😭
#help-38
wraith hinge
supple copper
,w characteristic polynomial of ((2,0,1);(2,1,3);(1,-1,0))
solid kilnBOT
wraith hinge
i got the same
supple copper
,w factorise -x^3 +3x^2-4x + 3
solid kilnBOT
wraith hinge
bruh
supple copper
Try this
wraith hinge
i dont get it
wraith hinge
yea but
i just dont wanna do it the normal way
we're meant to solve this question using characteristic equation
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@wraith hinge Has your question been resolved?
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@wraith hinge Has your question been resolved?
wraith hinge
.reopen
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✅
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@wraith hinge Has your question been resolved?
fair bison
Huh, this is weird. It doesn't seem like the polynomial f has much to do with the characteristic polynomial of A.
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Available help channel!
urban copper
,tex \begin{enumerate}
\item Find all values of $a > 0$ such that the area enclosed by the line $y = a$ and the graph of $f(x) = 4x^3 + 12ax^2 + a$ is equal to $3a^2$.
\end{enumerate}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
verbal stream
!status
trim joltBOT
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
1
vivid glade
just roleplay as someone finding the area as enclosed
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
vivid glade
,, \int a - \int b
this is wack, frankly
how do you normally find the area of a region enclosed by curves?
urban copper
mmm
its one definite integral minus the other definite integrals
problem is I dont have a range for the boundaries of the integral
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@urban copper Has your question been resolved?
urban copper
im not sure
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@urban copper Has your question been resolved?
sudden mist
First consider drawing a crude sketch of the curves in a graph
Maybe given certain values of a
Then what you want to find is the area enclosed, and as such the intersection between the curves (which are reasonable)
urban copper
okay
how to do it
I dont have access to calculator only pen and pencil
@sudden mist
sudden mist
Yeah so how would you usually sketch a curve in a graph?
sudden mist
When you say never? Do you mean you’ve never even plotted points before? As that might be a start
But say, y=a
This you can surely sketch given a reasonable value for a?
urban copper
im newbie
sudden mist
Why are you dealing with integrals when you’ve not even dealt with such things before?
urban copper
well y=a is a line that passes through a but idk
sudden mist
Or for that matter the undergrad role assigned?!
Hm, okay let’s try doing this without a graph
It will be a very big handicap for us
But let’s try I suppose
urban copper
sure
sudden mist
So we can straight off try and find the intersections for both curves
How would you do that?
urban copper
equate them
sudden mist
Right! Try and do so and show what you arrive at
urban copper
?
sudden mist
What?
urban copper
now what
sudden mist
Solve for x
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
sudden mist
**938c2cc0dcc05f2b68c4287040cfcf71**
I’m confused, how are you able to do this step and not see what to do next?
Well maybe I’m being harsh, but it’s not at all clear what you need help with here if you’re not willing to explain
urban copper
?
sudden mist
This was meant towards your deleted wolfram message
urban copper
,align &\implies 4x^3 + 12ax^2 + a = a \ &\iff x^2(12a + 4x) = 0 \ &\iff \begin{cases} x_1 = 0 \ x_2 : 12a + 4x = 0 \implies x = -3a \end{cases}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
sudden mist
Okay, so now that we know the intersections for the x-coordinate we can try to continue
Since a>0
We can make the reasonable assumption that the enclosed area must be wherever x is non-positive, since the question states there is one and we’ve just seen where these curves intersects
sudden mist
We can try to show this more carefully but I think it’s clear for our purposes
sudden mist
assumption?
We haven’t shown it yet, so I’m wording it like that
Furthermore, we can show that the curve y=a will be under the other curve
sudden mist
Which part?
urban copper
how
sudden mist
Yeah that’s what I was about to ask
Can you see how?
(Try and do it)
I’m not here to spoon feed you all the answers
sudden mist
Okay so we want to show that 4x^3 + 12ax^2 + a > a for x in between -3a and 0
urban copper
sure
urban copper
We can make the reasonable assumption that the enclosed area must be wherever x ...
?
sudden mist
In other words, 4x^3 + 12ax^2 > 0
Try and see if this true for the given x
And remember a>0
sudden mist
Oh Ok so you’re just trolling?
Lmao why’d you delete the message?
I think Im done helping here.
proper kernel
,tex \begin{enumerate}
\item Find all values of $a > 0$ such that the area e...
you here?
urban copper
ye
proper kernel
you first need to find out where y = 4x^3 + 12ax^2 + a and y = a intersect
do you know how to do that
proper kernel
you mean $4x^3+12ax^2+a=a$
proper kernel
@urban copper are you stuck
urban copper
yes
proper kernel
try factoring
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
,, 4x^3 + 12ax^2 = 0 \implies 4x^2(x + 3a) = 0
proper kernel
with that done, what are the solutions to the equation
@urban copper you have 4 * x^2 * (x + 3a) = 0
for them to multiply to 0, that would mean at least one of the factors is 0
that means 4, x^2, or (x + 3a)
urban copper
youre a bit preoccupied it looks like
?
proper kernel
you now have the equation 4x^2(x + 3a) = 0
what are the solutions to this equation?
proper kernel
so the intersections are at x1 = 0 and x2 = -3a
the area lies between these intersections
so which intersection is the left end of the area?
proper kernel
thats correct
the right end of the area ends at 0
so horizontally, this area spans from -3a to 0
vertically, this area is bounded by y = a and y = 4x^3 + 12ax^2 + a
which one is at the top?
(one of them will always be on the top and the other one always be at the bottom, so you can test out values of a and x to see which one it is)
@urban copper are you stuck again
urban copper
second one is on the top
we can graph them to be sure
but i only have access to pen and paper
proper kernel
I told you to test out values, did you do that?
urban copper
i did
proper kernel
did it confirm that the second one is on the top?
proper kernel
what do you think value testing is for in calculus?
urban copper
unsure tbh
proper kernel
youve seen that y = 4x^3 + 12ax^2 + a and y = a only intersect in two places
so that means, between x = -3a and x = 0, that one of them is always on top and the other is always on the bottom
compared to other methods, just placing in numbers for a and x will show which one is on top
proper kernel
so that means, between x = -3a and x = 0, that one of them is always on top and ...
this has proven that, when one value is tested, the entire region is confirmed
its also easier to just compare numbers than having to judge 4x^3 + 12ax^2 + a's behavior
are you with me so far?
urban copper
no
proper kernel
do you want to elaborate?
urban copper
sure
proper kernel
Im going to assume you didnt test the values properly
lets try some values like a = 1
that would mean y = 4x^3 + 12x^2 + 1 and y = 1
now we know the intersections are between x = -3 and x = 0
from using a = 1 with -3a and 0
I didnt use any theorems
dont delete your comments
thats bad practice to hide your mistakes
anyways, I can then test x = -1 to see which one is higher
y = 4(-1)^3 + 12(-1)^2 + 1 = -4 + 12 + 1 = 9
so we've seen that y = 4x^3 + 12x^2 + 1 is higher than y = 1 at a=1, x=-1
are you with me so far?
urban copper
x=-1?
proper kernel
it doesnt matter which value of x I test
so long as that value is between x=-3 and x=0,
this value would be within the region where the area should be
and in that region, either 4x^3 + 12x^2 + 1 or 1 is higher
proper kernel
so long as that value is between x=-3 and x=0,
you didnt let me finish, either, on why the value ultimately doesnt matter
I chose x = -1 out of these values because it is the easiest to calculate
are we ready to continue?
urban copper
okay
proper kernel
now lets use a property
not a theorem yet, just a property
urban copper
i was just confused because you used a = 1 and x = -1 but okay
proper kernel
dont read into it too much
urban copper
it was just too abrupt
proper kernel
everyone moves at a different speed, dont worry aobut it
urban copper
sure
proper kernel
anyways
in general, we know that both y = 4x^3 + 12ax^2 + a and y = a are both continuous
continuity promises some nice things, one of them being IVT
continuity is still there if I set a = 1, so y = 4x^3 + 12x^2 + 1 and y = 1
now we've seen that they intersect only at x = -3 and x = 0
they can't intersect anywhere else
still with me?
(I havent gotten to anything important yet, just asking)
urban copper
yes
proper kernel
now at x = -3 and x = 0, y = 4x^3 + 12x^2 + 1 and y = 1 are at the same height
(they intersect)
at x = -1, y = 4x^3 + 12x^2 + 1 is higher
lets assume that at x0, y = 4x^3 + 12x^2 + 1 is lower
with the additional restriction that -3 < x0 < 0
if this x0 is found, we would be able to prove that just testing one value isnt enough to rule the entire -3 < x < 0 region as "y = 4x^3 + 12x^2 + 1 is higher"
proper kernel
with the additional restriction that -3 < x0 < 0
now for this to happen, 4x^3 + 12x^2 + 1 would need to lower and dip below y = 1
however, that would mean it would need to cross y = 1
which can only happen at x = -3 and x = 0
so the only way 4x^3 + 12x^2 + 1 can be below y = 1 at x = x0 is if x0 is outside of -3 < x < 0
since x0 can never be inside -3 < x < 0, that means y = 4x^3 + 12x^2 + 1 is always above y = 1 in the region -3 < x < 0
do you understand so far?
urban copper
sure
proper kernel
not a yes
urban copper
how does ivt come into play
proper kernel
do you understand so far?
is this a "yes" or a "suuuure"?
urban copper
i agree
proper kernel
IVT never comes into play, because you didnt ask for a formal proof
proper kernel
however, that would mean it would need to cross y = 1
we can use IVT along with a few modifications to prove this line here
since continuity already is pretty clear visually, you dont need a proof to see how the proof would work
urban copper
how to prove
proper kernel
there you go
you need to keep that rigor throughout next time
now do you remember IVT?
urban copper
proper kernel
yep
now a more general proof shouldnt need to depend on a = 1
I chose it as an example, so lets let a > 0 instead
proper kernel
this clashes with the a being used in IVT though
so instead you can read that as "continuous on [b, d]"
now with a being general, we found the intersections of y = 4x^3 + 12ax^2 + a and y = a at x = -3a and x = 0
now as before, I test a value out and do x = -a
since a > 0, then -a < 0,
since 1 < 3, then a < 3a, so -a > -3a
hence -3a < -a < 0
so x = -a is inside (-3a, 0)
do you understand so far?
urban copper
so instead you can read that as "continuous on [b, d]"
sure
proper kernel
still not a yes
anyways lets see what happens if x = -a is tested
y = 4(-a)^3 + 12a(-a)^2 + a
y = -4a^3 + 12a^3 + a
y = 8a^3 + a
since a > 0, 8a^3 > 0
adding a to both sides, 8a^3 + a > a
y = 8a^3 + a is greater than y = a
so y = 4x^3 + 12ax^2 + a is higher than y = a at x = -a
do you get that?
proper kernel
you can also just say the word "yes"
urban copper
ye
proper kernel
because the word "sure" is a universal sign of sarcasm so you cant use it in a legit way unless said in person
in short, I dont hear the tone you use when you use it
proper kernel
so instead you can read that as "continuous on [b, d]"
now let b be -a and -3 < d < 0 (and also f(x) = 4x^3 + 12ax^2 + a)
the theorem here assumes that b < d, so Ill need to repeat this again for when d < b
we know f(b) = f(-a) > a
if f(x) dips below y = a, then f(x) would < a
so lets assume f(d) < a
now let k = a
by IVT, there exists a constant in [b, d] = [-a, d] such that f(c) = k = a
however if f(c) = a, then y = f(x) and y = a intersect at x = c
but since -3 < -a = b < c < d < 0, c is not -3 nor 0
this would prove that c is somehow a third intersection which cannot happen, since we've already seen f(x) only intersects with a at x = -3a and x = 0
(also if f(d) = a, then the same "third intersection occurs but with x = d, so f(d) = a cannot happen)
so by contradiction, f(d) > a
so f(x) > a for all x in [-a, 0)
do you understand?
urban copper
too much words and little latex gimme a minute to read
proper kernel
alr
latex would make this harder since you dont get to parse grammar
if you want, I can give you a latex version, but by the time I finish you'd probably get what I typed
urban copper
I would appreciate it i am not in a hurry btw and i enjoy math
proper kernel
can you read $\forall$ and $\exists$?
solid kilnBOT
mtt
urban copper
universal and existencial quantifiers?
proper kernel
you have them swapped (yea thats correct)
and theyre read "for all" and "there exists"
anyways the latex here is going to use a lot of these since theyre short
proper kernel
@urban copper here you go
proper kernel
yes, I said earlier that the IVT is a bit one-sided
it assumes b < d
I then would have to post a similar proof for when d < b
the proof is very similar in that other than changing d < b, none of the machinery changes, so its not worth posting two very similar images if focusing on reading one is good enough
when you understand this proof, Ill post the second one
I also said earlier to change [a, b] to [b, d]
proper kernel
<@1187916384341078070> here you go
it looks like theres an unfinished line in the middle there
proper kernel
<@1187916384341078070> here you go
removed them
you can see how similar these are
brb
proper kernel
@urban copper back, you here?
trim joltBOT
@urban copper Has your question been resolved?
urban copper
here
proper kernel
did you read the proofs?
urban copper
only first one
proper kernel
you still need time reading them?
urban copper
I think maybe math is not for me
proper kernel
formal proof latex is not for you
mostly because you werent taught formal proof latex
proper kernel
now let b be -a and -3 < d < 0 (and also f(x) = 4x^3 + 12ax^2 + a)
that is why you take english when you get it
it has grammar and words to connect nouns and phrases together
asking for a latex proof has no guarantees of being smoother than the words
urban copper
fair
proper kernel
now how familiar are you with IVT
urban copper
little to nothing
proper kernel
right, thats not good
to be able to understand how IVT is used, youll need to be familiar with what its saying
do you want to get into that or just take as a fact that f(x) > a for all x in (-3a, 0)?
urban copper
lets assume it
proper kernel
oh
then lets move on
we've verified that the top of the region is always y = 4x^3 + 12ax^2 + a and the bottom is y = a
it starts on the left at x = -3a and ends on the right at x = 0
this is enough to write an integral of the region in particular
do you know how to write that integral?
urban copper
sure
proper kernel
sounds good, try it out
urban copper
,, \int_{-3a}^{0} 4x^3 + 12ax^2 dx - \int_{-3a}^{0} a \ dx
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
proper kernel
you mean $\int_{-3a}^0(4x^3+12ax^2+a)\dd{x}-\int_{-3a}^0a\dd{x}$
solid kilnBOT
mtt
urban copper
sure
proper kernel
usually this is memorized as $\int_{-3a}^0((4x^3+12ax^2+a)-(a))\dd{x}$ instead
solid kilnBOT
mtt
proper kernel
so you directly consider it as the difference or gap between the top and bottom
urban copper
now what
proper kernel
now you simplify and solve that integral
urban copper
,align &\implies \int_{-3a}^0\left(4x^3+12ax^2\right)\dd{x} \ &\implies \int_{-3a}^0\left(4x^3\right)\dd{x}+\int_{-3a}^0\left(12ax^2\right)\dd{x}
wraith hinge
Hi
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
,align &\implies \int_{-3a}^0\left(4x^3+12ax^2\right)\dd{x} \ &\implies \int_{-3a}^0\left(4x^3\right)\dd{x}+\int_{-3a}^0\left(12ax^2\right)\dd{x} \ &\implies \left[ \frac{4x^4}{4} \right]{-3a}^{0} + \left[ \frac{12ax^3}{3}\right]{-3a}^{0}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
,align &\implies \int_{-3a}^0\left(4x^3+12ax^2\right)\dd{x} \ &\implies \int_{-3a}^0\left(4x^3\right)\dd{x}+\int_{-3a}^0\left(12ax^2\right)\dd{x} \ &\implies \left[ \frac{4x^4}{4} \right]{-3a}^{0} + \left[ \frac{12ax^3}{3}\right]{-3a}^{0} \ &\implies \left[0 - \left(-3a\right)^{4}\right] + \left[ 0 - 4a\left(-3a\right)^3\right]
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
proper kernel
thats correct
now simplify what you have down there
youre kidding me
please do that by hand
urban copper
,align &\implies \int_{-3a}^0\left(4x^3+12ax^2\right)\dd{x} \ &\implies \int_{-3a}^0\left(4x^3\right)\dd{x}+\int_{-3a}^0\left(12ax^2\right)\dd{x} \ &\implies \left[ \frac{4x^4}{4} \right]{-3a}^{0} + \left[ \frac{12ax^3}{3}\right]{-3a}^{0} \ &\implies \left[0 - \left(-3a\right)^{4}\right] + \left[ 0 - 4a\left(-3a\right)^3\right] \ &\implies -\left(81a^4\right) -4a(-27a^3)
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
proper kernel
for a timesave, you can split 81 into 3 * 27
urban copper
sure
,align &\implies \int_{-3a}^0\left(4x^3+12ax^2\right)\dd{x} \ &\implies \int_{-3a}^0\left(4x^3\right)\dd{x}+\int_{-3a}^0\left(12ax^2\right)\dd{x} \ &\implies \left[ \frac{4x^4}{4} \right]{-3a}^{0} + \left[ \frac{12ax^3}{3}\right]{-3a}^{0} \ &\implies \left[0 - \left(-3a\right)^{4}\right] + \left[ 0 - 4a\left(-3a\right)^3\right] \ &\implies -\left(81a^4\right) -4a(-27a^3) \iff 81 = 3 \times 27 \ &\implies -3(27a^4) -4a(-27a^3)
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
,align &\implies \int_{-3a}^0\left(4x^3+12ax^2\right)\dd{x} \ &\implies \int_{-3a}^0\left(4x^3\right)\dd{x}+\int_{-3a}^0\left(12ax^2\right)\dd{x} \ &\implies \left[ \frac{4x^4}{4} \right]{-3a}^{0} + \left[ \frac{12ax^3}{3}\right]{-3a}^{0} \ &\implies \left[0 - \left(-3a\right)^{4}\right] + \left[ 0 - 4a\left(-3a\right)^3\right] \ &\implies -\left(81a^4\right) -4a(-27a^3) \iff 81 = 3 \times 27 \ &\implies -3(27a^4) -4a(-27a^3) \ &\implies -3(27a^4) + 4(27a^4) \implies (27a^4)(-3+4)
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
proper kernel
@urban copper are you going to finish that
urban copper
okay
,align &\implies \int_{-3a}^0\left(4x^3+12ax^2\right)\dd{x} \ &\implies \int_{-3a}^0\left(4x^3\right)\dd{x}+\int_{-3a}^0\left(12ax^2\right)\dd{x} \ &\implies \left[ \frac{4x^4}{4} \right]{-3a}^{0} + \left[ \frac{12ax^3}{3}\right]{-3a}^{0} \ &\implies \left[0 - \left(-3a\right)^{4}\right] + \left[ 0 - 4a\left(-3a\right)^3\right] \ &\implies -\left(81a^4\right) -4a(-27a^3) \iff 81 = 3 \times 27 \ &\implies -3(27a^4) -4a(-27a^3) \ &\implies -3(27a^4) + 4(27a^4) \ &\implies (27a^4)(-3+4) \ &\implies 27a^4
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
proper kernel
27a^4 is correct
Im pretty sure you mistyped it when you put that into the calculator earlier
now you know the area is 27a^4
according to the problem, what must this area also be equal to?
urban copper
,tex \begin{enumerate}
\item Find all values of $a > 0$ such that the area enclosed by the line $y = a$ and the graph of $f(x) = 4x^3 + 12ax^2 + a$ is equal to $3a^2$.
\end{enumerate}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
,align &\iff 27a^4 = 3a^2 \ &\iff a^2 = \frac{3}{27} \ &\iff |a| = \sqrt{\frac{3}{27}} \ &\iff a = \pm \sqrt{\frac{3}{27}}
proper kernel
yep, now solve for a
not with a calculator
thats not ideal
try square rooting instead
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
proper kernel
whats 3 / 27?
urban copper
sure
,align &\iff 27a^4 = 3a^2 \ &\iff a^2 = \frac{3}{27} \ &\iff |a| = \sqrt{\frac{3}{27}} \ &\iff a = \pm \sqrt{\frac{3}{27}} \ &\iff a = \pm \frac{1}{3} \ &a > 0 \implies a = \frac{1}{3}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
proper kernel
yep
by coincidence*, when a = 1/3, the area is also 1/3
so a, the area, 3a^2, and 27a^4 are all 1/3
urban copper
by coincidence*, when a = 1/3, the area is also 1/3
how so
urban copper
.close
urban copper
,tex \begin{enumerate}
\item Determine all values of $x \in \mathbb{R}$ for which the series $\sum_{n=1}^\infty \frac{2^n x^n}{n + 5^n}$ is convergent.
\end{enumerate}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
cyan zinc
there's the 'formal' way and then there's the easy way. the easy way is to just see that 2x < 5
urban copper
formal I would prefer otherwise my homework solution might not be rigurous enough
<@&286206848099549185>
ionic pendant
you can use the root or ratio tests
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
,tex [
L = \lim_{n \to \infty} \sqrt[n]{|a_n|}.
]
The Root Test states:
\begin{enumerate}
\item If $L < 1$, then the series $\sum_{n=1}^\infty a_n$ converges absolutely.
\item If $L > 1$, then the series $\sum_{n=1}^\infty a_n$ diverges.
\item If $L = 1$, the test is inconclusive, and the series may converge or diverge.
\end{enumerate}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
,align a_n &= \frac{2^n \cdot x^n}{n + 5^n} \ \lim_{n \to \infty} \sqrt[n]{\left|a_n\right|} &= \lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n \cdot x^n}{n + 5^n}\right|}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
and we want it to converge
,, \lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n \cdot x^n}{n + 5^n}\right|} < 1
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
trim joltBOT
@urban copper Has your question been resolved?
urban copper
you can use the root or ratio tests
help me
shell shale
wat is x lol
winged hinge
squeeze thm
urban copper
wat is x lol
$x \in \mathbb{R}$
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
winged hinge
use squeeze theorem to find limit
white dagger
<@&286206848099549185>
I think ratio test is easier to do
urban copper
I want to practice ROOT TEST
desert sky
white dagger
I want to practice ROOT TEST
Well your function is always positive so you can remove absolute value
winged hinge
yeah use squeeze theorem to find the limit of the sequence in the root test
urban copper
Well your function is always positive so you can remove absolute value
wdym
can u elaborate
urban copper
right
white dagger
So you can reduce it to $\lim_{n \to \infty} \sqrt[n]{\frac{2^n |x^n|}{n + 5^n}}$
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white dagger
And I think you can split case to x >= 0 and x < 0 
urban copper
,, \lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n \cdot x^n}{n + 5^n}\right|} < 1 \ \lim_{n \to \infty} \sqrt[n]{\frac{2^n \cdot |x^n|}{n + 5^n}} < 1 \ \lim_{n \to \infty} \frac{\sqrt[n]{2^n \cdot | x^n |}}{\sqrt[n]{n + 5^n}} < 1 \ \lim_{n \to \infty} \left(\sqrt[n]{2^n} \cdot \sqrt[n]{|x|^n}\right) < \lim_{n \to \infty} \left(\sqrt[n]{n + 5^n}\right) \ |x| < \lim_{n \to \infty} \sqrt[n]{n + 5^n}
can u help or no
@white dagger
white dagger
<@793428843871731732>
$\sqrt[n]{x^n} = x$
urban copper
,, \lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n \cdot x^n}{n + 5^n}\right|} < 1 \ \lim_{n \to \infty} \sqrt[n]{\frac{2^n \cdot |x^n|}{n + 5^n}} < 1 \ \lim_{n \to \infty} \frac{\sqrt[n]{2^n \cdot | x^n |}}{\sqrt[n]{n + 5^n}} < 1 \ \lim_{n \to \infty} \left(\sqrt[n]{2^n} \cdot \sqrt[n]{|x|^n}\right) < \lim_{n \to \infty} \left(\sqrt[n]{n + 5^n}\right) \ 2|x| < \lim_{n \to \infty} \sqrt[n]{n + 5^n} \ 2|x| < \lim_{n \to \infty} \sqrt[n]{5^n\left( 1 + \frac{n}{5^n}\right)}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
,align &\lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n \cdot x^n}{n + 5^n}\right|} < 1 \ &\lim_{n \to \infty} \sqrt[n]{\frac{2^n \cdot |x^n|}{n + 5^n}} < 1 \ &\lim_{n \to \infty} \frac{\sqrt[n]{2^n \cdot | x^n |}}{\sqrt[n]{n + 5^n}} < 1 \ &\lim_{n \to \infty} \left(\sqrt[n]{2^n} \cdot \sqrt[n]{|x|^n}\right) < \lim_{n \to \infty} \left(\sqrt[n]{n + 5^n}\right) \ &2|x| < \lim_{n \to \infty} \sqrt[n]{n + 5^n} \ &2|x| < \lim_{n \to \infty} \sqrt[n]{5^n\left( 1 + \frac{n}{5^n}\right)} \ &2|x| < \lim_{n \to \infty} \sqrt[n]{5^n} \cdot \lim_{n \to \infty} \sqrt[n]{\left(1 + \frac{n}{5^n}\right) }
white dagger
,align &\lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n \cdot x^n}{n + 5^n}\right|}...
$\lim{\sqrt[n]{1 + \frac{n}{5^n}}}$
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white dagger
And this will make you go around the circle, though
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
And this will make you go around the circle, though
?
urban copper
And this will make you go around the circle, though
elaborate
white dagger
?
I think you need to use the log trick here which is: $\lim_{n \to \infty}e^{\frac{1}{n} \ln{(n + 5^n)}}$
solid kilnBOT
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white dagger
And use L'Hopital rule
urban copper
let me handle this
,align &\lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n \cdot x^n}{n + 5^n}\right|} < 1 \ &\lim_{n \to \infty} \sqrt[n]{\frac{2^n \cdot |x^n|}{n + 5^n}} < 1 \ &\lim_{n \to \infty} \frac{\sqrt[n]{2^n \cdot | x^n |}}{\sqrt[n]{n + 5^n}} < 1 \ &\lim_{n \to \infty} \left(\sqrt[n]{2^n} \cdot \sqrt[n]{|x|^n}\right) < \lim_{n \to \infty} \left(\sqrt[n]{n + 5^n}\right) \ &2|x| < \lim_{n \to \infty} \sqrt[n]{n + 5^n} \ &2|x| < \lim_{n \to \infty} \sqrt[n]{5^n\left( 1 + \frac{n}{5^n}\right)} \ &2|x| < \lim_{n \to \infty} \sqrt[n]{5^n} \cdot \lim_{n \to \infty} \sqrt[n]{\left(1 + \frac{n}{5^n}\right) } \ &|x| < \frac{5}{2} \cdot \lim_{n \to \infty} \sqrt[n]{1 + \frac{n}{5^n}}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
white dagger
,align &\lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n \cdot x^n}{n + 5^n}\right|}...
Wait
I mean: $\lim_{n \to \infty}\sqrt[n]{n + 5^n} = \lim_{n \to \infty}e^{\frac{1}{n} \ln{(n + 5^n)}}$
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winged hinge
squeeze theorem 1 < 1 + n/5^n < 1 + 5^n/5^n = 2 
white dagger
squeeze theorem 1 < 1 + n/5^n < 1 + 5^n/5^n = 2 <:cat_happycry:63640781314457601...
Or you can use this
😂
white dagger
squeeze theorem 1 < 1 + n/5^n < 1 + 5^n/5^n = 2 <:cat_happycry:63640781314457601...
Wait a minute what about n-th root
winged hinge
Inequality is preserved when u take n-th root and limit of a constant to nth root is 1
urban copper
can u help or not
winged hinge
do u know squeeze theorem
urban copper
y
winged hinge
u just can use it to justify why the limit is 1
urban copper
u show the latex i can wait
,align &\lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n \cdot x^n}{n + 5^n}\right|} < 1 \ &\lim_{n \to \infty} \sqrt[n]{\frac{2^n \cdot |x^n|}{n + 5^n}} < 1 \ &\lim_{n \to \infty} \frac{\sqrt[n]{2^n \cdot | x^n |}}{\sqrt[n]{n + 5^n}} < 1 \ &\lim_{n \to \infty} \left(\sqrt[n]{2^n} \cdot \sqrt[n]{|x|^n}\right) < \lim_{n \to \infty} \left(\sqrt[n]{n + 5^n}\right) \ &2|x| < \lim_{n \to \infty} \sqrt[n]{n + 5^n} \ &2|x| < \lim_{n \to \infty} \sqrt[n]{5^n\left( 1 + \frac{n}{5^n}\right)} \ &2|x| < \lim_{n \to \infty} \sqrt[n]{5^n} \cdot \lim_{n \to \infty} \sqrt[n]{\left(1 + \frac{n}{5^n}\right) } \ &|x| < \frac{5}{2} \cdot \lim_{n \to \infty} \sqrt[n]{1 + \frac{n}{5^n}}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
winged hinge
Basically it says if u have a_n < b_n < c_n and lim a_n = lim c_n, then lim a_n = lim b_n = lim c_n
lim of 1^(1/n) is clearly 1
urban copper
ok
winged hinge
lim of 2^(1/n) is also 1
urban copper
,w limit of n to infinity of (1+ n/(5^n))^(1/n)
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winged hinge
lim of 2^(1/n) is also 1
that's a little harder to prove if u don't already know it but uhhh the proof uses the binomial theorem
winged hinge
yeah I'm tryna remember the full proof lol
urban copper
how to find x in R s.t. this shit converges
white dagger
$|x| < \frac{5}{2}$ so $\frac{-5}{2} < x < \frac{5}{2}$
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938c2cc0dcc05f2b68c4287040cfcf71
white dagger
And you need to test whether the series convert at endpoint separately
white dagger
so $x \in \left(\frac{-5}{2}, \frac{5}{2}\right)$
Yeah, but maybe series converges at $x = 5/2$ or $x = -5/2$
solid kilnBOT
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white dagger
So you need to test it
urban copper
can u draw it and show me the latex
winged hinge
if a> 1 then a^(1/n) = 1 + b_n for b_n >= 0, so a >=1 + nb_n, so (a - 1)/n >= b_n >= 0, then squeeze theorem implies lim b_n =0 so lim a^(1/n) = 1
white dagger
**938c2cc0dcc05f2b68c4287040cfcf71**
Plug x = 5/2 and x = -5/2 into original series
And check whether it converges at those points
urban copper
Plug x = 5/2 and x = -5/2 into original series
can u draw it
white dagger
Let me latex it
$\sum_{n=1}^{\infty} \frac{2^n (\frac{5}{2})^n}{n + 5^n}$ and $\sum_{n=1}^{\infty} \frac{2^n (\frac{-5}{2})^n}{n + 5^n}$
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urban copper
right
second one is alternating
due to (-1)^n
can someone help or not
lets apply root test for left series
to determine if it converges or diverges
after that we can play with the second one, which looks a little more intimidating
white dagger
lets apply root test for left series
Bruh those two series are obviously
You don't need root test
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938c2cc0dcc05f2b68c4287040cfcf71
urban copper
so its 1/1
white dagger
Obviously it is 1 or you can check with L'Hopital
white dagger
Ok so it diverges at x = 5/2 so we exclude it
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
white dagger
,, \sum{n=1}^{\infty} \frac{2^n (\frac{-5}{2})^n}{n + 5^n}
?
The second one you can use alternating series test to see that it diverges. So the final result is $x \in (-5/2, 5/2)$
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urban copper
so its not inclusive
because both ends diverge
@white dagger
so are we done or ?
urban copper
The second one you can use alternating series test to see that it diverges. So t...
alternating series test
?
white dagger
so are we done or ?
Yes, we are done
urban copper
how to show it using leibnitz
white dagger
how to show it using leibnitz
Leibniz test?
urban copper
alternating series test
but we need to show it is decreasing sequence
can we use lagrange?
white dagger
but we need to show it is decreasing sequence
It has two condition
Another condition is: $\lim_{n \to \infty}a_n = 0$
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urban copper
sure
,, \sum_{n=1}^{\infty} \frac{2^n \cdot \left(-1\right)^n \left(\frac{5}{2}\right)^n}{n + 5^n} \ a_n = \frac{2^n \cdot \left(-1\right)^n \left(\frac{5}{2}\right)^n}{n + 5^n}
white dagger
$a_n = \frac{5^n}{n + 5^n}$
white dagger
We don't care about (-1)^n
urban copper
wdym
$\sum_{n=1}^{\infty} \frac{2^n (\frac{5}{2})^n}{n + 5^n}$ and $\sum_{n=1}^{\infty} \frac{2^n (\frac{-5}{2})^n}{n + 5^n}$
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
I was talking about the second series
I believe you are talking about the first one innit
@white dagger
white dagger
<@793428843871731732>
You simply it
Then it will become: $\lim_{n \to \infty} (-1)^n \frac{5^n}{n + 5^n}$
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white dagger
Use second condition of alternating series test
urban copper
oh it converges to zero
white dagger
No it diverges
white dagger
If you have alternating series: $\sum (-1)^n a_n$
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white dagger
Then for it to converges two conditions need to be satisfied:
1 is a_n need to form a decreasing sequence
2 is $\lim_{n \to \infty} a_n = 0$
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white dagger
Condition 2 is often easier to test first
urban copper
2 is $\lim_{n \to \infty} a_n = 0$
its not
it’s not zero
$\lim_{n \to \infty} (-1)^n \frac{5^n}{n + 5^n}$
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
white dagger
it’s not zero
So it diverges
white dagger
$\lim_{n \to \infty} (-1)^n \frac{5^n}{n + 5^n}$
$a_n$ here is $\frac{5^n}{n + 5^n}$
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white dagger
Meaning it diverges
white dagger
So -5/2 is excluded as well
white dagger
Then for it to converges two conditions need to be satisfied:
.
urban copper
So -5/2 is excluded as well
why
is that a minimum condition
for any series
i DONT GET IT
@white dagger
white dagger
i DONT GET IT
That is alternating series test
urban copper
but first one is not alternating
how can u
use alternating test for a NOT ALTERNATING SERIES
white dagger
but first one is not alternating
We are doing second one, though
white dagger
WHAT ABOUT FIRST ONE
Divergence test
urban copper
???
white dagger
If $\lim_{n \to \infty} a_n \neq 0$ then it diverges
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urban copper
If $\lim_{n \to \infty} a_n \neq 0$ then it diverges
sure
okay so we are done right
white dagger
okay so we are done right
Yes. $x \in (\frac{-5}{2}, \frac{5}{2})$
solid kilnBOT
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urban copper
sure
white dagger
You can test with Wolfram alpha to be sure
white dagger
can u do it or no
My computation time excessed
urban copper
,w SumConvergence[(2^n x^n)/(n + 5^n), n]
solid kilnBOT
white dagger
Ok so it is true
urban copper
,w SumConvergence[(2^n x^n)/(n + 5^n), 1]
solid kilnBOT
urban copper
,w SumConvergence[(2^n x^n)/(n + 5^n),n]
solid kilnBOT
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@urban copper Has your question been resolved?
proper kernel
I can confirm $x\in\qty(-\frac52,\frac52)$ is the answer, so whats your remaining question
solid kilnBOT
mtt
urban copper
.close
wraith hinge
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wraith hinge
This YouTube video says to factor a radical you should memorize the perfect squares which I have until 256
But when factoring how do I know the multiplication of those larger numbers
Do I start guessing with my calculator trying each perfect square
Like he shows 36 times 3 = 108
How would I know that
proper kernel
more practically, you dont memorize that that happens
what you do instead is repeatedly divide by potential factors until you break down the whole number
for example, I see 108 and its even
I try dividing by 2
108 = 54 * 2
108 = 27 * 2 * 2
27 is the third power of 3, I know that
108 = 3 * 3 * 3 * 2 * 2
108 = 3 * 3 * 3 * 2 * 2
then with this, you can see that pairs of the same number naturally form squares
3 * 3 * 2 * 2 = 3^2 * 2^2 = 6^2
so 108 = 3 * 6^2
so sqrt(108) = sqrt(3 * 6^2)
= sqrt(3) 6
wraith hinge
Wow thank you so much
proper kernel
np
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grand shard
Can someone help me in understanding the solution by providing a slightly more elaborate explanation?
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
but
if you take n^8 out of the sqrt is n^4
also
(n+1)^2 is to the power of two same with n^2
grand shard
I got the first step thanks but I'm trying to wrap my mind around why the 2nd step was needed because this kinda seems like it's changing the equation too far than the original, for the limit
urban copper
,, \frac{(n+1)^2}{n^2} = \left(\frac{n+1}{n}\right)^2
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
agree or disagree
grand shard
AH THANKS
I couldn't figure out such a simple thing in the entire problem smh
Thanks a lot man
😭🙏
urban copper
no worries, keep it hard
grand shard
.close
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atomic seal
hi
atomic seal
lim(x,y)->(0,0) (e^x-e^y)/x-y
i found this question somewhere
i asked my sir about it and he said its out of my syllabus yet i wanna know what the limit is
you can ping me
dusky gust
i found this question somewhere
i asked my sir about it and he said its out of ...
you assume t = x - y
(i haven't done anything like this but i think it is what question is really telling you to do)
e^y (e^(x-y) - 1) / (x-y)
so you have t and y
you get e^y = 1
and you know what (e^t-1)/t is
@atomic seal
dusky gust
you get e^y = 1
...
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urban copper
,tex \begin{enumerate}
\item Let ( f ) be a differentiable function that satisfies
[
x^2 f(x) = 3x + \int_{9}^{3x} t f\left(\frac{t}{3}\right) dt.
]
Find the Taylor polynomial of order 2 for ( f ) at ( x_0 = 3 ).
\item Find \( f \) differentiable such that
\[
f^3(x) f'(x) = 15 x^9 e^{2x^5}, \quad f(0) = 1.
\]
\item Find \( a > 0 \) such that the area of the region enclosed between the graphs of
\[
f(x) = a \sqrt{x} \quad \text{and} \quad g(x) = x^2 - 10x \quad \text{for} \quad 0 \leq x \leq 9
\]
is equal to 207.
\item Calculate the radius of convergence of the series
\[
\sum_{n=1}^{\infty} \left(\frac{n+1}{n+5}\right)^n \left(\frac{n+3}{2^n}\right) x^n.
\]
\end{enumerate}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
dusky gust
2xf + fx^2 = 3 + 3x f - 9f(0)
urban copper
can u draw it in latex
dusky gust
can u draw it in latex
cmd?
urban copper
do u have pen and paper
urban copper
ok, write the latex then
urban copper
fix RHS
solid kilnBOT
チラチーノ
dusky gust
,, f(x) = f(3) + f'(3)(x-3) + 1/2! f''(3)(x-3)^2 + ...
solid kilnBOT
チラチーノ
dusky gust
you need taylor polynomial of order 2
so you can ignore rest of it
you need f''(3) and f'(3)'s values
and f(3)
dusky gust
**チラチーノ**
put x = 0
you get f(3) = 1/3
,, \item 2xf(x) + x^2f'(x) = 3 + 3x f(x) - 3
\item 2xf(x) + x^2f'(x) = 3x f(x)
solid kilnBOT
チラチーノ
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
dusky gust
,, x^2f'(x) = x f(x)
solid kilnBOT
チラチーノ
dusky gust
now you put x = 3 here
,,f'(3) = (3 \times 1/3)/3^2 = 1/9
how to write multiply symbol
digital cloak
$\times$
solid kilnBOT
Pseudonium
dusky gust
ok thx
you get f'(3) = 1/9
you already got f(3) = 1/3
wait my calculations are bad
let me edit
urban copper
please check your calculations
solid kilnBOT
チラチーノ
urban copper
dusky gust
now looks good
solid kilnBOT
チラチーノ
dusky gust
**チラチーノ**
differentiate it again and put x = 3
you will get f''(3)
now just put there and you get the expression
(1st question solved)
solid kilnBOT
urban copper
,, f^{(2)}(3) = \frac{3f’(3) - f(3)}{9}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
,, f’’(3) = \frac{\frac{1}{3}- f(3)}{9}
,, f^{(2)}(3) = \frac{\frac{1}{3}- f(3)}{9}
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
f’’(3) = 0
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@urban copper Has your question been resolved?
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@urban copper Has your question been resolved?
carmine spade
winged hinge
f(3) = 1. $3^2f(3) = 33 + 0$ (integral from 9 to 3*3 = 9 is 0), so f(3) = 1
winged hinge
nah I'm on mobile lol
solid kilnBOT
992DuckEnthusiast (duck eater)
forest vortex
@urban copper
urban copper
<@1187916384341078070>
@forest vortex
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@urban copper Has your question been resolved?
fair bison
@urban copper what are you working on right now?
I'm confused, what part of the question are you stuck on
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spice granite
.close
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halcyon stratus
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halcyon stratus
I need solve all questions
halcyon stratus
I can solve first order equations
But I haven’t tried the second order
I know few methods
dark sundial
Have you read your book or lecture notes?
halcyon stratus
I’m perfectly good till first order
dark sundial
you have to think of functions that have a derivative that is of the same type as the original functions
the exponential satisfies this requirement
dark sundial
If you attended the class of second order odes
or read the sectian that explains it
halcyon stratus
I didn’t attend unfortunately
halcyon stratus
I’m tensed
dark sundial
just a second
halcyon stratus
4 total problems and the time is running
I believe 1B 1C 1D follow the same pattern?
clear cloud
Am I supposed to know this?
You can check for first order differential équations, you always need exp for things like y' = ay
dark sundial
but there's a catch
you only have 1 possibility for lambda in this equation
and there's a theorem that states
that a second order linear ode has 2 lambdas
halcyon stratus
After you found lambda, does the problem complete?
halcyon stratus
I feel overwhelmed
dark sundial
y0 and y'0
dark sundial
then all you can say is y(t)=Ae^(3t)+Bte^(3t)
dark sundial
I’m tensed
how much time left do you have?
dark sundial
it's more than enough
halcyon stratus
Really?
dark sundial
yes
halcyon stratus
But I don’t wanna waste ur time
halcyon stratus
Are u okay with helping me throughout the entire assignment?
dark sundial
I can help you with some of the questions, the others are just a repetition
halcyon stratus
then all you can say is y(t)=Ae^(3t)+Bte^(3t)
Where did A and B come from
halcyon stratus
I can help you with some of the questions, the others are just a repetition
Thanks bro
dark sundial
the ode is a linear one
so the solution is a linear combination of possible solutions
if you had 2 different lambdas you'd have y=Ae^(lambda1)+Be^(lambda2)
in the specific case of a repeated lambda, 3 and 3 in the first problem
you have to put a t, before the second solution
the independent variable
halcyon stratus
Kinda confusing tbh
dark sundial
A quadratic always has 2 solutions right ?
yes
but they are equal
the parabola touches the x axis in just 1 point
at the vertex
dark sundial
A quadratic always has 2 solutions right ?
36-4*9=0
halcyon stratus
Makes sense the root is zero is here
dark sundial
yes
dark sundial
you have to substitute lambda 1 and 2 in the answer I sent you
halcyon stratus
Thanks for this formula
dark sundial
you're welcome
halcyon stratus
I’ll solve bc and send u the solutions
dark sundial
differential equations is my favorit topic
halcyon stratus
What does graph the solution in interval mean
halcyon stratus
differential equations is my favorit topic
It’s cool yeah
dark sundial
What does graph the solution in interval mean
you have to plot the solution
and use y0 and y'0 to determine A and B
you have a system of linear equations
dark sundial
in this case you have 2 different roots
dark sundial
this is the correct formula for this case
the case where you have a 2nd order ode with 1 root on the polynomial equation is the special case
halcyon stratus
dark sundial
you've applied the wrong formula
dark sundial
use this one
dark sundial
It doesn’t have T?
just when you have just 1 lambda
halcyon stratus
dark sundial
substitute back A and B in the equation
and plot it
I'm helping you and another person, at discord too, in another language, at the same time.
halcyon stratus
No worries
dark sundial
This is way more exciting than being a teaching assistant, of students who don't even want to attend lectures.
halcyon stratus
This is way more exciting than being a teaching assistant, of students who don't...
Lmao I skipped so many classes
I hate summer classes
dark sundial
But you're asking for help now
dark sundial
I hate summer classes
My university went out of a strike recently. Our academic calendar is messed up.
halcyon stratus
My university went out of a strike recently. Our academic calendar is messed up.
That’s sad, can I copy the graph from a graphing calculator
dark sundial
That’s sad, can I copy the graph from a graphing calculator
I think so
halcyon stratus
Can u give me few mins
I need to copy 1a 1B to the answer sheet
4 more hours till deadline
halcyon stratus
6:33 pm in New York
dark sundial
@halcyon stratus finished copying?
halcyon stratus
Few more mins please I had some other work
dark sundial
ok
halcyon stratus
Sorry for making u wait
My fucking dasher canceled my order and I haven’t ate anything today
dark sundial
this is not healthy
go eat something
I'm at home
We in Brazil generally go to university at the city we live in, with our parents.
dark sundial
Then, solve it
halcyon stratus
Done copying
halcyon stratus
I’m moving on to c
dark sundial
I'm done with dinner
trim joltBOT
@halcyon stratus Has your question been resolved?
halcyon stratus
im having trouble with d
halcyon stratus
this is 16 not 6. and 4.5
dark sundial
ok
halcyon stratus
do u think we have time
halcyon stratus
(-2-i*sqrt(6))/2
its sqrt of 16
the complex roots are -1+2i
if u figured out 2,3,4 let me know
dark sundial
ok
halcyon stratus
,afk eating
dark sundial
calculating the 3a
halcyon stratus
C is solved
dark sundial
I belive that in 3a he is using Alpha and Beta instead of A and B as constants