help-10

[timid silo]

u dont have to do that

well do you know that quantities only having the same dimensions can be added or subtracted?

[obtuse pebble]

@dusky arrow Has your question been resolved?

[obtuse pebble]

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[pastel pilot]

I can't solve this quadratic equation, I've multiplied 4 and negative 3 to get -12 then found 6 and -2 combine to 4, after that I try and divide both 6 and -2 by 4 which cant work and im stuck with (4x+6) (4x-2) which isnt the right answer

[tender tusk]

why have u tried to divide by 4?

[pastel pilot]

because thats the coefficient of the first term

[tender tusk]

hmm, never heard of this method

[pastel pilot]

i dont use the quadratic formula

[ruby path]

is that even a quadratic

[safe haven]

it is

[ruby path]

I initially read that incredibly blurry index as 3

[tender tusk]

oh lol

[pastel pilot]

so how can I solve it?

[safe haven]

why not try to break 4x² down

[tender tusk]

from experience i just split 4x into -2x + 6x

and then further fractorise

[pastel pilot]

wouldnt that make it -4x?

[tender tusk]

no, -2x+6x is equal to 4x

[pastel pilot]

ah wait nvm

can you tell me what you'd do step by step then?

[tender tusk]

well from there just factorise

[ruby path]

btw the split that Galaxy did

[pastel pilot]

can you type it out please?

[ruby path]

do you get where it comes from?

[pastel pilot]

yeah u get it from 6x - 2x

[hot dawn]

(2x-1)(2x+3)

[pastel pilot]

this is my working

[hot dawn]

Expand that

Your leading coefficient term will always be the product ab from (ax+n)(bx+m)

So you definitely did some thing wrong

[pastel pilot]

this is how I've been doing the questions and it gets me the answer, im confused why it doesnt work for this one

[hot dawn]

I do t get your working

What is x12

[pastel pilot]

thats from 4*3

[tender tusk]

when solving a quadractic with a leading coefficient that isnt 1, u cant go straight to (x-a)(x-b)

[pastel pilot]

nono I did 4x*3 first to get 12, then found 6 and -2 combine to give you B, which is 4

[tender tusk]

right, that is correct

[ruby path]

I mean the rationale

Why did he not do 3x + x instead

[pastel pilot]

then divided those two numbers by the leading coefficient which is 4

[tender tusk]

then i assume u went (4x+6)(4x-2) straight after finding those 2?

[pastel pilot]

yeah because neither could be divided by 4

usually I'd try and divide

the answer for this is meant to be (2x-1)(2x+3)

[tender tusk]

this seems unorthodox

[tender tusk]

like i said, u cant go from finding 6, -2, straight to (4x+6)(4x-2), this is typically a shortcut method for when the coefficient of the leading term is 1

[pastel pilot]

^

if it was 4/4x it would become (x+1)

[tender tusk]

?

[hot dawn]

Buddy

If you want us to spot your mistake

You’re going to have to explain your “method” you have been using here

[pastel pilot]

I can try and show an example for another question I got right

[tender tusk]

that may have been a coincedence

try factorsing 6x^2-7x+2 using that same method, you'll see that it doesnt get u the answer

[hot dawn]

For a general quadratic ax^2+bx+c for integers a,b,c, you want to find factors (nx+k)(mx+l) such that kl=c, nm=a, and nl+km=b

That is the general method

[tender tusk]

just split 4x into -2x+6x and factorise

[pastel pilot]

let me try I guess

but I dont know how to factorize polynomials, we've only been doing upto trinomials

[tender tusk]

look for the highest common factor between the first 2 terms and the last 2 terms

[pastel pilot]

alr let me try

@tender tusk so far I've gotten 2x(2x-1) + 3(2x-1), what do I do next?

[tender tusk]

well u can pull a common factor yet again

[pastel pilot]

which is what?

[tender tusk]

what is similar about those 2 terms?

[pastel pilot]

2x and -1

@tender tusk

[tender tusk]

right

[pastel pilot]

then what?

[tender tusk]

well factor out 2x-1

[pastel pilot]

how do I do that exactly 😅

[tender tusk]

(2x-1)(2x+3)

notice that i have just pulled out 2x-1 from both terms

[pastel pilot]

shouldnt that make it 2x + 3 then 🤔

[tender tusk]

is that not what i have typed?

[pastel pilot]

I'm confused, if you pulled out 2x-1 wouldnt it become 2x+3?

[tender tusk]

when u pull out a factor, u dont just get rid of it

[pastel pilot]

what happens to it then

[tender tusk]

just look at the case 4x-2, i notice that they have a common factor of 2, as such i will pull out the common factor of 2 which leads to 2(2x-1), notice how the 2 i pulled out hasnt disappeared

[pastel pilot]

yeah

[tender tusk]

now compare to the original question

[pastel pilot]

okay so I get us factoring out 2x-1 from both to get (2x+3), but I don't get how the 2x-1 stays

.close

[obtuse pebble]

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[fluid snow]

Regarding the differential equation from before, @slim cove, which was

[fluid snow]

This is for dampened electromagnetic oscillations

Now apparently there are three cases

Oscillating case, creep case, aperiodic limit case

(maybe the translations here are wrong but I think you get the idea)

Though in all those cases, we have the same starting differential equation..?

How come

But different solutions

[slim cove]

The simplest way to view it if you think about Q = e^rt; then you would get a quadratic equation that looks like r^2 + (R/L)r + (1/LC) = 0. The equation will behave differently depending on whether the discriminant is positive, zero, or negative.

[fluid snow]

What if I just take

And solve it

Without thinking about all this

[slim cove]

If positive, then you get real solutions for r, so Q grows exponentially. If negative, then you get complex solutions for r, so Q is oscillatory.

[fluid snow]

The solution will be correct for all three cases or what?

[dark stirrup]

I believe critically damped is a special case iirc

[slim cove]

The validity will depend on the values of R, L, and C. For example if you have a solution of the form Q = e^-delta t cos(omega t), you had better hope that omega is a real number

[fluid snow]

Yeah

[fluid snow]

Oh

[slim cove]

That depends on the discriminant of that quadratic equation earlier being negative

[fluid snow]

Would it be a good strategy to first solve it without thinking about all this, then in the end, after you have a solution, analyze it

[slim cove]

Sure, but now that you know that there are three different cases for a second-order equation with constant coefficients, it's good to keep that in mind

I mean the best strategy would be to use complex numbers from the outset

[fluid snow]

This is for a physics presentation about electromagnetic oscillations in 11th grade and unfortunately in school we don't cover complex numbers, idk if it'd make sense for me to introduce them here

[slim cove]

It probably would tbh

[fluid snow]

My teacher said he doesn't care how long it'll take me, can be 60 minutes or 8 hours

So yeah, maybe I can introduce complex numbers

[slim cove]

Have you learned polar vs cartesian coordinates in school?

[fluid snow]

No

[dark stirrup]

$i^2=-1$, the rest is left as an exercise to the audience

[warm shale]

SWR

[slim cove]

I see

It might be worth it to briefly go over how complex numbers are useful for keeping track of amplitude and phase then

And write down Euler's formula

[fluid snow]

Do complex numbers really simplify matters that much?

[dark stirrup]

you can give a detailed solution to this differential equation without using complex numbers, but it's probably easier to just use imaginary numbers

[fluid snow]

We can use the Ansatz e^(lambda * t)

Always

But is that really this useful

[slim cove]

Yes

[dark stirrup]

yes

It's like the engineer's best friend

[slim cove]

It is better than guessing like three different Ansatzes

Lol

[fluid snow]

Well you gave a pretty intuitive explanation for why we need e^(lambda * t ) * cos before

So I might get away with that?

[slim cove]

The calculation is much messier because you have to use the product rule for derivatives

Plus it's not clear why you need three separate cases

If you guess e^(lambda t), the equation quite literally simplifies down to a quadratic, and the value of lambda has a clear qualitative meaning, with the real part representing the speed of growth and the imaginary part representing the speed of oscillation

[glossy yacht]

have you done integrating factor to solve first order des

[fluid snow]

No, nothing of this sort

[obtuse pebble]

@fluid snow Has your question been resolved?

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[frosty notch]

I’m lost

[timid silo]

if a curve passes through some point, then the point will satisfy the equation of the curve

[frosty notch]

Yeah

I plugged in 2/3x^3/2=1/(2/3x^3/2+1/18)

Where x=0 y=9

[obtuse pebble]

@frosty notch Has your question been resolved?

[cosmic fog]

ur integration is wrong

[frosty notch]

Um

[cosmic fog]

dy/root(y) = root x dx

[frosty notch]

Y^-1/2

Integration

[cosmic fog]

integral is 2 root y

[frosty notch]

2 times y^1/2

No

?

[cosmic fog]

yep

[frosty notch]

Right side is just

2/3 y^3/2

[cosmic fog]

u mean x

[frosty notch]

Ye mb

[cosmic fog]

so 2 root y = 2/3 x^3/2 + c

put 0,9

c = 6

so u get root(y) = (x^3/2)/3 + 3

square both sides

should be done

[frosty notch]

(1/3 x^3/2+c)^2 right

Or rewrite it as

1/3 x^3/2 squared +c since c is just a variable

Wait

Yeah c=6

.close

[obtuse pebble]

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[true sorrel]

I'm extremely close to solving this but I need some final tip to make the proper rearrangements

[true sorrel]

My solution is to compare any triangle to the case where the line l is perpendicular to the bisector. This special line intersects any other case at the points D and E

Angle bisector theorem givas

[ \frac{BP}{CP}=\frac{AB}{AC} ]

[warm shale]

Eruc

[true sorrel]

And some exploration with sine law gives that

$\frac{BP}{CP}=\frac{BD}{CE}$ since $\frac{\frac{\sin\angle CPE}{CE}}{\frac{\sin\angle BPD}{BD}}=\frac{BD}{CE}=\overbrace{\frac{\frac{\sin\angle CEP}{AC}}{\frac{\sin\angle BDP}{AB}}}^{\mathclap{\sin\angle CEP=\sin\angle AEP=\sin\angle BDP}}=\frac{AB}{AC}$

[warm shale]

Eruc

[true sorrel]

So $\frac{AB}{AC}=\frac{BD}{CE}$

[warm shale]

Eruc

[true sorrel]

And of course, AB=AD-BD, AC=AE+CE

And with this, I should be able to rearrange stuff to prove that

[\frac{1}{AB}+\frac{1}{AC}=\frac{1}{AD}+\frac{1}{AE}]

[warm shale]

Eruc

[true sorrel]

But how?

[obtuse pebble]

@true sorrel Has your question been resolved?

[true sorrel]

<@&286206848099549185>

[obtuse pebble]

@true sorrel Has your question been resolved?

[austere aurora]

am i missing something here, i don't see how line l will affect the lenght of AB and AC in the first place

[molten stirrup]

thats the point

you have to prove that 1/ab +1/ac is constant regardless of line l

[true sorrel]

I think the label is a bit confusing, l is the black line

[austere aurora]

that still doesn't affect AB and AC

[molten stirrup]

then prove it ;)

[true sorrel]

so as you "twist" l around P (which is fixed along with A), AB gets shorter/longer while AC goes the other way

[austere aurora]

no but isn't AB and AC fixed?

it doesn't vary according to l

[true sorrel]

Not if you draw another line l

[austere aurora]

i might be missing something here lol

[molten stirrup]

not rlly line l intersects with the angle A making 2 points

the twist is lien l is constantly spining around P

so it is asking to prove that ..

[austere aurora]

where did you find this question lol, also ABC is a triangle right

[molten stirrup]

not really, if it was or u assumed it was, i would still be correct in 1 case

but it constantly moving

[austere aurora]

ahhh ok

[pastel gyro]

Not a geometric way or a hint to the previous method but there is an analytic way of doing showing this. Consider the case $0 < m < 1$ and consider the two rays $y = 0$ and $y = \frac{2m}{1 - m^2}$. The bisector line has equation $y = mx$, so pick any arbitrary point $P = \left(x_0, mx_0 \right)$, and denote $A = \left(0, 0 \right)$. Choose $\alpha > m$ for the slope of the line that will pass $P$ and cross the legs of $A$ (or $\alpha < 0$, else the line won't intersect the two legs. In this argument, we will stick with the former, but adjusting is easy). So $y - mx_0 = \alpha \left(x - x0 \right)$ resulting in $y = \alpha x + \left(m - \alpha \right)x_0$. So by solving for the intersection points, we find $B = \left(\frac{\alpha - m}{\alpha}x_0, 0 \right)$ and $C = \left(\frac{\left(1 - m^2 \right) \left(m - \alpha \right)}{2m - \alpha \left(1 - m^2 \right)}x_0, \frac{2m \left(m - \alpha \right)}{2m - \alpha \left(1 - m^2 \right)}x_0 \right)$.

[warm shale]

PowerUp

[pastel gyro]

Thus $|AB| = \frac{\alpha - m}{\alpha}x_0$ and $|AC| = \frac{\left(m - \alpha \right) \left(1 + m^2 \right)}{2m - \alpha \left(1 - m^2 \right)}x_0$ and (ensuring correct signs): $\frac{1}{|AB|} + \frac{1}{|AC|} = \frac{\alpha}{\left(\alpha - m \right)x_0} + \frac{\alpha \left(1 - m^2 \right) - 2m}{\left(\alpha - m \right) \left(1 + m^2 \right)x_0} = \frac{\alpha \left(1 - m^2 \right) - 2m + \alpha \left(1 + m^2 \right)}{\left(\alpha - m \right) \left(1 + m^2 \right)x_0} = \frac{2}{x_0 \left(1 + m^2 \right)}$

[warm shale]

PowerUp

[pastel gyro]

which is indeed independent of $\alpha$ (i.e. the line passing $P$ and intersecting the legs). The argument can be adjusted for $m >= 1$.

[warm shale]

PowerUp

[true sorrel]

That's really cool! The reason I'm working on this is coincidentally because I've been stuck in the analysis swamp for so long that I badly need to brush up on my geometry skills 😁

[pastel gyro]

Yeah my geometry skills are so dull that I just instinctively search for alternative methods lol.

[true sorrel]

It really can, consider for example these two cases

[obtuse pebble]

@true sorrel Has your question been resolved?

[true sorrel]

Still looking for a geometric approach, <@&286206848099549185>

Haven't gotten much further than this #help-10 message

I feel like AD = AE is of much help, but I seem to be getting nowhere

[radiant bison]

Q?

[true sorrel]

Problem here #help-10 message, figure here #help-10 message

Green line AE is perpendicular to the bisector AP

[radiant bison]

Name of the triangle is ade?

[true sorrel]

ADE is the special case where the line l is perpendicular to the bisector

ABC is the general case

[radiant bison]

What's the triangle name figure is not clear?

[errant canopy]

Holy latex spam

[radiant bison]

B and c are points of intersection so they can't be edges of the triangle right

[true sorrel]

A triangle can be defined as 3 points, so the points A, B, and C do make a triangle

[true sorrel]

Maybe this makes it more clear?

[radiant bison]

But b and c are points on triangle ade correct?

[true sorrel]

ADE and ABC are two different triangles, or rather said two different cases generated by different lines l

If we move B and C, they become the same triangle

ADE is a special case of ABC

[radiant bison]

Ok got it now

Pd=pe

Ang bpd = epc

[true sorrel]

You might be rediscovering results shown here #help-10 message

[pastel gyro]

Let $AB = a, AD = b, AC = \lambda b, CP = t_1 CB, DP = t_2 DE$ where $a$ and $b$ are vectors and $0 < \lambda , \mu , t_1 , t_2 < 1$. We have $CP = t_1 \left(-\lambda b + a \right) = t_1 a - t_1 \lambda b$ so $AP = t_1 a + \left(\lambda - t_1 \lambda \right)b$. But we also have $DP = t_2 \left(\mu a - b \right) = t_2 \mu a - t_2 b$, so another form for $AP$ is $AP = t_2 \mu a + \left(1 - t_2 \right)b$. Since $AP$ is an angle bisector, then by the angle bisector theorem we have $\frac{AD}{AE} = \frac{b}{\mu a} = \frac{t_2}{1 - t_2}$ and $\frac{AC}{AB} = \frac{\lambda b}{a} = \frac{t_1}{1 - t_1}$. Thus we get the four equations $t_2 \mu = t_1 , 1 - t_2 = \lambda - \lambda t_1 , \left(\frac{b}{a} \right) \left(1 - t_2 \right) = t_2 \mu , \left(\frac{b}{a} \right) \left(1 - t_1 \right) = \frac{t_1}{\lambda}$. We wish to show that $\frac{1}{AB} + \frac{1}{AC} = \frac{1}{AE} + \frac{1}{AD}$, or written in the symbols above, $\frac{1}{a} + \frac{1}{\lambda b} = \frac{1}{\mu a} + \frac{1}{b}$. But we observe that $\frac{1}{a} + \frac{1}{\lambda b} = \frac{t_1}{\left(1 - t_2 \right) b} + \frac{1 - t_1}{\left(1 - t_2 \right)b} = \frac{1}{\left(1 - t_2 \right)b}$, and $\frac{1}{\mu a} + \frac{1}{b} = \frac{t_2}{\left(1 - t_2 \right)b} + \frac{1}{b} = \frac{1}{\left(1 - t_2 \right)b}$ which proves equality.

[warm shale]

PowerUp

[true sorrel]

Neat

Give me a minute to let it sink in

Impressive without needing to rely on any perpendicular shenanigans

[pastel gyro]

I forgot to write that $AE = \mu a$.

[warm shale]

PowerUp

[true sorrel]

Tracking along the path I already have seems frustratingly close

[pastel gyro]

Yeah, I just used the angle bisector theorem and compared vector components.

[true sorrel]

But I did do something wrong because I'm currently at [longer length] / [shorter length] = 1/2 🤔

Thank you @pastel gyro !

!solved

Or what is it again

.close

[obtuse pebble]

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[pastel gyro]

Nppp

[obtuse pebble]

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[fossil night]

is 1/infinity > 0 or = 0?

[fossil night]

iseee

i understand but i dont at the same time

[keen quarry]

whenever we deal with infinity, we always use limits since infinity isn’t a number and therefore cannot be used to perform addition, subtraction, multiplication, or division

using limits is the closest we can get to actually using infinity as a number

[fossil night]

ohhh

[keen quarry]

but attempting to use infinity like a number will always result in undefined

[fossil night]

isee

i get it now

well thanks

.close

[obtuse pebble]

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[neat oasis]

I've made an error somewhere, please help me find where I went wrong. I only need help with the first part; I haven't attempted to find R yet since I don't know the correct power series.

[neat oasis]

thats what i did first, i got the sum of x^(2n-1)

oh wait i see, i should've kept the integral around the series once i turned 1/(1+x^2) into the sum of x^2n

no

ok

.close

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[jagged summit]

in this proof i dont understand why in the 3rd line the negative of the sin disapppears

[grizzled shore]

i² = -1

[jagged summit]

yeah that would make sense...

damn didnt think of it that way

cheers

[grizzled shore]

Sometimes just distribute and see if it’s the same as the line above

[obtuse pebble]

@jagged summit Has your question been resolved?

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[tepid crag]

help pls

i don’t know how do get f(u)

what are the steps

[pine glacier]

let u =9x+10

[tepid crag]

du: 9

yeah

what would be next?

[pine glacier]

make x^2 in terms of u

[tepid crag]

so we are getting the u two times?

[pine glacier]

no

you use u=9x+10 to find x^2 in terms of u

[tepid crag]

so it would be du/ 9?

[pine glacier]

no

solve for x

but you want x^2

so square what you get

[tepid crag]

9^2?

i plug in du in x?

[pine glacier]

no

u=9x+10

solve for x

[tepid crag]

oh okay

but like i’ve never done that before

i’m just curious why this one

[pine glacier]

why make u=9x+10 ?

[tepid crag]

honestly idk i’ve just been following a process

[pine glacier]

its the same process just with an extra step

you make the usual u sub but you are left with x’s

so you need to make these in terms of u

[shy ruin]

just rewrite u = 9x + 10 by isolating x then substituting that in for x^2

[tepid crag]

okay

the reason we do this is to get rid of x^2?

[tepid crag]

,rotate

[tepid crag]

my work so far

what now

[pine glacier]

now you have x

but you need x^2

so square it

[tepid crag]

would this be right

[pine glacier]

yes

[tepid crag]

okie

[shy ruin]

then plug that back in to the original function

then you can expand it then split into separate integrals and solve

[tepid crag]

i’ve done this so far

[shy ruin]

looks good till the last step I got a bit confused

[tepid crag]

yeah it looks wrong

[shy ruin]

why not multiply out the sqrt(u)

just write it as fractional exponents

[tepid crag]

so like

this?

[tepid crag]

would this be right

[subtle sinew]

You should have parentheses here

[tepid crag]

oh oko

was my awnser still right or nah?

[subtle sinew]

This part?

No

Because of the missing parentheses

[shy ruin]

You need to distribute the u^(1/2) to all 3 terms

[tepid crag]

rightt

got this

war

one sec

[shy ruin]

how is 20u^1/2

[tepid crag]

the -20 is 3/2

i fixed it

[shy ruin]

yeah

then split that into 3 integrals and solve

[tepid crag]

okok

[shy ruin]

make sure to write du every time

You skipped it alot

[subtle sinew]

Also when you did the u sub, you had this where it's du = 9 dx meaning dx = du/9

So you forgot a 1/9

Outside of the integral

You forgot to sub that dx with a du

[shy ruin]

That too

[tepid crag]

oh right

okie i got the awnser finally

tysm

.close

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[open spoke]

can someone explain this DFA for me?

[obtuse pebble]

@open spoke Has your question been resolved?

[tame python]

what do you think the DFA does so far

[open spoke]

each consecutive 0 is odd length

[open spoke]

there is no substring 11

start at 0, end at 1

yeah thats all i have so far

it seems like alot of rules i feel like i could simplify it but idk how

[tame python]

are you trying to write a regex for this?

[open spoke]

sure

[tame python]

you know it begins with 0, so the regex has to begin with that

then after that its 01

but theres a varying even number of 0s, so (00)*

[tame python]

after an odd number of 0s followed by a 1, it can go back to the 0s any amount of times, which is a ()* or a ()+

after ending with a 1, it cant end with another 1, so its just that 1

do you want to try writing a regex for this then

[open spoke]

0 (00)* 1 (0001)*

[tame python]

0(00)*1(0(00)*1)*

so you can describe this as odd substrings of 0s with 1 at their ends

[open spoke]

how did u come up with that lol

nested kleene stars lmaoo

[tame python]

oh wait

I misread the DFA

whoops

(00)*1(0(00)*1)*

even 0s, then always odd 0s, all of which end in 1s

[open spoke]

???

1 is a counterexmamplke

[tame python]

another mistake

forgot kleene stars allow 0 of a copy

00(00)*1(0(00)*1)*

this should be it

[open spoke]

001

[tame python]

🤔

man Im not good at this

so the original I said was correct 0(00)*1(0(00)*1)*

I mismisread it

@open spoke anyways do you get what the DFA matches now

[open spoke]

i like this explanation

if its right atleast

[tame python]

this time we have the correct regex

[tame python]

if you see it, it cant really be seen any other way

[open spoke]

im still trying to figure out the nested kleenestar

never done that before

[tame python]

remember, each star is "some number of copies"

so the inner (00)* means 0, 2, 4, 6, 8, ... 0s

the 0(00)*1 means an odd substring of 0s followed by a 1

the (0(00)*1)* means any number of (odd substring of 0s followed by a 1)

could be e, or 01, or 010101, or 01000101, etc.

[obtuse pebble]

@open spoke Has your question been resolved?

[obtuse pebble]

@open spoke Has your question been resolved?

[obtuse pebble]

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[dapper robin]

how to find period of this function

step by step

idk how to find period:C

[velvet rover]

https://www2.clarku.edu/faculty/djoyce/trig/identities.html

[obtuse pebble]

#geometry-and-trigonometry - euclidean geometry, coordinate geometry, trigonometry

[dapper robin]

oh ok

ill see

[tulip prawn]

just make a graph

[dapper robin]

graph

like every time?

im talking in exam

[dapper robin]

in which topic is periods

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[timid silo]

Can you please provide me with resources so that I can solve these equations ?

[short spire]

I mean for 1) if you just stare at it long enough should be able to see what solutions exists, and for 2) since this is a true/false question you might be able to argue differently

[short spire]

In any case the resources you would probably need here is this: https://www.khanacademy.org/math/ap-calculus-ab/ab-differential-equations-new/ab-7-6/a/applying-procedures-for-separable-differential-equations

But again, you might be able to argue differently

So if this doesn’t work I suggest thinking differently

[timid silo]

How do I put the initial value condition ?

[short spire]

Oh you can see straight away that the statement is false, do you see how?

There’s only one solution, do you see which for 1)?

[timid silo]

The log function is not defined for "0"

[short spire]

Precisely, so going back to the original equation we can conclude there’s only one solution and it’s the so called “trivial” one, do you see it?

[timid silo]

Trivial means y(0)=0 right ?

[short spire]

I think in general you might need to reference a theorem like Picard–Lindelöf theorem but in this case I think it’s obvious enough

[short spire]

So, what I mean is the so called “trivial solution”, but that doesn’t matter; I’m only saying that because the solution is not at all complicated

A hint: it’s a constant function

[short spire]

This is a pretty big hint btw

[timid silo]

Y = 0

[short spire]

Yup!

And so we don’t even have to check 2)

Since we know the statement is false for 1)

Does that suffice help wise?

[timid silo]

Thanks 👍

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[sturdy steppe]

what is the shortest possible path to visit every square on a square grid from an optimal starting position, possibly a formula for an n by n grid
I looked online and all I could find was the version where they go to the middle of the squares
I am counting edges as visiting 2 squares at once
this is the best I could come up with so far

[brazen gorge]

optimal path
elaborate?

[sturdy steppe]

shortest possible distance traveled

[brazen gorge]

umm

[brazen gorge]

https://stackoverflow.com/questions/27305163/what-is-the-shortest-path-that-touches-every-tile-in-a-square-grid

don't all patterns travel the same distance?

if you need to go over every square then go through them exactly once and any path that does that has to have the same length

[sturdy steppe]

I looked online and all I could find was the version where they go to the middle of the squares

[brazen gorge]

i don't think your problem is that much different

what about in this case, what squares are counted as visited?

[brazen gorge]

i see

[sturdy steppe]

it is clearly a different problem when n is odd

[brazen gorge]

and here i assume this is the case

[sturdy steppe]

yes

to demonstrate

[brazen gorge]

yeah and the first one also visits some squares more than once

oh i got it i think

let me make a diagram

[sturdy steppe]

the outer part of this should be optimal, then for the 3 by 3 you need to visit 4 vertices of the middle square, should also be optimal

every turn is wasting a square that could be visited so the solution should be close to a spiral

wait no it's not

it gains that square back at the corner

both 8

[stray python]

i need help

[sturdy steppe]

this is occupied pls delete

[obtuse pebble]

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

[sturdy steppe]

for the even cases the spiral and the Us are equal

3(n-2)+2(n-4)...+2(n-(n+2))

[brazen gorge]

does this cover all squares you need to cover?

[sturdy steppe]

yes

[brazen gorge]

i am realising that it might be impossible for an odd by odd to cover each square exactly once

[sturdy steppe]

it is

well

if you want it to be the shortest possible

otherwise the spiral from center to center does it

[brazen gorge]

my idea was to do this: overlay an (n-1) by (n-1) grid and solve it using the same methods as the one where you go through the middles of squares

and avoid overcounting as much as possible

[sturdy steppe]

I thought of it as recursive, you need to visit these

then you are left with a smaller version of the same problem

so you should also be able to do this path

(n-2)/2 * 2 to connect the 2(n-2)+2(n-4)+... parts

[sturdy steppe]

which is eqv to this

connecting 2s in blue

[drifting wraith]

you can't improve this, so it's solved

[brazen gorge]

i think so

[sturdy steppe]

is it the same as this

11 vs 12

you already touch one of the vertices of the middle square in the 3 by 3 just from going back from the outer shell

[past sand]

This part is wasteful

[sturdy steppe]

yes you prefer the 3 by 3 over the row so you can avoid the triple counting U turn

[sturdy steppe]

to which you can always reduce the odd by odd grid with the spiral or this

let me see if the latter is the same as the spiral

it is in a 5 by 5

[drifting wraith]

that doesn't mean it can be done better

[past sand]

It literally does

[sturdy steppe]

here

[drifting wraith]

well fair

[past sand]

(same length as a spiral)

[sturdy steppe]

okay so odd should be optimal as long as you can reduce it to a final 3 by 3 without waste

traveling from the outer shell to inside it will take 2 no matter what

well if your shell reducing line ending is optimal

and it always can be

so we are done

.close

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[shell ridge]

I need help

[obtuse pebble]

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

[shell ridge]

Idk anything

[last pilot]

...?

[shell ridge]

Hi

[last pilot]

okay this question is fairly easy tbh

hello

uh

[shell ridge]

I need help with this

[last pilot]

what part do you not understand

[last pilot]

yes we know

[shell ridge]

Just how to do it

[brazen gorge]

volume is cross-sectional area * length

[last pilot]

^

or base times height

those are equivalent forms of the same thing

[shell ridge]

I tried what I thought it was but I am wrong

[brazen gorge]

what did you think it is?

[neat mantle]

think of the area as the base

then you have to multiply the base and the height

in this case, base = 18cm^2 and height is 3cm

[shell ridge]

12

[neat mantle]

its not 12

[shell ridge]

I'm confused asf

[brazen gorge]

why?

explain the thought process

[neat mantle]

,w 18cm^2 * 3cm

[shell ridge]

Base time height

[warm shale]

Results provided by WolframAlpha

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

Requested by clonesolopros

[shell ridge]

I thought it was 3x4

Idk wgy

Why*

Idk where I got the 4 from

[brazen gorge]

idk where you saw a 4 anywhere there

[brazen gorge]

yeah lmao

[shell ridge]

I'm stupid asf

[neat mantle]

8-1-3 =4 xd

[shell ridge]

Wb dis?

[last pilot]

same thing

length times area

hint: some of the lengths are simply present to confuse you

[shell ridge]

5×4?

?

[neat mantle]

18*7

[shell ridge]

7×18?

[neat mantle]

yes

[obtuse pebble]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[last pilot]

oops

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[neat mantle]

Yeah ik 😦

[shell ridge]

L

[last pilot]

then...

why did you give out the answer?

[neat mantle]

idk

[shell ridge]

Bottom number is 7 btw

[last pilot]

again.

dtg_dan.

[shell ridge]

Hi

[last pilot]

we have helped you solve two problems.

try this one yourself.

[shell ridge]

Ye?

But idk which to times

[last pilot]

please don't ask us to continually give out the answer

cross-sectional area times length.

[shell ridge]

There's a 8 in the middle below

[shell ridge]

But I idk how to properly do it

[last pilot]

then TRY

[shell ridge]

I thought I just timed it all

[last pilot]

if you make mistakes its FINE

[shell ridge]

I did bruh

The number is in the middle

[shell ridge]

What's the calculation

[last pilot]

...?

again

try

and tell me what you did

[shell ridge]

Like what do I do to get the answer?

[last pilot]

WE JUST EXPLAINED

sorry

[shell ridge]

I just times it all together

[last pilot]

im a little frustrated

i have some personal issues rn

mb

excuse me?

[shell ridge]

I'm calling myself it

[last pilot]

that... doesn't matter?

just because you can't solve a math problem

doesn't mean you are that

it's fine

we're all good at different things

[shell ridge]

Can u explain again

[last pilot]

okay

[shell ridge]

Or show the text where u alr explained it?

[last pilot]

it's okay

i'll explain

volume = cross sectional area times length (the length that is perpendicular to the cross-sectional surface)

cross sectional area is just the area that's shown

[shell ridge]

9 times 4?

[last pilot]

the side whose area yk is the cross-sectional surface

[shell ridge]

4?

[last pilot]

not quite

[shell ridge]

8l7?

7*

[last pilot]

try to calculate the area of the side first

it's a triangle

with base 9 and height 4

what is the area of the triangle

[shell ridge]

9×4?

[shell ridge]

7?

[last pilot]

no...?

dude

try working it out

dont just guess

[shell ridge]

I'd base is 9
Height is 4
It's left with 7?

[last pilot]

no...

look

[shell ridge]

Do I do 7×4?then ×9

?

[last pilot]

the base of the triangle has a length of 9.

[last pilot]

no...

[shell ridge]

Ok

[last pilot]

the base of the triangle has a length of 9

[long plinth]

How do you even get 7 by doing some calculation on 9 and 4???

[last pilot]

the height of the triangle is 4

[neat mantle]

The order of multiplication doesnt matter

[shell ridge]

.

[last pilot]

...?

[shell ridge]

Cuz the numbers r 4 9 and 7

[neat mantle]

Do you know how to calculate the volume of a triangular prism?

[last pilot]

my app crashed

[last pilot]

wdym

[shell ridge]

Nop

[last pilot]

okay

[shell ridge]

Showing tropo

[last pilot]

first off

[neat mantle]

Its the base of the prism x the height

[last pilot]

no...

that's wrong

it's the area of the triangle

[neat mantle]

where the base is a triangle*

[last pilot]

times the height

yes

[long plinth]

Nobody said the area of the triangle would be one of those three numbers, though?

[last pilot]

then that's correct

ah

[long plinth]

Sorry.

[last pilot]

is alr

lol

[shell ridge]

I'm getting confused

[last pilot]

o k a y

[long plinth]

Do you know how to calculate the area of a triangle?

[last pilot]

^^

[shell ridge]

Nope I forgot

[last pilot]

it is

the length of the base

times the height

times 1/2

[shell ridge]

9×4?x ½

[last pilot]

YES

@shell ridge do you understand what you have to do now

[shell ridge]

Kinda

[last pilot]

...

what

do you not

understand

[neat mantle]

it is important that you understand everything

[last pilot]

indeed

[long plinth]

Though that can be a somewhat anxiety-inducing way to state it ...

[last pilot]

??

[shell ridge]

I put 18 but it's wrong /:

[last pilot]

18

[shell ridge]

Now there's another new question

[last pilot]

is the area of the triangle

...

[shell ridge]

¿

[last pilot]

sorry to ask but tropo can you handle this for a bit

[shell ridge]

Nukehrs r 6 9 7

[last pilot]

i can't seem to explain this

(no offense, dtg; might just be my way of teaching)

[long plinth]

Okay, I'll take a shot.

[last pilot]

thank you

i'll watch

[shell ridge]

Prob cuz I'm stupid tho

[long plinth]

By the way, does the "watch video" button do anything that helps you? Some people feel they learn better that way than by reading text?

[last pilot]

no?

that's not true

[shell ridge]

I don't understand it

[long plinth]

Okay, let's try text then.

[last pilot]

it's quite all right if you don't understand a certain concept

you can't be judged as stupid just because you can't solve a maths problem

[long plinth]

So now we have a different triangular prism what we need to find the volume of.

[last pilot]

intelligence is far more than simply marks, or studies

i'll let tropo continue now

[long plinth]

First off, do the words "triangular prism" make sense to you in the first place?

[shell ridge]

No

Anyway I gtg for now bye and soz for the waste of tike

Time*

[last pilot]

it's okay

[long plinth]

How well do you understand the drawing in the questi-- oh, too late.

[last pilot]

i suggest you watch a youtube video on this @shell ridge

he's still here

[last pilot]

ffs, tropo

[long plinth]

Sorry?

[last pilot]

nothing

okay

ffs means smth else

than what i thought it to be

mb

i was googling it so i could explain it to you

and uh-

[long plinth]

Usually it's an exclamation of strong exasperation ...

[last pilot]

i did not know that

i thought it was like f's in the chat

...

mb

[long plinth]

"f's in the chat"?

Anyway, my guess when it's going as wrong as this is that it never became clear to Dan which three-dimensional shape we're talking about in the first place. (The problem just showed a triangle next to a weird quadrilateral, and what's up with that).

[last pilot]

...
yes

i use that phrase a lot

apologies once again

[long plinth]

Okay I didn't know what that was. https://knowyourmeme.com/memes/f-in-the-chat states it is "used to symbolize sorrow, compassion, or more commonly to sarcastically mock a FAIL or embarrasing situation" which sounds, um, somewhat ambiguous.

[warm canopy]

(originates from an old call of duty game where you "press F to pay respects" at a funeral)

[last pilot]

^

it was my way of saying you tried

sorry for the waste of time

[long plinth]

Anyway, looks like Dan has left. @shell ridge, when you get back feel free to open a new help channel.

.close

[obtuse pebble]

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[waxen swan]

can someone explain the Central Limit Theorem in simpler terms? I don't get what it means that the sample size is "sufficiently" large and uses another distribution, like I only learnt binomial and normal distribution

[waxen swan]

like what is (ii) talking about

[grizzled shore]

the central limit talks about the asymptotic distribution of the sample mean

when you have infinitely many samples, the distribution of the mean is normal

that's great and all, but you dont have infinite samples in real life

so...you just kinda pretend that 300 is sufficiently large

it's the sinx = x of statisticians

you wave your hand and pretend that the sample mean for "sufficiently large samples" is normally distributed

(which of course the theory DOES NOT say is true, you need to have infinitely many samples for this to be the case)

but we pretend

[waxen swan]

so anything => 30 falls under CLT

[grizzled shore]

well that's only a rule of thumb

some say 25, some say 30, it's really arbitrary

from the strict "trueness" of the statement, it doesn't know for any finite sample size

[waxen swan]

ah okay
what does the last sentence here mean though

[grizzled shore]

in (i) you use the central limit theorem

you used that the sample mean is (approximately) normally distributed

that's how you found the probability

now the central limit theorem doesn't say whether your original numbers need to be normally distribtued or not

i believe it only says they must have finite variance

yeah it only says we need finite variance (and hence also finite mean)

so regardless of the distribution of the mass of parrots, as long as it's finite variance, the sample mean will be normally distributed by CLT

so we dont actually need that the mass of parrots are normally distributed

[grizzled shore]

that's what (ii) is saying

[waxen swan]

so from what I understand now, i'm just closing an eye to the sample size basically?

[grizzled shore]

you look at the sample size

it's big

then you say by CLT the sample mean (mass) will be normally distributed (this is approximate, this isn't actually true, it's close enough though)

then you find the probability assuming the sample mean is normally distributed

nowhere have we assumed that the parrot masses themselves are normal, we did not need this assumption for CLT

[waxen swan]

omg it clicked

thank you

[grizzled shore]

(we do need that the parrot masses have finite variance for CLT)

but clearly, i mean the mass of an animal has finite variance

that is very reasonable to think

[waxen swan]

mean and variance are finite

[grizzled shore]

you can make examples (and they do often appear in practice) where the variances are not finite

[grizzled shore]

in which case this particular CLT wont work on it

there are many different versions of CLT 💀

[waxen swan]

I don't think my questions will go there so i should be fine

[grizzled shore]

no they wont, just something to note

so you are clear the assumptions are important

[waxen swan]

already struggling, did discrete random variables, normal distribution and sampling all in 1 day

[grizzled shore]

that's a lot, take some rest!

statistics is very unintuitive sometimes, it takes a while to get a hang of it

[waxen swan]

yeah after this i'm done
my mind is kinda spinning, tomorrow ill do practices

[waxen swan]

thanks for the explanation

.close

[obtuse pebble]

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[obtuse pebble]

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[dreamy forum]

Quick question: Is "f(x) tends to infinity" equivalent to saying that it diverges to infinity?

[drifting wraith]

i think so

[dreamy forum]

Ok because it is a term in a proof I am reading and I can't wrap my head around it

[waxen swan]

the values of f(x) go towards infinity like x^2

[dreamy forum]

Specifically it concerncs unbounded functions

Not every unbounded function tends to infinity or diverges to infinity that is a fact I think

It is mentioned in the proof of Ladner's theorem regarding the existence of NP-intermediate problems

[warm canopy]

that is true here is an example

[dreamy forum]

I am reading the proof of Arora-Barak

[dreamy forum]

I looked at something similar with the modulus operator wild graph that one

I think I understand then. Ty @drifting wraith @waxen swan @warm canopy

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[dawn vale]

so chain rule

[median dome]

not necessary and also idrk how you were planning on using chain rule here anyway

[dawn vale]

4x^(1/2) (x^2+2x^(1/2))

[median dome]

I think you meant to say product rule

[weary reef]

Product rule would work but there's a easier way

[median dome]

distribute the 4x^(1/2)

[dawn vale]

is this the answer?

[weary reef]

Nope

[weary reef]

Yeah that's right

[dawn vale]

how do you know when to use chain rule and product rule?

[weary reef]

You typically use product rule when differentiating the product of two functions. In this case, you could just distribute to get something easier to differentiate.

Chain rule is for nested functions. You apply chain rule implicitly almost every time you differentiate. If you used product rule for this question, then you definitely applied the chain rule as well

[dawn vale]

but its not really 2 different they are they?

like they are times together

[weary reef]

This is NOT a correct application of the product rule

[dawn vale]

k thank you

🫡

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[crisp ember]

i have managed to get this far

[plain vapor]

,rotate

[crisp ember]

but do not know where the brackets in whatthe answer is suppsoed to be come from

[plain vapor]

you approximated it a bit too much but your reasoning is right

you see, you get that n^k/k!, but there's also other terms of smaller value, like n^(k-1) etc...

and if you write that , and then factor out the n^k/k!, you get that

[crisp ember]

im still a little confused

[plain vapor]

i'll give you an example

(x)(x-1)(x-2)

the bigger term ofc is x^3

but you also have smaller terms with x^2 andx

when your x is big, yes they don't count much,

and the bigger of them is the x^2

so you write the approx like x^3+O(x^2)

[crisp ember]

ok so why are they adding and not timesing when everything was timesing before

[plain vapor]

and then factoring out the x^3 you get x^3(1+o(x^-1))

[plain vapor]

think about it

[crisp ember]

ahhhhhhh

ahaha i get it

[plain vapor]

when i expand the brackets it becomes a summation

[crisp ember]

makes sense when you think of it like a quadratic or polynomial

[plain vapor]

yep

so yeah every other term will be O(x^-1)

[crisp ember]

makes sense

thank u

[plain vapor]

np

[obtuse pebble]

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[eternal jay]

Is this right for question c?

[eternal jay]

and do i add 38 the original years for the answer

which would be 2033

[low ermine]

Yeah I think u are right, that's it

[obtuse pebble]

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[eternal jay]

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✅

[eternal jay]

is this right?

[bold bane]

The answer is "correct" but how you calculated the integral on the interval from [2,5] is not correct.

@eternal jay

[eternal jay]

it supposed to be positive 25/12 not negative i assume thats the mistake

[bold bane]

The technically correct way would have been to use int_2^5 (0 - f(x)) dx.

[eternal jay]

but this wokrs too no?

[bold bane]

Where 0 is the line y = 0.

[eternal jay]

would i get marked down if it was test?

or is this ok the other method is just better?

[bold bane]

It "works" but is not the correct way to calculate the value.

[eternal jay]

hmm ok anhyways thx

[bold bane]

Do you know how to calculate the area between two curves?

[eternal jay]

no i dont

[bold bane]

It's just int_a^b f(x) - g(x) where the curve f(x) >= g(x).

You are just subtracting the area of one from the other.

[eternal jay]

oohhh that makes sense

[bold bane]

Now with this particular problem, although it probably is never explained in class, the lower area is the line y = 0 which is the x-axis.

You are technically subtracting that area without realizing it.

[eternal jay]

yea i never heard of that until today

[eternal jay]

ur teaching me more than my teacher rn lol

[bold bane]

Because the curve on the interval [2,5] is below the x-axis, you need to subtract f(x) from the line y = 0.

[bold bane]

The latter intergral is how you would calculate the area so that you get a positive value.

And through integral properties, you can rearrange it this way.

You did the last step without understanding why you did it.

And it's important to understand the why.

[eternal jay]

wait this makes so much sense icl i didnt understand wat i was doing i was just doing it

thx a bunch

[bold bane]

yw

[eternal jay]

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[fluid snow]

Is this correct?

[bronze mica]

This seems reasonable

[fluid snow]

Apparently, it's this

That's a bit different to $f(-x + x_0) - y_0 = -f(x + x_0) + y_0$

[warm shale]

Kepe

[bronze mica]

No? It's equivalent

[fluid snow]

It's x_0 + x

Oh wait

[fluid snow]

You're right

Thank you!

[fluid snow]

But I can't expect f(-x + a) - b to be -x, right?

That's just for the toy example

[bronze mica]

No

[fluid snow]

Or can I?

[bronze mica]

At least I don't think you can

I mean for one it's not -x in this example, it's -x^3

[fluid snow]

oops I meant -x^3 yeah

[fluid snow]

Yeah I think we can only expect that if a cubic can be written as (x - p)^3 + z which is not always the case

Thanks

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[rough stump]

if z = 2 + 7i / 2 - 8i then 2 -7i / 2 + 8i = ?

A) z^-1 B) z(conjugate) C) -z

[obtuse pebble]

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.

[rough stump]

i do not have picture

sorry

[half silo]

simplify both

[rough stump]

i do

ohhh

wiat