help-17

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[jolly prism]

I struggle on drawing the tangent line of a point, would you have any tips, the question is asking to get the instantenous speed for 1/2 hour

[outer warren]

use a ruler

and adjust it until it looks like the curve is just bouncing off it around the specified location

[jolly prism]

would this be wrong?

[cursive comet]

Yea

[outer warren]

look how how your line is cutting the curve around that location

[cursive comet]

Bro is asking for help on how to draw a line

[outer warren]

using that as an example, you want something like that

[jolly prism]

alright got it thanks @outer warren

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[echo roost]

How do we resolve forces at perpendicualr and parallel

[distant smelt]

what do you mean at perpendicular and parallel?

[vocal sleet]

@echo roost Has your question been resolved?

[solid sage]

Resolve the forces in a manner where the 5g force is the hypotenuse, and extend the normal force down and you get your triangle

Using some basic angles, we can find that the top angle of the resultant triangle is also alpha

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[jolly dome]

I am getting e^1/2 and e^-2 as the ans

[daring bolt]

that's right

[vocal sleet]

@jolly dome Has your question been resolved?

[jolly dome]

but i tried entering it into the calci and it says its wrong

[daring bolt]

e^(-2) is wrong since ln(ln(x))=ln(-2) is undefined.

the other should be right

,w lim_(x to e^(1/2)) e^(2ln(ln(x)))-2+3ln(x)+ln^2(x)

[twin meteor]

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[daring bolt]

it works

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[wraith venture]

chain rule

[vocal sleet]

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✅

[wraith venture]

they just applied the chain rule to f(y) = y

-(a-b) = b-a
Figure out what they did

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[vast shale]

i dont really understand how the distance between P and L is |d + r cos theta|

[vast shale]

if dist(F,L) = d then dist(F, L) = dist(P, F)(cos(theta)) + dist(P, L) ==> d = dist(P, F)(cos(theta)) + dist(P, L)

dist(P, F) = r then d = |r cos(theta)| + dist(P, L) ==> d - |r cos(theta)| = dist(P, L) ??

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[inland cargo]

I dont see why if a_(l+1) = 0 implies that f1 f2... fn are linearly dependent

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[inland cargo]

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[steel oasis]

Hey

[steel oasis]

Can someone help me prove why this limit for a gradient is = 0

Python tells me as P approaches infinity, dz/dP = 0

[desert hornet]

!xy

[vocal sleet]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[steel oasis]

Where y,x and m are constants

[desert hornet]

,rccw

[desert hornet]

and you want the value of this gradient as p tends to infinity?

[steel oasis]

Yes

[desert hornet]

well the bottom has a p^2 term while the numerator has only p terms

maybe if you divide the numerator and denominator by p^2 you will see it more clearly

[steel oasis]

Hmmm

Did it and get this

Is it because anything divided by inf is 0

[steel oasis]

^

[desert hornet]

well the last term in the denominator should just be m

[desert hornet]

yeah, so you get 0/m which is 0

[steel oasis]

Why should it be m?

[steel oasis]

Could you show me proof?

[desert hornet]

mp^2/p^2 = m, no?

[steel oasis]

Oh mb

I read wrong

Okay thanksss

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[orchid flicker]

its a curv integral with the vektor F(r) =(-y^3, -xy^2,z) with y(x)=x^alpha and r_a(0,0,0) and r_b(1,1,0)

[vocal sleet]

@orchid flicker Has your question been resolved?

[orchid flicker]

obviously not, who made this bot...

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@orchid flicker Has your question been resolved?

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[sullen pilot]

quick question

[sullen pilot]

does a function only have a horizontal asymptote when x -> infinity?

[flat whale]

When else do you think a function would have a horizontal asymptote

[sullen pilot]

all i know is when the graph is infinite is when HA are used

but im just making sure

or is there any other way to get a HA?

[paper depot]

x → +∞ or x → -∞

so you can have up to 2

[vast shale]

Nope, if the domain of the function doesn't go all the way to infinity in at least one direction you can't get get a horizontal asymptote

[vocal sleet]

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[sullen pilot]

Ah okay i see, thanks guys

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[jade stone]

If a|b and b|a, then why does a=+/-b

[jade stone]

a|b translates to a divides b without a remainder

or a=bq

[hard atlas]

a=bq and b=am

so a=bq=amq

[jade stone]

I'm still having trouble understanding this

how does that make it +/-b

[hard atlas]

mq=1

so what can q be

[jade stone]

1

[hard atlas]

or?

[jade stone]

-1

right

if m is also -1

on a=bq

q can be 1 or -1

[jade stone]

what about this

why is |b|<=|a|

[hard atlas]

note that |xy| = |x| |y|

and |y| is always at least 1 for nonzero integers

[jade stone]

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[vast night]

Why does e^infinity = 0 in this case?

[silk osprey]

e^negative infinity is zero

[vast night]

Then have i done something wrong?

[silk osprey]

so as t goes to infinity

e^-t=0

[vast night]

oh its still -infinty?

or still -

[silk osprey]

wym the whole function will be 5000

the work on the right is right

on the left idk what ur doing

y is there an extra e^infinity

[vast night]

suppost to be like this

the extra i just to tell that e^infinity is 0 but thats wrong i guess

[silk osprey]

but it’s not 2 * 1/0

it’s 2* 1/infinity

and 1/infinity is zero

[vast night]

oh

so it is correct what ive done but e^infinity is infinity not 0 ?

[silk osprey]

because when u put the e in the denominator to make the exponent positive

u get infinity

but it’s in the denominator

so 1/infinity=0

[silk osprey]

yea

e^+infinity =infinity

e^-infinity=0

[vast night]

ah okay. probably why i had a hard time understanding why.

thanks

[silk osprey]

because e^-infinity=1/e^infinity=1/infinity=0

[vast night]

yeah thanks alot

[silk osprey]

👍🏻

[vast night]

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[jade stone]

how or what property of division happened here?

[jade stone]

as a matter of fact, what happened here?

some properties of division have been used, but I'm just confused how or what

[paper depot]

от 7 | (10x + y) следва, че 7 | (10x + y - 7x)

[jade stone]

момент, как?

[paper depot]

защото 7x се дели на 7

[jade stone]

а защо е -?

[paper depot]

защото са решили да направят така

целта ни е да докажем, че 7 | (x - 2y), нали?

[jade stone]

да

[paper depot]

това е един от възможните начини

не е единствен

[jade stone]

това тук ли е използвано?

[paper depot]

да

сборът от кратни на n също е кратен на n

и т.н.

[jade stone]

мхм

[paper depot]

не е сложна работа

ето ти още един начин да се докаже същото:

7 | (10x + y)
=> 7 | 5(10x + y)
=> 7 | (50x + 5y)
=> 7 | (50x - 49x + 5y - 7y)
=> 7 | (x - 2y)

[jade stone]

ясно

[jade stone]

а тук за използвали свойството b|(a+kb)

в дясно

за да извадят/прибавят достатъчни х и у

за да се получи х-2у

[paper depot]

вдясно по-скоро са разложили по хитър начин 15x+5y

за да има кратни на 7, които можем да махнем

[jade stone]

ясно

благодаря ти

.close

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[civic kite]

anyone know how to input log into the calculator

[civic kite]

I keep getting wrong answers

how do I make the base of the log

[silk comet]

what's your calculator?

what functions do you have available?

[civic kite]

SHARP scientific calculator EL-510R

can we go over the question?

so we want to make = u log2(x) right?

u = log2(x)

[silk comet]

sure, or you could take out 1/ln(2) as a constant before doing anything

[civic kite]

right I ended up doing that when I did the u sub

because when I did the change of base formula I saw I could take it out

if I do it after the u sub will it make things wrong?

[silk comet]

no

[civic kite]

ok so u = log2(x)

log(x) / log2

1/log2 * log(x)

derivative of log(x) is 1/x

so we have 1/log2 * 1/x du/dx

we solve for dx

dx = log2 * x du

x cancles out in fraction

so we are left with u * log2

we integrate u to be u^2 / 2 * log2

so we get log2(x) / 2 * log2

[silk comet]

that's the indefinite integral yep

[civic kite]

so then we just go

[silk comet]

but you forgot to square

[civic kite]

yea the *log2

so it would turn into log2^2(x) / 2

so it would be log2^2(6) / 2 - log2^2(1) / 2

for the definite integral right?

[silk comet]

yes

[civic kite]

this is where I have problems

because I can't put it into my calculator

and wolfram alpha as well

[silk comet]

well you can use the same change of base formula

log_2(6) = ln(6) / ln(2)

[civic kite]

YESSSSSSSSSSSSSSSSSS

so basically I will use ln every time on my calculator?

never use log?

[silk comet]

,calc log_2(6)

[twin meteor]

The following error occured while calculating:
Error: Undefined function log_2

[silk comet]

lol

,calc log2(6)

[twin meteor]

Result:

2.5849625007212

[civic kite]

should I never use log on my calculator?

[silk comet]

,calc ln(6) / ln(2)

[twin meteor]

The following error occured while calculating:
Error: Undefined function ln

[silk comet]

you can

it's up to you

[civic kite]

ok so

[silk comet]

the change of base formula works with any base

log_2(6) = log_10(6) / log_10(2)

[civic kite]

when I am putting log^2(2) into my calculator

do I do ln(2)^2

or ln^2(2)

[silk comet]

well ln^2(2) means (ln(2))^2

[civic kite]

its the same thing?

[silk comet]

go back to when you resubstituted the expression for u

u^2 / 2 * log2 -> (log_2(x))^2 / 2 * log2

so clearly you take log_2(stuff) and then square it

[civic kite]

so I have u * log2 right

u * log2 / 2

[silk comet]

what happened to u^2

[civic kite]

ok yea

u^ 2 / 2 * log2

so that becomes (log2(x))^2 / 2 *log2

[silk comet]

yes

and log_2(x) is ln(x) / ln(2) as you noted

[civic kite]

and what about the log2?

[silk comet]

on the outside?

[civic kite]

yes

[silk comet]

well go back to where it came from

[civic kite]

just ln(2)?

[silk comet]

it came from this

[civic kite]

I know

[silk comet]

so what do you think it should be

[silk comet]

yes

[civic kite]

I mean how do I calculate it

[silk comet]

it was ln all along

[civic kite]

when putting in calculator

[silk comet]

also wolfram can do log-base

[civic kite]

also when calculating the indefinite integrals

[silk comet]

,w log_2(6)

[twin meteor]

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[civic kite]

do I have to do the subtraction all in one go in a lot of brackets

or can I do one piece and then do the second piece

or will there be rounding errors?

[silk comet]

you mean definite?

[civic kite]

yea

my calculator saves the ANS

but I don't know how many digits it saves

[silk comet]

there will

[civic kite]

so will I get rounding error?

[silk comet]

probably enough if you use Ans

just don't round multiple times

[civic kite]

hahahahaha so I have to do the whole indefinitne calculation in like 20 brackets?

it's so hard to remember where the brackets are

[silk comet]

or you can use Ans

do it step by step

[silk comet]

yeah, it's an art

[civic kite]

I wish they didn't want rounding decimals its so silly

should just leave it in fractions and move on

[silk comet]

should probably do what you're asked

[civic kite]

Im putting this into my calcultor

It's soooooooooooooo long

even when I do piece by piece

this is seriously silly

I guess I will just do it on paper first with the brackets

and then copy that exactly

ok Tushar

I got it all bracketed

and I'm going to attempt to put it in calculator

can you tell me if you get same answer?

2.32

[vocal sleet]

@civic kite Has your question been resolved?

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[wicked elbow]

how do i know which z table i need to use to find k, ive looked at the answers and the first question uses a positive z table while the second uses a negative z table?

[regal bane]

I'm not sure what your tables look like

We could use this for the first question though

[wicked elbow]

yea im just not sure how im supposed to know which z table to use since they both use different ones

or how to actually get the answer tbf lmao

im reading that i need to do 1 - (z score) but not getting the answer its showing

figured out first question got -0.95 but unsure how to get the answer 0.18 for 3b <@&286206848099549185>

.close

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[modern geyser]

Can someone explain Concavity to me. I know its when the second derivative is increasing or decreasing but I can't tell when that happens on a graph. People say to use smily faces as an example but that doesnt make sense to me.

[spiral inlet]

It's when the second derivative is positive or negative, not increasing/decreasing

which means that it's when the first derivative (the slope of the graph) is increasing or decreasing

@modern geyser can you look at point A and tell whether the first derivative there is positive or negative?

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@modern geyser Has your question been resolved?

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[modern geyser]

.reopen

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✅

[modern geyser]

At A its decreasing. I mispoke, I know its when the first derivative is increasing or decreasing but I cant see where that is on the graph

[spiral inlet]

Yes, at A the first derivative is negative.

And look at this second point shortly after A, where I drew the second red line

See how the first derivative is positive there?

[modern geyser]

I do yeah

[spiral inlet]

So clearly the first derivative increased from A to that other point

The slope increased, from a negative number, through zero, to something positive

producing a smiley-face shape between

[modern geyser]

but at the max between B and C the derivative is 0 so the first derivative is becoming negitive. so how can the slope be increasing and decreasing.

[spiral inlet]

Between B and C, the slope is decreasing (which means the 2nd derivative is negative, and the graph is concave down)

It seems to start decreasing somewhere about halfway between A and B

[modern geyser]

so from the point between b and c and the point d is the first derivative getting more negtive if its going down or postive because its going towards 0.

[spiral inlet]

From here to here?

[modern geyser]

yeah

[spiral inlet]

At the top red point, the first derivative is 0.

After that, it immediately becomes negative, so the slope has decreased

but it looks like it starts increasing again just before D

See the blue smiles and the purple frown

[modern geyser]

ok I think I understand . I assumed both was happening at the same time. Thank you

[spiral inlet]

Ahh, yeah, no, you've got both happening on this graph in different places

[modern geyser]

I see thanks a lot

[spiral inlet]

Sure thing 👍

[modern geyser]

.close

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[somber garden]

structural engineering calculations i need help understanding the concept

[somber garden]

<@&286206848099549185>

[flat whale]

don't ping helpers without a question

[somber garden]

i posted a question

[flat whale]

????

what is there to answer

[somber garden]

what do you mean

[flat whale]

there is no question

[flat whale]

this is not a question

[flat whale]

https://www.dictionary.com/browse/question

[somber garden]

https://dictionary.cambridge.org/us/dictionary/english/arrogant

[flat whale]

if you're not gonna ask a question, don't waste a help channel

.close

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[mental wigeon]

Hi

[mental wigeon]

Csn u help with this

I subbed 2x as a

Idk what to do next

<@&286206848099549185>

???

Pls help

[vast shale]

i think this leads to a quadratic equation, so solve with that in mind

just fix up the placements and you'll see it

[mental wigeon]

So

-9a²+3a+2=0

Ignoring cos

For now

[vast shale]

yeah

[mental wigeon]

So

How do i get more solutions tho

I only got 48.2 408.2 109.5 and 469

<@&286206848099549185>

Ok i have to go to bed

This is not solved though

[vocal sleet]

@mental wigeon Has your question been resolved?

[vast shale]

no u have to sub (cos2x) with a

dont forget the '2' there

[mental wigeon]

?

I replaced 2x with a

Then replcaed cos with a as well

[vast shale]

so in this case it would be -1/3 = (cos2x) since u replaced (cos2x) with a

at least u should've, u forgot the 2 while calculating

[mental wigeon]

Yh

I got two base values for arc cos

109.5 and 48.2

Is that correct?

[vocal sleet]

@mental wigeon Has your question been resolved?

[vocal sleet]

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[sleek flame]

i’m turning it into double integrals not iterated

i got 1/2 as answer i actually don’t see what’s wrong :((

iterated integrals work for some reason

[dull bear]

getting 1/2 as an answer here?

[sleek flame]

YEAAA

no idea why

[dull bear]

,calc 0 * (1/2)

[twin meteor]

Result:

0

[sleek flame]

ohh

wait

no noo

why

am

i like this

oh

[dull bear]

You found the first integral as 1/2, the second one as 0

[sleek flame]

yea

1/2 times 0

is 0

[dull bear]

oh didn't see you writing this, thought you were asking why I meant what I said

There's me not reading

[sleek flame]

no worries at all

[dull bear]

(but it happens to us all )

[sleek flame]

i cant read either

thank you so much @dull bear

always answering my questions

[dull bear]

Always a pleasure

[sleek flame]

.close

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[ashen haven]

Need help with this question

<@&286206848099549185>

[willow bridge]

do you know the formula for the sum of cubes (non alternating) ?

[ashen haven]

I think its (n(n+2)/2)^2

Correct me if im wrong

And then sub in n=k-1

[willow bridge]

youre wrong

[ashen haven]

I meant +1

My bad

[willow bridge]

all good

[ashen haven]

^

[willow bridge]

why?

[ashen haven]

To get the sum of cubes up to k-1

[willow bridge]

but thats not what you want

an idea would be to sum the cubes (even an odd) and remove 2 times the even cubes

[ashen haven]

Wdym

[willow bridge]

1^3 - 2^3 + 3^3 - 4^3 + 5^3 - ... = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + ... - 2(2^3 + 4^3 + ...)

does that make sense

[ashen haven]

Oh i see

How do you write that in terms of k

[willow bridge]

there are k odd numbers

so how many numbers is there total

[ashen haven]

2k

?

[willow bridge]

2k - 1

because it ends with an odd number

[ashen haven]

Ohh i see

[willow bridge]

so there is k-1 even numbers

[ashen haven]

Then should i sub in n=2k-1

[willow bridge]

so you sum the first 2k-1 cubes and remove the even cubes (you still have some work to do for the even cibes)

[willow bridge]

yes

[ashen haven]

What about the k-1

[willow bridge]

the even cubes?

[ashen haven]

Yes

[willow bridge]

notice that 1^3 - 2^3 + 3^3 - 4^3 + 5^3 - ... = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + ... - 2(2^3 + 4^3 + ...) = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + ... - 16(1^3 + 2^3 + ...)

[ashen haven]

((2k-1)(2k)/2))^2

?

[willow bridge]

yes for the 2k-1 cubes

but you still have to remove twice the even cubes

[ashen haven]

How?

[willow bridge]

1^3 - 2^3 + 3^3 - 4^3 + 5^3 - ... = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + ... - 2(2^3 + 4^3 + ...) = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + ... - 16(1^3 + 2^3 + ...)

does that make sense?

• 16(1^3 + 2^3 + ...) is the important part

[ashen haven]

Like this?

[willow bridge]

no

you have to use the formula again

for the even cubes

2(2^3 + 4^3 + ...) = 16(1^3 + 2^3 + ...) = ? (use the formula)

(how many even cubes are there?)

[ashen haven]

(K(k+1)/2)^2

?

[willow bridge]

there are k-1 even cubes

because the sum ends on an odd number

[ashen haven]

Ohhh

So is it ((2k-1)(2k)/2))^2-((k-1)(k)/2))^2

As the final answer

[willow bridge]

almost

((2k-1)(2k)/2))^2-16((k-1)(k)/2))^2

because you removed them twice and factored out 2^3

[ashen haven]

Ohhhh

Forgot about that

[willow bridge]

all good

and then it simplifies a lot

[ashen haven]

I will do that part

[willow bridge]

you should get 4k^3 - 3k^2

[ashen haven]

Thank u very much

[willow bridge]

youre welcome

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[hard crest]

71 where do I even start 💀💀

[inner osprey]

normally when you have really a weird function in an integral you want to try a u-substitution of some sort

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[fair prawn]

can someone check my answer?

[vocal sleet]

Please don't occupy multiple help channels.

[fair prawn]

<@&286206848099549185>

can someone check my answr

[vocal sleet]

@fair prawn Has your question been resolved?

[fair prawn]

.close

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[west berry]

e^x=x+5

[west berry]

Can you solve for x

Please

[round plover]

stuff of that form generally can't be solved exactly (unless you use special functions like the lambert-w function), but you can calculate them numerically

[vocal sleet]

@west berry Has your question been resolved?

[west berry]

What is that thing

[west berry]

And where I can learn that

[round plover]

do you just need a number for x

[west berry]

No I want to learn lamber w functions

[round plover]

it's a part of complex analysis, which I don't do, so you'll have to ask someone else

[west berry]

I was doing just basic calculas

Why the hell they give complex analysis thing on that

Bruh man I am screwed

[round plover]

what's the original question

[west berry]

Like it says 2 function

e^x

And another one is x+5

So they want the area bounded by those two curves

So I just try to solve for x so I can have the limits

[round plover]

can you send a screenshot of the problem if possible

[west berry]

55

55

[round plover]

you could still do it numerically I guess

the integral itself isn't difficult

[west berry]

Numerically means

[round plover]

calculate the intersection points using approximation methods

newton raphson, etc.

or just plugging numbers into your calculator until you get something close

fwiw, the values are x=-4.99322 and x=1.93685 (approximately)

[west berry]

Nice 👍

Thanks for the help

.close

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[hot mica]

Idk how to get c)

[vocal sleet]

@hot mica Has your question been resolved?

[mighty nacelle]

,rotate

[mighty nacelle]

ok

so the first thing i would do is find the midpoint of AC

[hot mica]

Which is 6 6

[mighty nacelle]

ok, then you want to find the distance from the midpoint to a vertex

[hot mica]

Oh I get it

Distance from midpoint to vertex is the same no matter the vertex

[mighty nacelle]

yeah

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[obsidian bison]

what should i do next?

it doesnt seem i can solve this integral

also can someone explain when i shouldn't use this integrating factor method?

because there are some differential equatation that it's supposedly impossible to find the integrating factor for

but i can't imagine one of them

[dull bear]

So you chose to substitute u = arctan(x) right? How did you get the du from there?

@obsidian bison

[obsidian bison]

yes

wrong sub

i was able to solve it

but does this method works for any first order linear differential equations?

@dull bear

[dull bear]

Well at least in theory it should, though there may be cases where you don't get a nice integral out of it

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[willow whale]

For a) is the factor 6^x2 (3 + 5^x5) ?

[desert hornet]

looks good

[willow whale]

This is an example I was given

[pine snow]

looks good

[willow whale]

Oh thank you !

With b) how do I know what indices is correct for the factor number ?

I know it's 4 but is it 4^x2 ? And why would it be 2 ? Apologies I'm not sure on how to word it

[pine snow]

we can first look at the x's

which x has the lowest power

[willow whale]

6

[pine snow]

nice

first factor found

next is the numbers

if theyre even numbers u can take a factor of 2 out

its just a trial n error until u can find the common factor

[willow whale]

Ohhh ok so they don't need to also be a factor ?

I'm just subtracting them ?

[pine snow]

no i mean like

2(10x^6 + 16x^7 -12x^9)

thats what i mean

with the 2

[willow whale]

In that instance what am I doing with the indices ?

As I have a + + then - ?

[pine snow]

U dont need to pay that much attention as you will factor out positive factors

so the indices won't change a bit

But if you were to factor out a negative factor

It would reverse the indices : - - +

Do u get the hang of it?

[willow whale]

I think I understand a little

[dense fox]

One doubt... #help-20

[pine snow]

its like this - * - = +

negative * positive = negative

.

[dense fox]

#help-20

[willow whale]

I will try and work b) out and see what I get

Thank you !

[pine snow]

Gl i suck at helping but aight

Most people would use positive indice as the factor number

[willow whale]

Haha no I think you helped ! I essentially need an indices there so I have x2 and just subtract that. That's what I have on my examples

[pine snow]

but negative works the same and essentially correct too, although the teacher prolly wont accept it

[willow whale]

Ah see I didnt get told that

I was going by highest factor number

Like these are a few we did

[pine snow]

i was jus confused by indice

i thought it was exponent

or was it?

[pine snow]

quite vague

on how to do

not gonna lie

[willow whale]

Ha yeah it was a quick power point and I've never done any algebra before.

Gonna take me awhile for this 😂

[pine snow]

Yeah

Make sure to memorize that well

[vocal sleet]

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[marble kettle]

hi i dont understand why there the denominator for c is squared. can someone explain please? thanks

[paper depot]

this is what you do when you have a repeated factor in the denom

you include fractions with denominators equal to powers of that factor, up to the power that appears in the original

so e.g. if the demon were (x-2)(x-1)^3 you'd have B/(x-1) + C/(x-1)^2 + D/(x-1)^3

[marble kettle]

but is there a reason why we need to do this?

[paper depot]

well imagine a simple case

(5x+7)/x^2

you would not be able to express this with only A/x or only A/x^2

[tepid hound]

you have to write before (x-1) and then (x-1)^2 , if it was (x-1)^3 you have to write first (x-1) then (x-1)^2 and then (x-1)^3

[paper depot]

i said exactly this above

[tepid hound]

yes ive seen

[marble kettle]

but how is this related to having to writting the demoninators in all powers?

[paper depot]

mmmh

not sure i can give a convincing explanation beyond this lol

you'd have to go look for a proof of PFDs in the general case

[marble kettle]

oh ok

thanks anyways

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[winged vapor]

is this right

[dry yacht]

Looks good, but remember to keep the plus sign in between y^2(x-2) and (x-2)(x-5).

[winged vapor]

oh right
thx

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[storm owl]

hey

[storm owl]

i got a study guide review can someone give me the answers?

i just need to double check

w another sheet

[outer warren]

no, we don't do that here.

if you have issues with specific problems
post them (ideally one at a time), show what you tried and the concerns you have.

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[unique lagoon]

I was wondering how I could get up to learning complex analysis, I have been self studying single variable calculus for about a year now, though I still am not sure if I know everything yet, I was also wondering what resources are good for getting there and how long it would take. I haven't taken linear algebra either.

[sly sierra]

have you tried reading an introductory complex analysis text?

[unique lagoon]

Not yet, and I should probably learn multivariable calculus before that, right?

[sly sierra]

not necessarily, you don't need much multivariable calculus

[unique lagoon]

Well, is there any part of multivariable calculus that may be required or that I would benefit from?

[sly sierra]

understand the basic ideas of vectors and partial derivatives
it's useful if you know what differentiability of a function of two variables means
maybe green's theorem if the book decides to use that as a launching point for proving the basic theorems

[unique lagoon]

Is there any other prerequires of complex analysis, like should I do real analysis first or-?

[sly sierra]

you don't necessarily need a full real analysis course, but it depends on the emphasis of the complex course

definitely good to be comfortable with concepts like convergence, continuity, differentiability, power series, open and closed sets in the plane

oh and line integrals

[unique lagoon]

Okay, where do you suggest I learn these?

[sly sierra]

some intro textbooks will teach you what you need as you go

brown and churchill for example has quite modest requisites

[unique lagoon]

Okay, great!

[sly sierra]

you could always try reading a book like that and if you find something you need background on, go and learn that in a targeted way
that would be more efficient than doing a full multivariable calculus and/or real analysis course first

[unique lagoon]

how long do you think a full multivariable calculus / or real analysis course would take btw

[sly sierra]

for self-study?

[unique lagoon]

Yeah.

[sly sierra]

well for example, mit has a good multivariable course on youtube
https://www.youtube.com/playlist?list=PL4C4C8A7D06566F38
36 lectures, about 50 minutes each

[unique lagoon]

Yeah, I was thinking about taking 18.02SC

along with their real and complex analysis courses

and 18.06SC for linear algebra

[sly sierra]

what are these 16.02sc, 16.06sc, looks like MIT-style numbering but aren't their math courses all 18.xx?

[unique lagoon]

Sorry, 18.xx

yeah you're right haha

[sly sierra]

ah got it

yea i can't vouch for all of them, but i've watched many of the multivariable ones at the playlist above (18.02), they're very good

i think they have accompanying problems/exams/solutions somewhere as well

haven't looked at those though

[unique lagoon]

Good to hear, I would expect that the school that hosts integration bees would do well at teaching calculus haha

[sly sierra]

yea that lecturer (denis auroux) is really good, explains things very well

[unique lagoon]

Would you say it is comprehensive?

[sly sierra]

and i assume gilbert strang for linear algebra, he's good too

[sly sierra]

yea, as far as i can tell it's a complete coverage of all the standard topics for multivariable calculus

[unique lagoon]

Haha, I find that funny since my schooling has longer classes everyday and still if you watch one video a day in theory you could learn it faster than school can teach in one semester, or even half of one.

[sly sierra]

does your school spend time going over a lot of example problems? that could explain why they take longer

the mit guy does a few but he expects you get most of your practice doing homework on your own

[unique lagoon]

Honestly yeah, most of the time is dedicated to example problems, still, seems understanding is poor in my class.

Then again I am in the lowest level math class in my high school

[sly sierra]

might be good to supplement with online lectures, maybe your teacher is just doing a poor job of explaining

or, sometimes it just takes some time to wrap your brain around new knowledge

[unique lagoon]

I already am fairly fluent in all of algebra 1 and some of 2

So it doesn’t affect me much, but a lot of my class is struggling.

I actually have a math final today, though I’ll do good even though I don’t study for that level of mathematics

Though I do want to take all of my school’s math classes over the summer on a charter school to show my school I can actually do math or something

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[sleek flame]

what does it mean when it says contradicts the minimality of q, is that saying q is not the smallest such denominator

[silk comet]

yes

[sleek flame]

ok

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[hoary cypress]

if z be a complex number satisfyng |Re(z)| + |Im(z)| = 4, then find range of |z|

[vocal sleet]

Please don't occupy multiple help channels.

[hoary cypress]

can someone pls help me with thiis question

and my previous one

they are both different questions

[desert hornet]

!status

[vocal sleet]

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

[desert hornet]

uhm

[hoary cypress]

but its been an hour and. a half since i asked my prev one and noone responded so i thot its not unfair to ask a new question

[desert hornet]

looks like the bot is not responding

anyway, what have you tried?

[hoary cypress]

i draw |x| + |y| = 4

its a square

wiith vertices
0,4
0,-4
4.0
-4,0

[desert hornet]

correct

[hoary cypress]

but idk what to do after that

[desert hornet]

where is the value of x^2+y^2 maximum?

[hoary cypress]

idk

[hoary cypress]

is it a circle inscribed in the square

and max is circle circumscribng the squre

am i right

[desert hornet]

yes

you're right

[hoary cypress]

.close

[vocal sleet]

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[vocal sleet]

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[dusty kiln]

helo

[dusty kiln]

i have to calculate local extrema of it

and

i calculated partial derivatives:

and i know max is probably 4 and min is 0 but how can i get that

[vast shale]

do you know like the formula you get

from calculating the determinant of the Hessian

[dusty kiln]

but to make hessian matrix i need critical points

and idk how to get these

i can only get tan(x)=tan(y)

[vast shale]

set both equations equal to 0

[dusty kiln]

i did that

but

i font know how to solve it

to get points

[vocal sleet]

@dusty kiln Has your question been resolved?

[pallid wren]

hi im new and i have a question so when a twine has a strength of 4 tex, which means 1 km of this thread weighs 4 grams. Calculate the diameter of the thread in mm if the material density is 1.44 g / 2 square centimeters?

[vocal sleet]

@dusty kiln Has your question been resolved?

[gloomy wadi]

Maybe you can use the difference formula of the sine if you factor 2

[dusty kiln]

what formula

[gloomy wadi]

sin(x-y)=sin(x)cos(y)-sin(y)cos(x)

[dusty kiln]

yeah i tried

but anyway no idea what to do then

[gloomy wadi]

The first equation should be 2 sin(y-x)=0 and the second 2 sin(x-y)=0, since sin(y-x)=-sin(x-y), your critical points are the points (x,y) such that x-y=k*pi, with k an integer

[dusty kiln]

o yeah i have done that

rn

but im thinking

[gloomy wadi]

Did you find the hessian? Because if you have done that, the rest is easy!

[dusty kiln]

i dont think the hessian works here

bcus det will be 0

but i put the numbers in function

and got 4 and 0 so i guess i just need to prove that its max andmin for that function and that makes sense

[gloomy wadi]

It works:
The hessian is:
\begin{equation}
H_f=\begin{pmatrix}
-2\cos(x-y) & 2\cos(x-y)\
2\cos(x-y) & -2\cos(x-y)\
\end{pmatrix}
\end{equation}

[twin meteor]

cristorenzo99

[gloomy wadi]

Do you know what a positive or negative matrix is?

[dusty kiln]

im not sure

i know that if the determinant is 0, i can say is it max/min or not

i cant *

[gloomy wadi]

Try to sub x-y=k pi and cos(k pi)=(-1)^k

So you have two hessian matrices, one is {-2,2;2,-2} and the other is {2,-2;-2;2}

Strange

[dusty kiln]

yeah

but what can i do with that

[gloomy wadi]

Maybe I have an idea

If you expand the squares, use cos^2 x+ sin^2 x=1, and use the difference formula for the cosine (cos(x-y)=cos(x)cos(y)+sin(x)sin(y)) you should get 2+2*cos(x-y)

And maybe it's easier because, if you set t=x-y, you get 2+2cos(t). Now you know that the cosine is minimum at kpi, with k odd and maximum at kpi, with k even

[dusty kiln]

yeaah soit will be enaugh to prove that its max and min ?

[gloomy wadi]

I think it will, the points x-y=k pi with k odd are minimum points and the points where x-y=k pi, with k even, are maximum points.

[dusty kiln]

okay thank i think i get it

./close

.close

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[hazy trout]

1. why does the set of all linear combinations of n linearly independent vectors span R^n? What is the intuition behind this? 2) Also is basis a "stricter version" of a span in that all the vectors must be linearly independent of one another? 3) Does the basis of n vectors have to span R^n entirely?

[wraith venture]

k linearly independent vectors in a vector space V span a k dimensional subspace of V, by definition of them being a basis of it
If k=n, the only n-dimensional subspace of R^n is R^n, so they must span R^n.
Bases are the fundamental idea behind dimension.
In a way, they are stricter than spanning sets since all bases are spanning, but not all spanning subsets are bases

[hazy trout]

I can understand intuitively in R^2 if we have the two unit vectors: (0,1) and (1,0) how they span the entire x-y plane through linear combinations of them. But what if we have two linearly independent vectors like (3,4) and (7,9)? How can we intuitively know that it spans the whole plane?

[hard atlas]

linearly independent means they point in "different directions"

which is all you need to span the plane

[wraith venture]

which works well for 2 vectors
But for > 2 vectors, this can be misinterpreted

[hard atlas]

you can go everywhere if you can go in two different directions

[wraith venture]

they're 2 linearly independent vectors since one isn't a multiple of the other => they span a 2D space, but R² is 2D => they span R²
Or find the formula explicitly by solving the system

Oh I forgot the "intuitively" part
This may not be super intuitive

[hazy trout]

Regarding your last point. just so I understand. If we have 3 vectors: (1,0), (3,0), and (2,2). the set of vectors (1,0) and (2,2) are a basis and a span of R^2. The set (1,0) and (2,0) span R and they are not a basis because not linearly independent?

[wraith venture]

yes

[hard atlas]

they dont span R

[wraith venture]

though they span R x {0} technically, not R

[hard atlas]

they span a subset of R^2 which looks like R

[wraith venture]

they span a 1-D space, isomorphic to R (you'll see what that means later)

[hazy trout]

What is the difference between R and spanning "R^2 which looks like R" / "R x {0}"?

[hard atlas]

well (1,0) isnt even in R

so how could it span it

[hazy trout]

Wait, sorry why is (1,0) not in R?

[hard atlas]

well is (1,0) a real number?

[hazy trout]

Ohh I see

So is R x {0} like a coordinate, but the y is not changing?

[hard atlas]

the set Rx{0} is {(x,0): x in R}

[wraith venture]

(see cartesian product)

[hazy trout]

is there a case where 2 linearly independent vectors don't span R^2? I still don't really get intuitively why just because 2 vectors point in different direction/linearly independent they span the whole R^2space?

[wraith venture]

n=2

since they're 2 lin-indep vectors, they span a 2-D space, by definition right

What are the 2D subspaces of R² ?

[hazy trout]

Just so I understand, that's equivalent to the set of coordinates where x ranges from -inf to inf, but the only value y can take on is 0, and because it's a coordinate, it exists as a subset of R^2?

[hard atlas]

well if we wanna be technical, (1,0,0) and (0,1,0) are two linearly independent vectors which dont span R^2

[hard atlas]

yeah

[wraith venture]

also known as the x-axis

[hazy trout]

what's a subspace? I assume it's the whole x-y plane?

[hard atlas]

a subset that is also a vector space

[wraith venture]

a subspace of V is a subset of V that is a vector space

[hard atlas]

(with the same operations on it)

[wraith venture]

(over the same field)

[hard atlas]

(implied by the scalar multiplication being the same operation)

[hazy trout]

Ok, so is the subspace of R^2 just a plane?

[hard atlas]

the only 2d subspace of R^2 is R^2 itself

aka the plane, yes

there are a lot of 1d subspaces. all the lines going through the origin

[wraith venture]

{0} is also a subspace

[hazy trout]

Ok so maybe, I just can't get pass the by definition part. I might just fail to see intuitively why it's by definition that any 2 linearly independent vectors span a 2-D space

I can see it intuitively for some like the unit vectors, but why is it true for all of them

[hard atlas]

2 dimensions by definition means "spanned by 2 linearly independent vectors"

[hazy trout]

Hmm I see

Is there a way to reduce any two linearly independent vectors into (0,1) and (1,0)? I feel like if that's possible, it would make sense to me

[hard atlas]

well what do you mean by "reduce"

for some more intuition. if you have your two linearly independent vectors, that means you can travel on the lines spanned by those

first you can go along the first line

and then you can go parallel to the second line

and because they point in different directions, you really can go everywhere you want to

I think 3b1b has a nice animation for that somewhere in his videos

[wraith venture]

everyone should watch the whole playlist anyways

4 years later, it's still fundamentally stuck in my mind, it's still how I see things
They're just not quite pink and blue

or green and red

but they move all the same

[hazy trout]

Wait, linear combination doesn't not need to be scalar multiples right?

Like a*(v1,v2) + b*(v3, v4), a and b can take on any rational value?

[hard atlas]

well some scalar multiple of the first plus some scalar multiple of the second

any real value

just like for (1,0) and (0,1)

[hazy trout]

ohhhh, that makes a lot of sense now. So we can take infinitesimally small distances in either direction, to span the entire plane

Ok thank you for the help!

.close

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[zealous cradle]

If you have a closed hydraulic system with two pistons at the same elevation. Pressure stays the same for both pistons? So would applying 1kN of force on a smaller piston apply 10kN of force on a piston with 10 times the area?

[vocal sleet]

@zealous cradle Has your question been resolved?

[vocal sleet]

@zealous cradle Has your question been resolved?

[vocal sleet]

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[upper flower]

Not reallly sure about this, you started in the right direction, but after that i dont really understand what you were trying to do

I think if you put = sign between them then it would be much more readable

[warm vale]

I checked the answer in "Symbolab" and it said the answer is correct, however it doesn't work when submitted for the assignment.

[upper flower]

hmmm

is the exponent -3?

or 3?

cant really see it tbh

[warm vale]

Yes the exponent of the whole equation is -3

[upper flower]

okay

let me try to do it and see if iget the same thing

[warm vale]

Okay thanks Gabe

[upper flower]

Thats weird i got the same answer as you did

idk then tbh

[vocal sleet]

@warm vale Has your question been resolved?

[vocal sleet]

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[upbeat fractal]

How do I solve 3-5?

[upbeat fractal]

Ok I got 1 and 2, 1 has a real solution so it’s true and 2 doesn’t so it’s false

stuck on 3-5 :(

<@&286206848099549185>

o hello

[iron parrot]

The question is asking if for all real x, if $x^2 > 4$, then is $x > 2$

[twin meteor]

Dork9399

[iron parrot]

Can you come up with a counterexample

Or just try simplifying x^2 > 4

[upbeat fractal]

x>+-2

[iron parrot]

Is that the same as x > 2?

[upbeat fractal]

no :0

oo

so like for example -3^2 > 4 but -3 not > 2 so its a counter

[iron parrot]

Yes

[upbeat fractal]

ty

what about 4?

[iron parrot]

Assume that y is just a constant, and try to solve for x (in terms of y)

[upbeat fractal]

so i turn it into y=

[iron parrot]

no x =

[upbeat fractal]

oh

ok

i got

x = 1/ (y^2 + 1)

so i try to find a counterexample where x is real but theres no real y?

or is it the other way around

[iron parrot]

Can you manipulate the right hand side to make any real value?

[upbeat fractal]

wdym, is it not already?

[iron parrot]

Just take x to be a real value, and try to find some y that will satisfy

Remember that reals arent just positive integers

they can be irrational

negative

or both

[upbeat fractal]

Ok

ok it works for x=1 and y=1

nvm

it doesnt lol

it works for x=1/2 and y=1

[iron parrot]

Try more extreme values

not necessarily big, but like square roots, or negative numbers

[upbeat fractal]

ok ill do -5

x=

uhh

y= sqrt(-6/5)

thats not a real number right

since its the sqrt of a neg