This is more an exact question
#help-38
lunar parcel
lunar parcel
Okay I see now, so in the end it is actually continuous because the domain (0; x]
lunar parcel
So maybe I just need to say because the domain is from 0 than it continuous, otherwise it wouldn't be
tough iron
Yeah
lunar parcel
Okay, but where i get the domain from? Do I say that i just take it from integral notation, does it actually indicates the domain?
so from the integral form 0 to x I can say that the domain should be [0;x], but because 1/0 cannot be so we say that domain is (0;x]?
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@lunar parcel Has your question been resolved?
lunar parcel
<@&286206848099549185>
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soft granite
Hey guys i need help with this one:
What is the greatest power of 15 divisible by 41 x 42 x...x176?
soft granite
i have some calculations done but idk if its right, those numbers are so high that i cant prove it with calculator
vagrant prism
Hey guys i need help with this one:
What is the greatest power of 15 divisible b...
is it $41\cdot 42 \cdots 176$?
solid kilnBOT
artemetra
soft granite
sorry...
vagrant prism
use \
soft granite
41 x 42 ... x176
vagrant prism
**artemetra**
that's what i meant here lol
soft granite
my asnwer is 15^34
lunar parcel
literal tears of blood
vital quail
Q?
soft granite
What is the greatest power of 15 divisible by 41 x 42 x...x176?
lunar parcel
I really don't know how to do it but I would say do something like 176! - 40! so you would get what is the value of $41\cdot 42...176$ and then maybe $15^x = 176! - 40!$ so it should be divisible? but im really not sure, dont trust me bro
soft granite
my solution is 15^34 BUT, i cant prove it bc those numbers are so high
solid kilnBOT
Džiugas
soft granite
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real sparrow
Need help with this quaratic equation question. I tried making the cases but it aint working
onyx heart
Note that -|b| < 0 and |b| > 0
onyx heart
This implies -|b| < |b|
wet shard
yh
next briar
The way I see it is if a < -|b| as the question says, then obviously a + |b| < 0, and a + |b| is more positive than a - |b| but the first one is less than zero so the second one has to be as well… do you follow that?
Kinda following from what Robin is saying
wet shard
ok ok
next briar
Because the value of |b| is positive and a + |b| < 0 so it must follow that a - |b| < 0, it’s MORE negative if you think about it I guess!
onyx heart
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wet shard
.close
wary mist
Question, for which a∈ℝ is the linear linear system of equationslinear system of equations
a) solvable
b) clrearly solvable (only one solution and not infinite solutions)
For following:
wary mist
I figured out that if a=1 there are infinite solutions and that -2 is not solvable, I am not sure about other numbers tho
Wolframalpha shows me this but I am not sure if I understand why it behaves that way and how to explain it in my own words
I tried to use Gaussian Method to solve this with the a's but it seems it won't work that way
maiden hare
Gaussian elimination should work but you can just notice that
ax+y+z = x+ay+z = x+y+az
Taking each equation and simplifying you get
(a-1)x = (a-1)y = (a-1)z
So a = 1 gives you 0 = 0 = 0, which means x, y, z can take any value
Any other value for a means x=y=z
Now it's only a matter of solving ax+x+x=1
wary mist
So far I understand this part. Where it gets funky it when I try to find a number with a definitive solution, which seems to be everything besides 1 and -2 (even pi lol)
maiden hare
Or (a+2)x=1
Yes, every value for a except -2 will result in (a+2)x=1 having a single solution
wary mist
I get it now, thank you. It is like a mistake in a movie or a painting. Once you see it, you can't unsee it
maiden hare
wraith hinge
Need help
nimble stone
nimble stone
are you able to graph f(x)
wraith hinge
Yes
nimble stone
show me
wraith hinge
.close
sacred sandal
How do I find the limit of the sequence as n approaches infinity? I can’t figure out how to do this one
sacred sandal
I tried starting with u substitution with u being the square root of x but I don’t really understand how to do it
zinc ginkgo
Bound the integrand from above or below by something that only depends on n then take the limit
sacred sandal
Wait so are you saying to split it into 2 integrals, I’m confused
zinc ginkgo
Wait so are you saying to split it into 2 integrals, I’m confused
No I'm not
If you want to show a_n diverges, you bound it from below by something that diverges
And similarly for convergence
zinc ginkgo
So you set n to a value?
No
zinc ginkgo
Wait can’t I just do a comparison test for this one?
Try it
sacred sandal
Ok I did the comparison test and I’m pretty sure that it diverges
Here also I have another question
So it’s the integral of 1/(1+x^2) from n to infinity
And so I end up getting arctan(infinity) - arctan(n)
So it’s pi/2 - arctan(n)
Is it just 0 because n is also going to infinity?
fringe plaza
Calculate the result of
lim ( 1-cos(3x) / x^2 )
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Calculate the result of
lim ( 1-cos(3x) / x^2 )
fringe plaza
Looks like arabian to me
I mean, I don't even know how to start solving this even though I'm familiar with the concept of limit
crude blaze
wraith hinge
What’d I do that was rude?
fringe plaza
Nah
wraith hinge
I’m asking for help not the answer
west whale
what is the triple integral that defines the volume that is inside z=x^2, y=1-z^2, y=z-1 and x>=0 y>=0
west whale
how do i make it so the triple integral ends with dz dy dx (or dz dx dy)
i tried this, but its wrong
answer should be 68/105 u^3
ed plzz help
:c
brisk solar
I’m trying
west whale
thanks :D
west whale
i figured it out when it starts with dx and dy but dz idk
west whale
my homework says you can
west whale
it says i can either do it with dzdydx or dzdxdy
brisk solar
Oh
west whale
i don't see how that helps us tho
brisk solar
z-1 to 1-z^2
west whale
what if we split it in 2 integrals
brisk solar
Yeah I think that’s what ur meant to do
brisk solar
What’s the z value where the blue and green intersect
It’s 1 ok
Wait is it I’m having a stroke
Can you do the question if u can choose the order
Or is it just the if its dzdydx or dzdxdy that’s an issue
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@west whale Has your question been resolved?
west whale
Or is it just the if its dzdydx or dzdxdy that’s an issue
it says i must do it like this ^^^
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gloomy sphinx
Hey how would I go about solving this?
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Hey how would I go about solving this?
zinc ginkgo
limit test
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@gloomy sphinx Has your question been resolved?
gloomy sphinx
So would it converge to 5/64?
zinc ginkgo
So would it converge to 5/64?
no
if a series converges, then the limit of the terms must go to zero
gloomy sphinx
Oh ic ty. Is that test the one that can't confirm convergence?
gloomy sphinx
Thought so, thank you. So what test would you have to use on this one who's limit equals 0
gloomy sphinx
Large n?
zinc ginkgo
n=4,5,6,7, ..... the first finite number of terms don't affect the convergence of the series
you only care about when n is large
whatever definition of "large" your mind thinks is probably fine
gloomy sphinx
Wait so you ignore the -3n and do an integral test for n to the -2?
zinc ginkgo
you could yes
gloomy sphinx
So it converges to 1/4?
Wait is it just the geometric test that can find the sum of a series?
zinc ginkgo
So it converges to 1/4?
the integral test doesn't tell you the value the series converges to
zinc ginkgo
Wait is it just the geometric test that can find the sum of a series?
no
geometric series is of the form sum a * r^n for fixed a and r, and index n
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Available help channel!
errant bronze
How does lim k->infinity of frac{(k+1)e^(-k+1)}{ke^-k} equal 1/e
edgy cedar
Well try simplifying the expression first
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@errant bronze Has your question been resolved?
errant bronze
I did here #help-38 message
I do not understand properties of numbers ig bc somehow this is supposed to break down and cancel out to 1/e
I keep getting e(k+1)/k with what I did
meaning I messed up somewhere on the simplification
edgy cedar
Ah
edgy cedar
Here the numerator should be e^{-k-1} right
errant bronze
ye
edgy cedar
So you should get (k+1)/ek
errant bronze
wtf
edgy cedar
Which should be clear is 1/e as k tend to inf
errant bronze
You're saying e^{-k+1} = {k+1}/{e^k}
errant bronze
$$\frac{e^{-k+1}(k+1)}{ke^k} = \frac{k+1}{ek}$$
errant bronze
???? how is it -k-1
edgy cedar
Because it’s -(k+1)
edgy cedar
a_k = 9ke^-k
a_(k+1) = 9(k+1)e^{-(k+1)}
Literally replacing k with k+1
U with me?
errant bronze
Okay I
$$\frac{9(k+1)e^{-(k+1)}}{9ke^{-k}} = \frac{(k+1)e^{-k-1}e^k}{k}
Why didn't it go ☠️
edgy cedar
$$\frac{9(k+1)e^{-(k+1)}}{9ke^{-k}} = \frac{(k+1)e^{-k-1}e^k}{k}$$
solid kilnBOT
$Pure$
edgy cedar
Right yeah
edgy cedar
Lhop is overkill but sure
errant bronze
okay how wwould you do it
knotty oriole
(k+1)/k=1+1/k
errant bronze
forgot the e
errant bronze
ye
knotty oriole
Oh yeah you can just pull it out of the limit tho
errant bronze
can you throw in the e please i'm getting lost in the sauce
(k+1)/ke
turns into e(1+1/k)
?
edgy cedar
Take 1/e out of limit
edgy cedar
Same thing it’s just
errant bronze
TY AND SOLVED 💙
edgy cedar
Terrible latex but you get the point
errant bronze
How does $$(\frac{n}{n+1})^n = \frac{1}{e}$$
solid kilnBOT
🌸 Katsune
edgy cedar
Ah sorry misread
errant bronze
Ye my latex just now assuredly is correct, my problem is understanding how that is the case
edgy cedar
For this you’d need to rewrite that thing as e^(log…)
errant bronze
what
edgy cedar
Like $$\left[\frac{n}{n+1} \right]^n = e^\log{\left[\frac{n}{n+1} \right]^n}$$
solid kilnBOT
$Pure$
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
errant bronze
Why is this? I have never seen this before.
I do not understand this property of e or log exists?
edgy cedar
Well you know that e^log(x) = x
You know there might be an easier way to do this
Let me think for a bit
errant bronze
Is e^ln(x) = x as well?
knotty oriole
When you rewrite your limit in this way you can apply continuity of e^x
edgy cedar
$$\lim e^{n\ln{(\frac{n}{n+1})}} = e^{\lim n\ln{(\frac{n}{n+1})}}$$
edgy cedar
When you rewrite your limit in this way you can apply continuity of e^x
You can just consider the limit of the exponent as doordootee said
And this is a common limit
knotty oriole
Like doing lim(e^blahblah)=e^lim(blahblah)
edgy cedar
I know my latex skills need improvement 😔
errant bronze
This looks cracked
not intuitive at all
looks like some kind of property i'm unfamiliar with
I don't even know how you did the 2nd part to the 3rd part
solid kilnBOT
$Pure$
knotty oriole
The log property is because e^x and ln(x) are
knotty oriole
**$Pure$**
This part is by continuity of thr exponential function
Because lnx and e^x are inverses ln(e^x)=e^lnx=x no matter what value you choose for x.
(Assuming the expressions are defined for your choice of x)
Same idea as composing sine and arcsine or any other function with its inverse.
errant bronze
Uhhh I'm watching it rn but 2 minutes in he gets to actually proving the inverse of our current topic https://www.youtube.com/watch?v=FIUG2TcYUIg
edgy cedar
Intuitively $$\left(\frac{n}{n+1}\right)^n = \left(\frac{n+1}{n}\right)^{-n} \
= \frac{1}{\left(\frac{n+1}{n}\right)^n}$$
solid kilnBOT
$Pure$
knotty oriole
The trick pure is talking about usually also uses lhopital.
errant bronze
so like I asked (x/(x+1))^x but the inverse ((x+1)/x)^x would be (1+1/x)^x or simply e
Oh hmmm
where does the e come into that
knotty oriole
With the trick pure wants to use you apply the limit to the exponent
errant bronze
$$\lim_{\infty} (\frac{n+1}{n})^n = e$$
knotty oriole
The e is already there and the limit of the exponent is a more elementary limit you can work out
You are missing a limit.
You can just leave off the inf stuff since we all know the limit is tending towards infinity here
errant bronze
ok
knotty oriole
lim(f) type notation is common for that sometimes
solid kilnBOT
🌸 Katsune
edgy cedar
$$\lim_{n \to \infty} \left(\frac{n+1}{n}\right)^n = e$$
knotty oriole
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$Pure$
knotty oriole
,rotate
solid kilnBOT
errant bronze
arabic?
edgy cedar
Lmao
edgy cedar
knotty oriole
,rotate
solid kilnBOT
edgy cedar
There we go
knotty oriole
This
The final linit at th bottom can be done with lhopital
Should be -1
So the whole limit will be e^(-1)=1/e iirc
errant bronze
sigh unfortunately i think this is too advanced for my current understanding of properties of e, logs, and limits
knotty oriole
Ah fuck god damn it my pic cut off the lim on the lhs lol
errant bronze
like it's one thing if i made an algebraic error or procedural error something like the stuff i was asking earlier but this looks completely alien to me
knotty oriole
sigh unfortunately i think this is too advanced for my current understanding of ...
It should not be
edgy cedar
sigh unfortunately i think this is too advanced for my current understanding of ...
This should be undergrad lvl or even lower
knotty oriole
Look x=e^lnx for ANY value of x because the exponential function and the natural log function are inverses of each other.
They undo each other
errant bronze
This should be undergrad lvl or even lower
im american 
knotty oriole
That is the first line here
errant bronze
lmao sorry but that made me laugh
knotty oriole
Note that the limit sign is missing on the lhs due to cropping.
The second line is by continuity of the exponential function.
Recall continuity lets you move a limit inside of a function.
errant bronze
e^limit(anything) = e
edgy cedar
No?
knotty oriole
From here though please:
One way to define e is like that or we can use the same trick I'm showing
To evaluate that limit
errant bronze
I don't understand how the left side of the image transforms into the right
You said the lim part is necessray for that
knotty oriole
Which image?
edgy cedar
From here though please:
So you want to know why this is true?
errant bronze
Yes
There's some property of e that I'm unaware of that makes this true, i'm guessing
edgy cedar
It uses the same trick
solid kilnBOT
edgy cedar
$$ \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^n = \lim_{n \to \infty} e^{\ln{\left(\frac{n+1}{n}\right)^n}}$$
Omg this sucks
solid kilnBOT
$Pure$
edgy cedar
Does this line make sense to u
errant bronze
No. Is that the true form instead of $$e^{\lim_{\infty} ln \left(\frac{n+1}{n}\right)^n}$$
solid kilnBOT
🌸 Katsune
knotty oriole
It's from the fact that the natural log and the exponential are inverse functions.
edgy cedar
Are you aware that $$x = e^\ln{x}$$
solid kilnBOT
$Pure$
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
errant bronze
Yes
knotty oriole
This is that for a funky x
errant bronze
So just raising all this from e, makes it e?
I thought that and asked this:
Okay what I meant was
given lim(any function)
you could just do this: e^ln(lim(anyfunction) and get e
was what i was getting from all this
knotty oriole
no
errant bronze
yeah it's not right xwx
knotty oriole
It's just that we have a particular function where this works.
errant bronze
This is messing me up so bad 😭 like this randomly inserted e and ln and then the function was put into the power https://cdn.discordapp.com/attachments/908078008609431674/1170925504203866225/370444508644769823.png?ex=655ad05a&is=65485b5a&hm=c77b416f14e4aab6c219b6f64241a209cb397247a4fe7dddb8b250c29683a1dd&
put the entire limit in the power, add ln, and have e as the uhhh whatever that's called
knotty oriole
**$Pure$**
Compile Error! Click the <:errors:504266556054962196> reaction for mo...
It's this property here
errant bronze
the thing that is being raised
knotty oriole
For the first line
But x=((n+1)/n)^n
For the second line it's continuity of the exponential function.
edgy cedar
It may seem random but this is just a trick to do this sort of limit youll get better at spotting these eventually
knotty oriole
Continuity is the property that lets you do stuff like lim(g(x))=g(lim(x))
g here is the e^x function
edgy cedar
Wait so does this make sense to you
errant bronze
Yeah ig I just have to review properties of e and ln some more :S
Oh god but it still has to equal e so that entire thing on the right somehow equals e
knotty oriole
Answer their question tho
edgy cedar
forget about that and just tell me if this makes sense
errant bronze
no it doesn't, like the right side is supposed to just be e, how did everything else go away
So the right side is = e, but how did everything but the e go away then?
e^ln(any function) = any function
knotty oriole
You're looking way too far ahead.
edgy cedar
Wait so does this make sense to you
So you understand why the left hand side is equal to the right hand side?
edgy cedar
Okay great
$$\lim_{n \to \infty} e^{\ln{\left(\frac{n+1}{n}\right)^n}} = e^{\lim_{n \to \infty} \ln{\left(\frac{n+1}{n}\right)^n}} $$
solid kilnBOT
$Pure$
edgy cedar
Does this make sense
errant bronze
No. The lim went into the power for an unknown reason.
knotty oriole
That's by continuity of the exponential function.
errant bronze
Can you link a reference to that property? I'm not seeing anything for "continuity of exponential functions."
knotty oriole
You should have covered continuity in calc 1?
errant bronze
I know about continuity with respect to derivatives and integrals as in the function is continuous along the entire domain?
It seems you're using the word in a different way that I do not know >w<
errant bronze
I love Paul's notes >w< I'll read this more later...
knotty oriole
Continuity isn't a property of derivatives and integrals so much as a property of functions.
Some functions are continuous and others are not.
In this case exponentials happen to be continuous is all.
knotty oriole
Continuity is the property that lets you do stuff like lim(g(x))=g(lim(x))
Practically what that means is when we have to take a limit of a continuous function we can move the limit inside and outside the function in the sense described here
Fact 2 is typically how continuity is presented in calculus 1.
edgy cedar
Seems like a good oll epsilon delta proof
knotty oriole
So, there's a bit of a subtlety here that we're taking a limit of a sequence rather than a real valued limit I suppose.
There's a sequential characterization of limits here that applies which is covered in real analysis that makes the thing we're talking about acceptable.
But I forget how it is justified in calc 1 and calc 2.
errant bronze
I'm here :3
errant bronze
So, there's a bit of a subtlety here that we're taking a limit of a sequence rat...
Series ye?
edgy cedar
Just looking at sequential continuity and it holds for any sequence so we have the result by the sequential characterisation of limits
Well I’ll go to bed it’s almost 4am
@knotty oriole can help you with the rest
errant bronze
Omg go to bed >W< ty so much Pure
edgy cedar
Np
errant bronze
I'm gonna review that continuity later but I want to give a shot at my next problem nyow :c
knotty oriole
Was looking at pauls onlime math notes to see if they explained the continuity thing I'm talking about but they skipped it.
knotty oriole
Just looking at sequential continuity and it holds for any sequence so we have t...
The issue is probably that this is not covered in calc 1 or 2
We def did this in real analysis but idk for calc.
knotty oriole
Series ye?
Not really a series thing
You have this trick in analysis called the sequential characterization of limits that lets you convert a limit of some function with x approaching a to a limit of a sequence approaching a as n approaches infinity.
Which gives you the continuity stuff I mentioned.
errant bronze
Okay, yeah the American system probably shouldn't be blamed, the student must take responsibility I guess...
In our system, analysis is not lower division nor does is it mandatory for any students going thru the calc sequence
quite frankly I think that's actually rare if not only seen in upper div
it's not even offered in my college i'm p sure
knotty oriole
I found what I need I think
knotty oriole
Okay, yeah the American system probably shouldn't be blamed, the student must ta...
In some countries people don't do calc they just immediately do analysis instead. Which is sort of different.
More of a proofs emphasis
knotty oriole
it's not even offered in my college i'm p sure
Fwiw I am american. I'm in analysis rn.
Did calc at cc. There was no analysis at my cc either.
In uni tho analysis is a kind of standard math major course.
Okay lemme post some pics
Ignore my highlighting and annotations.
This justifies swapping limits using continuity
But not yet for sequential limits.
errant bronze
Oof no okay
knotty oriole
errant bronze
I would need to see that in an Orgo Chem Tutor video
reading it like that -- I can't decipher it
knotty oriole
The pink footnote here is how this book lets us move towards limits tending toward infinity
knotty oriole
I would need to see that in an Orgo Chem Tutor video
The bit in yellow explains it kinda slick.
You should have theorems like this way back when you covered continuity in calc 1.
There's another bit here for exponentials so gimme a sec. Not done yet.
errant bronze
Calc 1 was just derivative techniques 😭 i can't remember anything else
and like graphs and stuff and concavity
no for goddess sake i'm cracked
knotty oriole
You should have covered continuity
It might have been kinda brief tho
Probably right after limits
Continuity of the exponential is exactly one of those issues that's a little subtle and could have got swept under the rug. Probably not continuity itself tho.
errant bronze
Do you know about the Ratio Test :c
knotty oriole
Yeah
errant bronze
Continuity of the exponential is exactly one of those issues that's a little sub...
Chances are it was like one small section that went over everyone as we just jumped into properties of derivatives
$$\lim_{\infty} 3\left(1+\frac{1}{n}\right)^{n^2} = 3\lim_{\infty}\left(\frac{n+1}{n}\right)^{n^2} = \frac{\frac{1}{n}+\frac{1}{n^2}}{\frac{1}{n}} = \frac{n}{n+n^2} = \frac{n}{n(1+n)} = \frac{1}{1+n} = 0$$
OKAY there
solid kilnBOT
🌸 Katsune
knotty oriole
Chances are it was like one small section that went over everyone as we just jum...
Yeah here I would just maybe look up some videos on continuity and continuity of the exponential function. The book I sent you earlier was my old calc book. Continuity of exponential gets punted around because there's an elegant way to define it by establishing the log as an integral and then establishing all the exponential props after producing it as the inverse function of the log.
errant bronze
So I ended up with 1/(1+n) meaning limit goes to 0, but then I got this:
knotty oriole
But that kinda means you have all these exponential function properties you just accept on faith until you know integrals.
errant bronze
knotty oriole
Gimme a sec to read
errant bronze
Oml the n=2, what does that mean 😭
wanton rune
it's definitely gonna be something eulery
errant bronze
Yes the root test ALSO I just made progress
knotty oriole
Yeah
errant bronze
Hold on here's what I did
knotty oriole
It will be more e stuff
errant bronze
$$\lim{\infty} 3\left(1+\frac{1}{n}\right)^{n^2} = 3\lim{\infty}\left(\frac{n+1}{n}\right)^{n^2}$$
solid kilnBOT
🌸 Katsune
wanton rune
i was going to say e-y but that didn't sound right
errant bronze
Take the root so ((n+1)/n)^n
knotty oriole
The n=2 at the bottom of the sum just mean the sum starts counting from n=2 instead of 1.
errant bronze
So that literally is 3lim\inf (e)
throw the 3
it's e
and the n=2 is inconsequential
3 also irrelevant
wanton rune
ye
errant bronze
Whenever I see the ((n+1)/n)^n now I will think "that is just e"
wanton rune
limit as it goes to infinity yea
errant bronze
kind of weird the root test is literally just "okay take the root, and then the limit"
wanton rune
mhm
knotty oriole
The n=2 at the bottom of the sum and the 3 here are not related
knotty oriole
Whenever I see the ((n+1)/n)^n now I will think "that is just e"
It's not though.
The e involves a limit.
The limit of a thing and the thing itself are different ya know?
But you can do a similar e and log trick with the root test here to find this limit
knotty oriole
**🌸 Katsune**
Fwiw this is not the correct way to start with the root test
The whole thing needs to be inside the nth root
At the start
wanton rune
that's why the 3 goes away :)
knotty oriole
If you do that you can't just pull the 3 from the limit because it ends up having an nth root on it.
knotty oriole
that's why the 3 goes away :)
It does end up being inconsequential, but starting from the wrong limit here is a poor way to start.
errant bronze
The whole thing needs to be inside the nth root
What?
knotty oriole
The root test requires you place your a_n formula inside of an nth root.
knotty oriole
**🌸 Katsune**
Your limit here is almost right but the nth root is missing
errant bronze
Take the root so ((n+1)/n)^n
This
So basically it just becomes
$$3\lim{\infty}\left(\frac{n+1}{n}\right)^{n}$$
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🌸 Katsune
wanton rune
he's saying if your $a_{n}=3\left(1+\f{1}{n}\right)^{n^{2}}$
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Moosey
wanton rune
then it ALL needs to be in the n-th root
wanton rune
well, |a_n| specifically, but there's no negatives so it won't matter
solid kilnBOT
🌸 Katsune
wanton rune
you can put [n] in front of \sqrt{}, i.e. \sqrt[n]{3}
knotty oriole
The n is the variable of the limit
errant bronze
wanton rune
and ^
errant bronze
Root test: put all of this in an \sqrt{n}{all this}
the n^2 becomes n, that's it
3 is irrelevant
knotty oriole
The 3 becomes 3^(1/n) it can't be pulled out of the limit because of the 1/n power
wanton rune
^
knotty oriole
It ends up disappearing like moosey said tho
wanton rune
(it becomes 1)
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Moosey
errant bronze
$$\lim{\infty}\sqrt[n]{3}\left(\frac{n+1}{n}\right)^{n}$$
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🌸 Katsune
knotty oriole
Yeh
errant bronze
Uhhhh okay
knotty oriole
Now you can do the e to the natural log stuff again
wanton rune
knotty oriole
Oh shit I forgot something from way earlier I was gonna send
wanton rune
ye :)
errant bronze
wanton rune
(you can do some log to see that pretty much at a glance)
knotty oriole
so n goes to in finity the sqrtn3 goes to 1
If you apply the e to the natural log stuff me you and oure did you can show this
Okay I found it
,rotate
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knotty oriole
This is my old ass calc book again
It's describing pretty much exactly what we've been doing.
This is briggs calculus. Your book or whatever probably has a section on these kinds of indeterminate forms.
Should be near the parts on L'hopital like this one.
wanton rune
b-but every form is just 0/0 or infty/infty in disguise :(
knotty oriole
They had asked for references on this stuff earlier moosey
wanton rune
ahhh
knotty oriole
They're kinda uncomfortable with x=e^lnx type manipulations and then moving limits into the exponent by continuity of exponentials.
The latter problem is kind of annoying because continuity of the exponential sometimes gets punted around until you can define natural log via integrals.
errant bronze
limx = e^(ln*limx)
knotty oriole
That works if limx exists and there aren't weird extra variable limits preventing you from moving the limit into the log.
Excuse my messy board and poor writing.
We can't move the limit directly into the log for our actual limit in your problem because we have the n outside the log and under the 3
But continuity of the exponential still lets us put the limit in the exponent of the e.
And the limit can be worked out still.
The 3/n disappears since 1/n goes to 0.
The nln(1+1/n) limit can be handled by l'hopital iirc.
errant bronze
$$\lim_{\infty} e^{\frac{ln3}{n}+nln(1+\frac{1}{n})}$$
knotty oriole
Yesh
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🌸 Katsune
errant bronze
ln3/n goes to 0
then bc we have e to the ln, drop all the interior stuff down
NOOOO it still doesn't make sense
knotty oriole
Which part?
errant bronze
There's still an n in the exponent!!!!!!!!!!!
knotty oriole
Wdym?
errant bronze
and I still don't get how it becomes e!!!! it's just doing this:
So the only way I'm understanding this is: if it looks kind of like this, THROW EVERYTHING OUT and go with e!!!
knotty oriole
I don't think that always works.
solid kilnBOT
🌸 Katsune
errant bronze
Okay so as n goes to infinity, throw the first ln3/n out
this SHOULD be like
lim n(1+1/n)
And this somehow = e
but there's an n there still
knotty oriole
No you lost the natural log and the e and stuff
knotty oriole
We still have continuity of exponential.
What we want is to move the limit up into the power of e
Because we know for theblinit we are working on that we can apply l'hopital to the limit in the power
,rotate
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knotty oriole
Like this
After the ln(3) stuff is removed
At the end here we can apply l'hopitals rule to finish the limit.
errant bronze
Thank you for nyow, I'll take a look at it later, but I gotta go rn >w< signing up for courses for next semester in 10 minutes LOL, gotta beat everyone out on the register buttons after midnight hits.
knotty oriole
Good luck!
errant bronze
yay i got em :3 okay i want to sleep now, but i'll have to think about this for a bit 
wraith hinge
by the way l'hopital isn't really needed
you can simply use the limit definition of e
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modern narwhal
whats a simple graph with degree sequence
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whats a simple graph with degree sequence
trim lichen
!status
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
how many vertices does this graph have?
modern narwhal
7
modern narwhal
its gonna be like
n1 -> n2
....
n1 -> n7
trim lichen
that node is connected to everything else.
so draw that.
then figure out how to make the rest of the nodes have degree 4.
modern narwhal
if I connect a node to itself
is that degree 1 or 2
cuz if its 1 i think i got something that works
trim lichen
idt you can have loops here
but no a loop would add 2 to the degree
you dont need that here tho.
modern narwhal
wait lemme show you how I drew it and lmk if its proper
rose grotto
occult symbolism lookin headahh
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modern narwhal
does that work
modern narwhal
did you draw that btw or find it online
modern narwhal
how do you use it, i dont see a degree sequence input
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orchid sonnet
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orchid sonnet
Hello can you help me to understand the rule behind it why I was able to but the -2 like that
nimble stone
what do you mean?
are you talking about the factorisation?
(a+b+c)q+dq=q((a+b+c)+d)=q(a+b+c+d)
nimble stone
yeah they just factorised out the e
orchid sonnet
Wow. Okay.. thank you so much. I’m dealing with one more similar problem so I will try to apply it there and if I don’t do well I will return
nimble stone
no problemo
orchid sonnet
no problemo
So I basically did the same thing
But how do I find convex and concave from this? Do I have to take x^2 out?
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bold dagger
Quick question, what's the formula for calculating the ratio of rates in modeling problems?
bold dagger
.close
wraith hinge
is this the correct way of solving this inequality?
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is this the correct way of solving this inequality?
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random token
is this the correct way of solving this inequality?
No
That’s how you’d solve it if the absolute value signs were around the whole thing
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frank walrus
mild scroll
i have to demonstrate them using inductions , i have a basic knowledge of how to work with it but i simply cant get to a right answer.
vagrant prism
show your steps
mild scroll
let me take a photo of it
it seems like i've lost that paper but i'll rewrite it real quick
then im lost
wispy kiln
what does the three dots mean
mild scroll
divisible
wispy kiln
i hav no clue but you can write $g^n + 3 = 4a$ and if you multiply both sides by $g$ you can get $g^{n+1} + 3g = 4ag$ then add $3$ you get $g^{n+1} + 3 = 4ag - 3g + 3$ however im not sure how to prove $4ag - 3g + 3$ is divisible by 4
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Kyrum
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mild scroll
i hav no clue but you can write $g^n + 3 = 4a$ and if you multiply both sides by...
that's a 9..
wispy kiln
o
i think u wrote 4 then?
well if its 9 i think u can do the same thing i did above and perhaps g^n+1 + 3 will evaluate to some expression divisible by 9
and i think u could do the same thing for all 5 questions
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mild scroll
i think u wrote 4 then?
sorry , completely forgot about this but it's 9^n
sour minnow
hi
maiden parrot
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maiden parrot
Dont understand how to get the answer
tepid rock
Dont understand how to get the answer
rpm is revolutions per minute
You do know a revolution is 2pi radians, yes?
maiden parrot
Yes
maiden parrot
rpm is revolutions per minute
Mhm
tepid rock
So if it is 80,000 revolutions, how many radians is it?
trim lichen
@maiden parrot how many radians make one revolution?
maiden parrot
There should be pi in your answer lol.
So 160000pi
tepid rock
Yep, that's right.
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manic jackal
If Z is a standard normal random variable and it is defined that U = |Z|, then U is called a folding standard normal random variable. Express F_U(t) as a function of N_Z(t)
manic jackal
The function of density is always non negative, I'm not sure what it is like to apply an absolute value
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@manic jackal Has your question been resolved?
manic jackal
Not sure which of the two is the correct one
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fathom ivy
The function of density is always non negative, I'm not sure what it is like to ...
density yes, but domain isn’t
idea is that if you say sample X=-1 from Z then U=|-1|=1 instead
manic jackal
density yes, but domain isn’t
idea is that if you say sample X=-1 from Z then U=...
Domain is now x≥0 right?
fathom ivy
yes, so neither of your proposed densities work since it’s clearly nonzero for some negative x
manic jackal
yes, so neither of your proposed densities work since it’s clearly nonzero for s...
This should be it, right?
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spice fulcrum
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spice fulcrum
Ive expanded the original expression into $-\frac{a^2+b^2}{b^2}$
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leo
spice fulcrum
I think A is the correct answer, but I need help proving that equation
old slate
I think A is the correct answer, but I need help proving that equation
4zz¯/(z + z¯) = 4(a+bi)(a-bi)/2(a^2) = (a^2 - (bi)^2)/a^2.. (1)
now -(bi)^2 = (b^2)(i^2) ,by def i^2 = -1 so that is equal to, b^2 = (Im(z))^2.. (2)
from (1) and (2) also Re(z) = a
(a^2 - (bi)^2)/a^2 = 1 - {(bi)^2/a^2} = 1 - {Im(z)/Re(z)}^2
Sorry about the mess, I haven't learned how to type in LaTex yet 😔
zinc urchin
The answer is D but I cant understand the concept being asked in this question
nimble stone
which part confuses you
zinc urchin
h(x) triples for every decrease in the value of x by 1
i tried plugging in some numbers for x
nimble stone
means if h(0) is 12 then h(-1) is 36 and h(1) is 4
zinc urchin
oh so i test that with the answer choices?
nimble stone
you could recognise its similar to a geometric sequence (kind of)
if you know about those
zinc urchin
oh yea its starting to make sense
this was actually kinda easy, i couldnt make the connection
hey i got one more question, its kinda similar but involves transformations
the ans is C and i understand x-2 means i gotta shift it to the right
but how do i get to 175 as the y intercept
i could do it with desmos but i just wanted to understand how its done
nimble stone
would you be able to explicitly write r(x-2)
from there its just simplification
using exponent laws and whatnot
zinc urchin
plugging x-2 in the function and then putting that in exponent of 1 and 5?
nimble stone
that wouldnt be that helpful but you can ig
i was thinking more so of using $a^{b+c}=a^b \cdot a^c$
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AℤØ
nimble stone
zinc urchin
.close
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midnight narwhal
hi
maiden hare
No, the 4 stays with the (k+1)
thorn trout
you factor it as (k+1)^2(k^2+4)
midnight narwhal
okay
maiden hare
you factor it as (k+1)^2(k^2+4)
Pretty sure that's a superscript 3
wind crown
help me to do this
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help me to do this
tough iron
Have you ever done nodal analysis
wind crown
wowowowo thnks
tough iron
Its much better than anything i can explain to you if you've never seen it before
wind crown
ya i think i kinda get it is this correct?
while waiting imma do another question first imma ask again after this
haloo
@tough iron rawr
tough iron
Heyo
Well i dont know im not really in the space now to solve it myself, but you got online circuit simulators online to check
tough iron
LTspixe
wind crown
tough iron
Give me a sec
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@wind crown Has your question been resolved?
vale python
sure
lets look at each item
A first
as x goes to + infity (infinity to the right)
what happens to f(x)
why?
as you are going to the right with x
what is happenning to y
nope
chill
wait
look at the graph
do you understand what x—>+infinity means?
think of it this way
when x goes to a very big positive number
right
mhm
when x is a very big positive number
what is y?
no
the more you go to the right in your graph (as x grows towards infinity, or a very big number)
whwre is your y going
look at the graph, dont guess
what is y at x=2
look at the x axis
yes!
do you know what an x axis is?
but im asking ab the x axis!
you got it partially right. when the x axis is 0, the y axis is -6, can you see that?
look at the y coordinate at that point
here
when x is 2
the graph is at 0
its y coordinate
try the question again now that you understand it
x—>infinity+ means the more you go to the right on this x axis
x—>infity- to the left
yes!!!
now what about when you go to a big big number on x
and a big negative number
what is happening to the graph
vale python
👏🏽
vale python
positive or negative infinite
but you said it gets bigger
so you mean positive, dont make it confusing
yes
can you see that if you go all the way to the left on your x axis your y goes all the wy up
if you go all the qah to the right in your x axis your y goes up aa well
trail obsidian
Basically if it gets 'Bigger' downwards it's negative infinity and 'Bigger' upwards it's positive infinity
vale python
so whats the answer
look it over
wdym?
the question is asking the behavior on the extreme left and right
there is no “end behavior”
ok
x—> infinity+
do you get what this means?
it means when x is going to a big number, in the way the arrows are pointing
infinity- is the other way
vale python
as x is growing, what is happening to the function?
in terms of its y value
yes!
so as x grows, f(x) grows can you see that?
so if one goes to infinity+ where is the other
when x goes to the right (to +infinity) where does f(x) go? up (to +infinity) or down (to -infinity)
so when x—> +infinity, f(x)—>?
good
thats half of your answwr
yes
now look at it to the other side
when x goes to -infinity (to the left), what happens to f(x) up (+infinity) or down (-infinity)?
yes
you look if the functtion is going up dor down
in its left most point
perfect
mhm
you need to review graphs!
trail obsidian
Yeh
fringe rover
Hey so I got a question
trim joltBOT
Hey so I got a question
fringe rover
What do you do if there is a tie between the mode?
nimble stone
then there are multiple modes
fringe rover
So I just put 0?
nimble stone
the number of sibling is more likely to be the x
fringe rover
nimble stone
if y is frequency, ie number of people with said number of siblings, then youre doing it wrong
nimble stone
yeah y is frequency
the data set is actually
0,0,1,1,1,1,2,2,2,2,2,2,3,3,4,5
not 1,1,2,2,4,6
nimble stone
why wouldnt you
fringe rover
Yeah I was right
Alright but also is the mean correct?
Or should I round it by the nearest decimal and just put it at 2
nimble stone
,calc (4+12+6+4+5)/(2+4+6+2+1+1)
solid kilnBOT
Result:
1.9375
1.9 is fine
nimble stone
the average of 1,2,5,9,x is 4
the average of 2,3,y is 6