help-13

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[hot crag]

my man

please stop pinging us

[solemn fern]

i only pinged once bro

i waited 15 mins

[hot crag]

...?

a. it'd been 5 minutes since you posted when you pinged us the first time
b. it'd been 13 minutes since you posted when you pinged us the second time
c. the factoid states both wait 15m and ping ONCE

[solemn fern]

is it really that strict? 2 mins?

if thats the case then sorry for the 2 mins

[hot crag]

*sighs*
ykw leave it

[solemn fern]

ok so are you the helper?

[hot crag]

i help as per my discretion, sorry

[solemn fern]

and what does that mean?

[hot crag]

...?

[solemn fern]

bro im new to this how do i get help

[hot crag]

just wait okay

[solemn fern]

alright

<@&286206848099549185>

[hot crag]

yep, good

that works see

[solemn fern]

omg u were there the whole time waiting for it

[hot crag]

(now again i cant help, this is one of my weak points so im sorry)

[hot crag]

nah lol

[solemn fern]

BRUH

[hot crag]

sorry :(

someone else will cm dw

[solemn fern]

oh im sure they will

i give up on life
who said humans are better than ai
i will ask an ai
and if it gets it wrong, at least i didnt lose half my brain cells waiting

[hot crag]

no no no dont do that

[cedar kiln]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[flint plinth]

we have an understanding

[cedar kiln]

@solemn fern Has your question been resolved?

[solemn fern]

Should Mhyd1427 wait for 30 mins and not get an answer tho😢

I squeezed my brain and google and wolfram alpha to get an approximation 66 square units

[flint plinth]

you're trying to calculate this?

[solemn fern]

Im pretty sure

But i think i mightve made a mistake forming it

[flint plinth]

it doesn't look very friendly

[solemn fern]

Really isnt

Even wolfram alpha had to approximate it

So this means that there was a mistake forming the integral

But its the only way it seems

[flint plinth]

the thing inside the square root is a perfect square

but i don't know how much that helps

$$x^8 + \frac{1}{2} + \frac{1}{16}x^{-8} = \left(x^4 + \frac{1}{4}x^{-4}\right)^2$$

[wraith dagger]

Bungo

[solemn fern]

Or maybe the question is assessing my ability to form integrals and use approximation methods maybe?

[flint plinth]

possibly, what's the context, is this a calculus course or what

[solemn fern]

Yep it is calculus

I have the question posted i think here

Here

This was my attempt

[cedar kiln]

@solemn fern Has your question been resolved?

[cedar kiln]

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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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[burnt glacier]

hi! thers a way to solve this limit using euler?

[cedar kiln]

@burnt glacier Has your question been resolved?

[cedar kiln]

@burnt glacier Has your question been resolved?

[versed siren]

you mean by using euler constant?

[burnt glacier]

umm i know the answer is going to contain the euler constant but theres a limit pattern that we can write as euler

this one

[upper ruin]

You can use the fact that $x$ can be written as $x = e^{\ln(x)}$

[wraith dagger]

Alberto Z.

[burnt glacier]

yes thats the solution , but i wanted to know if we can use this rule instead

[versed siren]

Idk, it will be easier tu use x = e^lnx

[upper ruin]

It doesn't seem to easily apply here

[cedar kiln]

@burnt glacier Has your question been resolved?

[burnt glacier]

Is there any way we can do it?

Youre right. I know its not ideal but i would like to figure this out

[upper ruin]

I don't think so, or at least I can't figure it out

[pastel vault]

I would suggest using the most appropriate strategy for the question

Rather than trying to think you can apply one strategy to all of them

[upper ruin]

I agree

[cedar kiln]

@burnt glacier Has your question been resolved?

[burnt glacier]

i agree as well but my professor did somehow managed to solve it this way

i tried to solve as he did but i only made it halfway

this were i got stuck at

he started like this but from this point im kinda stuck

[pastel vault]

That's crazy

So he didn't take the logarithm of that

[burnt glacier]

a crazy man do crazy things i guess lol

but it did got me curious about it

[burnt glacier]

how can we pull this off from this point?

[pastel vault]

Fair, your question is going to be sitting for a while though

[burnt glacier]

Thanks. How did you managed to put lnx in the numerator?

[raw gulch]

etc

[burnt glacier]

can you try it with this method?

[raw gulch]

there is initial mistake here, so I'm not going to continue that, since: (1+x)^1/x --> e, when x->0, and in your case, x - 1 does not go to zero when x ->0, clear?

[raw gulch]

this way, you can't solve it,

you need this:

but not what you wrote in your paper

[raw gulch]

to obtain the required result, I suggest you to use what I've written first

unless you figure out some sophisticated substitution of the variable

[burnt glacier]

But thats what i did eventually

I “added zero” to the limit

/doubled it by 1

[raw gulch]

yes, but x - 1, must go to zero if x -->1

and in your case

x - 1 goes to -1

so you cant replace it with e

[burnt glacier]

oh so it only works if x—/—>0 ?

[raw gulch]

your expression must go to zero,

not mandatory x

but all this expression behind 1

you see ?

x - 1, here, goes to zero

i understand what you wanted to do, sure, but in your case, it failed

do you know the oroginal answer ? if not i can tell you

then you can verify your calculations

[cedar kiln]

@burnt glacier Has your question been resolved?

[gentle pecan]

Are we permitted to use LHopitals Rule?

If so, ths limit isn't too bad to find

[cedar kiln]

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[fresh loom]

any help here?

[cedar kiln]

@fresh loom Has your question been resolved?

[fresh loom]

.close

[cedar kiln]

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[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[naive ember]

can someone explain wtf is happening here?

[cedar kiln]

@naive ember Has your question been resolved?

[cedar kiln]

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[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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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.

[raw yoke]

minor problem; they never taught us how to do questions where you need to find the line of intersection

I've got the cartesian equations for each

-4Xx1-3Xx2-5Xx3-7=0

3Xx1+5Xx2+x3+8=0

[cedar kiln]

@raw yoke Has your question been resolved?

[main needle]

here's an example video https://www.youtube.com/watch?v=Q68F0Tr0-q8

you want to eventually get to equations like x1= ... + ...*x3 and x2= ... + ...*x3

just have to cancel out some rows in a matrix first

[pastel vault]

Find the normal vectors of both planes using the cross product

Then the line of intersection must be normal to the two normal vectors you get

If you draw a diagram you can see why

Oh right once you get your line, sub x, y, z into ax + by + cz = d for any one of the planes

That's how you find a point the line passes through

[cedar kiln]

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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.

[crimson sedge]

HI

[crimson sedge]

i dont understand this,

im new to sin / cos stuff

if anyone could explain this, it would mean SO MUCH

[tepid wraith]

this not trigonometry...

[crimson sedge]

o..

IM SORRY

I DONT UNDERSTAND IT

i dont know it

[tepid wraith]

i don't see A and B

[crimson sedge]

oh

[tepid wraith]

why you doing calculus without trig knowledge

[gusty forum]

probably meant P and Q
i can tell the underpaid teacher energy just from that ss

[tepid wraith]

lol

for P x is 0 because it's on y axis

[crimson sedge]

okay

[tepid wraith]

now plug in x=0 in y=x^2-4x+6

[crimson sedge]

so i sub in?

oh okay

[tepid wraith]

ye

[crimson sedge]

0^2 - 4(0)+6

like this?

[tepid wraith]

yes

what do you get

[crimson sedge]

so its 6

[tepid wraith]

what's 6

[crimson sedge]

P

wait

WAIT

Y

[tepid wraith]

yes

y coordinate of P is 6

now what would the point be

[crimson sedge]

wait

do i sub in the y?

y = 6

[tepid wraith]

no

[crimson sedge]

oh okay

[tepid wraith]

you got x and y

for P

[crimson sedge]

0, 6

[tepid wraith]

yes!

[crimson sedge]

(0 , 6)

OH OKAY

SO I KNOW THAT

B AKA Q

IS

LIKE

X = 5?

[tepid wraith]

yes

[crimson sedge]

omg okay

wait

5^2 - 4(5) + 6 = the y for q

[tepid wraith]

please talk in a singular or few sentences

[crimson sedge]

okay sorry

[tepid wraith]

yes

[crimson sedge]

25 - 20 + 6 = 11 ?

[tepid wraith]

yes

write coordinate

[crimson sedge]

y = 11 therefore, coordinates for q is (5 , 11)

[tepid wraith]

yep

now do you know integration?

[crimson sedge]

dy/dx etc ?

[tepid wraith]

that's differentiation dear but ill be good if you know differentiation

[crimson sedge]

shucks T.T

[tepid wraith]

you don't know integration?

[crimson sedge]

I just learnt it pretty sure..

[tepid wraith]

or differentiation

[tepid wraith]

good

what's a definite integral used for

[crimson sedge]

um is it unique values

[tepid wraith]

huh

it's used for area under curves

[crimson sedge]

ou

okay let me write that down rq

[tepid wraith]

do you know how to integrate?

[crimson sedge]

not really.. T.T

[tepid wraith]

for example $\int x^n dx=\frac{x^{n+1}}{n+1}+C$

[wraith dagger]

The Prophet Of The Damned

[crimson sedge]

OH YEAH I KNOW THAT

[tepid wraith]

great

that's all you need

[crimson sedge]

when it goes to the opposite its like n-1 rightt?

[pastel vault]

Yeah so what's the integral of x^2?

[pastel vault]

I don't know what you mean here

[tepid wraith]

same

[crimson sedge]

sorry, maybe i got mixed up with a different topic for that

[tepid wraith]

answer dis

[crimson sedge]

[tepid wraith]

plug x=2 into this

[crimson sedge]

okay

[tepid wraith]

bruh

[crimson sedge]

sorry

[tepid wraith]

you'll learn dw

[tepid wraith]

what you get

[pastel vault]

Not x = 2

Plug n = 2

Very important difference

[tepid wraith]

mb

i forgo lmao

[crimson sedge]

okay wait

x^3/3 + c

?

[tepid wraith]

yep

[crimson sedge]

how did you know it is to plug n=2

[tepid wraith]

now what's the integral of -4x

[pastel vault]

Cause when n = 2, x^n = x^2

[tepid wraith]

because x^2 compare it with x^n n=2

[pastel vault]

Ah actually, what's the integral of x

Think about what the exponent is if you just have x

[crimson sedge]

^0

[pastel vault]

I feel they need smaller steps than this

[tepid wraith]

-4 factors out that's why

[pastel vault]

Nope, x^0 = 1

[pastel vault]

I know we'll get to that later

[tepid wraith]

man

[crimson sedge]

oh ya whatever^0 is 1

[pastel vault]

Yeah basically cause you have x^0 = x^(2 - 2) for example
= x^2 / x^2 = 1

You can have any other number instead of 2 and it will still cancel

[tepid wraith]

kim why did you decide not to learn calc before trig

[crimson sedge]

oh okayy

[pastel vault]

Okay so now what's the exponent of x

[tepid wraith]

x=0

[crimson sedge]

i dont know anything

[tepid wraith]

learn things in order dude

[crimson sedge]

by the way, is this in O levels syllables?

[tepid wraith]

huh

[crimson sedge]

nevermind

[tepid wraith]

answer this rn

don't get distracted

[crimson sedge]

n?

[pastel vault]

O levels I think doesn't have integration

I need to check, anyways

[pastel vault]

Nope, x^n is just x^n

[crimson sedge]

okay im not done for 😭

[tepid wraith]

in regular ol' x he meant

[tepid wraith]

the exponent of x^n is x^n

[crimson sedge]

oh, i dont quite understand that sorry

[tepid wraith]

now answer

[pastel vault]

Honestly take your time

[crimson sedge]

I don't understand this, i meant

[tepid wraith]

what is the exponent of x in x^1

[crimson sedge]

exponent of 1 equals to the base, so is it x

[tepid wraith]

no x is in the exponent would mean x^x

[crimson sedge]

OHHHH

[tepid wraith]

if i say $x^n=x^1$ what is the value of $n$?

[wraith dagger]

The Prophet Of The Damned

[crimson sedge]

1

[tepid wraith]

yes

[tepid wraith]

so plug n=1 into #help-13 message

[crimson sedge]

x^2/2 + c

how do I tell from the question, when to plug in a number?

[tepid wraith]

great

[tepid wraith]

ask yourself this question

replace 1 with the power that's given

[crimson sedge]

okay

[tepid wraith]

may i now?

[pastel vault]

I was helping someone else

Yeah go ahead

[tepid wraith]

ok do you know an integral is a linear operator

[pastel vault]

Let's say the green function is f(x)

And the blue function is k * f(x)

So the area under the green function is the integral of f(x)

[pastel vault]

Too complicated, use simpler words

[crimson sedge]

no .. 😭

[tepid wraith]

definite integral

[pastel vault]

Yep yeah

[pastel vault]

Okay so what must the area under the blue function be?

[tepid wraith]

id use my mother tongue for that

[crimson sedge]

wait huh?

[pastel vault]

I should rephrase

If the green function is f(x) and the blue function is k * f(x)

How much larger is the area under the blue function compared to the green function?

[tepid wraith]

by what factor

[crimson sedge]

squared? :D .. sorry i dont know

[pastel vault]

Okay, so do you see that all we're doing to the shape under the function

[crimson sedge]

yeah

[pastel vault]

We're just stretching that shape in the y-direction

And we're increasing it by a factor of k

k times larger

[crimson sedge]

k is constant?

[pastel vault]

Yep k is a constant

If we have k, we mean it's a constant

Like if we have x, y, or z, we mean it's a variable

[crimson sedge]

how can we tell which one it is?

[pastel vault]

You just have to read and get exposure to more maths

[crimson sedge]

okay

[pastel vault]

It's like how you get better at understanding the events in a book or how characters develop if you read more

[crimson sedge]

ooo yeah thats true

[pastel vault]

Yep

So because the integral is just the area under the graph

[tepid wraith]

definite integral*

[pastel vault]

The integral of k f(x) dx = the area under k f(x) = k * the area under f(x) = k * the integral under f(x)

[pastel vault]

I know I'm simplifying things

$\int k f(x) \ dx = k \int f(x) \ dx$

[wraith dagger]

south

[pastel vault]

You can have a variable such as x in your limits of integration

And it's still true

[pastel vault]

That's an indefinite integral since you will get a function of x, not a number

[tepid wraith]

do you know sum rule for differentiation?

[crimson sedge]

wait can this be explained in a simpler term?

[pastel vault]

What didn't you understand about that explanation?

[tepid wraith]

i think we need a transaltor

[crimson sedge]

no 😭 ..

[pastel vault]

No they just need things broken down into really small steps

You need patience

[tepid wraith]

you skipped precalc and differentiation

oh god

[tepid wraith]

it was a joke chill

[crimson sedge]

sorry, at what age do people learn this usually.

[tepid wraith]

im 13 idk

[crimson sedge]

im cooked

[tepid wraith]

south is in uni

[tepid wraith]

and asian

[pastel vault]

Wait for real

Nahhhh man

You can't trick me now

[crimson sedge]

im 15 and i dont know this...

im SO DONE

[tepid wraith]

but in india it's 15-16

[pastel vault]

It's not about age actually

[tepid wraith]

yes for real real

[crimson sedge]

thats true

[pastel vault]

underage discord, sus

[tepid wraith]

whenever 11th or 12th grade

[tepid wraith]

13 is allowed bro 😭

13 not against tos

[pastel vault]

Nah I mean your account was created in... yk

[tepid wraith]

i turn 14 in may

[tepid wraith]

🤫 🧏

[crimson sedge]

wow man

im actually envious of people that are hella good in maths like this

[pastel vault]

Everyone has different talents

[tepid wraith]

get back to topic

[pastel vault]

I suck at drawing

[tepid wraith]

im not 😔 \

[pastel vault]

Okay I might explain the next thing then

So if you have f(x) + g(x)

[crimson sedge]

okay

[pastel vault]

The area under f(x) + g(x) is just the sum of very thin rectangles

And the height is f(x) + g(x) ofc

[tepid wraith]

knowledge doesn't imply success in math

[pastel vault]

I'll pull up a picture

[tepid wraith]

practice does

[tepid wraith]

practice is a real one

[pastel vault]

Like these rectangles in practice would be so thin, their width would be pretty much 0

So the area would be exact

[tepid wraith]

imma ignore the riemann sum on the right

[pastel vault]

Yeah like ignore the formula, it is relevant but like it's not worth going into

[tepid wraith]

not rn

[crimson sedge]

oh i see

[tepid wraith]

but it's a jumpscare when limit of a sum

[pastel vault]

Yeah the height of these rectangles is going to be f(x) for the particular x value

If you let the rectangles have width almost-0

Then every x will have a rectangle corresponding to it

So the 'height' is just f(x)

And the width we call dx, dx stands for a very small number on the x-axis

[crimson sedge]

okay

[crimson sedge]

wait

is that for every qn

[pastel vault]

Yep, so if we use the same widths of the rectangles for f(x) and g(x)

[pastel vault]

Yes

The height of a function is just f(x) for that value of x

[crimson sedge]

can you give an example question

[pastel vault]

Basically this explains why $\int f(x) + g(x) \ dx = \int f(x) \ dx + \int g(x) \ dx$

[crimson sedge]

then explain the answer (u dont have to if its too yea)

[wraith dagger]

south

[pastel vault]

The height of the function on the left is f(x) + g(x)

And the height of the right hand side is also f(x) + g(x)

And all the widths are the same

[crimson sedge]

how do you tell

[pastel vault]

So the combined area of all the rectangles is the same -> the integrals are the same

[pastel vault]

We just make it happen that way

[crimson sedge]

okay

[pastel vault]

You can always choose some points on the x-axis of f(x)

And make those points the same for g(x)

[crimson sedge]

OH YEAH

like any coordinates?

[pastel vault]

The assumption is that these are the areas from x = a to x = b, for both of them, sorry

[tepid wraith]

this implies $\int af(x) + bg(x) \ dx = a\int f(x) \ dx +b\int g(x) \ dx$ when combined with the scaling example

[wraith dagger]

The Prophet Of The Damned

[pastel vault]

Yeah, so $\int a f(x) + b g(x) \ dx = \int a f(x) \ dx + \int b g(x) \ dx = \cdots$

[wraith dagger]

south

[pastel vault]

Combining the two rules together

Okay now we can do your question, the area of region A1 which is the region under the parabola

That's $\int x^2 - 4x + 6 \ dx = \int x^2 \ dx + \int -4x \ dx + \int 6 \ dx$

[wraith dagger]

south

[tepid wraith]

we're speedrunning calc 1

[pastel vault]

Using the f(x) + g(x) rule for 3 functions

Like (f(x) + g(x) + h(x)) = (f(x) + g(x)) + h(x), break it into 2 first

And then break it again

$= \int x^2 \ dx + (-4) \int x \ dx + 6 \int 1 \ dx$

[wraith dagger]

south

[pastel vault]

Using the k f(x) scaling rule

$= \frac{x^3}{3} + (-4) \frac{x^2}{2} + 6x$ plus a constant

[wraith dagger]

south

[pastel vault]

(We add a constant cause if you differentiate an integral, you get the original function back)

(So if you differentiate a constant, you get 0 and you get the original function back as well)

With me so far? This is a lot I know

[crimson sedge]

well.

if im quite honest, ive never been taught these formulas

[pastel vault]

True

[tepid wraith]

underpaid teacher alert

[crimson sedge]

😭

[pastel vault]

We're almost there I promise: we can turn this into a number which tells us the area

[crimson sedge]

ive never been taught this in year 10

idk why

[tepid wraith]

this is the homerun

[pastel vault]

Now we can actually just sub x = 5 in, that gives us a number we can call A
And sub x = 0 in, that gives us a number we can call B

Then the integral from 0 to 5, or the area under the parabola is just B - A

[tepid wraith]

it's year 11/12 stuff usually

[crimson sedge]

oh i see

[crimson sedge]

OH

[pastel vault]

Like I know we're subbing in individual numbers

But the area from x = 0 to x = 5

Is just (area from x = -1 to 5) - (area from x = -1 to 0)

Or (area from x = -10000 to 5) - (area from x = -100000 to 0)

You get the idea

[crimson sedge]

what im getting is that A1 is the integral of the graph? so we plug x=5 in the equation and A2 is area under PQ - A1 ?

[pastel vault]

Yeah so A2 is the integral of the equation of the line - parabola

In other words, the area under the line, minus the area under the parabola

That's A2

[crimson sedge]

okay

[pastel vault]

But actually, if you notice, the area under a line is just a trapezoid

[crimson sedge]

can u highlight wheres the parabola

[pastel vault]

Let me scroll up

[crimson sedge]

okay

[pastel vault]

I highlighted the parabola in blue

[pastel vault]

Hopefully you can see this is true

[tepid wraith]

we need slope for that

[pastel vault]

Yeah so of course the first thing would be to figure out the equation of the line

[pastel vault]

So that we can minus the parabola and get a function

[tepid wraith]

handwriting is shit, sorry

[tepid wraith]

delta x is 5-0

you can see its 5 units

now delta y would be the difference of the y coordinates of p and q

now remember what y coordinate of p and q are

[crimson sedge]

(0,6) (5, 11)

so

6 and 11

[tepid wraith]

yez

the difference of 11 and 6 is?

[crimson sedge]

5

[tepid wraith]

so delta y is?

[pastel vault]

Or if you know the slope already, just tell us

[crimson sedge]

is it 11/6 ?

[tepid wraith]

no

[crimson sedge]

wait 6/11

[tepid wraith]

it would be the difference

[crimson sedge]

oh i thought changes in Y = Y1/Y2

[crimson sedge]

5 then?

[tepid wraith]

yep

[crimson sedge]

okay

[tepid wraith]

what was delta x

[crimson sedge]

5

[tepid wraith]

so what will be the slope

@crimson sedge

[crimson sedge]

then

minus a1 for the final answer?

[tepid wraith]

bruh

ok

[crimson sedge]

is it wrong.........

[pastel vault]

Yeah I was saying the area under a line is a trapezoid

It's correct for when you have a line

[crimson sedge]

the change in Y as x increases

so uh

[tepid wraith]

nvm you got it champ

[tepid wraith]

nah but looking at solution isnt

[crimson sedge]

this is so stressful 😭

[tepid wraith]

practice

gtg

@pastel vault take

[pastel vault]

You need a break

Like honestly just come back to this tmw

And then you can practice more if you want to

[crimson sedge]

okay

thank you so much again

[pastel vault]

No worries

[cedar kiln]

@crimson sedge Has your question been resolved?

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[civic coral]

How do I tell whether the cross product of OA and OB is parallel to (1,-2,2,)

[coral jewel]

cross product of two vectors is the normal vector of the plane containing those 2 vectors

it doesnt make sense for a vector to br parallel to a point

[civic coral]

Emmm

(1,-2,2) is vector

[carmine dock]

If vectors u, v are parallel
then we can write u=kv where k is a scalar

[cedar kiln]

@civic coral Has your question been resolved?

[broken yoke]

If you are asking about if 1i-2j+2k is a parallel to OA X OB then 1i-2j+2k=k(OA X OB) here k is a constant. In other terms if two vectors are parallel it means they are multiple of one-another.

[cedar kiln]

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[oblique lynx]

in my signals and systems class, one TA uses such notation

$$
\langle f(t),g(t) \rangle = \int_{-\infty}^{\infty}f(t)g^*(t)dt
$$

Is this property/notation possible to use only in Hilbert spaces? I thought that this notation is used for inner product.

[wraith dagger]

konxmok

[oblique lynx]

also can somebody please help my understand, why is Hilber space so special? Why dont we use like some normal vector space. We touched it a bit in my quantum mech class but i dont se much why is it used (i dont have strong background in functional analysis since i study engineering)

[crimson delta]

well hilbert space is basically vector space + inner product

and yes thats precisely the notation for an inner product

[crimson delta]

(you also need that the space is complete)

[oblique lynx]

but isnt inner product also defined on normal vector space?

[crimson delta]

and together with that inner product they are a hilbert space

[oblique lynx]

also do you think that you may can point me to some literature that would help me understand the basics of spaces? I always had rather hard time with this topic and it will come handy for me

[crimson delta]

when we say something is a hilbert space, we say that it is a "normal" vector space and it has an inner product

basic of spaces?

[oblique lynx]

oh okey

[crimson delta]

not sure what thats supposed to mean

[oblique lynx]

sorry ofr the formulation, in lin alg we defined some abstract spaces, then here in physics and signals they throw hilbert spaces and i am just confused by the topics, so i thought that i will try to understand it more deeply (even when its not actually needed for any of my classes)

[crimson delta]

for vector spaces any linear algebra book should be good. for specifically banach/Hilbert spaces any functional analysis book should give some information

[oblique lynx]

as we are engineers, they dont really "explain it", they just say that it works good f.e. in the Hilbert space

[oblique lynx]

okey, thats what i thought, we have special non-mandatory class on functional analysis and measure theory, i will take it next year

also i have one more question, i have to calcuate correlation functio R12 of these two signals

I know the cor. func. is defined as

$$
R_{12}(\tau) = \langle s_1(t+\tau),s_2(\tau) \rangle =\int_{-\infty}^{\infty}s_1(t+\tau)s^*_2(t)dt
$$ but i am not really sure how to construct the integral with the tau shift, can somebody help me construct it? (i am able to calculate it, just not to costruct it)

[wraith dagger]

konxmok

[oblique lynx]

.close

[cedar kiln]

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[somber juniper]

i need help with upper and lower bounds theorem. i just learned that a polynomial can have multiple upper and lower bounds. when the question asks us the upper bound what we're supposed to tell?

[somber juniper]

like it makes 0 sense

i dont understand this theorem at all. i know this in general but everday i learn that i know smthn wrong abt this theorem

1. how do i find the upper or lower bounds for a polynomial?
2. also learned that the greatest value we find using rational root theorem is the upper bound and the lowest value we find is the lower bound. Is this True or False?
3. i just learned that a polynomial can have multiple upper and lower bounds. when the question asks us the upper bound what we're supposed to tell?
4. im so confused what this theorem even means. as i know upper bound mean we can have no roots greater than this and this works with the same (but in opposite) with the lower bound

these are my questions abt this theorem can someone pls hel me with these

[idle tusk]

wdym by multiple upper/lower bound?

[somber juniper]

i used a precalc calculator and it told me this

hold on

[crimson delta]

well if for example -10 is a lower bound, then clearly -20 also is a lower bound

its a "worse" bound, but its still a bound

[somber juniper]

okay i think i dont even know what does bound mean

okay pls lets start from the beginning im abt to go crazy bc i've been trying to finish this damn lesson for the past 6 days

and i have been trying to find the stupid bounds of a polynomial for the past 3 hours

what is this theorem

i dont get it

[carmine dock]

Do you mean the bounds of the roots of the polynomial?

[somber juniper]

im this close to quit math and major in social sciences and become unemployed fır the rest of my life

[somber juniper]

i dont know

well yeah i guess

idk what the question is even asking okay

it says

Find integers that are upper and lower bounds for the real zeros for the polynomial

i got

$$x^3-3x^2+4$$ and i found -1 as lower bound and 4 as upper

[wraith dagger]

ates

[somber juniper]

i also got $$2x^3-3x^2-8x+12$$

[wraith dagger]

ates

[somber juniper]

i have been trying to find the upper and lower bounds of this one for 3 hours

[somber juniper]

i keep finding that -6 and -12 is the lower bound

which one im supposed to give as an answer

[carmine dock]

either one is fine i think

Do you know what is an upper/lower bound?

[somber juniper]

i knew it but im so confused that i forgot whats even that

[somber juniper]

im not sure whether i know it correctly or not

[carmine dock]

Ok so consider a set A
If a real number p is greater than or equal to any elements of A
then p is a upper bound for A

[somber juniper]

then this means we got infinity numbers of upper bound?

[carmine dock]

yep

[slate lintel]

yeah technically 3552256783362 is an upper bound

butttt there's probably a specific answer they're looking for, maybe the lowest possible integer that works as one?

[somber juniper]

then we cant answer to the questions stating a specific upper/lower bound

[somber juniper]

oh

the lowest?

[carmine dock]

Yeah maybe they are asking for the lowest upper bound

[somber juniper]

how do i find upper bound or lower bound? I use rational root theorem usually but this time it didnt work for the second polynomial

$$2x^3-3x^2-8x+12$$

[wraith dagger]

ates

[somber juniper]

this one

[carmine dock]

Can you show your work

[somber juniper]

alr wait a sec

here

12 is upper 6 is upper as well

so i just try each of these numbers and find the lowest ?

[carmine dock]

Maybe the question had specified
what is the original question

[somber juniper]

im doing question 78 atm

[carmine dock]

You can write any one of them in this case

[somber juniper]

okay so u see i wrote -1 as lower and 4 as upper

it says its incorrect

and it doesnt give answer to others

see

no 78

and the only one i did is incorrect

but -4 is also lowest here

like agh

-4 is the lowest lowest bound

[carmine dock]

Your answers are valid

[somber juniper]

it says -2

[carmine dock]

this is really weird

[somber juniper]

i think they're not i miscalculated them by putting 1 instead of 0

im redoing it and -4 is also a lowest bound

[somber juniper]

see?

i've been dealing with this for the past 6 days

once it says A is correct and on the next page it says A is incorrect

[carmine dock]

For 77, the roots are -1,2,2
So even though 3,-2 is a possible answer
4,-1 should also be correct

[somber juniper]

is -4 also correct?

for 77

also someone said the highest value we got when we're dealing with rational root theorem is the upper bound and the lowest value is the lowest bound

[carmine dock]

if you are talking about lower bound, yes

[somber juniper]

is this correct

[somber juniper]

yeah

[somber juniper]

i tried this and it works as well

so basically the highest value is the upper bound?

<@&286206848099549185>

[crimson sedge]

What the problem

[somber juniper]

is the highest value we find using rrt the upper bound?

[crimson sedge]

?

[somber juniper]

check out here

number 2

[somber juniper]

rrt as in rational root theorem

[crimson sedge]

Ohh

[cedar kiln]

@somber juniper Has your question been resolved?

[somber juniper]

.close

[cedar kiln]

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[cedar kiln]

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[calm elm]

i'm trying to draw something and my page is 24.6cm and that length is supposed to represent 31m

[calm elm]

i can't for the life of me figure out how to divide the length of the paper into meters or whatever

so when i draw a point on my paper that's supposed to be say 1.3m, i want to know how many cm or mm that is

[stoic gale]

Only use 20 cm of the page, each cm represents 2 meters or something along those lines

[calm elm]

that's a good idea, i'll do that

thanks

[stoic gale]

Nws

[cedar kiln]

@calm elm Has your question been resolved?

[cedar kiln]

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[tired delta]

Anyoneee

[nocturne reef]

!status

[cedar kiln]

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

[nocturne reef]

Sidenote, I wish my handwriting was as neat as yours lmao

[tired delta]

I don't know where to begin.

[nocturne reef]

Okie dokie, so do you know how to find extrema (maxima/minima) using the derivative?

[tired delta]

yess yess

[nocturne reef]

Okay, so how do we do that? ANd what do we get if we differentiate f(x)?

[tired delta]

by using power rule igg

[nocturne reef]

Yeah exactly, so give that a go and let me know what you get for f'(x)

[tired delta]

its 2x - 5

[nocturne reef]

Good, so what is 2x-5 equal to at a maximum/minimum?

[tired delta]

umm dont know 😭

[nocturne reef]

haha okay, so the rule is that your extrema are when f'(x) = 0

So 2x -5 = 0

So what's x?

[tired delta]

5/2

[nocturne reef]

Exactly, now they'll want it as a coordinate so we need to find the y value when x = 5/2

How do we do that?

[tired delta]

by puting the value of x

[nocturne reef]

Yeah you sub x back into f(x)

So what do we get for the y value when we sub x=5/2 in

[tired delta]

0

[nocturne reef]

I came out with -1/4 lol

$(\frac{5}{2})^2 -5 \cdot \frac{5}{2} + 6$

[wraith dagger]

TayBee

[nocturne reef]

Don't substitute into f'(x) as we know that it equals 0 when x=5/2

Substitute it back into your original function, f(x)

[tired delta]

ohh ok ok

[nocturne reef]

Okay, so our extreme point is at (5/2, -1/4)

Now is it a maximum or a minimum?

[tired delta]

maxima igg

[nocturne reef]

What does a quadratic with a positive x^2 term look like?

[tired delta]

ax^2+2x+c=0 ??

[nocturne reef]

Yeah but I mean like if you draw it on a graph, what shape is it

[tired delta]

u maybee

[nocturne reef]

Yeah exactly a u

[tired delta]

yess yess

[nocturne reef]

So is the extreme point (at the bottom of th eu) a max or min\

[tired delta]

can u send a picc i cant get it

[nocturne reef]

Positive quadratic looks like that right?

So is the green dot the maximum or minimum value of the function

[tired delta]

yess

ig its the origin ??

[nocturne reef]

Origin is irrelevant here, it's the minimum - the line represents all the values the function can take right? So that green dot is the smallest value the function can ever take

This is your specific functiuon

[tired delta]

yess yess so its absolute minimaa

[nocturne reef]

Exactly

So that's the first question done, we have a minimum at (5/2, -1/4)

As a sidenote, we can also work out if something is a max or min with the second derivative, f''(x) - differentiate f(x) twice basically. Then substitute in the value you got for x when f'(x) = 0.

If f''(x) > 0 then it's a minimum and if f''(x) < 0 then it's a maximum.

But a quadratic is something we're all familiar with so I don't thinkkkk they would need you to do that here

If you did

f''(x) just equals 2

So it's always >0 therefore your extreme point is a minimum

But anyway

Part b

What do the words domain and range mean?

[tired delta]

set of input

igg

[nocturne reef]

Yeah so the domain is the input

The range is the output

[tired delta]

yess

[nocturne reef]

The domain is basically asking the question of are there any values of x that would make f(x) undefined if you put it in

So like if you had a fraction, it would be undefinied if the denominator was equal to 0, or if you had a square root, it would be undefined if what was inside the square root was negative

But here

Are there any values of x you could put in that would cause a problem?

[tired delta]

noo

[nocturne reef]

Good, so our domain is just $x \in \mathbb{R}$

Latex has failed me

X is a real number lmao

Or alternatively (-infinity, +infinity)

Depending on how you've been taught to write domains

[tired delta]

ohh ok ok

and wb rangee?

[wraith dagger]

TayBee

[nocturne reef]

There we go XD

And range is what we get out

So what possible values can y be?

Think back to what we worked out re. the minimum and what the graph looks like

[tired delta]

2,1,0

[nocturne reef]

So our minimum y value is -1/4 remember?

[tired delta]

yess

[nocturne reef]

But as you can see it can be anything higher than that

So y is any number that is greater than or equal to -1/4

Does that make sense?

[tired delta]

yes yess

so it can be 0

ohh so 1?

[nocturne reef]

Ignore that I got muddled XD

We'll write our range as:

$y \ge -0.25$

[tired delta]

ohh ok

[wraith dagger]

TayBee

[nocturne reef]

because if you look at the graph, the lowest y value we have is -0.25

And from there it just increases

[tired delta]

yess

[nocturne reef]

But it increases forever

To infinity

So that's our range

Does that make sense? Thats q b done lol

[tired delta]

so infiity is our range ??

[nocturne reef]

No our range is y >= -0.25

That's how we write a range

[tired delta]

ohh ok ok

[nocturne reef]

The range is just an inequality representing what values y can take

Using interval notation

You could write

[-0.25, +infinity)

[tired delta]

ohh ok ok

[nocturne reef]

So part c is just putting it all together

[tired delta]

wait a min i will be right back

[nocturne reef]

When you draw a function you should include

1. x intercepts
2. y intercepts
3. extreme points/points of inflection
4. asymptotes
5. The general shape

[nocturne reef]

Np

There are no asymptotes because it's just a quadratic, we know our extreme point (the minimum), we also know the general shape, so that's 3 4 and 5 done

All we now need to work out are the x and y intercepts

So how do we do that?

[tired delta]

now we can put the values

[nocturne reef]

Yeah, so our y intercept is when x=0 and our x intercepts are when y=0

So starting with the easy one, what's our y intercept?

[tired delta]

-1/4

[nocturne reef]

No that's our minimum

So we go back to our function

x^2 - 5x + 6

And we substitute in x=0

What do we get if we do that?

[tired delta]

oh yes

-1/2'

[nocturne reef]

$y = 0^2 -5(0) + 6 = 0 - 0 + 6$

[wraith dagger]

TayBee

[nocturne reef]

What is 0-0+6 XD

[tired delta]

hmm?

[nocturne reef]

On your calculator, 0^2 - 5(0) + 6

Should just give you 6

So that's our y intercept

[tired delta]

yes yess

[nocturne reef]

You can see it labelled here

So now for our x intercepts

We substitute in y=0

$x^2 -5x + 6 = 0$

[wraith dagger]

TayBee

[nocturne reef]

So solve that, and what values do we get for x

[tired delta]

2

?

[nocturne reef]

Yes good 2 is one

But there are two solutions

What's the other one?

[tired delta]

its 0

[nocturne reef]

Nope

Factorise the quadratic or use the quadratic formula

Or use the solver on your calculator if that's allowed lmao

[tired delta]

-2

[nocturne reef]

(x-2) is one of the brackets but that gives x=2 which you've already found

What's the other one

What two numbers multiply to +6 and add to -5

[tired delta]

2 times 3?

[nocturne reef]

-2 times -3

So your two brackets are (x-2)(x-3)

Which means x-2 = 0 or x-3 =0

So x = 2 or?

[tired delta]

yess yess

they an be both

can*

[nocturne reef]

Yes, x=2 is one solution

What's the second one?

[tired delta]

x=3

[nocturne reef]

So there's are your two x intercepts

(2,0) and (3,0)

So label everything on your graph and that's the question done

Good work haha

[tired delta]

thanks sir or friend

whichever u r

whats ur age btw?

[cedar kiln]

@tired delta Has your question been resolved?

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[summer lintel]

A symmetric matrix $A$ is called positive definite if $0 < x^T Ax$.

\textbf{Lemma 6.37}

For a symmetric matrix $A$ of size $k \times k$, the following statements are equivalent:

\begin{enumerate}
\item $A$ is positive definite.
\item All eigenvalues are positive.
\item There exists a constant $C > 0$ such that $C \cdot \lVert x \rVert^2 \leq x^T Ax$ for all $x \in \mathbb{R}^k$.
\end{enumerate}

\textbf{Proof:}

First we show that (1) $\implies$ (2).

$\lambda \cdot \lVert v \rVert^2 = \lambda v^T v$ and $\lambda v^T v = v^T A v$ and $\lambda v^T v = v^T A v$.

Why is this true?

[wraith dagger]

// 1800

[pastel vault]

Recall that $||v||^2 = v \cdot v = v^T v$

[wraith dagger]

south

[pastel vault]

You're just writing the first v as a row vector v^T

[summer lintel]

why is v*v = v^t v

[pastel vault]

And then $\lambda v^T v = v^T \lambda v = v^T A v$

[wraith dagger]

south

[pastel vault]

If v = [v1 v2 ... vn]^T

Then the first entry of the row vector v^T is v1, which gets multiplied by the first entry v1 in the column vector

Similarly you get v2 * v2

So it's v1 * v1 + v2 * v2 + ... + vn * vn

[summer lintel]

so

[pastel vault]

Should be a row vector dot a column vector

You get the idea

[summer lintel]

dot product of v is v*v = v_1^2,v2^2,v3^3 etc

[pastel vault]

The dot product gives you a number, a scalar

So it's just v_1^2 + v_2^2 + ... + v_n^2

[summer lintel]

yeah that is what i meant

you add them all uip

[pastel vault]

Cool

[summer lintel]

$|v| = \sqrt{v_1^2 + v_2^2 + \ldots + v_n^2}$. Squaring both sides of this equation gives us $|v|^2 = v_1^2 + v_2^2 + \ldots + v_n^2$.

[wraith dagger]

// 1800

[pastel vault]

And yes, this is correct

[summer lintel]

and the dot product of v with itself is equal to the v^t dot v?

[pastel vault]

Yes

[summer lintel]

ok

what about the next step

[pastel vault]

This

[summer lintel]

yeah

[pastel vault]

So lambda is just a constant: we can order it anywhere

[summer lintel]

how does lambda turn in A

[pastel vault]

By definition, lambda v = A v if lambda is the eigenvalue of the matrix A

It's a definition

[summer lintel]

ok

[pastel vault]

So if $Av = \lambda v = \lambda I v$, multiplying by the identity matrix which does nothing

[wraith dagger]

south

[pastel vault]

$Av - \lambda I v = 0 \implies (A - \lambda I) v = 0$

[wraith dagger]

south

[pastel vault]

See where that comes from now?

[summer lintel]

but then we have a v too much

if lambda is equal to A v

[pastel vault]

I didn't say that

lambda v = Av

[summer lintel]

so what is the definition of lambda

[pastel vault]

lambda is the special constant

Such that lambda v = matrix A multiplied by vector v

And v is a special vector such that when you multiply it by matrix A, all it does is scale the vector along a line

And it doesn't move it away from that line

https://www.youtube.com/watch?v=PFDu9oVAE-g

[cedar kiln]

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[split egret]

Are they both 136

?

[hot crag]

theta, yes. gamma, not necessarily

mb: gamma, no

[split egret]

Then what is gamma and how to find

[hot crag]

i mean

it's fairly obvious, cointerior angles

[split egret]

So it's 124

[hot crag]

no...?

don't reply ping me

please

[split egret]

Ok sry

Wdym no...?

[hot crag]

is alr

[split egret]

?

[hot crag]

gamma is cointerior with 54 deg

what do cointerior angles sum to?

[split egret]

Yeah so it's 126

?

[hot crag]

yse

you said 124

126 is right

!done

[cedar kiln]

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[hot crag]

@split egret

[split egret]

Ok thanks bye

.close

[cedar kiln]

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[cedar kiln]

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[fathom star]

I know the power reducing formula for cos^2x: 1+cos2x/2

do the 2's turn into 3's for cubed?

1+cos3x/3

[fathom star]

hm, maybe I missed something in the book then. Dont recall seeing this, but I have a example for sin^4x. 🤔

[fathom star]

Do you just have to remember each formula individually or is there a conversion for squard to another power?

[buoyant latch]

I just google it

[fathom star]

If only I could google during test 😅

[crimson delta]

well cos^3=cos*cos^2 so you could use the one for ^2 twice

would take longer obviously

combined with double angle

easier to just remember

[fathom star]

Not sure im understanding. wouldn't doing it power reducing for ^2 be for ^4?

[crimson delta]

forget it

made a mistake in my head. would be even worse than I thought

[fathom star]

would you do the power reducing for cos^2x then multiply the result by cosx?

[crimson delta]

yes but then you have cosx*cos(2x) and then other stuff you have to worry about

[fathom star]

Still not sure what im supposed to be doing then

[crimson delta]

dw. just remember the formula for ^3

😛

[fathom star]

Not even sure why this is on the practice test based on what Im reading about it

.close

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[modern prism]

Need help with this

[pastel vault]

!status

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[sand meadow]

is any option right?

i think for bijection null space should be 0

[cedar kiln]

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[undone reef]

Pyramid SABCD base - rhombus. The length of the rhombus side is 15 cm, and the length of the larger diagonal is 24 cm. The side walls of the pyramid form an angle equal to 45 degrees with the base (Which is a rhombus). Prove that OE = 7.2cm (O - Base center, E base edge center)

[cedar kiln]

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.
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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.
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[deft saddle]

try drawing it out first

[undone reef]

I have the drawing it was given with the task

but its in another language so i didnt do a photo, i rewrote it

[deft saddle]

send anyway

[undone reef]

i think im onto something but i dont know how to turn cos alpha = 3/5 into degree

Photos quality really bad