help-49

[lyric charm]

yeah

[radiant lagoon]

for the given norm

then how can I construct an example to show the space is incomplete

[lyric charm]

the idea is that you might have functions whose values become smaller but whose derivatives explode, and that's how your limiting function ends up outside C^1.

er

ok not exactly "limiting function ends up outside C^1"

ok sorry i messed up a little

give me a function that belongs to C^0 but not C^1

[radiant lagoon]

ln x

[lyric charm]

on [0,1]?

[radiant lagoon]

no

|x-1/2|

[lyric charm]

ok

now try to cook up a sequence of cont-diff functions that converges to this one

[radiant lagoon]

$\sqrt{\left(x-\frac{1}{2}\right)^{2}+\frac{1}{n}}$

[grand pond]

Xyrokryen

[midnight plank]

@radiant lagoon Has your question been resolved?

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[midnight plank]

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[flint anchor]

i found fxx fyy and fxy but im kinda lost in D=fxx*fyy-fxy^2

[flint anchor]

because we'd have -6x*((3xe^y)-(9e^3y))-(3e^y)^2

is that correct?

[fallow scarab]

Show how you got this D

[flint anchor]

well fxx=-6x fyy=(3xe^y)-(9e^3y) fxy=3e^y

D=fxx*fyy-fxy^2

[fallow scarab]

Try factoring D

[flint anchor]

for fyy=(3e^y)(x-3e^2y)

[fallow scarab]

That's not D

[flint anchor]

oh D

i mean i could just find solution for D

for the given values

(-18x^2)(e^y)+(54xe^3y)-9e^2y

@fallow scarab sorry if i shouldnt mention you

[fallow scarab]

?

[flint anchor]

like this?

[fallow scarab]

Yea just stop pinging me

[flint anchor]

sorry

[fallow scarab]

This isn't factored

Factoring is like 3+9 = 3(4)

[flint anchor]

i know

but to get here i factored

-6x *((3xe^y)-9e^3y))

[midnight plank]

@flint anchor Has your question been resolved?

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[flint anchor]

can someone help idk where to start

[twilit field]

what is property 10

[twilit field]

is this stewart

[flint anchor]

yes

[twilit field]

can you please show property 10

[flint anchor]

this is the first time the book mentions it

[fallow scarab]

Doubt

[twilit field]

this is property 10

[fallow scarab]

Lmao

[flint anchor]

ok

[twilit field]

can you solve it now

[flint anchor]

i'll try

[midnight plank]

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[flint anchor]

so im kinda confused with this problem

[flint anchor]

idk where to start

[gaunt nimbus]

okay

lets consider the max value of sqrt{4 - x^2y^2}

since x^2y^2 always >= 0

-x^2y^2 always <= 0

therefore

4 - x^2y^2 <= 4

for area of the first one

its just

a circle

second one

use the fact that -1 <= sinx <= 1

area is triangle

[flint anchor]

wait

[flint anchor]

i dont get this part

where did u find that

[gaunt nimbus]

do u agree that a^2 >= 0

for any a

[flint anchor]

why

[gaunt nimbus]

parabola

range of parabola

[cold oracle]

my math teachers send my class this

Is there a set whose size is strictly between that of the integers
𝑁
N and the real numbers
𝑅
R?

idk how to solve

[flint anchor]

occupied channel @cold oracle

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[cold oracle]

sorry

[quartz hornet]

its okay!

[flint anchor]

well yeah this makes it better but in a test im not allowed to use desmos

[quartz hornet]

if i square a negative

what will i get

[flint anchor]

+- smth

or wait

[quartz hornet]

square man

a=-5

[flint anchor]

yeah my bad

[quartz hornet]

a^2?

[gaunt nimbus]

its a known inequality

[quartz hornet]

yeah do that for any number

for any real number a , a^2 > 0

[flint anchor]

wait

-5^2 like this u mean?

u get a positive

[flint anchor]

so i just have to know that

[quartz hornet]

(-5)^2

[flint anchor]

yeah 25

[quartz hornet]

(-1)^2 (5)^2

(-1)^2 =?

[flint anchor]

1

[gaunt nimbus]

u should know that if u're doing calc 3...

[quartz hornet]

i mean yeah-

[flint anchor]

i suck 😭

[gaunt nimbus]

did u take ap calc bc

anyhow

do u agree tho that x^2y^2 = (xy)^2 >= 0

[flint anchor]

whats ap and bc

[flint anchor]

yeah

[gaunt nimbus]

gud

now

-x^2y^2 <= 0, right?

negative changes sign

[flint anchor]

for both x and y?

[gaunt nimbus]

like from this inequality

[flint anchor]

kinda contradicting

[gaunt nimbus]

${(xy)^2 \geq 0 \implies -(xy)^2 \leq 0}$?

[grand pond]

k

[flint anchor]

x can be both - and + right

[gaunt nimbus]

yes

any possibility

[flint anchor]

yeah so i agree

[gaunt nimbus]

right

so

adding 4 to both sides

${4 - x^2y^2 \leq 4}$

[flint anchor]

yeah

i agree again

[grand pond]

k

[gaunt nimbus]

i made a mistake 🥀

[gaunt nimbus]

this

[flint anchor]

still agree

[gaunt nimbus]

ok so

the maximum inside the square root is 4, right?

[flint anchor]

question tho

[gaunt nimbus]

then what is the maximum after taking the square root

in other words

what is M?

[flint anchor]

in the exercise it says (xy)^2 <=1

or no sorry

nvm

[gaunt nimbus]

x^2 + y^2 leq 1

[flint anchor]

so why did we look for this like is it a standart procedure for this type of exercise?

for (xy)^2 => 0

[gaunt nimbus]

its a great way to find max/min

[flint anchor]

ahh ok

[gaunt nimbus]

oh

then

what is M

[flint anchor]

isnt it the minimum?

[gaunt nimbus]

why do u think its the minimum

[flint anchor]

well because of this

the lowest we can go is 4

if it was 0 for x^2 + y^2

[gaunt nimbus]

lowest or highest

[flint anchor]

lowest but i feel like thats wrong and idk why

[gaunt nimbus]

the expression is always less than or equal to 4

so 4 is the maximum inside the square root

[flint anchor]

oh yeah true

[gaunt nimbus]

now

to the minimum

we have to be wary of our constraint

x^2 + y^2 <= 1

for the minimum

we want the highest value of (xy)^2, right?

[flint anchor]

hmm

im lost

im sorry im just slow 😭 bare with me

[gaunt nimbus]

ok so

our f(x,y) is ${\sqrt{4 - x^2y^2}}$

[grand pond]

k

[flint anchor]

yes

[gaunt nimbus]

notice how (xy)^2 brings the value down as it increases?

[flint anchor]

yeah

i just tried it

i mean goes both ways

because -2 for x and y would give the same result as +2

[gaunt nimbus]

remember our constraint

x^2 + y^2 = 1

[flint anchor]

<= but yeah

ok

[gaunt nimbus]

ok

so

do u agree with me that the maximum value is on this line

[gaunt nimbus]

.

[flint anchor]

ok so -1 0 1

[gaunt nimbus]

doesnt have to be that

[flint anchor]

yeah i know im just saying

[gaunt nimbus]

ok so

from x^2y^2

and x^2 + y^2 = 1

we can rewrite the final eq as

x^2(1 - x^2)

[flint anchor]

why = 1

or are u doing it wrong

[gaunt nimbus]

the maximum (xy)^2 lies on the border of the area

[flint anchor]

all i know now is this, beyond this im confused

[gaunt nimbus]

oh

do u understand tho

that the maximum is 4

[flint anchor]

yeah

i get that

[gaunt nimbus]

at point (0,0)

ok

now to the min

[flint anchor]

yes

[gaunt nimbus]

to find min, we must maximumze x^2y^2

[flint anchor]

yeah

[gaunt nimbus]

to do that we think of the possible xy values

because of our constraint

[flint anchor]

they are infinite tbh

[gaunt nimbus]

x^2 + y^2 <= 1

[flint anchor]

oh yeah

[gaunt nimbus]

it should lie on the edge

cuz being inside of the circle decreases our xy value

[flint anchor]

like 1 and 0

[gaunt nimbus]

yes, like 1 and 0

we know that it is on the border

but where on the border?

to do that

[flint anchor]

so we want 2 values that are close to this

hence 1 and 0

[gaunt nimbus]

1 and 0 is an example of a point that is on the border

but isnt the max value

[flint anchor]

yeah

[gaunt nimbus]

to find it

we express x^2y^2 in another way

[flint anchor]

it had to do with this right

[gaunt nimbus]

recall that x^2 + y^2 = 1 means that y^2 = 1 -x^2

[gaunt nimbus]

ye, our x and y are restricted by this

[flint anchor]

no i meant the 1 and 0

[gaunt nimbus]

so what we seek to maximise is now x^2y^2 = x^2(1-x^2)

[gaunt nimbus]

they are within that inequality, yes, but arent the correct answer

[flint anchor]

yeah

[flint anchor]

ok

[gaunt nimbus]

do u know how to maximise x^2(1-x^2) from calc 1

[flint anchor]

ok how did we go here

[gaunt nimbus]

y^2 = 1-x^2

[flint anchor]

yeah

how does that give us x^2y^2 = x^2(1-x^2)

[gaunt nimbus]

substitute y^2

[flint anchor]

with what

[gaunt nimbus]

1-x^2

[flint anchor]

ohhh

i see what u did

[gaunt nimbus]

ok

so

do u know what x value maximises x^2(1-x^2)?

[flint anchor]

1?

[gaunt nimbus]

using calc 1 techniques

[flint anchor]

i cant recall ngl

[gaunt nimbus]

x^2 - x^4

derivative

2x - 4x^3

find critical point set = 0

2x(1-2x^2) = 0

2x(1-sqrt{2}x)(1+sqrt{2}x) = 0

critical points are 0, -1/sqrt{2}, 1/sqrt{2}.

in the question, we are considering only for x >= 0

therefore

x is either 0 or 1/sqrt{2}

cant be 0

evidently

so lets check if 1/sqrt{2} is our max

do a second derivative test

2 - 12x^2

gives - 4

so it is the max

${x^2y^2}$ is maximum at ${x = \frac{1}{\sqrt{2}}}$

[grand pond]

k

[gaunt nimbus]

can u find what y is?

from our

x

[flint anchor]

where did we get x^4

[gaunt nimbus]

$x^2(1-x^2) = x^2 - x^4$

[grand pond]

k

[flint anchor]

damn

1-2x^2 this?

@gaunt nimbus sorry if im bothering

[gaunt nimbus]

wdym

[gaunt nimbus]

thats fine. ur not bothering

[flint anchor]

here

[gaunt nimbus]

factorisation of 2x - 4x^3

[flint anchor]

can you write this as latex please

[gaunt nimbus]

[ 2x(1-\frac{1}{\sqrt{2}}x)(1+\frac{1}{\sqrt{2}}x) = 0 ]

[grand pond]

k

[gaunt nimbus]

,w 2x(1-2x^2) = 0

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by nuclearinteraction

[midnight plank]

@flint anchor Has your question been resolved?

[gaunt nimbus]

u good m8?

[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[grim sierra]

(algebraically)

[lyric charm]

no algebraic solution

but also WHAT is going on with those black shapes.

[grim sierra]

i just graphed it to make it easier to under stand 🙂

[olive matrix]

i am so not doing this right now

[grim sierra]

can you show me out the steps

[midnight plank]

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as this action is irreversible, and this channel may abruptly lock.

[random fox]

Slove this

Ignore the four options above the question

[midnight plank]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[wary ferry]

Did I do this right

[desert siren]

Did you do the desmos thing

[wary ferry]

I don't get the function

[desert siren]

Oh ic ic

Lmao w (5x^2 - 4x)/(x+1)^4 =0

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by executor.1011

[desert siren]

Your critical point at 4/5 seems a little off

The limits all seem right tho

[desert siren]

Better

Lmao w (-10x^2 + 22x- 4)/(x+1)^4 = 0

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by executor.1011

[wary ferry]

lmao w?

[desert siren]

Yeah ok the graph seems right

[desert siren]

Lmao is my personal prefix lol

[wary ferry]

oh

[desert siren]

Use ,w if you wanna use it

[wary ferry]

,w then the problem

oh

i guess it works

[desert siren]

Graph seems correct

Any other doubts?

Within this q

[wary ferry]

this just seemed a bit wonky since i couldnt find y values

[desert siren]

Its fine u got it

Well almost except for the crit pt

But yeah

[wary ferry]

well i may have overlooked that part a little

[desert siren]

Nice work

[wary ferry]

all the limits look good?

[desert siren]

Lolol fr i dont even look at inflection pts while drawing

Just the crits

[wary ferry]

i should start doing the crits first

[desert siren]

Yes the limits look good

[desert siren]

Yeah

[wary ferry]

could you help me with the next problem

[desert siren]

K

Man this is nice hw i like it

I used to have hw like this when i was learning these topics

But i never did it

Cos i was too lazy

[last slate]

bro this is literally so easy

[wary ferry]

I'm brain dead tho

[desert siren]

Don't.

[last slate]

i am not exaggerating, in india even a 8th class can student do this , (8th class student is like 13 yo)

[desert siren]

<@&268886789983436800>

[last slate]

am sorry am i violating any server rule

[warm hazel]

If you can't say anything helpful, say nothing.

[solid iris]

do not belittle anyone over difficulty of a problem

[kind ledge]

😭😭

[desert siren]

!status

[midnight plank]

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

[wary ferry]

setup

[round parcel]

So, the volume formula for this is V = lwh, where l is length, w is width, h is height.

[last slate]

just curious , is this college level mathematics in usa? also just solve it by using l=2b

[wary ferry]

calc in college

[round parcel]

So, we can start filling things in.

[last slate]

this is calculus?

[round parcel]

The volume is 30. The length is 2w.

V = lwh
30 = 2w^2h.

So, what is h in terms of w?

[warm hazel]

Leave the channel, please.

[wary ferry]

why is 2w to the power of 2h

[round parcel]

It's to the power of 2.

Exponents come before multiplication in PEMDAS or BODMAS.

(30 = 2w^2h)

[grand pond]

Chai T. Rex

[wary ferry]

oh it seemed like the h was in the exponent the way it was typed

[sudden yacht]

Nope

The way it was typed seemed like it was not in the exponent

Otherwise there would have been a pair of brackets, i.e. w^(2h)

[round parcel]

So, solve that for h. What do you get?

[wary ferry]

15/w^2

[round parcel]

OK, good.

Now we have everything in terms of h.

Now we need to look at the sides.

Oh, sorry in terms of w.

The bottom is the length times the width.

So, the bottom is lw, which is 2w^2.

[last slate]

can we not calculate h by taking diagonal of the box?

[round parcel]

What is the area of one of the sides?

[last slate]

l

and width is 2l

[wary ferry]

xh or 2xh

[round parcel]

Good.

Now put it in terms of x.

We have the bottom as 2x^2.

[wary ferry]

what

[round parcel]

Well, you said that h = 15/x^2.

[wary ferry]

yes

[round parcel]

You said that one side is xh.

Fill in 15/x^2 for h.

[wary ferry]

wym put in terms of x

[round parcel]

I mean the only variable should be x.

h shouldn't be there.

[last slate]

how ca he eliminate x

when value of x is not given

[round parcel]

<@&268886789983436800> Shark is a troll that was told to leave.

[last slate]

there are 2 variables in this question

how is this troll , see it yourself

value of x is not given

[daring dagger]

this channel is occupied by someone else right now nevermind didn't see previous context

[desert siren]

for someone who claims this is easy you sure are frickin it up

[formal blade]

That's not helpful either

It's best to assume they're just watching the channel interested in the problem despite the earlier comments.

[desert siren]

Hm alright i apologise

[wary ferry]

im lost rn

[round parcel]

You said xh is one of the sides, right?

And h = 15/x^2.

So, what you can do is to substitute 15/x^2 for h, like this: x(15/x^2).

Does that make sense?

[wary ferry]

15/x?

[round parcel]

Right.

You said the other side was 2xh.

What do you get after you substitute 15/x^2 for h in that?

[last slate]

there is no such side as xh its x and 2x sir

[warm hazel]

I see you can't help yourself but comment on this, so let me help you resist the urge to keep interrupting.

[wary ferry]

how are the sides and that equation combined?

im lost

[round parcel]

We'll see that in a minute.

[wary ferry]

im confused what im doing

[round parcel]

Whenever you have a = b, you can replace a with b.

Which is what we're doing here.

[wary ferry]

so h=hx?

[round parcel]

We have h = 15/x^2.

No, h = 15/x^2.

[round parcel]

We can go back to this ^

You said the two side areas are xh and 2xh.

[wary ferry]

yes

[round parcel]

You also said that h = 15/x^2.

So, we're going to use substitution, which means we can replace one side of an equation with the other.

So, we can see h = 15/x^2 is an equation with h on one side.

We can see that xh is the area of one of the sides.

[wary ferry]

but h is the hight and hx is an area

[round parcel]

Right.

We want it to only have one variable, though.

Right now that area has two variables in it.

[wary ferry]

are you trying to tell me hx=15/x^2

[round parcel]

No.

We got that h = 15/x^2, not hx.

Now that we know that, we can change h to 15/x^2.

So, if we had 7h or something, that would be 7(15/x^2).

If we had h^2 + h, that would be (15/x^2)^2 + (15/x^2).

What I'm doing there is replacing h with 15/x^2.

Does that make sense?

[wary ferry]

little

[round parcel]

Let's try a simpler example.

Let's say that x = 2 and y = 3.

What's x + y?

[wary ferry]

5

[round parcel]

Right, and you used substitution there.

x + y
2 + 3
5

You replaced x with 2 because x = 2.

[wary ferry]

yes

[round parcel]

You replaced y with 3 because y = 3.

You can also do that on more complicated things.

[desert siren]

Oh thats a good way of explaining imma steal that

[wary ferry]

how is this related

[round parcel]

Well, that's what I'm talking about with hx.

We know that h = 15/x^2, right?

[wary ferry]

oh i substitiute in 15/x^2 for h then the x tags along

[round parcel]

Right.

hx = (15/x^2)x = 15/x

Now we only have one variable with that one.

[wary ferry]

same with the othe

[round parcel]

Right, what do you get for 2hx?

[wary ferry]

30/x

[round parcel]

OK, good.

Now we want to figure out prices.

The base (the bottom side) is $15 per square meter.

The area of the bottom is 2x^2.

So, what's the price of the base in terms of x?

[wary ferry]

900

wait

450

[round parcel]

How did you get that?

Oh, I see.

No, that's not the way to do that.

x isn't the area.

[wary ferry]

just a guess

i saw a number and a variable and plugged em in

[round parcel]

If you have a price per area and an area, you do price/area * area.

That'll give you the price.

(\frac{\text{price}}{\text{area}} \cdot \text{area} = \text{price})

[grand pond]

Chai T. Rex

[round parcel]

Does that make sense?

The areas cancel.

[wary ferry]

i know that]

[round parcel]

OK, so we have the price/area as $15/m^2.

We have the area as 2x^2.

So, 15 * 2x^2 = price.

price/area * area = price.

Does that make sense?

[wary ferry]

na

this is probably not right

[round parcel]

No, x is a length, right?

So, you can't fill in $15 for x because that's a price, not a length.

[wary ferry]

oh

unfortunate

[round parcel]

They give us two prices per area.

$15/m^2 for the bottom.

$4/m^2 for the top.

m is meters, which is a length.

So, m^2 is an area.

[round parcel]

Right.

Now you used price/area * area to get price.

There's a small mistake.

Remember that there are four sides.

[wary ferry]

Ok

[round parcel]

You have only two there.

[round parcel]

Good.

Now we have the price in terms of one variable.

[round parcel]

OK, and you can also add the first two together.

[wary ferry]

Ok

[round parcel]

So, we have the price as 360/x + 30x^2.

[wary ferry]

yes

[round parcel]

Now we want the cheapest, so we want to minimize the cost.

So, how can we find minima using calculus?

[wary ferry]

dirive

[round parcel]

Right, we can differentiate it.

[ancient bear]

this is calc?

[wary ferry]

yes

[round parcel]

So, what's the derivative of 360/x + 30x^2?

[ancient bear]

dang 9th grade gonna be easy

[round parcel]

OK, good.

Now how can we use the derivative to give us some minima?

[wary ferry]

find x

[round parcel]

Well, we can find x, but we need an equation for that.

Right now we don't have an equation (with =), we only have an expression.

What happens when you reach a local minimum is that it comes in from the left going downwards, then it goes off to the right going upwards.

,w plot x^2

[grand pond]

Results provided by WolframAlpha

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Requested by chai.t.rex

[round parcel]

Like the minimum there has the graph coming in from the left going downwards.

Then it has the graph going on to the right going upwards.

[wary ferry]

yes

[round parcel]

So, the derivative tells you the slope, right?

[wary ferry]

yes

[round parcel]

On the left of the minimum, the slope is negative.

On the right of the minimum the slope is positive.

What's the slope at the minimum?

[wary ferry]

0

[round parcel]

Right.

So, the derivative will be zero.

So, we can do -360/x^2 + 60x = 0.

[wary ferry]

thats what i ment when i said find x

[round parcel]

Oh, OK.

So, what do you get for x?

[wary ferry]

to be honest im blanking on algebra,lol

[round parcel]

OK, so first get rid of denominators.

How can you do that?

[wary ferry]

multiply by x2

[round parcel]

Right, so multiply both sides by x^2. What do you get?

[wary ferry]

this seems wrong but i got cube root of 6

[round parcel]

Yes, that's right.

So, x = cbrt(6).

[wary ferry]

yes

now sub

?

[round parcel]

Yes, substitute that into the price formula we got.

[wary ferry]

is that the same as 6^(1/3)

[round parcel]

Yes.

[wary ferry]

just checking

[round parcel]

1/n as an exponent is the nth root.

[wary ferry]

297.1734524

is my math mathing

[round parcel]

How did you get that?

Oh, wait.

[round parcel]

Yes, that's correct, but unless they want you to put it in the form of a decimal, you should do (90 \cdot 6^{\frac23}).

[grand pond]

Chai T. Rex

[desert siren]

Lol i didn't expect the answer to be so cursed looking at the question

[wary ferry]

tbh i dont know if i would be able to do a similar question

[round parcel]

OK, so we got the minimum price.

We just need the dimensions (don't forget to put m for meters).

[desert siren]

Just resolve this twice after finishing rn

You'll get the hang of it

[desert siren]

One time looking at soln one time without

[wary ferry]

how do i find this

[round parcel]

OK, so we had this ^

So, the length is 2x, the width is x, and the height is 15/x^2.

[wary ferry]

ye

oh

so that

answers are in the top right

[round parcel]

How did you get 8.25 for the height?

Oh, I see.

Remember that it's 15/x^2.

Not 15/x.

@wary ferry

You can also check by multiplying all three together to see if you get the volume of 30 the problem has.

[midnight plank]

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[wary ferry]

I see

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[night hawk]

!status

[midnight plank]

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

[night hawk]

3

In the book, were told to assume that x, y, and z are nonnegative, but I got z = -6.

btw when I write circle(number, number) -> ..., it means that i’m using those equations to get a new equation

and an arrow pointing into a circle with a number means ill later refer to that as equation ...

[midnight plank]

@night hawk Has your question been resolved?

[night hawk]

<@&286206848099549185>

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[viral dagger]

for all $n$, prove that $\sqrt[n]{\sqrt{3}+\sqrt{2}}+\sqrt[n]{\sqrt{3}-\sqrt{2}}$ is irrational

[grand pond]

skissue.in.a.teacup

[viral dagger]

status 1

[midnight plank]

@viral dagger Has your question been resolved?

[fallow scarab]

did you try recurrence relation and induction?

[topaz epoch]

I remember doing this for case n = 100, I'm not sure this is exactly right, it may help though

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[midnight plank]

@molten bay Has your question been resolved?

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

<@&268886789983436800>

[proven cairn]

g(z)=10 z^3 and h(z)=e^z -z+1/z-4 since in rouche theorem |g(z)| > |h(z)| and f(z)=g(z)+ h(z) , f(z) and g(z) are analytical function

f(z) and g(z) have the same number of zeroes (counted with multiplicity)

[molten bay]

How will you check it satisfied the condition of

|g(z)|>|h(z)|

[proven cairn]

This function has a singularity (pole) at 𝑧=4 because of the term 1/z-4 and then |z|<4
​

you can check it?

The choice of 𝑔(𝑧) and ℎ(𝑧) in Rouche Theorem depends on the value of ∣𝑧∣

[proven cairn]

I took this in order to find number of zeroes

[proven cairn]

this is true for |z|=2,3

[midnight plank]

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[molten bay]

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[fleet smelt]

is this chanel open

to use?

[midnight plank]

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[tired spruce]

I need help if anyone knows how to use desmos. I need to graphic a harmonic series (sum of 1/n or wtv) but I also need to show a bar graph of all individual squares under the function and kinda show how they add up and round off? does anyone know how to do that

[dreamy lichen]

you could use 0 <= y <= 1/N {N <= x <= N+1} or sth like this

[tired spruce]

can you please show me how to do this but with all the boxes adding up together?

thats the part I cant really figure out

[dreamy lichen]

wdym by adding up?

like stack them on top of each other?

[tired spruce]

yes

[dreamy lichen]

hmm

I suppose that's doable

[tired spruce]

like its supposed to show that if you keep adding individual blocks it will follow the curve of it

but idk how bro

[dreamy lichen]

oh logarithm curve

Maybe I could use 2 colors to distinguish the individual boxes?

so thaat its not just one big red box

[tired spruce]

either way is fine

[dreamy lichen]

okay

ill use these 2

hold on

the way im doing it will take ages to compute

im thinking of doing it recursively

is this what you meant?

[tired spruce]

yes how did you do this

[dreamy lichen]

it takes ages to render though, computing that sum is terrible

maybe i could define new sequence

hold on

[tired spruce]

the first 10 is fine but does L have to be the end of the sequence?

[dreamy lichen]

https://www.desmos.com/calculator/olg8p3csvn

idk if its actually quicker

[dreamy lichen]

You can define L to be as high as you need it to be

just make it even or else it will break i think

[tired spruce]

ill prob just do 10 I just need a screenshot of a sectino of the graph

im just gonna change the variables that shouldnt be an issue right cuz im using n as a list of 10k terms to generate the series as a dotted curve

[dreamy lichen]

yeah, it shouldnt

[tired spruce]

wait is this simplier if its one color?

[dreamy lichen]

oh you dont even need N

[dreamy lichen]

Possibly

[tired spruce]

like

https://www.desmos.com/calculator/ufyuseeewk

I just needed something like this

but with series of 1/n

[dreamy lichen]

this is how it looks with one color

it doesnt look as bad as i thought it would

oh because im using a different way than i originally intended

makes sense

[tired spruce]

is that just with the one box instead of subtracting two ?

[dreamy lichen]

Yeah

instead of spliting it to N1 and N2, its just N

[tired spruce]

is there anyway I can see the equation for that?

[tired spruce]

yea understood

sorry for overcomplicating it to you

[dreamy lichen]

sure

and this is H

H is basically harmonic sum up to N

[dreamy lichen]

this curve fits it even better

ln(x+0.5) + 0.577... (mascheroni)

[tired spruce]

yes this works perfectly thank you

[tired spruce]

yea I read that somewhere but im only in calc 1 its alright

[dreamy lichen]

the sum is between ln(x) + mascheroni and ln(x+1) + mascheroni, so ln(x+0.5) + mascheroni is quite a good approximation

[tired spruce]

ok tysm

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[steep condor]

why does order get reversed when we transpose a product of matrices

[dawn dagger]

Maybe this is something worth looking into
https://math.stackexchange.com/questions/1279861/why-intuitively-is-the-order-reversed-when-taking-the-transpose-of-the-product#:~:text=the intuitive reason is that,for traveling along the paths.

I even saw some nice visuals.

[grim vector]

Yep

[steep condor]

cool lol

idk what are feynman path integrals tho

will learn all that i guess

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[unkempt canopy]

Explain like im 5

If dx= 1/10
And x= 100
Shouldnt x+dx= 100+10/100 or just 100+10?

And what If dx = 1/100 instead of 1/10

Would it still look like 100+1?

[glass lantern]

100 + 10/100 isn't 100+10

get your arithmetic checked,

and 100+1/100 would be 100.001

[unkempt canopy]

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[twilit field]

I started off by finding the eigenvalues of T

which were ±i

so I thought that x^2+1=0 is a polynomial satisfying this

So the minimal polynomial is x^2+1

[obtuse basin]

Yeah, that’s the minimal polynomial

[runic hamlet]

no way thats how axler intended this

you are supposed to use the construction from the proof

[twilit field]

oh

okay, I'll try that then

thanks

[twilit field]

thanks

[runic hamlet]

I dont know how many results he covered about the relationship between the eigenvalues and the min poly

[twilit field]

this is how axler cover the construction

There is also "Eigenvalues are the zeros of the minimal polynomial"

which has been covered

[runic hamlet]

thats not enough to recover the min poly

[twilit field]

this too

[runic hamlet]

you need to argue that the eigenvalues arent repeated in our case

but still, I very much doubt that you are supposed to do this with the eigenvalues

[twilit field]

okay, I'll try constructing it similar to the proof

[twilit field]

The proof way would require me to construct tha matrix instead and work with that, right

I suppose this method works too

the Matrix is $\begin{bmatrix} 0&-1\1&0 \end{bmatrix}$

[grand pond]

wai

[twilit field]

so T(e_1)=e_2

T(e_2)=-e_1

T^2(e_1)=T(T(e_1))=T(e_2)=-e_1

this gives is the minimal poly is x^2+1

I think I'm getting it

thanks

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[molten bay]

if |z|=1 then how can we check |e^z|<=e?

[dreamy lichen]

try using z = a + bi

and think about what |e^z| becomes

[molten bay]

e^a.1

a is real value

[dreamy lichen]

if |z| = 1, what values can a have?

and thus what values can e^a have?

[molten bay]

sqrt(a^2+b^2)

a^2+b^2=1

[dreamy lichen]

just imagine it

imagine |z| = 1

which point will maximize the real part? The a

[molten bay]

yes it is circle at orifign

[dreamy lichen]

the a is the x-coordinate

[molten bay]

which point means?

[dreamy lichen]

which x-coordinate can appear on the circle

[molten bay]

(1,0)

[dreamy lichen]

This is the point with largest a, yes

[molten bay]

yes

[dreamy lichen]

and that point is gonna have magnitude e^1

when you do e^z on it

[molten bay]

ohh i got it

thanks

[dreamy lichen]

it can be also shown that the lower bound is e^(-1)

or 1/e

since the leftmost point is (-1, 0)

[dreamy lichen]

np

[molten bay]

i see

thanks

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[molten bay]

I'm applying argument principal here but what will be roots of this polynomial?

[midnight plank]

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[neat silo]

solve over reals
$$\sqrt{6x-2} = 2 + \sqrt[3]{x+5}$$

[grand pond]

acgn

[tawdry laurel]

h m m

i would begin by letting x+5 = a^3

,rccw

[tawdry laurel]

and guess a root

@neat silo

i have the solution

[tawdry laurel]

do you know how to proceed from here?

[midnight plank]

@neat silo Has your question been resolved?

[neat silo]

3

[tawdry laurel]

wat

[neat silo]

Yes

[tawdry laurel]

wait

[neat silo]

x = 3 is a sol

[tawdry laurel]

it works?

okay

mine was 2, it also works

[tawdry laurel]

ok

[grim vector]

a = 2 yeah

but x = 3

[tawdry laurel]

ah mb

yes

[neat silo]

Ty

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[dusty portal]

Just wanna check that my proof is correct

[dusty portal]

By definition, right?

[robust swallow]

when i multiply 4b^4(a^4) do i get 16b^4 a^4 or 16a^4 b^4??? or is it same thing

[tawdry laurel]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tawdry laurel]

your last two products are equivalent btw

[robust swallow]

alr

[dusty portal]

waaaaaaah

Oh shit, didn't even notice you were green

[runic hamlet]

just replace all your z's by 1

|z| < 1 after all

[tawdry laurel]

yay

[dusty portal]

Mkay

I'm sorry if I'm being paranoid but that is sufficiently small |z|, right?

Oh wait

💀

Haha I think I screwed up the proof a tiny bit

$|p(z)|=|a_0+a_1z\ldots a_nz^n|\le|a_0|+|a_1z|\ldots|a_nz^n|=a_0+a_1|z|\ldots a_n|z|^n\le a_0+a_1\ldots a_n\le K$ for some $0<|z|<1$ and $K\ge\sum_{k=0}^n a_k$, $K>0$, which means that since $|z|$ is sufficiently small that $p(z)=O(1)$ as $z\to 0$ by definition.

[grand pond]

;(

[dusty portal]

And a K will always exist, since the sum is finite

Yipee

I'm a little stuck on the second question though

How do I show that $a_0+a_1|z|\ldots a_n|z|^n\le K|z|^n$ for sufficiently large $|z|$?

[grand pond]

;(

[dusty portal]

Oh wait

😈

I can show that

$\frac{a_0}{|z|^n}+\frac{a_1}{|z|^{n-1}}\ldots a_n\le K$ for sufficeintly large $|z|$

[grand pond]

;(

[dusty portal]

AHAHHAAH

ezpz

Since $|z|$ is large (and hence 0<$\bigg\vert\frac1{z}\bigg\vert^n$<1) we have that every term will tend to zero and hence all we really need is $\frac{a_0}{|z|^n}+\frac{a_1}{|z|^{n-1}}\ldots a_n\le K$ which is now trivial!

[grand pond]

;(

[dusty portal]

Yeah we can just let $K\ge\sum_{k=0}^n a_k$, $K>0$ again can't we

[grand pond]

;(

[subtle blaze]

why does it say for some 0 < |z| < 1

shouldn't it be for all such z's?

[runic hamlet]

or, replace every |z|^k by |z|^n

[brittle grotto]

uh, I may be going crazy, but why is it fine to pull the $a_i$ out of the norms? it can be the case that $a_i<0$, right?

[grand pond]

00100000

[dusty portal]

Convienience

[runic hamlet]

|a_i| always

[dusty portal]

I forgor 💀

[subtle blaze]

it sounds like your inquality only holds for some z's and some K's

but you need for all z's for the limit to work

[dusty portal]

But |z| is smol?

[subtle blaze]

you said your inequality holds for some 0 < |z| < 1

what if it's only 0.5

and it doesn't hold for any other z's

[dusty portal]

Ohhh

[subtle blaze]

then you don't have the result that p(z) is in O(1)

[dusty portal]

Bad wording 💀

I meant to say for all z with the property 0<|z|<1

[subtle blaze]

no

for all z's

[dusty portal]

Alright

[subtle blaze]

that's the same statement as before

[dusty portal]

I think I just jumped ahead with the wording 😔

$|p(z)|=|a_0+a_1z\ldots a_nz^n|\le|a_0|+|a_1z|\ldots|a_nz^n|=|a_0|+|a_1| |z|\ldots |a_n| |z|^n\le |a_0|+|a_1|\ldots|a_n|\le K$ for all $z$ such that $0<|z|<1$ and $K\ge\sum_{k=0}^n|a_k|$, which means that since $|z|$ is sufficiently small that $p(z)=O(1)$ as $z\to 0$ by definition.

[runic hamlet]

|a_i|

and just choose K= sum |a_i|

[dusty portal]

By the same logic as the previous exercise we have that $|p(z)|\le |a_0|+|a_1| |z|\ldots |a_n| |z|^n$. Since $|z|$ is large (and hence 0<$\bigg\vert\frac1{z}\bigg\vert^n$<1) we have that every term containing $\frac1{|z|}$ is small and hence $\frac{a_0}{|z|^n}+\frac{a_1}{|z|^{n-1}}\ldots a_n\le K\implies |a_0|+|a_1z|\ldots |a_n| |z|^n\le K|z^n|$ for $K\ge\sum_{k=0}^n |a_k|$. Therefore $p(z)=O(z^n)$ as $z\to\infty$, by definition.

[dusty portal]

Yeah I know

Working on it

[runic hamlet]

why does the term a_0/z^n + ... + a_n matter

you are pulling it out of nowhere

[dusty portal]

Wdym

Ah shit

[grand pond]

;(

;(

[brittle grotto]

isn't the first part of the problem just an equivalent statement to ``for all $a_i$, there exists some $C$ such that $\limsup_{z\to 0}\frac{a_0+\ldots+a_nz^n}{C}<\infty$"? Then, since $\limsup$ is linear and each $a_iz^i$ approaches 0 as $z\to 0$, blah blah blah

[grand pond]

00100000

[dusty portal]

whar

whar 💀

[brittle grotto]

I might be going crazy

I've been going crazy a lot recently

so take my fuckery with a grain of salt 💀

[dusty portal]

Ok

[midnight plank]

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[twilit field]

Am I tripping or is factorisation the best way to solve this

[twilit field]

nvm

$T^5+2T^4-7T^3-6T^2+5T+4=0$.\
Applying $T^{-5}$ to both sides \
$1+2T-7T^2-6T^3+5T^4+4T^5=0$

[grand pond]

wai

[runic hamlet]

why is that the min poly of T^-1

[twilit field]

As the min poly of T is of degree 5 , T^{-1} has 5 eigenvalues as well, and thus is of degree 5

[runic hamlet]

no

[twilit field]

As T is invertible, all eigenvalues are non-zero

[runic hamlet]

T can have up to 5 eigenvalues

it doesnt need to have 5

[twilit field]

hmm

[twilit field]

suppose it had a lower degree,applying T raised to the additive inverse of the most negative power to both sides, we'd get a poly of lower dgree than 5

like T^{-3}+T^{-2}=0 (say). applying T^3 to both dies we'd get T+I=0, violating the minimality

[runic hamlet]

write that down properly

[twilit field]

okay

Suppose the minimal polynomial had a lower degree than 5. say $m$. Applying $T^m$ to both sides of the minimal polynomial , we'd find that the minmal polynomial of T is of lower degree than 5, resulting in a contradiction

[grand pond]

wai

[runic hamlet]

too imprecise

let m(x)=sum a_i x^i

[twilit field]

Let $m(x)= a_ix_i$ be the minimal polynomial of $T^{-1}$. Suppose $m<5$. We thus have $\sum_{i=1}^{m} a_iT^{-i}=0$ . Applying $T^{m}$ to both sides gives us $\sum_{i=1}^{m} a_iT^{m-i}=0$.This would correspond to a polynomial of degree $m$ , violating the minimality of the polynomial of $T$

[grand pond]

wai

[runic hamlet]

awful

[twilit field]

what's wrong

[runic hamlet]

so much

have you even read what you produced

[twilit field]

yes, just realised

Let $p(x)= \sum_{i=0}^{m} a_ix_i$ be the minimal polynomial of $T^{-1}$. Suppose $m<5$. Letting $x=T^{-1}$, We thus have $\sum_{i=0}^{m} a_iT^{-i}=0$ . Applying $T^{m}$ to both sides gives us $\sum_{i=0}^{m} a_iT^{m-i}=0$.This would correspond to a polynomial of degree $m$ . However $m<5$, and the degree of the minimal polynomial is given to be $5$, this is a contradiction. Thus the minimal polynomial of $T^{-1}$ must be of degree 5 or more.

Is this better

[runic hamlet]

no

[twilit field]

What's wrong?

[runic hamlet]

.

literally right at the start is garbage

the name m is already taken

you cant use it for the degree

[grand pond]

wai

[runic hamlet]

x^i

to which polynomial does the last expression correspond

[twilit field]

Which expression

[runic hamlet]

\sum_{i=0}^m a_i T^{m-i}

[twilit field]

p?

[runic hamlet]

it needs to be q(T) for some polynomial q

its not equal to p(T)

[twilit field]

so it corresponds to a new polynomial q

[runic hamlet]

yes

[twilit field]

Let $p(x)= \sum_{i=0}^{m} a_ix^i$ be the minimal polynomial of $T^{-1}$. Suppose $m<5$. Letting $x=T^{-1}$, We thus have $\sum_{i=0}^{m} a_iT^{-i}=0$ . Applying $T^{m}$ to both sides gives us $\sum_{i=0}^{m} a_iT^{m-i}=0$.This would correspond to a polynomial,$q$, of degree $m$ . However $m<5$, and the degree of the minimal polynomial of $T$ is given to be $5$, this is a contradiction. Thus the minimal polynomial of $T^{-1}$ must be of degree 5 or more.

[grand pond]

wai

[twilit field]

Is this better

[fallow scarab]

x^i, not x_i in the beginning

[twilit field]

Let $p(x)= \sum_{i=0}^{m} a_ix^i$ be the minimal polynomial of $T^{-1}$. Suppose $m<5$. Letting $x=T^{-1}$, We thus have $\sum_{i=0}^{m} a_iT^{-i}=0$ . Applying $T^{m}$ to both sides gives us $\sum_{i=0}^{m} a_iT^{m-i}=0$.This would correspond to a polynomial,$q$, of degree $m$ . However $m<5$, and the degree of the minimal polynomial of $T$ is given to be $5$, this is a contradiction. Thus the minimal polynomial of $T^{-1}$ must be of degree 5 or more.

[grand pond]

wai

[midnight plank]

@twilit field Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[runic hamlet]

but what is the polynomial q

[runic hamlet]

if p(x)= sum a_i x^i, then q(x) = ?

what are the coefficients of q

you have to know these