Does x grow faster than ln(x)?
#help-13
lyric jungle
crimson sedge
how to know that
im rlly bad at maths
For every real number
x > ln x
they solved it like that
how do i continue from where they started
crimson sedge
oh
so i take x above and x in den
then its =1 1
1*
thanks man
i have exams in 13 days
official exams gonna be super hard
so gotta be prepared
.close
leaden notch
Hey
leaden notch
is there a mathematical way to write a number with unknown digits where the math problem is dependent on using the digits?
crimson sedge
like $x_0+10x_1+10^2x_2+10^3x_3+\cdots$?
wraith daggerBOT
Toby
crimson sedge
eg, 523=500+20+3
crimson sedge
yup
leaden notch
oh man thanks! so is there like a short hand version for this?
like for decimals, i used to think i could do a.b and then if the number was 2.5 a = 2, b = 5
crimson sedge
you could write it like $x_2x_1x_0$ if you are very clear what you are saying in context
wraith daggerBOT
Toby
crimson sedge
just make it clear that you are concatenating and not multiplying
crimson sedge
yeah
neat meteor
How to show that the product of two independant standard normally distributed variables is chi squared distributed
neat meteor
Nk * Nl with Nk,Nl ~ N(0,1)
cedar kilnBOT
@neat meteor Has your question been resolved?
dire geode
lots of ways to do it
https://stats.stackexchange.com/questions/45854/pdf-of-the-product-of-two-independent-random-variables-normal-and-chi-square
neat meteor
Is there a way without solving hellish integrals or transformations?
dire geode
nope
rose stream
cedar kilnBOT
rose stream
How do i factor this
crimson delta
,w x^5-5x+3
wraith daggerBOT
crimson delta
you don't
rose stream
Um
dire geode
x(x^4 - 5) + 3 = 0
x(x^2 - sqrt(5))(x^2+sqrt(5))+3=0
rose stream
In the solution section of the pdf that we are given, the factors are (X²+x-1)(x³-x²+2x-3)
I want to know the method
How did they do it
Also is there any formula for factoring quintic eq?
crimson delta
there is no general formula to find the roots of a quintic only involving the common operations and roots
rose stream
Welp so breaking the eq down and adding, subtracting and dividing stuff in
dire geode
crimson delta
there are some algorithms to factor polynomials over the rationals but they aren't easy
in your case maybe they started with the factored form and multiplied it out
dire geode
Pretty sure Wolfie does some simpler version of that paper
crimson sedge
you could probably show that it has no integer linear factors by hand by using IVT. Then perhaps trying to expand a quadratic*cubic and comparing coefficients might give you insight into coefficients
rose stream
you could probably show that it has no integer linear factors by hand by using I...
💀 that would take an eternity
crimson delta
yeah that could be the way
otherwise feel free to look into stuff like the berlekamp-zassenhaus algorithm lul
crimson sedge
factoring polynomials is harder than factoring integers I think
crimson delta
according to wikipedia it actually isn't lul
crimson sedge
oh
rose stream
factoring polynomials is harder than factoring integers I think
Very
crimson delta
crimson sedge
interesting
crimson delta
maybe just over rationals/integers/finite fields. but still interesting
crimson delta
yeah sure but who comes up with that shit. that's the kinda stuff you come up with after you know the solution
crimson sedge
you need to be lucky and guess the factor
rose stream
you need to be lucky and guess the factor
Yep that could be a way, but in this case, its near impossible
crimson sedge
x^2+x+1 is the second degree two polynomial I would try if I was told that it had a degree two factor (the first being x^2+1)
rose stream
You gotta have a supa computer in your head if you wish to do this via trial and error
crimson sedge
indeed
rose stream
But anyhow, will ask teach to explain this out, wonder how his reaction will be after he's not able to solve his own problem.
Will also keep yall updated if i learn something new.
Thanks for your time guys!
.close
arctic marsh
Sorry people, in this exercise I have to find the conditions of existence of the function. What did I do wrong?
crimson sedge
.reopen
A table that is 8 centimeters long and a bug is positioned on the 3rd centimeter from the left edge. Itrandomly moves 5 centimeters either to the left or right. Calculate the average amount of moves the bug makes.
crimson sedge
The probability that it moves left or right is equal to 0.5
So I made a table of the amount of moves it makes and the probability and basically saw that for n moves the probability it achieves that many moves is 2^(-n)
I wrote it out till n=6 and then took the expected value and saw it was converging to 2 moves
Is this approach correct? Is there a more robust way to calculate the value of convergence?
crimson sedge
Is this approach correct? Is there a more robust way to calculate the value of c...
This is certainly the approach I would've went with. But please go further, how can you calculate the exact average amount of moves the bug makes with this?
hmm would I have to find the value of the sum of n/2^n as n approaches infinity?
Currently, the calculation for the probability looks like $\sum_{n=0}^{\infty}\frac{n}{2^n}$
wraith daggerBOT
Labyrinth
crimson sedge
Phew, I would've bet I messed up the LaTeX somewhere
Looks great!
Oh wait can I just use the convergence formula of a geometric series
a/(1-r)
crimson sedge
Oh wait can I just use the convergence formula of a geometric series
Yes... but no
Sadly, this isn't a geometric series
Oh
How can you tell?
Because it isn't of the form a+ar+ar^2+ar^3...
It would be if you replaced that n at the top with a constant
Ahhh right
Forgot about that
So then is it an arithmitic series?
Well, an arithmetic series grows linearily, no?
Does this one look linear?
Ahh right it does
It looks like a mix between an arithmetic series and a geometric series
so this is not one
Oh I didn't know that was possible lol
In theory you can have all sorts of crazy series
Well, I'm really not sure how to give you the intution for that, but there's formulas for these series too
I don't think they have a name though
👁️ 👄 👁️
The idea basically is to go from $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ then use some clever Math to transform it into the sum we just saw
wraith daggerBOT
Labyrinth
crimson sedge
Let me try my best to walk you through it
$\frac{n}{2^n}$ can be written as $n(\frac{1}{2})^n$
wraith daggerBOT
Labyrinth
crimson sedge
So, you want to transform $x^n$ into exactly that. First, do you know of anything that could bring that n to the front?
wraith daggerBOT
Labyrinth
crimson sedge
Yeah
And that is...?
...What's your idea?
@crimson sedge Uhm?
So if I set x=2 and then multiply it by (n/2^(2n))?
...What?
Oh wait sorry I misunderstood
I could take the natural log of x^n to get nlnx
Okay, why not
Basically, you want an n at the front, as a product, and another n above, as an exponent. Know of anything that could do that?
Something almost like $x^n$ => $nx^n$
wraith daggerBOT
Labyrinth
crimson sedge
oh I could take the derivative of x with respect to n
With respect to n?
Just to be clear, is it d/dx or d/dn?
with respect to x
d/dx
Yeah, alright, I was getting confused
So, now you have $nx^{n-1}$, how do you change that n-1 into a n?
wraith daggerBOT
Labyrinth
crimson sedge
divide by x so it ends up being n*(x^n/x^1)
crimson sedge
divide by x so it ends up being n*(x^n/x^1)
Okay, so, how do you turn $nx^{n-1}$ into $nx^n$?
wraith daggerBOT
Labyrinth
crimson sedge
Multiply it by x
Great
Now, it seems that you've already realised that x should be $\frac{1}{2}$, so try to turn $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ into the sum you had at the beginning
wraith daggerBOT
Labyrinth
crimson sedge
Do whatever you need to do on the left and remember to do the same on the right
Just did it on my side, and my answer matches with the internet's answer so it seems as if I did it right
I get x^(1-2n) / (x-1)^2
I first took the derivative, then multiplyed it by x, then divided it by x^2n
on both sides
crimson sedge
I first took the derivative, then multiplyed it by x, then divided it by x^2n
I don't understand that last part... Why divide by x^2n?
Oh, I see now. It would make sense if you were trying to achieve $\frac{n}{x^n}$
wraith daggerBOT
Labyrinth
crimson sedge
so I could make the x^n be 1/x^n
Yeah
But you could just keep $nx^n$ and choose x to be 1/2
wraith daggerBOT
Labyrinth
crimson sedge
Whatever, as long as it works
Oh true..
That's a smart move
*whispers* I only just learned all of this online and then started teaching it to you in a way that allows me to say I didn't just spit out the answer, this is something I found on the internet and would've never thought of myself
So, if you do it the "normal" way, you should get $\frac{x}{(1-x)^2}$
wraith daggerBOT
Labyrinth
crimson sedge
*\*whispers\** *I only just learned all of this online and then started teaching...
😂
crimson sedge
**Labyrinth**
Yeah
Now just plug in x=1/2 and you get your answer
Phew! What a long one
I learned something as well
Ahh I see it now, thank you so much!
Could you potentially apply this to any series
Like take a similar convergent series and manipulate it till it turns into your desired formula
.close
cedar kilnBOT
Available help channel!
vernal pike
$y^{\prime \prime }-2y^{\prime }+3y=6\sin 3x-6\cos 3x$
cedar kilnBOT
$y^{\prime \prime }-2y^{\prime }+3y=6\sin 3x-6\cos 3x$
wraith daggerBOT
AuHasard
vernal pike
What's the general solution for this?
$y_{h}\rightarrow r^{2}-2r+3=0\rightarrow r{1} =1-i\sqrt{2} ,\ r_{2}=1+i\sqrt{2}$
Is this correct?
steel stag
How did you get 1+-2i ?
vernal pike
$y_{h}=Ce\cos (2x)+C_{2}e\sin (2x)$
wraith daggerBOT
AuHasard
vernal pike
$r=\frac{2}{2} \pm \sqrt{\left( \frac{2}{2} \right)^{2} -3} \rightarrow r=1\pm \sqrt{-2}$
steel stag
But specifically about the values for r1 and r2, how did you get 2
wraith daggerBOT
AuHasard
wraith daggerBOT
Social Capital Gainer
vernal pike
yes true
wraith daggerBOT
AuHasard
vernal pike
Which one
steel stag
,,y(x)=e^{px}(A\sin(qx)+B\cos(qx))
wraith daggerBOT
Social Capital Gainer
steel stag
For $r=p\pm iq$
wraith daggerBOT
Social Capital Gainer
vernal pike
$y(x)=e^{x}(A\sin (\sqrt{2} x)+B\cos \left( \sqrt{2} x\right) )$
wraith daggerBOT
AuHasard
wraith daggerBOT
Social Capital Gainer
vernal pike
Now you need to try a particular integral of the form
Wdym?
steel stag
The part of the general solution that gives the right hand side of the differential equation
vernal pike
$6\sin 3x-6\cos 3x$
wraith daggerBOT
AuHasard
steel stag
**Social Capital Gainer**
That's what we need to get once this has been put into
,,y''-2y'+3y
vernal pike
once this has been put int
I don't understand
wraith daggerBOT
Social Capital Gainer
steel stag
This is the left hand side of the ODE
vernal pike
Yeah
steel stag
We got the complimentary function
When that is put into the ODE, it gives 0
We now need to find the other part of the general solution, which will produce the right hand side once it is put into the ODE
So for
,,y=a\sin(3x)+b\cos(3x)
wraith daggerBOT
Social Capital Gainer
vernal pike
**Social Capital Gainer**
are you saying this is y_p?
steel stag
Yes
You need to chose a function which is of the same form as the right hand side of the ODE
In this case we have sines and cosines which have 3x in their argument
So we need to put the most general possible combination of sines and cosines with 3x in the argument, as this does not change during differentiation.
So that will be
steel stag
,,y=a\sin(3x)+b\cos(3x)
This
There are other cases you need to know, such as if the right hand side of the ODE is exponential, linear or quadratic
vernal pike
$y^{\prime \prime }-4y-12y=3e^{5t}\rightarrow y_{p}=Ae^{5t}$
wraith daggerBOT
AuHasard
vernal pike
But since we're given the coefficients already, which is 6, why is this step necessary?
steel stag
Yes exactly, since the RHS is of the form of an exponential with argument 5t, then we try the most general exponential with argument 5t
This is because the right hand side of the ODE is not y_p
If you put 3e^(5t) into the left hand side, you will not produce 3e^(5t) again. Therefore 3e^(5t) is not a solution of the ODE
It needs to be a different function which produces 3e^(5t) after it has been differentiated and put into the left hand side of the ODE
We need to find what this function is, but it will be similar to the right hand side of the ODE,
So we try Ae^(5t) and solve for the value of A
I'll do that very quick to show you
,,y_p'=5Ae^{5t}
wraith daggerBOT
Social Capital Gainer
steel stag
,,y_p''=25Ae^{5t}
wraith daggerBOT
Social Capital Gainer
wraith daggerBOT
Social Capital Gainer
steel stag
Notice how now our choice of Ae^(5t) lets us cancel the e^(5t)
Since it didn't change during the differentiation
,,25A-20A-12A=3
wraith daggerBOT
Social Capital Gainer
steel stag
,,A=-\frac{3}{7}
wraith daggerBOT
Social Capital Gainer
vernal pike
$y_{p}=-\frac{3}{7} e^{5t}$
wraith daggerBOT
AuHasard
steel stag
Yes, now if you put that into the LHS of the ODE, you will get 3e^(5t), therefore this is a solution
Since it satisfies the differntial equation
wraith daggerBOT
AuHasard
steel stag
Yes, so for you, you want to get the first and second derivatives of that form of y_p, I'll leave you to do that
vernal pike
$y^{\prime }_{p}=3a\cos 3x-3b\sin \left( 3x\right)$
wraith daggerBOT
AuHasard
vernal pike
is this correct?
vernal pike
$y^{\prime \prime }=-9a\sin \left( 3x\right) +9b\cos \left( 3x\right)$
wraith daggerBOT
AuHasard
steel stag
Check your sign on the cos
vernal pike
$y^{\prime }=-\sin \left( x\right) \rightarrow y^{\prime \prime }=-\cos (x)$
wraith daggerBOT
AuHasard
steel stag
Mhm
vernal pike
$y^{\prime \prime }=-9a\sin \left( 3x\right) -9b\cos \left( 3x\right)$
wraith daggerBOT
AuHasard
vernal pike
$y^{\prime \prime }-2y^{\prime }+3y=\left( -9a\sin \left( 3x\right) -9b\cos \left( 3x\right) \right) -2\left( 3a\cos \left( 3x\right) -3b\sin \left( 3x\right) \right) +3\left( a\sin \left( 3x\right) +b\cos \left( 3xx\right) \right)$
wraith daggerBOT
AuHasard
steel stag
Yes
And this is equal to 6sin(3x)-6cos(3x)
Let me now tell you another principle we will use here
If we have
,,a\cos(x)+b\sin(x)= 2\cos(x)-5\sin(x)
wraith daggerBOT
Social Capital Gainer
steel stag
Then a must be 2 and b must be -5
This is called equating coefficients
It works because sin and cos are linearly independent functions, so we can't obtain an expression for
,, 2\cos(x)-5\sin(x)
wraith daggerBOT
Social Capital Gainer
steel stag
Using just Asin(x)
We need the cos(x) term to be able to represent this expression
Like in vectors
steel stag
**AuHasard**
Anyway, we want to rewrite this in a way that will let us get the a and b
How do you think we will do that?
vernal pike
equating to 6sin3x - 6cos3x
steel stag
Yes, so can you do that for me,
vernal pike
by equating the coefficients
steel stag
Get two expressions involving a and b, which will be equal to 6 and -6
vernal pike
$\begin{gathered}\ -9b\cos \left( 3x\right) +3b\cos \left( 3x\right) =-6b\cos \left( 3x\right) \ -9a\sin \left( 3x\right) +3a\sin \left( 3x\right) =-6a\sin \left( 3x\right) \ -2\left( 3a\cos \left( 3x\right) \right) =-6a\cos \left( 3x\right) \ -2\left( -3b\sin \left( 3x\right) \right) \end{gathered}$
wraith daggerBOT
AuHasard
steel stag
I'm not quite sure what's happening here tbh
vernal pike
$-6b\cos \left( 3x\right) -6a\sin \left( 3x\right) -6a\cos \left( 3x\right) +6b\sin \left( 3x\right) =6\sin \left( 3x\right) -6\cos \left( 3x\right)$
wraith daggerBOT
AuHasard
vernal pike
simplified
vernal pike
group terms
steel stag
Yes
Factor out the cos(3x) and sin(3x)
Remember to include y as well in the left hand side
vernal pike
$-6b\cos \left( 3x\right) -6a\sin \left( 3x\right) -6a\cos \left( 3x\right) +6b\sin \left( 3x\right) =6b\left\boxed{( -\cos \left( 3x\right) +\sin \left( 3x\right) \right)} -6a\left\boxed{( \sin \left( 3x\right) +\cos \left( 3x\right)} \right)$
wraith daggerBOT
AuHasard
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
steel stag
We actually need to do it the other way around
Like this
,,(-9a-2(-3b)+3a)\sin(3x)+(-9b+6a+3b)\cos(3x)=6\sin(3x)-6\cos(3x)
All the sin coefficients and all the cos coefficients
wraith daggerBOT
Social Capital Gainer
steel stag
group terms
Is that making sense
vernal pike
$\boxed{(-9a-2(-3b)+3a)}\sin(3x)\boxed{+}\boxed{(-9b+6a+3b)}\cos(3x)=6\sin(3x)-6\cos(3x)$
wraith daggerBOT
AuHasard
vernal pike
I didn't understand how you got these
steel stag
So I used
$y^{\prime \prime }-2y^{\prime }+3y=\left( -9a\sin \left( 3x\right) -9b\cos \left( 3x\right) \right) -2\left( 3a\cos \left( 3x\right) -3b\sin \left( 3x\right) \right) +3\left( a\sin \left( 3x\right) +b\cos \left( 3x\right) \right)$
wraith daggerBOT
Social Capital Gainer
steel stag
I factored out the cos(3x) from wherever I could and also the sin(3x)
There are 3 cos and 3 sin terms once the brackets are expanded
Just group all the sines together and all the cosines together
vernal pike
Ok I've been on this question for one hour now, better if I work with easier problems.
steel stag
We are almost done, doing this over discord is slow, I'm sorry
It's just a lot of algebra, it isn't any harder than the last one
vernal pike
$-6b\cos \left( 3x\right) -6a\sin \left( 3x\right) -6a\cos \left( 3x\right) +6b\sin \left( 3x\right)$
wraith daggerBOT
AuHasard
vernal pike
$6\sin(3x)-6\cos(3x)$
wraith daggerBOT
AuHasard
vernal pike
we want to get that
steel stag
Yes
$\boxed{(-9a-2(-3b)+3a)}\sin(3x)\boxed{+}\boxed{(-9b-6a+3b)}\cos(3x)=6\sin(3x)-6\cos(3x)$
So equating coefficients, we get
-9a+6b+3a=6
And
-9b-6a+3b=-6
These are two simultaneous equations we can solve for a and b
And thats it
vernal pike
$$-6b\cos \left( 3x\right) -6a\sin \left( 3x\right) -6a\cos \left( 3x\right) +6b\sin \left( 3x\right)$$ $$\rightarrow$$
$$y = \left( -6b-6a\right) \left( \cos \left( 3x\right) \right) -\left( 6a-6b\right) \left( \sin \left( 3x\right) \right)$$
wraith daggerBOT
Social Capital Gainer
steel stag
Yes, but you have forgotten to use y
wraith daggerBOT
AuHasard
steel stag
Remember, we need to put y'', y' and y into the ODE
steel stag
Ok
So now when we equate coefficients with 6sin(3x)-6cos(3x), we get
-6a+6b=6
And
-6a-6b=-6
vernal pike
$\underbrace{\left( -6b-6a\right) }{should\ equal\ -6} \left( \cos \left( 3x\right) \right) -\underbrace{\left( 6a+6b\right) }{should\ equal\ 6} \left( \sin \left( 3x\right) \right)$
wraith daggerBOT
AuHasard
vernal pike
$$-6b-6a=-6\rightarrow -6\left( a+b\right) =-6\rightarrow a+b=1$$
$$-6a+6b=6\rightarrow 6\left( -a+b\right) =6\rightarrow -a+b=1$$
wraith daggerBOT
AuHasard
vernal pike
What did I do wrong here?
steel stag
Uh, something has gone off here, a tip for these sin and cos questions use the abbreviations c and s in your working to save space. It doesn't matter here, but it might help
vernal pike
ok
steel stag
Lemme check
vernal pike
$y^{\prime \prime }-2y^{\prime }+3y=6\sin 3x-6\cos 3x$
wraith daggerBOT
AuHasard
vernal pike
$$-6b\left( c\left( 3x\right) \right) -6a\left( s\left( 3x\right) \right) -6a\left( c\left( 3x\right) \right) +6b\left( s\left( 3x\right) \right)$$
$$\rightarrow$$
$$y=\left( -6b-6a\right) \left( c\left( 3x\right) \right) -\left( 6a-6b\right) \left( s\left( 3x\right) \right)$$
wraith daggerBOT
AuHasard
vernal pike
$$\begin{gathered}y=\left( -6b-6a\right) \left( c\left( 3x\right) \right) -\left( 6a-6b\right) \left( s\left( 3x\right) \right) =6\left( s\left( 3x\right) \right) -6\left( c\left( 3x\right) \right) \ y=\underbrace{\left( -6b-6a\right) \left( c\left( 3x\right) \right) }{=-6} -\underbrace{\left( 6a-6b\right) \left( s\left( 3x\right) \right) }{=6} \ \ \ \ \ \ \end{gathered}$$
wraith daggerBOT
AuHasard
vernal pike
$y=\begin{cases}\left( -6b-6a\right) \left( c\left( 3x\right) \right) =-6&\ \left( -6a+6b\right) \left( s\left( 3x\right) \right) =6&\end{cases}$
wraith daggerBOT
AuHasard
vernal pike
or wait
steel stag
Ye, but because we equated coefficients, we dont have the c and s
wraith daggerBOT
AuHasard
steel stag
Yes
vernal pike
$y=\begin{cases}-6(b+a)=-6&\rightarrow b+a=1\ 6(-a+b)=6&\rightarrow -a+b=1\end{cases}$
wraith daggerBOT
AuHasard
steel stag
Yep
vernal pike
$\ b=1-a\rightarrow -a+\left( 1-a\right) =1\rightarrow -2a=0\rightarrow a=0$
$\rightarrow b=1$
wraith daggerBOT
AuHasard
steel stag
Put a and b into the expression for y_p and add together y_c and y_p
To get the general solution
vernal pike
Can you show me
vernal pike
asin3x - bcos3x
vernal pike
$y_{p}=a\sin \left( 3x\right) -b\cos \left( 3x\right) \rightarrow a=0,\ b=1\rightarrow -\cos \left( 3x\right)$
steel stag
y_p=cos(3x)
vernal pike
b not 6
wraith daggerBOT
AuHasard
steel stag
Yes, but no minus, it was a plus in what we originally used
vernal pike
why plus
steel stag
**AuHasard**
Here
vernal pike
$y^{\prime \prime }-2y^{\prime }+3y=6\sin 3x-6\cos 3x$
wraith daggerBOT
AuHasard
vernal pike
Don't we just want to replace the coefficients with variables?
steel stag
It's a minus in the RHS of the ODE, but that doesn't matter
vernal pike
And leave the signs intact?
steel stag
Yes, just like we found yp for the exponential one was -3/7e^(5t)
We started with Ae^(5t)
And found A to be negative
It is easier to add everything and a or b will turn out negative if they need to be
steel stag
That's ok!
It's because we need to have the most general possible sum of sin and cos
Which is asin(3x)+bcos(3x)
The a and b end up being set to what ever they need to be
Like with vectors
vernal pike
I see, so b = -6 in the original DE?
steel stag
You have a unit vector along the x and y axis
So b is still 1
In the original expression for yp that we used which was
vernal pike
$y^{\prime \prime }-2y^{\prime }+3y=6\sin 3x-6\cos 3x$
wraith daggerBOT
AuHasard
steel stag
asin(3x)+bcos(3x)
vernal pike
if we uppose that right side is asin3x+bcos3x
would b then be = -6?
written in the form asin3x+bcos3x
steel stag
Yes, if you assume the right side of thr equation is the solution and put it in, it will almost never satsfy the equation
To be clear
We are looking for a function y_p such that when it is substituted into the DE, it will produce the right side
To find this y_p, we try a function which we know will work given the form of the right side of the equation
Since the right side of the equation involves sines and cosines
We know that y_p must also be composed of sines and cosines
Since the right side of the DE has 3x in the argument of the sines and cosines, then we also need to put 3x in our solution so that we can end up being able to equate coefficients
You can't equate sin(2x) coefficients with sin(3x) coefficients
They need to be the same
So we try
asin(3x)+bcos(3x)
And put that into the DE, and find the a and b we need
vernal pike
To find this y_p, we try a function which we know will work given the form of the right side of the equation
I.e. substituting the coefficients of 6sin3x - 6cos3x
We know that y_p must also be composed of sines and cosines
Yes
and b was 1
but now we want y = yh + yp?
steel stag
Yes exactly
vernal pike
What was our yh…
steel stag
,,y_h=e^x(A\sin(\sqrt{2}x)+B\cos(\sqrt{2}x))
wraith daggerBOT
Social Capital Gainer
steel stag
We know that putting this yh into the DE will give 0.
I want you to put yp=cos(3x) into the left side of the DE and see what you get
vernal pike
**Social Capital Gainer**
yes
vernal pike
I want you to put yp=cos(3x) into the left side of the DE and see what you get
of this? #help-13 message
vernal pike
Why would we do that
steel stag
To check if our solution is correct
vernal pike
,w second derivative of cos(3x)
wraith daggerBOT
vernal pike
,w derivative of cos(3x)
wraith daggerBOT
vernal pike
,w prove -9cos(3x) - 2(-3(sin(3x))+3(cos(3x)) = 0
steel stag
Aaaaaa
steel stag
Wolframalpha is giving funny info
Nope it is correct
It gave everything except simplifying the answer which is what we wanted
vernal pike
there
steel stag
Yeah -9cos(3x)+6sin(3x)+3sin(3x) = 6sin(3x)-6cos(3x)
And that is the right hand side of the DE
So our solution for yp is correct
vernal pike
it says -9cos(3x)-6(cos(3x) lol
steel stag
Eh whatever, we are correct anyway
We don't need wolframalpha for that
-9cos(3x) - 2(-3(sin(3x)) +3(cos(3x) is not equal to 0
It is equal to the right side of the DE which is what we wanted
wraith daggerBOT
steel stag
$y^{\prime \prime }-2y^{\prime }+3y=6\sin 3x-6\cos 3x$
wraith daggerBOT
Social Capital Gainer
steel stag
We wanted a y which works in this eqn and we found it
Yay
,,y=e^x(A\sin(\sqrt{2}x)+B\cos(\sqrt{2}x))+cos(3x)
wraith daggerBOT
Social Capital Gainer
steel stag
The cos(3x) produces the right hand side of the DE, and the other thing produces 0
So together they are the most general solution
vernal pike
,w simplify -9cos(3x) - 2(-3(sin(3x))+3(cos(3x))
wraith daggerBOT
vernal pike
Should I try simplify it myself?
steel stag
Yeah, absolutely
steel stag
If you want
steel stag
Yaay
vernal pike
i have to go now, but thanks for the elaborate explanation
cedar kilnBOT
@vernal pike Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
slim jackal
slim jackal
<@&286206848099549185>
cedar kilnBOT
@slim jackal Has your question been resolved?
south dawn
I hope someone can help. It seems that something is missing from this word problem. Is that all the information given?
I guess she needs to earn extra 5000 plus the tax, lol
cedar kilnBOT
cedar kilnBOT
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minor ginkgo
Here is my question which has proven to be a tough challenge for me. Let's say I wanted to place a certain number of tiles around a circle. I know the radius of the circle as well as the angle between each tile edge. For this example, I will say I have a circle with a radius of 1 foot and tiles at 30 degree increments (12 tiles). What I want to find is the approximate length of each tile to form as close to a square as possible. I thought that using the chord for one side of a trapezoid would help?
minor ginkgo
Here is a diagram
cedar kilnBOT
@minor ginkgo Has your question been resolved?
cedar kilnBOT
@minor ginkgo Has your question been resolved?
lament kraken
Here is a diagram
cant you just do area of the sector - area of triangle
minor ginkgo
Could you explain?
lament kraken
uhh nvm its more complicated than i thought
cedar kilnBOT
@minor ginkgo Has your question been resolved?
vast wolf
Find how many arrangements of the letter of the word UNIFORM are possible of the m is somewhere to the right of u.
buoyant latch
this feels kinda bruteforcey to me
vast wolf
Yeah it is a little
ember python
this feels kinda bruteforcey to me
Idk not completely
vast wolf
It says it's half if that helps
buoyant latch
i dont like it when you have to bruteforce
buoyant latch
wait shouldn't it just be half of 7c2
fair geyser
yes out of 5040 arrangements, you take half
buoyant latch
half the time u is on the left of m
vast wolf
Nope
buoyant latch
half its the other way round
high coyote
7!/2
buoyant latch
oh
ember python
wait shouldn't it just be half of 7c2
🤦♂️🤦♂️🤦♂️🤦♂️
vast wolf
Yeah haha
ember python
Somebody else came up with this idea on a problem someone else posted like a month ago
vast wolf
Cheers categorist
high coyote
7! arrangements where in half of them M to the right of U, so 7!/2
buoyant latch
no i was gonna do 7c2 and then do 5!
vast wolf
Ohh
buoyant latch
cos 5! ways to arrange the rest 5 letters
buoyant latch
with a /2 somewhere in there
vast wolf
But yeah just using the mental approach is faster
buoyant latch
but 7!/2 is even better than 7c2/2*5!
cedar kilnBOT
cedar kilnBOT
Available help channel!
grave stone
can anyone give me a hand with 4 quickly
cedar kilnBOT
can anyone give me a hand with 4 quickly
wooden harness
what
this question doesn't seem to make too much sense for me
oh wait nvm it does now
if the question means "for ever increase in price by $0.05, they sell 5 less pucks per week"
then it's solvable
@grave stone does that help a little?
dapper steppe
(6000-5x)(2+0.05x)=52000
grave stone
The problem with this is that I have no clue what to do when I get my equation set up
wooden harness
alright so
grave stone
$\left(6000-5x\right)\cdot \left(2+0.05x\right)=52000$
wraith daggerBOT
Golden
grave stone
Yes I keep getting that
wooden harness
well it's the right equation
dapper steppe
and then you just look for the roots
wooden harness
^
grave stone
zeros?
wooden harness
yea
dapper steppe
yes
wooden harness
using the quadratic formula might help
though if you expand everything out and multiply evewrythign by 4
the coeff of x^2 will be 1
maybe ther'es a neat way to factor it
oh yea
(x-1000)(x-160) = 0
well more precisely it's -1/4 (x-1000)(x-160) = 0
grave stone
so the zeros are (1200,40)
wooden harness
nah it should be x=1000 or x=160
wooden harness
nah I factored it
you can quadratic formula it of course
but the number's a bit big so I'm scared
anyway
You have $(6000-5x) (2+\frac{x}{20}) - 52000 = 0$. After expanding it out and factoring everything, you will have $-\frac{1}{4}(x-160)(x-1000) = 0$
wraith daggerBOT
Azzurala
wooden harness
So either $x-160 = 0$ or $x-1000 = 0$
wraith daggerBOT
Azzurala
grave stone
just wondering whrer is the x/20 coming from
wooden harness
0.05x = 1/20 x
grave stone
So I expand the brackets and simplify it into factored form>?
wooden harness
yea
or you can quadratic formula it
I haven't tried it, but the numbers looks big
and they might get unwieldy
not that they might, they will get unwieldy
grave stone
my brain is fried from trig so do you have to make it to standard form and then use quadratic formula?
wooden harness
yea
grave stone
$(-100x^2+290x-40000)=0$ is what I simplified it into. I take what out of it now?
wraith daggerBOT
Golden
wooden harness
uh
wraith daggerBOT
Golden
wooden harness
idk how you got 100x^2
grave stone
you said expand it through no?
wraith daggerBOT
Golden
grave stone
**Azzurala**
so what one is what now
wooden harness
and the bit inside should look recognisable enough to factorise
ok let me
do the working out on paper
@grave stone
grave stone
I got it!
wooden harness
nice
grave stone
yes
wooden harness
so there are 2 possibilities
either the price was increased by 0.05(160) or the price was increased by 0.05(1000)
and the number of sales were either 6000-5(160) or 6000 - 5(1000)
grave stone
is itguessing game or is there logic to this
wooden harness
**Golden**
how did you make this question in the first place
grave stone
ive gotten everything but that
I thought you were askin wehre the question came from
grave stone
I understand the equation but someone else beat me to the equation
wooden harness
ah
yea then
you can see taht in that equation
(amount sold) * (price) = revenue
amount sold = 6000 - 5x
price = 2 - 0.05x
grave stone
2.00
wooden harness
so you just plug in x=160
you get 1 set of solutions
plug in x=1000
you get another set of solutions
there are 2 possible cases here
wooden harness
happy to help
grave stone
.close
regal fog
How do I transform the exponential form (5³)⅐ into radical form?
crimson sedge
How do I transform the exponential form (5³)⅐ into radical form?
$(5^\frac{3}{7})?$
wraith daggerBOT
sqrt(-1) is approx -30
bitter reef
How do I transform the exponential form (5³)⅐ into radical form?
$k^{\frac{a}{b}} = \sqrt[b]{k^a}$
wraith daggerBOT
Doggo
regal fog
Thanks.
cedar kilnBOT
@regal fog Has your question been resolved?
stray otter
y=3x^2(2-x)
Find the values of (d^2)y/d(x^2) when dy/dx=0
cedar kilnBOT
y=3x^2(2-x)
Find the values of (d^2)y/d(x^2) when dy/dx=0
@stray otter Has your question been resolved?
stray otter
<@&286206848099549185>
lyric jungle
y=3x^2(2-x)
Find the values of (d^2)y/d(x^2) when dy/dx=0
Calculate dy/dx
And find values of x that satisfy dy/dx = 0
lyric jungle
And?
stray otter
x=0?
lyric jungle
Yes.
stray otter
ok
lyric jungle
So what's the value of dy^2/dx^2 at those values of x?
stray otter
that would be 12-18x so do we just substitute the answers we just got into our equation?
lyric jungle
Yes.
hallow breach
simplify the expression 1+1.0075+1.0075^2+1.0075^3+...+1.0075^59
correct to 2 decimals
shut reef
$a^x+a^y=a^{x+y}$
hallow breach
how do i apply that to this expression
elder pewter
Use the geometric series sum to n formula
elder pewter
$a^x+a^y=a^{x+y}$
Also what?
cosmic steppe
Wait what
cedar kilnBOT
hallow breach
wait but how to i simplify it im confused guys
simplify the expression 1+1.0075+1.0075^2+1.0075^3+...+1.0075^59
correct to 2 decimals
elder pewter
I gave you a hint
hallow breach
idk how to simplify it
cedar kilnBOT
Available help channel!
elder pewter
Can you read?
cedar kilnBOT
Can you read?
elder pewter
I gave you a hint
hallow breach
ok sorrry omg
elder pewter
Use the geometric series sum to n formula
Right here
hallow breach
why do u have to be so aggressive im just asking for some help
but im sorry
i struck a nerve with you
lilac totem
Lol the squidward pfp suits you @elder pewter
elder pewter
Because I said use the hint
And you said idk how to simplify
Like that's not a response
Do you understand the hint I gave?
If not I can help
But this requires you to try
hallow breach
oh ok i get it now im sorry
i read the question wrong
im a dummy
sorry for tilting you
@elder pewter
inner quest
phi is the standard normal distribution
upper abyss
In terms of θ1 or θ2?
I imagine Φ is the normal cdf function, which means its derivative is just the normal pdf
e^(blah²)
inner quest
both
inner quest
I imagine Φ is the normal cdf function, which means its derivative is just the n...
yes it is the standard normal cdf but i dont see how to do it
its the derivatvie of both terms of ln(phi(theta_1/sqrt(theta_2)))
upper abyss
what do you mean exactly? I would really appreciate it 🙂
That was me referencing the normal pdf formula. Know that, for sure, as that's the derivative of Φ(x)
This question is all chain rule, otherwise.
inner quest
I really cant figure the method out even with chaining
for the first one i.e theta_1 what would you say it results to?
upper abyss
Here's two questions you need to consider:
- Are you comfortable with the chain rule?
- Can you take the derivative of ln and Φ alone?
inner quest
let me answer the second one first: d/dx(ln(x)) is just 1/x
for the cdf it is just the density i guess
First question is a long time ago, I usually dont have to deal with chaining but have been for the last 2 hours..
inner quest
Here's two questions you need to consider:
- Are you comfortable with the chain ...
I hope that is okay; I really did not think I would have to deal with derivatives in this course.
inner quest
Here's two questions you need to consider:
- Are you comfortable with the chain ...
Please let me know if that is something you can assist me with
native cosmos
cedar kilnBOT
elder pewter
First add fractions
native cosmos
ahan
i could find the roots in complex form but ive to do it without calc
@elder pewter
elder pewter
Now use the relationship between coefficients and roots
native cosmos
Now use the relationship between coefficients and roots
a + b = - coeff of x / coeff / x^2?
ty
native cosmos
a + b = - coeff of x / coeff / x^2?
this would mean a + b = -2/3
but we need the value of this
cedar kilnBOT
@native cosmos Has your question been resolved?
cedar kilnBOT
@native cosmos Has your question been resolved?
dense girder
Help
quartz vale
first find a quadratic equation with these zeroes
dense girder
Wym
quartz vale
a quadratic equation, with the zeroes 0 and 0.6
dense girder
Ya
dense girder
Aos is 3
dusty hazel
So the x coordinate of the vertex isn't exactly hard to figure.
dense girder
(x - a)(x - b) is the simplest quadratic equation with zeroes a, b
O did that
dusty hazel
0 and 0.6, what lies halfway through the two numbers?
dusty hazel
...
dusty hazel
Bruh.
dense girder
Sorry
dusty hazel
What lies halfway between the two.
dense girder
3
dense girder
3
dense girder
How would i fing a
dense girder
Not sure
quartz vale
aos?
dusty hazel
Oh a(x-h)^2 + k?
dense girder
Ya
quartz vale
I nedd to put into vertex form
we know, find an equation
quartz vale
oh alr
dusty hazel
Still writing a general expression will help regardless.
You can proceed from there, and it's simpler so yeah.
crimson sedge
The zeros ar 0,0 and 0,6
considering that the solution is in terms of a u can find the vertex by the taking the mean 6+0/2=3 also as one of the roots is 0 notice that the intercept is 0
there is no legimitate solution
which is not in terms of a
most books assume a=1
kind of
native cosmos
how do you find ab
thats my question
crimson sedge
most books assume a=1
actually i am confused as well now 
cedar kilnBOT
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cedar kilnBOT
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unkempt aspen
cedar kilnBOT
unkempt aspen
Hey I'm having some difficulty attacking this last one.
How would I do this without L'Hopital's rule?
applying it twice gives me 16/(1/x^2)
is that the correct answer?
plucky pawn
you can divide everything by x
unkempt aspen
Well let's say I divide by the highest x which is what you normally do
you get 4 on top, 3/x and lnx/x^2
I don't know how to reduce lnx/x^2
plucky pawn
no it must be on the bottom highest x
unkempt aspen
is that actually the rule?
I wasn't aware of that, just that it was the highest x
that would still leave me with what to do about lnx/x
balmy apex
applying it twice gives me 16/(1/x^2)
how did you get that result?
unkempt aspen
An answer of 1/infinity kind of doesn't make sense because the exponential growth of nlogx is not that fast?
unkempt aspen
how did you get that result?
L'Hopital rule twice
plucky pawn
didnt you want to do it without lhopital
balmy apex
<@&268886789983436800>
plucky pawn
XD
unkempt aspen
wait I need the product rule don't I
calm sierra
ty
balmy apex
wait I need the product rule don't I
indeed
plucky pawn
is that symbolya
plucky pawn
nice nice! I wish I could use it