help-0

[thorn tapir]

Yeah

This is 100%

[steady basin]

what about the 0 column

[thorn tapir]

Try your own method on the RREF

RREF means Row Reduced Echelon Form

[steady basin]

yh ive done that

what im saying is

u said

no of columns is 4

but what about the 0 column

shouldnt number of columns be 5

[thorn tapir]

That is not part of the original matrix lmao

[thorn tapir]

No

the number of column is basically of the original matrix B

[steady basin]

ok

but when u do RREF

do u consider

the inhomeginity

as well

ie the added column

[thorn tapir]

You don't have to always add the 0 column to do RREF, it can be done on any matrix anywhere

[steady basin]

ie the RHS of the augmented matrix

[thorn tapir]

When you augment the b vector and then do RREF, then you're doing something called Gaussian Elimination of System of Linear Equations

You're removing variables by making their coefficients 0

[steady basin]

okok

[thorn tapir]

When you do row operations it's technically adding equations to each other

[steady basin]

ok

[thorn tapir]

Essentially the elimination method of solving equations

[steady basin]

ight kl

i think we r come

chartbit has also given me their agreement as well

thanks mate

[thorn tapir]

You're welcome

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[gray walrus]

A pound of coffee beans makes 50 cups of coffee when brewed according to the instructions (4 cups = 1 qt). How many liters of coffee can one kilogram of coffee beans produce?

[lunar palm]

do you know the unit conversions required for this question?

[gray walrus]

L/kg

[lunar palm]

more specifically you'll need the conversion formulas from
pound --> kilogram
&
cup --> litre

do you know them?

[gray walrus]

yes yes

[lunar palm]

convert the 1:50 ratio with those formulas

i.e 1 pound in kg:50 cups in liters

[gray walrus]

okok got it

thank you thank you

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[alpine sable]

I'm messing up somewhere in the second iteration. If someone could help me please

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[chrome depot]

Let x1, x2 be solutions to:
(5-x) : (x+1) = (x-1) : (x+a)

[chrome depot]

I have to find parameter a such that x1^2 + x2^2 = 26

How do I go about this?

[weary wyvern]

Write it as fractions then cross multiply

[lone heart]

@chrome depot Has your question been resolved?

[chrome depot]

I have managed to solve it myself in the mean time

Here is the solution:

[weary wyvern]

Didn't ask

[chrome depot]

Kind of rude but okay

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[plain hill]

about this

i find the sin(alpha)

because i do that sin^2(a) = 1/1+ cot^2 (a)

and i find that sin^2(a) = +/- 1/16

and after i do that this was equal at -1/16

and so sin(a)= -1/4

but about cos(a)

what i should do??

<@&286206848099549185>

[exotic belfry]

sin²+cos²=1

or cot = cos/sin

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[somber spade]

2x + 3y = 45
x + 4y = 35

[somber spade]

whats the best way of doing this?

[sharp thorn]

Multiply the second equation by 2

And subtranct

2x+3y=45
2x+8y=70

Solve from here

[somber spade]

okay that actually was really easy

is there other methods to do when you cant multiply?

or divide

[sharp thorn]

Draw lines and take the intersection

Don't worry there won't be much cases without the elimination mehod

[somber spade]

how would i draw lines and take intersection?

[sharp thorn]

Graphing calculato

[somber spade]

ah right

[sharp thorn]

You enter both equations

[somber spade]

is there another way which you can do on paper(

😅

[sharp thorn]

Idk any other methods sorry

[modern sedge]

2x+3y=45
2x=45-3y
x=(45-3y)/2

x + 4y = 35
x=35-4y

x=x
35-4y=(45-3y)/2
And solve for y

or the substitution

[sharp thorn]

Oh yeah I substitution

You have to convert x in terms of y and substitute or vice versa

[modern sedge]

x+4y=35
x=35-4y

2x+3y=45
2(35-4y)+3y=45 and solve for y

[sharp thorn]

Yep

[somber spade]

not really understanding any of this 😅

[sharp thorn]

Try my method

[somber spade]

i was able to do your first method easily

but i want to learn at least 2 methods on paper in case of a test

[modern sedge]

The first method always works for the equations in form of ax+by=c

[sharp thorn]

The two equations are
2x+8y=70
2x+3y=45

After subtract
5y=25
So y is 5

Then take the value of y and substitute it in any of the equation you'll get x

[somber spade]

yeah this was easy to understand

[somber spade]

this is what i didnt understand

[modern sedge]

Try this method instead

[sharp thorn]

It's substitution method

[modern sedge]

the one I posted earlier is a bit worse

[sharp thorn]

It's very hard to teach in chat .. maybe contact your teacher or smtg

[somber spade]

cant understand this either

[modern sedge]

x+4y=35, that's first equation
x=35-4y by subtracting 4y from both sides

2x+3y=45 that's the second equation
2(35-4y)+3y=45 since x=35-4y we can replace it with it

[sharp thorn]

MY English sucks already and I'm in phone so that's the maximum i can do

[modern sedge]

Which line dont you understand?

[somber spade]

after the second equation

[modern sedge]

Since x=35-4y is the result from first equation. This means that x is SAME as 35-4y, and since it's same you can replace x with 35-4y and you can also replace 35-4y with x

equal means same

If a=b and b=c, then a=c

[somber spade]

right let me try and do it

alright i understood it this fae

far

[somber spade]

how do i find the value of x and y from this??

[modern sedge]

2(35-4y)+3y=45

this has just one variable so it can be solved trivially

once you solve it you can plug it's value e.g. to x+4y=35 and get x

[somber spade]

uh

i didnt understand a thing about that

[sharp thorn]

We are converting x in terms of y

[somber spade]

could someone draw it on paper?

and explain whats being done step by step?

[sharp thorn]

After conversion we get
x=35-4y

We take this value of x and substitute in one of the equations

That's it

[somber spade]

yes i understood that

but how do i go from that

to x=5

[sharp thorn]

Im sorry I don't have paper now

So wait

Send me the 2 equations once sgn

[somber spade]

2x + 3y = 45
x + 4y = 35

[sharp thorn]

Okay so

Acknowledge me with every step ok,

Let me take the second equation

x+4y=35

Now let me take the 4y to the other side

So
X=35-4y

[somber spade]

yes

[sharp thorn]

Understand?

[somber spade]

yup

[sharp thorn]

Now we have a value for y

[somber spade]

x*

[sharp thorn]

Sorry for x

[somber spade]

yeah i understood this far

[sharp thorn]

That is x=35-4y

So let's take the first equation again

[somber spade]

yeah

[sharp thorn]

X+4y=35

Or wait ig I have to take the other equation

[somber spade]

first equation is 2x+3y= 45

[sharp thorn]

Yeah this

So we have a x in this equation right?

[somber spade]

yeah sp 2(35-4y) +3y = 45

[sharp thorn]

Yes exactly

[somber spade]

yes so after this i need assistanse

[sharp thorn]

Multiply 2 inside

And tell me

[somber spade]

what do you mean?

[sharp thorn]

After multiplication 2(35-4y) is 70-8y

This you got?

[somber spade]

yes

yeahyeah

[sharp thorn]

So mpput this into the equation

70-8y+4y=45

[somber spade]

70-8y+3y=45?

[sharp thorn]

Yes sry

[somber spade]

its fine

[sharp thorn]

So 70-5y=45

[somber spade]

so 70-5y=45

[sharp thorn]

Got it?

Yes

[somber spade]

yup

[sharp thorn]

Now subtract 70 on both sides

-5y=-25

[somber spade]

-5y = -25

yes

[sharp thorn]

Negative cancels negative

5y=25

Y=5

Got it?

[somber spade]

negative cancels negative?

i thought that was only in multiplication

[sharp thorn]

For your understanding let me put it this way

You multiply by(-1) on both the sidds

Sidds

Sides

Got it?

[somber spade]

yes

[sharp thorn]

Yea that's it

[somber spade]

but does thst happen automatically when its negative on both sides of equal?

or do you actually have to do (-1)

[sharp thorn]

Yes it does

[somber spade]

ohh

[sharp thorn]

You don't have to

[somber spade]

ohh thank you learnt something new

[sharp thorn]

You can just straight away cancel two negative signs in equality considering there is no addition or subtraction

for example -ax=-c
Also means that
ax=c

You can substitute any values for a x and c to check this yourself

[somber spade]

makes sense but also doesnt make sense to me

but yeah

i understand it

[sharp thorn]

Just cancel 2 negatives that's it

[somber spade]

yeah thanks

[sharp thorn]

Not an overkill

You're welcome

[somber spade]

wanted to learn 2 methods of doing it so i can double check my answers on my tests

[sharp thorn]

Yes

For double checking

You will get x and y nah

Just substitute those values in all the equations and make sure all of them end up true

[somber spade]

.close

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[native temple]

How to deal with earworms? I want quiet in my heart yet my mind thinks of a song I like even when I want quiet…

[spring plover]

Wrong server/ channel.

[native temple]

This is about studying though

[calm plank]

Bro what

[sharp thorn]

Not maths

[spring plover]

Does not belong here.

Take it to other channels on this server meant for discussion that isn't math help.

[calm plank]

But close this question first tho

[modern sedge]

.close

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[modern sedge]

#chill or #discussion

[lone heart]

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[alpine sable]

hey, im not really sure how to do hypotenuse leg proofs, would this be the right way to go about it? i dont know if theres anything specific that i need to do special to them or whatnot

[alpine sable]

(and then obviously do the triangle proof and stuff saying that the triangles are congruent)

[calm plank]

what is your question exactly?

[alpine sable]

is this the proper way to prove congruent triangles using HL

i thought there was some special angle stuff you need to do or something but im not sure

[gusty gorge]

can you state the HL theorem?

[lone heart]

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[lone heart]

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[lofty flame]

Summation (I=15, n=29+5, k = i^2 + 5i +4 )

Can anyone check if the answer is correct and tell me please how can I verify it

[lone heart]

@lofty flame Has your question been resolved?

[lofty flame]

<@&286206848099549185>

[vestal quail]

Tips on how to actually become good at math😭😭

[tacit bobcat]

You can verify it with calculators

[lofty flame]

I want to verify the whole summation I need to redact the solution

[tacit bobcat]

First example the first summation is verified this way:

,calc sum(map(range(1,5), f(x) = 7*x-3))

[ocean seal]

Result:

90

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@lofty flame Has your question been resolved?

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[quartz panther]

,rccw

[quartz panther]

how do i find the missing side?

[fringe sleet]

pythagoras

but first you need to find this length

so look for something symmetries so you can find it

[quartz panther]

2+5x - 2+x?

so @fringe sleet its just 3x

[tacit arch]

what is "it"

the red line?

[quartz panther]

the answer

for the missing side

coz sqrt(3x^2 + 6x^2) = 3x

right?

[tacit arch]

show your full work

[quartz panther]

yeah ive done something wrong i think

so c

• sqrt of a^2 plus b^2

and the missing side is 6x

i think

coz $2+5x-(2+x) = 4x$

[ocean seal]

MathematicsPractice

[quartz panther]

@tacit arch

sorry for ping im real desperate

[tacit arch]

missing parentheses here

oh you fixed it

[quartz panther]

the answer for the perimeter is somehow 14x + 4

i dont understand how that is the perimeter

[tacit arch]

where are you plugging in 4x here?

[quartz panther]

huh?

[tacit arch]

this is the red line

[quartz panther]

i messed up the working before sorry

[tacit arch]

or are you calculating something else

[quartz panther]

the red line is 4x

[tacit arch]

then where's your missing side/

[quartz panther]

sqrt 3x + 4x

sqrt 3x^2 + 4x^2

[tacit arch]

https://www.google.com/search?q=pythagorean+theorem

[quartz panther]

9x^2 + 16x^2

sqrt

is 5x

cool

.close

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[quartz panther]

.close

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[peak mist]

Hello what do I do to complete the ratio test'

[peak mist]

.flip

[lament forge]

,rotate

[lament forge]

well it looks like you should be able to simplify this expression quite a bit

[high rapids]

Radius of convergence is 1/ lim -> infinity |a_n+1/a_n| right?

[peak mist]

does this simplify it completley

[peak mist]

yes that is correct i did that

just need to know if the above simplification is correct

[high rapids]

Hmm what happened to the 5^n?

[peak mist]

its clear my simplifying is my weak point

would you mind walking me through how you owuld simplify it

[high rapids]

You could write 5^(n+1) as 5^n * 5

[peak mist]

got it

thanks

.clsoe

.close

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[thin dagger]

ok so do you guys know how to solve this?

[high rapids]

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[thin dagger]

1

[high rapids]

So which point is the relative maximum?

[thin dagger]

the top left point is the maximum

ik that but

you have to write it a weird way

and Im very scared of getting it wrong

its not interval notation or anything but

I think you write an ordered pair then the x

but idk which

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[warped topaz]

https://gyazo.com/c4377ab68cb27d0d12d4a88263cf42bf.png

[warped topaz]

I got that b = 4

But idk how to get a

[spring flower]

how do i ask for help?

[tacit arch]

!help

[lone heart]

Please read #❓how-to-get-help

[warped topaz]

Btw in this question that 2 means log to the base 2

its written weird

[serene junco]

Find a point on the graph whose coordinates are integers

[warped topaz]

(-3,0)

[serene junco]

well yeah, I'm guessing that's how you found b=4, right?

[warped topaz]

Yeye

[serene junco]

See if you can find another one, without any 0 coordinates

[warped topaz]

(-2,6)

[serene junco]

then since you have b already, you should be able to use that to find a

[warped topaz]

Ok 1 moment I think I can get it

A = 6?

That's all I had to do?

[serene junco]

yep 👍

[warped topaz]

Oh damn

Oke thank you!!

❤️

[serene junco]

np

[warped topaz]

. close

[serene junco]

❤️

[warped topaz]

.close

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[thick depot]

prove $(a + b)^2 − (a − b)^2 = 4ab$

[ocean seal]

Timsaay

[calm plank]

what have you tried so far?

[calm plank]

you made a mistake in your algebra

[thick depot]

where?

[calm plank]

when expanding (a-b)^2

remember that the minus sig applies to the whole expansion of (a-b)^2

not just the first term

if you put parenthesis around the expansion it should do the trick

but you still have to fix a bit

[thick depot]

You mean $-(a-b)^2 = -(a-b)-(a-b)$

[ocean seal]

Timsaay

[thick depot]

?

[calm plank]

not quite

so when you expand -(a-b)^2

it should be $-(a^2-ab-ab+b^2)=-a^2+2ab-b^2$

[ocean seal]

Browi

[calm plank]

get it?

[thick depot]

so the - is actually -1 multiplying it?

[calm plank]

yup that's the same thing

[thick depot]

Okay that makes sense

thank you

[calm plank]

no problem

now the rest of the proof should be easy

[thick depot]

Does this look better?

.close

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[ornate ginkgo]

Why the 1/2

[worn fox]

Differentiate tan(x/2) and see

[ornate ginkgo]

If i do that way it makes sense due to chain rule

So i just add the coefficient to the denominator whenever i see stuff like this?

[worn fox]

Yeah you want to undo the chain rule so to speak

[ornate ginkgo]

The reverse process is kinda hard to PROCESS.

[worn fox]

(This is just what you do when you u sub in an integral)

[ornate ginkgo]

Thank you tho

.close

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[misty sparrow]

Hello, what is the trick when knowing if 8 is divisible by a large number? (for ex. 460) Because it will consume too much time if I'm going to add 8 manually.

[amber obsidian]

Divide the large number by 8. If there is NO remainder, then the larger number is divisible by 8.

[misty sparrow]

Okay, got it! Thank you very much! I was thinking the same thing though I just want to make sure. closing this thread.

.close

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[alpine sable]

huh

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[alpine sable]

idk where to start

[calm plank]

well start by looking at the graphs

so immediately you can see that there are 2 different kinds

the ones trending more and more upwards, and the ones becoming more and more flat

[alpine sable]

ok through pattern recognition i'd say

49. A
50. y
51. c
52. D

[calm plank]

not exactly

try plugging in some values

so for example in 50

[alpine sable]

ok

[calm plank]

try plugging in 2

[alpine sable]

for x ?

where did you get 2

[calm plank]

wait actually plug in 4

it doesn't matter what you plug in

but choosing something easy to calculate is preferrable

so sqrt(4) is much easier than sqrt(6) for example

because sqrt(4) is just 2

now if only 1 graph has f(4)=2, then that must be sqrt(x)

[alpine sable]

?

can you show the work

[calm plank]

i haven't really done any work

i just concluded that if f(4) is not 2, then the function cannot be sqrt(x)

because sqrt(4) is 2

[alpine sable]

ok

[lone heart]

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[sinful field]

somebody help

[wet nest]

1600 ohm

[sinful field]

how come

[wet nest]

V=IR
12=IR
I=12/R
I_2 = 60/R
Req. = 400R/400+R
V=I_2Req
12= 60/R × 400R/400+R
Solve it to get R

[sinful field]

why not I=12/400

and I_2=5I

[wet nest]

Bruh, it is connected in parallel, voltage remains same, current gets divided
I_2=5I is true but I=12/400 is false

Req. = R1R2/R1+R2

[sinful field]

ohh right

i see

[lone heart]

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.

[sinful field]

okay thank you

.close

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[torpid storm]

Hello guys

[torpid storm]

Just a bit confused and need help

Doing this exercise on finding the limits of function by their graph

Here’s my answers, but apparently I got one wrong.

Which one tho? And why?

[limpid turret]

Let's start at the first

What is the value of f(-4)?

[torpid storm]

Sorry late reply but my answer for that is 5

Of wait is it at -5 instead??

[limpid turret]

Something seems off here

[limpid turret]

yes

[torpid storm]

I didn’t see the graph is wrong?

Ooohhh

[limpid turret]

Yeah that's weird

you got tricked

[torpid storm]

Man I only have one more attempt for the quiz

I thought I did something wrong, I mean I did, but I didn’t expect it like this lol

.close

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[sinful field]

.close

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[unkempt dew]

find exponentiation

[lone heart]

@unkempt dew Has your question been resolved?

[unkempt dew]

no

[lone heart]

@unkempt dew Has your question been resolved?

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[unborn stag]

unsure how to convert the theta portion to polar coords

cuz i get r = 3

[raven rover]

Try drawing A on the plane

That’ll probably help you visualize your bounds

[unborn stag]

i have drawn a circle of radius 3 with a line 3x going through it. they intersect at (3/sqrt(10), 9/sqrt(10))

[raven rover]

Great

So how can you get the angle this line makes with the horizontal

[unborn stag]

i was thinking tan(y/x) --> tan(3) but i don't feel very confident about it

[hidden heart]

how does it intersect at 2 positive points

surely it intersects at negative (x,y) and positive (x,y)

[unborn stag]

i formatted it wrong

[raven rover]

True, but what exactly is A as drawn in the plane?

[unborn stag]

i was only talking abt the positive coord

[raven rover]

Careful with how the problem cuts off the region in a particular area in the plane

[raven rover]

Yup

[hidden heart]

a circle with the lines y=3x and y = 0

[raven rover]

Yes great

So it’s a sector, right?

[hidden heart]

my b i read it as the x coordinates are lmao

[raven rover]

It’s a slice of a pie

[unborn stag]

yeah

[raven rover]

Great, so using polar cords, you know you have r going from 0 to 3, right?

[unborn stag]

yeah

i have that

i was stuck on theta

[raven rover]

You want to get every point from those close to the center to those on the edge

I see

So for θ

You want to find the θ which describes the line y = 3x

Then you go from where y = 0 (θ = what here?), to this y = 3x line

You want to describe both such lines using one value of θ Hera for each, these θ values will be your bounds

Think of them like where you make your pie slice

[unborn stag]

at y = 0 theta would be 0, but how would i describe y = 3x with theta, would i use trig?

[raven rover]

Yup!

Try drawing that triangle 🙂

[unborn stag]

alright, will try it now

yeah, i mean i keep getting tan(3)

which doesn't feel correct

im forgetting that its inverse tan right?

should be going from 0 to arctan(3)?

[hidden heart]

i believe so since tan(theta) = 3 so theta = atan(3)

[raven rover]

I think so 👍

[unborn stag]

im going to try integrating now

[raven rover]

👍

Something like this I think is what you’re looking at

[unborn stag]

yeah, i did that but integrated r first, i got 27.3375 as a final answer which im going to check rn

i got it

[raven rover]

Nice

[unborn stag]

thank you so much for the help

[raven rover]

Of course man

[unborn stag]

it was very concise

[raven rover]

👍❤️

[unborn stag]

much appreciated!

[raven rover]

I try my best 🙂

[raven rover]

Take care my friend

[unborn stag]

you too!

[raven rover]

Good luck with multivar

👍👍

[unborn stag]

.close

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[nimble trail]

how do heaviside functions with coffiecients work?

[nimble trail]

i.e. H(3x-5)

like when do they start and so forth

[lone heart]

@nimble trail Has your question been resolved?

[raven rover]

Well you’ll have to split up this function by case, one for when the input is positive/negative

[unkempt dew]

.reopen

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[steady swallow]

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30 degrees. Determine all the angles of the triangle

[hidden heart]

have you drawn a diagram

[steady swallow]

nope

wait

imma draw

[wraith robin]

how to find x

[hidden heart]

open ur own channel

[lone heart]

Please read #❓how-to-get-help

[steady swallow]

nvm

its 3x + 30 = 180

[hidden heart]

yeah

always draw diagrams when its a shapes problem

it helps a lot

[steady swallow]

ok

sure

.close

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[median oar]

ok how does this P(X=i) and P(X=x) relate to each other?

(1-p)^x is the probability of x fails (x because we have P(X=x))

p^k is the probability of k successes

what's that binomial at the front?

i feel like it has to do with the ordering of each of the k independent geometric distributions

but im not sure why it's k+x-1 choose k-1

[merry depot]

the number of ways for the successes to appear in the k+x-1 trials (the last trial is always a success, so it's fixed there)

[median oar]

so like if i had 4 successes, last one is a success, and we want to find the probability we failed 5 times to get the 4th success

it'll be like

F F F F F S S S S
F F F F S F S S S
F F F S F F S S S
etc

but the last S is always a success so it doens't matter

so from k-1 successes

OH

k+x is the number of total trials

-1 because that last one is always a success

and we choose k-1 success because the last one is always a success

aha thank you very much

[merry depot]

np\

[median oar]

.close

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[rose raven]

Is this sufficient

[tacit arch]

Yes that looks fine. Limit test is a little faster

[lone heart]

@rose raven Has your question been resolved?

[rose raven]

Cheers. I also have a limit test proof, but lecture content only covered ratio test and partial sums

This is the ambiguous case of the ratio test for this given question

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[thorn tapir]

A continuous injective map f exists such that f : [0,1] -> [0,1]^2.

Prove that f([0,1]) is nowhere dense in [0,1]^2.

[thorn tapir]

What I proved is this map isn't bijective, as no homomorphism can exist between these two spaces.

Now, for me to show it is nowhere dense, we have to show for any point in f([0,1]), there exists a epsilon > 0 such that B(epsilon, x) is not contained in f([0,1]).

This is because the definition of nowhere densebess says interior of closure is empty

f([0,1]) is compact and hence closed

So basically none of the points in the image are interior points

How do I approach this

[lone heart]

@thorn tapir Has your question been resolved?

[thorn tapir]

<@&286206848099549185>

[thorn tapir]

.close

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[raven haven]

#real-complex-analysis

[lone heart]

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[bronze bone]

can someone help me? the answer key is 1999+ 1999/2000

[spring plover]

Can you write the general term of this series?

[bronze bone]

what series do you mean?

<@&286206848099549185>

[flint pecan]

try writing the general term for sqrt(1+1/1^2+1/2^2), sqrt(1+1/2^2+1/3^2),....

[bronze bone]

how ? i don't know what the general term of it

can u help ?

[wild trail]

As in what would a general term look like?

or say what would the kth term be?

[flint pecan]

or the nth term (whatever you're used to)

[bronze bone]

like this ?

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[bronze bone]

.reopen

[lone heart]

✅

[bronze bone]

<@&286206848099549185> pls help me 😭

[vital kite]

what's the matter?

[raven dagger]

K doesn't go to 2000

[spring plover]

One small mistake, your upper limit should be 1999.
Also, now we need to get rid of this square root to proceed further. So try and create a perfect square inside the square root.

[astral briar]

hmmm

[astral briar]

expand the inside

[lone heart]

@bronze bone Has your question been resolved?

[exotic belfry]

$\sum_{i=1}^{1999}\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}=\
\sum_{i=1}^{1999}\sqrt{\frac{k^2(k+1)^2+(k+1)^2+k^2}{k^2(k+1)^2}}=\
\sum_{i=1}^{1999}\sqrt{\frac{(k^2+k)^2+k^2+2k+1+k^2}{k^2(k+1)^2}}=\
\sum_{i=1}^{1999}\sqrt{\frac{(k^2+k)^2+2(k^2+k)+1}{k^2(k+1)^2}}=\
\sum_{i=1}^{1999}\sqrt{\frac{(k^2+k+1)^2}{k^2(k+1)^2}}=\
\sum_{i=1}^{1999}\frac{(k^2+k+1)}{k(k+1)}=\
\sum_{i=1}^{1999}(1+\frac{1}{k(k+1)})=\
1999+\sum_{i=1}^{1999}\frac{1}{k(k+1)}=\
the\ rest\ is\ yours.$

[ocean seal]

ThM

[astral briar]

yes

[lone heart]

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[coarse trail]

Idk how part a works

[minor needle]

hint:
300p - 10p(16 - p)

[coarse trail]

Oh bro I'm stupid thanks

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[visual scaffold]

I need help to calculate the fourier trigonometrical series

[visual scaffold]

Also the exponential and armonic series but for now the trigonometric series

[naive valley]

how far did you get?

[visual scaffold]

Pretty much at the start, I think it's an even signal so the sine terms are 0

[naive valley]

it can't be even, otherwise the value at say 0.001 would have to be the same as at -0.001

but they have opposite signs

[visual scaffold]

Then it's odd?

[naive valley]

(or replace 0.001 with some small number that is less than tau/2)

yea, it's odd, you can check

[visual scaffold]

well in that case I need to find the sine terms

right?

[naive valley]

yep

so start by writing out the integral

[visual scaffold]

2/T integral from 0 to T/2 x(t)sin(kwot) dt

[naive valley]

looks mostly right, isn't it just sin(wt)?

[visual scaffold]

I'm not sure on the interval

[naive valley]

not sure what kwot signifies

oh wait we're doing fourier series not transforms

[visual scaffold]

yea

[naive valley]

$2\pi k t / T$ isn't it?

[ocean seal]

Bungo

[naive valley]

if i recall correctly

maybe your wo means $\omega_0$ and that equals $2\pi / T$?

[ocean seal]

Bungo

[visual scaffold]

yes

[naive valley]

cool

ok agree then

so now consider what the value of x(t) is

it's a piecewise function so you probably will break into two integrals

one for the part with t < 0 and the other for the part with t >= 0

[visual scaffold]

On the graph, there's also a tau constant

[naive valley]

yea

that will give you one of the endpoints for each integral

i.e. instead of integrating all the way to T/2, you will only go to tau/2

since x(t) is zero beyond that

[visual scaffold]

alright, give me a minute

so it's 2/T * [ integral from 0 to tau/2 of x(t)sin(....)dt + integral from tau/2 to T/2 of x(t)...]

actually, shouldn't the interval be from -tau/2 to tau/2?

in which case I break it into 2 from -tau/2 to 0 and 0 to tau/2

@naive valley

[lone heart]

@visual scaffold Has your question been resolved?

[visual scaffold]

There is also a gap before the signal repeats itself, should I include that in the integral?

[gusty gorge]

that's not a gap; that's part of the signal

[visual scaffold]

hm so the integral goes from -T+tau/2 to T-tau/2?

[gusty gorge]

you can pick any integration bounds, as long as they cover one period of the function

[visual scaffold]

i don't really understand

could you give me an example of some valid bounds for that signal

[bronze bone]

thanks a lot

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[obtuse basin]

I need help with this question part b)

[obtuse basin]

Could anyone help with part b). Im new

[lone heart]

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[modern sedge]

Distance is area under graph of velocity

[obtuse basin]

ik but

im stuck

[heady pollen]

the shape that we have is a triangle

[modern sedge]

What was your result in part a)?

[obtuse basin]

my answer for part a is 20m/s

[gray isle]

split the shape into two triangles by drawing a vertical line at t=60

[obtuse basin]

k then what next?

[obtuse basin]

k

[gray isle]

you can determine the area and hence distance value you seek from doing (area) addition/subtraction

[modern sedge]

Will it be just 2 triangles?

[obtuse basin]

Im still confused. I started off by calculating the area of the triangle ( part a) and got 20m/s as V when t = 90s so i used the same formula and tried t = 60s and my answer was 600m but the textbook says the answer is 750m i dont know how to arrive at that answer

[modern sedge]

You know that the whole area is 900 units, so if you subtract the area of violet triangle you should get the area between t=0 and t=60

[modern sedge]

Show your work, idk what formula did you use so I cant tell why your answer is incorrect

[gray isle]

using t=60 gives you the area of at triangle with base 60s,
which the shape you want is not

[obtuse basin]

i used vt/2 (1/2 x b x h)

[gray isle]

the figure with the red border is not a triangle

[obtuse basin]

ohhhhhh

my bad

[modern sedge]

Try this trick it seems easiest to me

[gray isle]

you should at some stage determine the v at that blue point

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[cloud scroll]

So I'm trying to work with a measure theory claim that if a function $f: \mathbb{R} \to \mathbb{R}$ such that the set $f^{-1}(\lambda) = {x \in \mathbb{R}: f(x) = \lambda}$ for any $\lambda \in \mathbb{R}$ is measurable, then $f$ is measurable.
Right now I think this statement is false because if I want $f$ to be measurable, then I need to show that the set $f^{-1}([-\infty, a))$ is measurable for all $a \in \mathbb{R}$. The interval $[-\infty, a)$ has uncountably many points, so if I want to construct this interval, I need to take an uncountable union that looks like
$$\bigcup_{\lambda \in [-\infty, a)}f^{-1}(\lambda)$$

[ocean seal]

_Cluster

[cloud scroll]

This is an uncountable union

so surely it can't be measurable can it?

[mossy laurel]

The problem here is that you cannot prove it general, you need a counterexample

[cloud scroll]

that's right

could I please have a hint to point me in the right direction?

[mossy laurel]

It's really hard to build a function or set that is not measurable

So you might suppose that the exists a set that is not measurable (which is true)

And you might try to build a counterexample using the indicator function of that set

[cloud scroll]

when you mean "there exists a set that is not measurable", are you talking about $f^{-1}([-\infty, a))$?

[ocean seal]

_Cluster

[cloud scroll]

otherwise what set are you referring to?

[mossy laurel]

You do not start with a function f, you build it

So first you take a set that is not measurable

And you build f from that set

[cloud scroll]

is there an example where I can see this?

[mossy laurel]

Vitali sets for instance

[cloud scroll]

I've never heard of those, so I guess I have to search them up

how's this? https://math.stackexchange.com/questions/137949/the-construction-of-a-vitali-set

okay, here's my idea. I'm going to define f such that

$$f(x) = \begin{cases}0 & x \notin [0,1]\text{ or } x \notin V\
1 & x \in V\end{cases}$$

where $V$ is the Vitali set over $[0,1]$, referenced in the stack exchange post above.

I'm going to be very dumb and guess this first

[ocean seal]

_Cluster

[cloud scroll]

@mossy laurel

sorry for the double ping, tex bot doesn't properly render pings

I claim that f is not measurable

oh, but the set $f^{-1}(1)$ violates the assumption of the claim since $f^{-1}(1) = V$

[ocean seal]

_Cluster

[mortal trellis]

you have to use (un)countability somewhere

that if you union uncountably many measurably sets you dont necessarily get a measurable set again

[cloud scroll]

you're right, but I am still a noob at measure theory so do you have any examples of this?

[mortal trellis]

I am gonna write a trivial union. $X = \bigcup_{x\in X} {x}$

[ocean seal]

Denascite

[cloud scroll]

okay, but how is $X$ not measurable? It could be an interval from the borel set

[ocean seal]

_Cluster

[mortal trellis]

it could be

but it could also not be

I said nothing about X

[cloud scroll]

you're right

how about the rationals then? they are measurable with measure 0, my bad

[lone heart]

@cloud scroll Has your question been resolved?

[cloud scroll]

Can I please have another hint? I'm having trouble figuring out what you're suggesting with X. Also I'm not sure how to use vitali's set to resolve the $f^{-1}(1)$ issue

[ocean seal]

_Cluster

[mortal trellis]

what if you instead managed to make it so that f^{-1}([0,1]) = V

but so that all elements in [0,1] are only hit once

[cloud scroll]

hang on, so are you suggesting that I define $f$ as follows:
$$\begin{cases}
0 & x \notin V\
x & x \in V
\end{cases}$$

[ocean seal]

_Cluster

[mortal trellis]

0 would not be good here cause that is contained in [0,1]

[cloud scroll]

you're right

maybe a different output would be better

like 2

[mortal trellis]

actually no you still run into some problems with that

[cloud scroll]

yeah, because I think $f^{-1}(2) = \mathbb{R} - V$

[ocean seal]

_Cluster

[mortal trellis]

yes

[cloud scroll]

hmm, what if I did the following instead

$$\begin{cases}
2x & x \notin V\
x & x \in V
\end{cases}$$

[ocean seal]

_Cluster

[cloud scroll]

oh, I need to account for the non-representative elements as well

$$\begin{cases}
2x & x \notin V \text{ and }x \notin [0,1]\
x & x \in V\
x + 1 & x \in [0,1] - V
\end{cases}$$

[ocean seal]

_Cluster

[cloud scroll]

scuffed but the idea is that

$f^{-1}(\lambda)$ for any $\lambda$ results in a singleton set which is measurable and has measure 0

[ocean seal]

_Cluster

[cloud scroll]

and $V \subseteq [0,1]$. So if we do $f^{-1}([0,1])$ we get $V \cup ([1,2] - (V + 1))$ which is clearly not measurable because both sets are disjoint and $V$ is not measurable.

[ocean seal]

_Cluster

[cloud scroll]

@mortal trellis could I please have your opinion on this?

[mortal trellis]

well scuffed is a word

give me a min

[cloud scroll]

sure thing mate

[mortal trellis]

ok I think it works

not what I had in mind tho

[cloud scroll]

what did you have in mind?

[mortal trellis]

take a bijection V->[0,1] and R\V ->R\[0,1]

[cloud scroll]

you can do that?

[mortal trellis]

why not

oh right you can probably run into CH problems

[cloud scroll]

it's harder for me to understand because I think of V as being smaller than [0,1]

[mortal trellis]

didnt actually think about that

[cloud scroll]

yeah, continuum hypothesis

[mortal trellis]

well still uncountable

but yes good point

ok then mine just doesnt work

[cloud scroll]

that's okay, it was still interesting to think about

and thanks for sticking with me

hope to talk with you again sometime

.close

[lone heart]

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[fallow rampart]

can anyone teach me how to solve these kind of stuff

[lunar palm]

squaring both sides is usually a start

your aim is to get every x out of a square root

[fallow rampart]

yeah

[lunar palm]

to do that for equations like this, you need to:
-square each current root (to get x-3 out of sqrt(x-3) for example)

and
-only be left with square roots involving a quadratic equation at the very least

if you have sqrt(x^2 +2x +4) you could say = sqrt((x+2)^2) = x+2, for example
or in the form sqrt(quadratic) = f(x), you could just square both sides for a polynomial with no square roots.

the idea is when squaring you'll instead only have one term of sqrt( ) which is handier for algebraically simplifying

squaring both sides will allow you to do this

give it a shot

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[lilac stratus]

Could anyone help me with this question? I'm not sure how to proceed

[echo socket]

Is it me or you can simply deduce the general formula a_n = 1/2 + 1/4 + ... + 1/2^n from that?

And so the limit is basically an infinite geometric sum

[lilac stratus]

yeah right, as soon as I posted it I tested it and yeah that's it lol, thanks anyway

.close

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[rotund laurel]

prove (x+y)(x-y)= (x-y) is x=y

[rotund laurel]

i cant under stand this image

[lone heart]

@rotund laurel Has your question been resolved?

[nimble fern]

which part of the image?

[rotund laurel]

@nimble fern

[nimble fern]

i see

do you know what is factorization?

[lone heart]

@rotund laurel Has your question been resolved?

[nimble fern]

hello there!
where are you stuck?

[lone heart]

@rotund laurel Has your question been resolved?

[rotund laurel]

hey i figured it out

ever seen a reply this fast ?

.close

[lone heart]

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[nimble fern]

nice! good

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[fading loom]

hello

[echo socket]

Let's say the side of the square WXYZ is some number a

Then AD = BC = a, right?

[fading loom]

yes

[echo socket]

Also, AB = DC = 13 + 8 - a

This information is enough to solve for the perimeter

[fading loom]

hm

[echo socket]

This is true because DZ is simply 8 - a and YC is 13 - a

And DC = DZ + ZY + YC = 8 - a + a + 13 - a = 13 + 8 - a

[fading loom]

huh

[echo socket]

Which bit is confusing you?

[fading loom]

this

[echo socket]

We can break the side length DC into three segments

DZ, ZY, YC

Clearly, DC is the sum of those segments

So DC = DZ + ZY + YC

[fading loom]

ohh

[echo socket]

And, like I said, DZ = 8 - a and YC = 13 - a, meaning we can plug those in

And get DC = 8 - a + a + 13 - a

-a and +a cancel out, leaving us with DC = 8 + 13 - a

[fading loom]

uhuh

[echo socket]

Okay, let's now solve for the perimeter

Perimeter is just the sum of the lengths, so, in this case, it's a + a + 21 - a + 21 - a

As you can see, all of the a's cancel out

So the perimeter is just 42

[fading loom]

OHHH

OKAYY THANK U SO MUCH

I GET IT NOW

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[lone heart]

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[marsh trout]

How can I show that there is a C > 0 such that :

[marsh trout]

$\mathbb{E}(X \mathbf{1}_{{X<C}}) > 0$

[alpine sable]

(\mathbf for bold face)

[marsh trout]

thx

[ocean seal]

Hmz

[mortal trellis]

what do you know about X?

[marsh trout]

it's a random variable

nothing more

[mortal trellis]

then this doesnt seem true

[marsh trout]

why ?

btw C is a constant

[mortal trellis]

X=-3

[marsh trout]

oh no

sorry

I forgot something

$E[X] > 0$

[ocean seal]

Hmz

[marsh trout]

in fact what bothers me is to calculate the expectation of a random variable multiplied by an indicator

[mortal trellis]

well for an integral you would essentially cut off the part corresponding to above the line y=C here

[cunning locust]

e * x = unresolved daddy issues and prob std's. you dodged a bullet there

[last ether]

What the fuck

[marsh trout]

I didn't understand what you two said to me

[marsh trout]

and how am i doing here ?

[mortal trellis]

I would probably just do a proof by contradiction

assuming you have something like monotone convergence theorem or something?

[marsh trout]

yes i have

but here i didn't have sequence of variables that converges

[mortal trellis]

make one up

always good to think about the sequence 1,2,3,4,...

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[lone heart]

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[lone heart]

@silent sage Has your question been resolved?

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✅

[lone heart]

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[lone heart]

@silent sage Has your question been resolved?

[spice eagle]

a2 + b2 =????????????

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[slate monolith]

1/chance

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[cinder spindle]

question d?