards a number lets say 2, so it can converge where its not = 0
also sorry for the late reply
pine wren
also, how did we figure out only from that that it is not = 0?
hallow knot
But the limit of the an is different then the result of the serie
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pine wren
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pine wren
also, how did we figure out only from that that it is not = 0?
because 1^n = infinity?
hallow knot
They use a famous limit result
pine wren
But the limit of the an is different then the result of the serie
how is it different if it goes towards a number? I know its not exactly the number te limit goes towards but we simplify it so that it is, no?
hallow knot
Limit (1 +1/n) ^ n = e
hallow knot
how is it different if it goes towards a number? I know its not exactly the numb...
If a serie goes towards a number , its necessary that the limit of the general term goes to 0
But that Alone doesent garantee that the serie converges
Cause i will give you an example
pine wren
If a serie goes towards a number , its necessary that the limit of the general t...
whats a general term? like f(x)?
pine wren
ok
hallow knot
For example
If you have 2 + 1/n
General term an = 2 + 1/n
Do you think the series of this converges or diverges?
When "n" gets very very large
2 + 1/n goes to 2 right ?
pine wren
diverges because 1/n is divergent
pine wren
2 + 1/n goes to 2 right ?
but without knowing that 1/n diverges I would guess that lim n->inf 1/n = 0 so the lim of 2 + 1/n would be = 2 and there for it would be convergent
hallow knot
Its 2
You are right
So when gets very large
Like n = 10000000 and above that
Its like you are summing 2
Forever
So it diverges
pine wren
wait so serie 1/n isnt divergent? because our teacher wrote dow that its a harmonical serie and that it diverges
hallow knot
1/n is
Cause
An goes to 0 its not the only condition
But if it doesent go
You know it diverges
pine wren
You are right
oh sorry tought you ment that its convergent
pine wren
ok
hallow knot
The limit as you said goes to 2
So when n is bigger
Its like summing infinite times 2
So you will get infinity
pine wren
but still im confused since wouldnt 1/10000 and going towards infnity be = 0?
hallow knot
You are confused about the limit of 1/n ?
pine wren
but still im confused since wouldnt 1/10000 and going towards infnity be = 0?
i know its divergent because our teacher wrote it down but I dont get the logic because this is my current logic in my head, or is this for limit of 1/n and not for the serie?
am I confusing the limit of 1/n for the serie of 1/n?
pine wren
am I confusing the limit of 1/n for the serie of 1/n?
or are they the same?
hallow knot
Now i am a little bit confused i am sorry i am not that good at English
But from my understanding
You are asking
If there is any difference
Between the limit of the general term an ( in this case 1/n ) and the number that the serie with an converges too
pine wren
wait I also have a language barrier since im from slovenia, is a serie this?
hallow knot
With a general term it is yes
In front of that symbol put a general term like 1/n , n+4/n and you have a serie
pine wren
With a general term it is yes
oh because n can be something else thats why you say with a general term?
or do you mean an when you say general term?
oh ok
I understand, ok
hallow knot
Let me go to my computer wait a sec
pine wren
Between the limit of the general term an ( in this case 1/n ) and the number tha...
then yes this is what bothers me, since in my mind they are the same thing
thank you for trying to help me so far I appreciate it
hallow knot
wait I also have a language barrier since im from slovenia, is a serie this?
Ok first of all do you understand what that symbol means ?
pine wren
yes, for an = 1/n it would be 1/1 + 1/2 + 1/3 ... + 1/n
pine wren
not exactly the definition but simplified, yes
hallow knot
1/100 = 0.01
pine wren
yes
hallow knot
Its not a big number
pine wren
thats why I am confused since doesnt the limit go towards a 0
pine wren
oh sorry I meant n is a big number
hallow knot
To 0
pine wren
or goes towards one
but an goes to 0
yes
n goes towards infinity and an goes towards 0
hallow knot
Yes
pine wren
ok
hallow knot
As you see from the first
pine wren
so whats the difference then?
hallow knot
Terms of the serie
1/1
- 1/2 + 1/3
Already that is different then 0
One thing is the limit of general expression
pine wren
ok fair but doesnt the symbol for the serie say it goes towards infinity?
pine wren
ok fair but doesnt the symbol for the serie say it goes towards infinity?
this one
hallow knot
One thing is what the general term aproaches other is what the sum of that general term aproachesr
n goes towards Infinity
1/n the general term no
That only means you are summing infinite terms
You are using the general term infinite times
1/1 + 1/2 + 1/3 + 1/4 ... + 1/Infinity
pine wren
this one
but doesnt this symbol of an mean that its = 1/1 + 1/2 + ... + 1/n, emphasis on + ... + 1/n
so we are summing it an infinite number of times
OHHHHHH
ok
I get it
hallow knot
Yessss
hallow knot
Exacly
pine wren
ok I understand the difference now
pine wren
how did we figure out that its divergent out of whats in the red box?
but I still dont understand this
how we directly knew that its not = 0
pine wren
but thats for an not an^-1
pine wren
since for b) we want to know if the serie an^(-1) is konvergent
hallow knot
Goes to e
hallow knot
Ok
hallow knot
No
pine wren
or e^infinity, since -n^2-1 goes towards infinity
pine wren
oh god Im confused again
pine wren
wouldnt that mean that this is = 0?
pine wren
The limit goes to -1
how though, what I sent last is from b) from my original question
hallow knot
But they got a different thing
pine wren
But they got a different thing
for a) you mean?
pine wren
yes
hallow knot
You should get e
hallow knot
I will try to do it
hallow knot
Did you found the mistake?
pine wren
the under part of the division
hallow knot
I can try to do it
pine wren
I cant take it into 2 parts, thats the problem
pine wren
only the upper part of the division
hallow knot
An^-1 its just e
pine wren
yea true
hallow knot
But you can do it for training
pine wren
ok I understand now, thank you very much for the help, do you have a paypal?
pine wren
come on for a coffee
hallow knot
Np you can add me if you want
pine wren
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umbral folio
Through which concrete, easy as he says, division process does he derive the infinite series from the first expression?
safe radishBOT
Through which concrete, easy as he says, division process does he derive the inf...
main mural
infinite geometric series i assume
he, in fact, specifies that in the paragraph below
so heisst diese Reihe eine geometrische Reihe
umbral folio
I know. But how do I divide and what to reach the series? Not with mere knowledge, but real calculation.
main mural
$\frac{a}{\alpha + \beta z} = \frac{a/\alpha}{1 + \frac{\beta z}{\alpha}} = \frac{a/\alpha}{1 - \left( - \frac{\beta z}{\alpha} \right)} = \ \frac{a}{\alpha} \cdot \left( 1 + \left(\frac{- \beta z}{\alpha}\right)^1 + \left(\frac{- \beta z}{\alpha}\right)^2 + \left(\frac{- \beta z}{\alpha}\right)^3 + \cdots \right)$
flat frigateBOT
artemetra
dreamy cliff
Oh
umbral folio
woah thank you: so he divides by alpha?
umbral folio
then you do a trick with the minus, but how does 1/-(-1beta*z/alpha) get me (1+….) meaning: how do you get it on top?
Why am I allowed to devide by alpha, am I not changing the whole equation therewith? So you just divide by alpha to get 1/1-q and then use your knowledge that this structure resembles a geometric series? But that is not what Euler does, because he says through iterative division he derives at the series.
main mural
Why am I allowed to devide by alpha, am I not changing the whole equation therew...
pretty much
i don't know how Euler derives the geometric series
umbral folio
Thank you anyways. He says it is apparent how through iterative division you derive from the first expression at the second..xD I do not see it either.