#help-0
1 messages · Page 201 of 1
is there a particular part that you're stuck on?
how old are you?
@green solstice how old are you?
I am 13
as you say.
13 year old with a 6 year old account...
I stuggle with multiplying above 10
well she is 13 now so. nothing unkosher in the present
aight fair
if you were asked to, say, multiply 20 * 6, could you do it
yeah I started discord quite young
yes because I can do 6 x 10 and then do it twice
but if its like 27
I get confused
and it takes me a long time to figure it out
I havent heard of it
thanks
shame, bc that's what is happening.
but then again, 21 groups of 6 is the same as 20 groups of 6 plus 1 group of 6 -- you might be able to get this even without it being phrased with the scary word "distributive law"
do you understand this?
yes that's what i said
okay yeah I get it now
Our prof just write it in the board
@silk fox open your own channel
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@vale wigeon is there a fast way to figure out two double digit multiplication in your head?
like 53 x 27 for example
two-digit by two-digit?
meh
i can't say i've ever needed that much in the way of mental math
oh, well whats the best way you think to approach that question
mental or on paper
like in general
I think learning abacus may help you to figure out large number multiplication in you head
"advanced" and "difficult" are both subjective
whats abacus
yeah I was kind of thinking that
it would probably be dificult for me but easy for you
It's a counting ancient tool.
it's a matter of how much you have had to practice with it i guess
this thing?
Yep
the only method i can suggest is a "factor breakdown" method:
for example if have 28x15, you can break it down into (2x14)x(3x5)
you can rearrange it to be 14x3x2x5 which you can make to be (14x3)x(2x5)
this equals (42)x(10)=420
this does mean you have to know how to divide fast or know the factors of numbers
this is also unhelpful for prime numbers
that sounds interesting
thank you
also abacus math is really cool, if you get super good at it, you look superhuman
how does abacus work
you learn how to use an abacus, then your muscle memory kicks in and you can do the math with an imaginary abacus
woah
some people can get some big numbers
but thats beside the point for now, start small and make your way up
@green solstice Has your question been resolved?
in my head i do 53 * 10 * 2 + 53 * 10/2 + 53 * 2. basically doing smaller simpler calculations and adding them up
but doing mental math for large numbers isnt very useful. learn a method to do it by pen and paper and thats all you need really
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i'm not sure how to even begin with this problem
figuring out the number of rows and columns for A can be a start
how?
first columns. you are multiplying with a 2x1 vector so for matrix multiplication to even be legal how many columns should A have?
2?
yes
and the 'output' is a 3x1 vector
av+bw is
so from that you can determine the rows of A
ye it would be 3 rows right?
how do you calculate that
its just matrix multiplication
oh right nvm
ye
the b's have to match to be legal
you can go through giving the elements in the 3x2 matrix variable names and multiplying it with (a b) to solve for the values
or just revise in your textbook what matrix multiplication looks like
maybe that will give you an aha moment of what values it should be
wdym?
one sec, ill try it out to see if im crazy or if it works
算盘 in china
btw
ye
it literally is the definition of matrix multiplication
oh ok
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@errant dagger Has your question been resolved?
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For what values of N does $\lim_{x\rightarrow\pi} \frac{d}{dx}(|\sin x |^{\frac{1}{N}x^2})$ exist?
Quixz
I took a derivative, graphed it in desmos, and immediately got stuck
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how is ln n > 1
because n>3>e,and ln (e) = 1
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can anyone help me with this
simple geometry but i cant get no.8
In a parallelogram, such as this, the diagonals gets divided into 2 equal parts no?
by that i mean like...
For line DEB
DE = EB
cmiiw
yes
but the problem is, im trying to solve for the exact measurements for up and pl which is ul cut in half by trying to get x and y
so that means up and pl are of the same length
x=1, y=2 gives 5 for both up and pl
but y=7x-3 contradicts it
there are lots of x and y that satisfy x+2y = 3x+y
0,0
i think you should start by substituting y = 7x - 3 for the lengths of UP and PL
thats what i did
OH WAIT
up = ul?
ahhh
ok
thats what we got, 1,2 and 0,0
but the other statement said y=7x-3
i think i got it
0.6 and 1.2?
sounds right to me 👍
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"Given the equation (a-2)x-2y-1=0, find for what value of a represents a straight line:
a) parallel to the x axis
b) parallel to the y axis"
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
4
Show your work, and if possible, explain where you are stuck.
i mean point a) was pretty easy i guess
a-2=0
a=2
b) is what i'm confused on
-2y=0
?
?
open your own channel
how
then what should i do now then*
this is a bonkers question
whats the form of the equation for a line thats parallel to the y axis
its gotta be x = c or something like that right
the given equation distributes out to ax - 2x - 2y = 1 so i imagine you want to cancel out the y term... is it legal to put a = 2y/x?
no
why not
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i know that the theorem is
if a <= b and b <= a, then a = b
but how do i use
like what are the steps in a proof for this
do i just count out the elements
and say that each one has a correlating value
0 -> 3
1 -> 4
2 -> 5
therefore mod(a) = mod(b)
?
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Hey guys
this is a small excerpt from a paper im reading
so the author takes the inverse Fourier transform of the top equation (equation 10)
but, the results of which in the next equation dont make sense
how does taking the fourier transform of u(k, t) somehow transform it into u(k, t+dt)?
this isnt a typo, because he goes on to take full advantage of it
and this "transformation" is critical to the paper but never explained
the final result works, meaning that it is correct, but i cant possibly fathom how
@obtuse parrot Has your question been resolved?
@obtuse parrot Has your question been resolved?
Out of curiosity, what is k is in this context?
wavenumbers, theyre precaculated in this context
Ah ok. I can't speak to the nitty-gritty, but the Fourier transform is a subset of Laplace and just converts from the time domain to the frequency domain. τ is actually just a constant (in general, maybe not here) https://en.wikipedia.org/wiki/Time_constant
Also, notice that the function on the left-hand side of the equation, u, has a "hat" in (10) but not in (11) - this denotes the Fourier transform of u, so to find u you just take the inverse. Applying the inverse Fourier to the left-hand side leaves you with just u. Applying it to the right-hand side translates the whole thing from the frequency domain into the time domain, so now you have to introduce time into the equation. I'm not sure why it's delta-time as opposed to just t here. But I think, from a quick google, the inverse of the right-hand side will be a convolution of of the two terms with respect to t.
In physics and engineering, the time constant, usually denoted by the Greek letter τ (tau), is the parameter characterizing the response to a step input of a first-order, linear time-invariant (LTI) system. The time constant is the main characteristic unit of a first-order LTI system.
In the time domain, the usual choice to explore the time resp...
Keeping in mind I learned the basics of this in a controls class 5 years ago :|
@obtuse parrot Has your question been resolved?
wrong place @alpine sable, you gotta take a channel in the available pool
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trying to calculate further Information gain, already found out odor is the best classifier but now i need to find the next attribute for cases where odor is "okay" and somehow my information gain turns out to be -0.02 which ISNT nonnegative so it cant be the answer. what did i do wrong
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@spare crane Has your question been resolved?
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if sin(pi/7) = 0.434 what is sin(6pi/7)
What's pi - pi/7?
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What're the bounds for this integral problem?
I messed up and had theta going from 0->2pi and r from 0->pi/2
But I'm thinking it should be theta from 0->pi and 0->1
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am I correct that this cant be solved conventionally?
you need to use algebra right
like you can't simply do 3a+b / 3
you have to let PQ = xa + yb
then solve it that way
or is there a shorter way
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can you help me write the equation
Apply transformations to a standard hyperbola
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Hello
Don't occupy a help channel for no reason
Do you have a question?
@alpine sable Has your question been resolved?
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I'm struggling with the following integral:
$$\int_{-2}^1\left(\frac{1}{x}-3 e^{-x}+\sin (\pi x)\right) d x$$
dgh
We've not been taught integration by parts for this problem. How do I solve it?
Integration by parts is for functions being multiplied so that isn't needed
I would split the 3 terms into 3 integrals and solve each separately
1/x is just ln|x|
the other 2 are u-sub
$$\int_{-2}^1\left(\frac{1}{x}-3 e^{-x}+\sin (\pi x)\right) d x$$
$$=\int_{-2}^1\frac{1}{x}dx - \int_{-2}^1 3 e^{-x} dx + \int_{-2}^1 \sin(\pi x)dx$$
dgh
Like this?
Yeah
To compute the definite integral $\int_{-2}^1\left(\frac{1}{x}-3 e^{-x}+\sin (\pi x)\right) d x$, we can apply linearity and split it into three separate integrals: $\int \sin (\pi x) d x-3 \int e ^{-x} d x+\int \frac{1}{x} d x$.
\\
The first integral can be solved by substituting $u=\pi x$ and using the standard integral $\int \sin (u) d u=-\cos (u)$, which gives us $-\frac{\cos (\pi x)}{\pi}$.
\\
The second integral can be solved by substituting $u=-x$ and using the exponential rule $\int a ^u d u= \frac{ a ^u}{\ln ( a )}$, with $a = e$, which gives us $- e ^{-x}$.
\\
The third integral is a standard integral, which gives us $\ln (x)$.
\\
Plugging these solved integrals back in, we get $\int_{-2}^1\left(\frac{1}{x}-3 e^{-x}+\sin (\pi x)\right) d x=\boxed{-\frac{\cos (\pi x)}{\pi}+\ln (x)+3 e ^{-x}}$
dgh
So now we just evaluate the definite integral between -2 and 1?
\begin{align*}
&\left(-\frac{\cos (\pi \cdot 1)}{\pi}+\ln (1)+3 e ^{-1}\right)\bigg|_{-2}^{1} = \
&\left(-\frac{\cos (\pi \cdot 1)}{\pi}+\ln (1)+3 e ^{-1}\right) - \left(-\frac{\cos (\pi \cdot (-2))}{\pi}+\ln (|-2|)+3 e ^{2}\right) \
=& \left(-\frac{\cos (\pi)}{\pi}+0+3 e ^{-1}\right) - \left(-\frac{\cos (-2\pi)}{\pi}+\ln (2)+3 e ^{2}\right) \
=& \left(\frac{1}{\pi}+3 e ^{-1}\right) - \left(\frac{1}{\pi}+\ln (2)+3 e ^{2}\right) \
\approx& \boxed{-21.12}
\end{align*}
dgh
Btw, I was supposed to solve this without a calculator @fierce prairie
cos(2π), cos(π) is fine…
That can be left as well
I don't think they wanted the numerical answer
Leet me check
@fierce prairie It says the right answer is "Integral doesn't exist (divergent)"
Because of the 1/x?
Hmm
I think so
I honestly haven't seen one like this but symbolab is also saying it diverges
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The answer is 2/0 or 0 right?
you should factor
well because the Limit equals to 0/0 when you substitute 0 in for z we can do Hopetals rule
Ya i guess the right move would be to Factor right?
You have already covered LHospital's Rule?
i see a common factor
But if you did this way, you don't get 2/0, I don't think. I'm doing it mentally but pretty sure it's not 2/0
Ya know, factoring, actually, I agree that's all you need
you have to do something special since its complex right
I agree with factoring too but you did LHospital's wrong
Zeros won't hurt you after you factor out common factor in top and bottom
Ooh...top will be odd power of z though
Does that matter?
I think the limit is same from LHS and RHS anyway
Welp i guess i over complicated myself lol, thanks guys
or youd need the limit to be the same approaching from any path
...of 0. What z approaches
right but isnt z complex
im just assuming
so theres no lhs or rhs of 0, theres a region surrounding it
How do you mean if it's not complex?
is it not complex?
well then why did they use z 
This is like a Monty Python sketch now
DOES THIS SHOP HAVE ANY CHEESE AT ALL???
$\lim_{x \to 0} \frac{z+1}{z^2 - 1}$
$\lim_{z\to0}\frac{z+1}{(z+1)(z-1)}$
jan Niku

Disorganized
Oh, yep: then @remote heron
1/(z-1)
So LHsopital doesn't need to come into this
@grave niche
Sure it can, but usually when you ask a teacher or a student if you are expected to use LH, they arent
its a sin to use the hospital rule where you dont have to
We had to learn how to prove LH before we could use it
So LH is basically a cheat code?
im pretty sure in order to use LH you need to get 0/0 or ♾️ /♾️ in a limit right
(Not the best meme)
analysis can kill
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How would I graph this?
this is in general form
@tired nexus Has your question been resolved?
place y alone
in ur case
y = 3x - 9
now put any value of x in your equation u would get the value of y and then plot in the graph
for e.g
if x=2
y = 3*2 - 9 = -3
(x,y) = (2,-3)
similary follow this process 2 or three times
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Im confused on how to integrate this
I got until here but then when i integrate this it just gives me 1 which isnt the answer
$\int u^{-1} \dd{u} = \ln|u| + C$ though
Ann
so where are you getting just 1 from
power rule cannot be used when exponent is -1
Oh
Wait what
But what happens to the numerator then?
where?
the numerator is just 1 if you rewrite it
$u^{-a}=\frac{1}{u^a}$
ℕαv
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Oh shit yeah thx
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What is the application of the unit circle in the real world, and what is the reasoning behind the specific unit of 1? Also are we limited to just the 2 special triangle cases where they are labelled on the unit circle. Or can we use any value for the angles and thus find the proper lengths of its sides and it stuff? I know its good for trigonometric rations.
Real-world application is always subjective to someone's skill-set
unit circle is used for trigonometry
So you're asking what the real-world applications to trig are, which is a heckin' lot
If by 2 triangle cases you mean 45-45 and 30-60, then no
Those are just the easiest to memorize
We use 1 because it's literally a "unit". It makes the math easiest
If it was anything besides 1, we'd have to divide or multiply to account for the non-unit ratio
Or can we use any value for the angles and thus find the proper lengths of its sides and it stuff
I don't know what you mean by this
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how do I even start doing this question ?
I know how to differentiate it, but how do I form that equation
?
?
I kinda understand but can you like show me how to do it ?
Compute first and second derivative of the y they give you above, and substitute those into the diff. eq. and check whether you reach an identity or not
real chads solve the diff equation first
Integratiosns?
Bro am doing it right now
Partial fraction
what
ITS CONFUSING
@thick beacon Has your question been resolved?
Thats it
I don’t know how to further differentiate and I tried to figure out a way to make it form the equation given but so far I don’t really know
dont take e^2x in common before differentiating again
in second last step itself you get two functions in addition both are similar to the original function so just differentiate them separately and then add together
i can understand your frustration from the fact that you are doing it by assuming the functions being multiplied first
then differentiate individually and then apply the product rule
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It would be of great help if someone could explain how the last step came to be. Appreciate it.
The question is:
,w d/dx arctan(ax+b)

So if you adapt it you can find such antiderivatives
but it's 3/4
ummm, what?
,w integrate 1/((x+a)^2 + b)
You can do this, slowly, with a lot of u-substitution
it added 3/4 to change the limit from 0 to pi/4 to 0 to 1
,w derivative 1/(x^2+1)
xd
ohh
,w derivative arctan(x)
i guess that was what it was intended for since its worth 6 marks
This is the key thing you need to know
after a butt-load of u-subs, you can get here and solve your integral
xd. Alright thanks, i'll try
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hello

Seems like a tough problem
thanks for offering
but it seems like the answer was hidden in plain sight
all i did was watch a 2 minute video and i understood lmaoo
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so this was the ques
can we simply cancel out the term which is present inside the mod and the denominator ?
why not ?
as the modulus function makes everything +ve , so why can't we just cancel it out
?
Yeah but the inside of the absolute value is negative
So you'll get 4(2 - x)
Then you can cancel if you wish
You actually can lol
i already got the ans 😁
i'm wasn't able to figure out how to solve the mod fucntion thingy
if i cancel it i get 16, not -16
.
how do u know that we will get 4(2-x) and not 4(x-2) ?
Because the term inside the absolute value is negative
So we multiply it with -1 to make it positive
how do we know that its negative ?
by putting x = 2-h ?
is there any other way ?
ok
Then you see for yourself that it's negative
the root of 4x - 8 is 2
If you put 1.99999 in the absolute value term
ah yes as 4*(number less than 2) < 8
Yes
how did you come to that conclusion ?
but it also extending in the positive side ?
how would i be able to plot such a graph so fast ?
You're interested in a number slightly smaller than 2
Not greater than
yea 💀
so it's literally just an upward sloping straight line passing through 2
oh ok so i would plug in the values of x which are less than 2
Yes
so is 4x-8 = 0 ?
Why so?. Since for 2- you move from right to left , that means you should foucs on greater number , right? Or am I wrong?
2- is approaching from the left
Okie sorry.
.?
when you said plot a graph, 4x-8 would be equal to 0 right ?
i said plot a graph to see that it's negative on the left
it should be a 1 second thing you do in your head to check how to evaluate the absolute value
i'm dumb 
well now you know
but for plotting a graph we need something in the right hand side right ?
y-x=0
like this sort of a thing
is anything stopping you from plotting the graph of the function f(x) = 4x - 8
like it's just to determine the sign of the values inside the absolute value
you don't need to get caught up on this
i haven't been taught to do this
i am from somewhere where we are taught trigo without graphs 💀
ig i will just stick with the 2-h method 💀
sure
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here's how I'd do it
- find angle MOA
- OAM
also reminder that the whole shape is a regular octagon as mentioned in the question
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how would you approach
v(t) = e^(-s)
I know that s(0) = 0
meaning
v(0) = e^(-0)
v(0) =1
Differentiate both sides of that with respect to t and get a(t) = e^(-s) * ds/dt
ok
Hm, we still need s(t) though
I suppose we need to solve the differential equation then
ds/dt = e^(-s)
ye
e^s ds = dt
e^s = t + C
You can solve for C by plugging in t = 0 and S = 0, yielding C = 1
So you end up with s(t) = ln(1 + t)
The rest is solving for value t such that the second time derivative of s is equal to -4
ok so then you get derivative of s(t) as v(t) as 1/(1+t) right
then derive that again for a(t)?
Yes
what is the derivative of 1/(1+t) then
nvm got it
what is the differential equation?
@echo socket e^s ds = dt
e^s = t + C
HOW DID you get that pleae
please
could you please @ me when your back
@weary void Has your question been resolved?
Integrate both sides
Sorry for the late response
oh
so you integrate e^s ds to get e^s
and integral of dt is t + c
ok i think i get it
thank you btw
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what class is it for? calculus?
yup
so I don't know if this is the best way to solve it
but you're looking to find a substitution $u = x^n$ for some $n$.
cwatson
take the derivative of that, which gives you "du = ?? dx", where the question marks you can fill in
definitely not, one could just compute the integral directly without substitution...
you also know that 7u^2 du should equal 28x^11 dx. so you can combine the 2 equations
true but they're looking for the substitution itself. it can be done in your head but I'm trying to explain how I went about it
@errant dagger Has your question been resolved?
i expanded it and equated it and i got it 🙂
u = x^n was perfect starting idea ty
i was very lost
.sloe
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Is this wrong?
i dont understand where the cos(x) in the denominator came from.
They factored out sec x = 1/ cos x
both the sec(x) ?
You don't factor something there's only one of
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can I say
log_x (0) = -∞
for x > 1
and
log_x (0) = ∞
for x < 1
or not
The limit of something doesn’t have to be defined at that point
It is entirely possible that the limit of a function is not equal to the function at that point
What does this mean sorry?
Consider the function f(x) = 1 for x<0 and x>0
And f(x) = 0 at x=0
The limit as x->0 of f(x) = 1
Even though f(0) = 0
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@alpine sable Has your question been resolved?
@alpine sable Has your question been resolved?
@alpine sable Has your question been resolved?
Put the above expression under the sum, and use the linearity of summation to split sums and get constant outside
in the end it ends up being a sum of k over k which is easy to do
he probably won't follow it from the first time since it's all words so I will keep an eye for questions
oh ok
Just wanna show him how to do it rather than giving the answer
yes that is the way
Mohamed Mohsen
you can plug that expression in here $\sum_{k=1}^{n}f(a+k \times \frac{b-a}{n}) \times \frac{b-a}{n}$
tell me if you follow up to this point
Mohamed Mohsen
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Hey, does the least squares methods still give a valid answer even if ker(A) != {0}
Cause I know there's no unique solution, but I wonder if it still gives you something
@prisma lantern Has your question been resolved?
What is A
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✅
Its a matrix
and how does A relate to least squares?
Well if you have a system of linear equations
like Ax = b
and b isn't in the image of A, i.e. there is no solution for Ax = b
then you can find a vector x* thats the closest it possible can come to solving Ax = b
riemann, i just got the tex preamble
are you the one who made that? i saw your name and profile picture there
And I know that if ker(A) has more than just 0
See "No linear dependence": https://en.wikipedia.org/wiki/Ordinary_least_squares#Classical_linear_regression_model
riemann came up with the idea and organised everything, snow made it possible
crazy
Hey so @alpine sable I'm not so good with statistics, whats the notation here?
Is it probablility that rank(X) = n is 1?
yup
Cool
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I don't understand their formula for the second derivative (in blue)
I understand that are are differentiating with respect to x but dx/dy is in terms of t
so we must do something.. but I don't get why they do that
@violet ore Has your question been resolved?
it's an application of the chain rule
Yes but I don't see how to derive it
if I have a parameterized curve x = f(t), y = g(t) then I accept dy/dx = [dy/dt]/[dx/dt] but the second derivative I just don't see
So they're doing (d/dx)[dy/dx] = [(dy/dt)/(dx/dt)] but how do they reduce ?
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am trying to develop a working understanding of the integral test, comparison test, and p series...this was a question and answer on my study guide but I'm having trouble following
Can someone help me understand from the first step. I think they applied u substitution firstly but I'm lost on what they did afterwards..
@high wolf Has your question been resolved?
Is it not via u substitution?
So that turned into integral of 14/u^3? But how did he get -7/u^3 from that?
Can we start from the beginning? So we have the original function given to us, and we must apply substitution on 2+ root x. Shouldn't that give us instead 7/(root x)*u^3?
Sorry, these sorts of functions are tricky for me to understand
we multiply root x by u^3?
What do we need to multiply u by?
ok.
Any advice?
Azka Aska
Is the above for u = 2+ root x?
thien
So this is the process, correct?:
How exactly did it become -7/u^2 though...?
Alright, I'm fixing to solve it.
Still not sure how, but it's fine. Somehow, we get -7/(u^2)...then we plug our u back in and get this: now how do we apply this integral test?
How about this. What are they doing in these few steps?
^
How are the getting 7/9 though?
What is going on?
Wait how do you get 1?
We have b there not 1?
Azka Aska
Wouldn't it be -7/9 though?
Aren't we meant to subtract upper bound from lower bound?
Okay, you're right. Sorry. So here we were doing integral test. But remind me, why do we set 7/9 less then infiniy?
What do I do with that information though?
How can I use it to determine if there’s convergence or divergence?
How do I know that this integral converges though?
What does this mean? That it’s finite?
Does it have to do with the 7/9?
How do we know it’s not infinite?
Now I’m understanding. But what about the other function?
The /7/(2+ root(b))^2?
This makes it of sense. May I ask more questions?
^
Give me one moment I am heading back to my computer to get the questions
I gave up here.
Im at the point where I have 14 * integral of 1/(u^3)du
I'm confused on the next step to integrate
yes
like, add one to exponent and multiply
Can you visualize the steps for me? Functions like these are weird to me.
Like how you did here
Azka Aska
Like this?
Got it. I can understand it. My other question is related to another problem:
im doing it now..let me try and see
So does this also converge?
does it not go to 0 like the problem before?
oh wait!
Is it because it's a fraction and this isn't a fraction?
so it won't go to 0 in this case?
As long as b is in the denominator or a variable is in the denominator..?
when we plug infinity in, it goes to 0?
But when it isn't in denominator it goes to infinity?
How does it depend? Can you explain a little further. 😄
Okay, are you still available?
Can we go over some more of these types of problems? They're called integral tests?
do these involve p-series as well?
Can we do this?
Okay, working on it rn
Does the process look right?
🙂
And it converges because -1/ln(inf)) goes to infinity because infinity in the denominator as b goes to infinity?
like you said before?
Got it. Are you familiar with the comparison tests?
Can we do a problem? I'm not good at comparison tests at all
Can we do this?
No problem!
I sent friend request. 🙂
@high wolf Still need help?
Yes. Can we work on comparison tests?
As n approaches infinity, what do you think the numerator acts like?
Same question for denominator
wouldn't it also be infinity?
Then I do not know.
3nsqrt(n) or 3n^(1.5)
Or even n^(1.5) will work
The denominator acts like n^2
So as n approaches infinity, the series acts like $\sum_{n=1}^{\infty} \frac{n^{1.5}}{n^2} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
Sup?
So you can use the limit comparison test with this series
where did you get 1.5 and 2?
@high wolf From the original series
3nsqrt(n) + 1 acts like 3nsqrt(n) = 3n n^(0.5) = 3n^(1.5)
n^2 + 4 acts like n^2
So your series acts like 3n^(1.5) / n^2 = 3/n^(0.5) = 3/sqrt(n)
you can get rid of the 3, doesn't matter, so the series you are interested in is 1/sqrt(n)
Are you still available?
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yo whats up with the coefficent for x =3
if the planes are parallel dont they have the same normal?
yeah thats what i did
so your plane should be 3x -3y + 2z = something not 6
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Please simplify
simplify this plese
What is the coefficient of x² ?
uhh
1
You need to simplify or solve for x?
simplify
Take Lcm in denominator
Do you need to like factories it ?
No rizz so i call it factories
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how do I know that the force at right end will be W-A or A-W
I'm solving other subject, but I thought this is simple maths
The force needed to make it stable ?
yes
because if not then the thingy will tip
Including rotation?
yes rotation too
Yes exactly. Unless A = W/2 you can't have a force on the right end that won't lead to nonzero torque
And keep it vertically stationary
if I say we have to check the direction of rotation after finding reaction at right side
it's not actual force but we are just considering it
Then you also look at the sum of torques
what if I write this. is this right
Is the sum of forces 0 ?
This is supposed to be a Galilean referential, in which the 3 laws of Newtonian mechanics apply. What does the first one say ?
Conclude
can I talk in terms of shear force or moment if you know those things
this is not mathematics, but I was trying to make it a simple question which would come under mathematics
It becomes math when you apply the principles and laws that turn your system into a bunch of equations
we are taking a section of a body, I drew only the section, and then we are finding the force at right side (from where we have cut the section)
it's fine anyway. close channel if I don't reply
You want the resulting force to be 0
So A+x - W = 0
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This may be a silly question, but I'm still confused with this kind of notations, I know this is the second derivative but why it's not written dy^2/dx^2 ?
just notation
$\frac{d^2y}{dx^2}=\frac{d(\frac{dy}{dx})}{dx}$
Pro_Hecker
we just write (dx)^2 as dx^2 for convienience
ok thanks, after all it boils down to notation
Having y² also seems to imply we are differentiating the function y²
Rather than differentiating the function y, twice
In the end it's just quirky Leibnitz notation
It looks like youre just multiplying the differential operators
thinking about it that makes sense
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Wassup guys
yo
it doesnt seem like people are very keen to help over here
pls respect the language
Did you send the question?
what are you struggling with?
No
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Whats the reason for r1 and r2? Because i know it equals 40 and r1 and r2 =20 but whats the reason
is RS angle bisector?
Whats that?
The dots suggest that it is I think
Yes
what do you need then?
The reason for r1 and r2
what is statement?
for what
a) <PRQ = 40
b)r1=20and r2=20
c)both a and b


