help-19

[woeful rivet]

why did he add 2k+1)

when he already had 2k-1

[proven beacon]

They assumed that for some natural number k it is true

Now they will prove that it is true for k+1

[woeful rivet]

why did they add the 2k+1 in the last time

[proven beacon]

1 + 3 + 5 + ... + (2k-1) <- Assume is true for k
1 + 3 + 5 + ... + (2k-1) + (2(k+1)-1)

[woeful rivet]

wait no why do they have 2k-1 when they 2(k+1) - 1 = 2k+1

i thought u sub in n = k+1

[proven beacon]

2k-1 is still part

k+1 is the next term, k is still part

[woeful rivet]

oh

okkk

okok

make sense

[static talon]

basically add on to the thing that u assumed is true and reorganize the RHS of the equation and make it so that it looks almost the same as the one u assume is true

hope that helps

[quartz adder]

Hey in this quadrilateral how is F the mid point of DB prove

[static talon]

hey u cant do that u need to open a new one

[proven beacon]

Im going to work through the problem you sent (1 + 3 + 5 + ...)

[quartz adder]

Guyssss

[proven beacon]

1). Prove it is true for a base case

[static talon]

go use #help-3

this is already occupied

[proven beacon]

1 = 1^2

1 + 3 = 2^2

Seems legit to me

2). Assume it is true for n=k

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

I will assume this is true.

3). Prove it is true for n=k+1

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

[woeful rivet]

um..

[static talon]

why is it k-1

[proven beacon]

Why k - 1

[static talon]

u put n=k+1

[proven beacon]

[1 + 3 + 5 + ... + (2k-1)] + (2k+1) = (k+1)^2

k^2 + 2k + 1 = (k+1)^2

[woeful rivet]

oh.

[proven beacon]

(k+1)^2 = (k+1)^2

Therefore yes it is true that 1 + 3 + 5 + ... + (2k-1) = k^2

For natural numbers

[woeful rivet]

is this it..

[static talon]

yes

and (k+1)(k+1+1)/2

[woeful rivet]

since lhs = rhs then k+1 is also true

?

[proven beacon]

Yes

[woeful rivet]

oh ok

[static talon]

ehh

[woeful rivet]

...

[static talon]

i would say

compare to n=k

[woeful rivet]

let me try the next question

[static talon]

and for n=k u got k(k+1)/2

[woeful rivet]

OMG

WTF

[static talon]

the next one should theoretically be (k+1)(k+1+1)/2

which is what u got

[woeful rivet]

um.. lets do question c!

[proven beacon]

question c is a little like the first question

is just that its the sum of squares instead

[woeful rivet]

is it best

to leave in

factored form

[static talon]

let me write down my way of solving the 1+2+3+4+...+n and see if it makes more sense

[woeful rivet]

bruh do i have to expand

all of it

[static talon]

does this makes sense

[woeful rivet]

yer

[woeful rivet]

for this one do i have to expand all of it

[runic swift]

factor the (k+1)

if you expand everything you're gonna have to factor a cubic

which is not fun

[woeful rivet]

okk

[proven beacon]

ok i got a solution for question b but its long

lo

[woeful rivet]

um..

did i factor this right..

[static talon]

looks good

[runic swift]

cool

keep going

[static talon]

minor mistake for my part i forgot to add "So it is true for n=1" that might deduct a point in the exam

[static talon]

yes

ur /6 disappeared on the last step

oh wait

[runic swift]

i dont think you factored that correctly

[woeful rivet]

omg i did it wrong

...

[runic swift]

4 3 is 12

[woeful rivet]

but its non monic

so i use

AC method

[runic swift]

your constant term is 6

but you factored into 4 and 3

and 4 3 doesnt make 6

[woeful rivet]

i thought ac method was like this

wait

[runic swift]

ok now that is correct

[odd edge]

@woeful rivet Has your question been resolved?

[odd edge]

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[quaint token]

Hello I would not say I specifically have one question but I am going to have continious as I am asking for assistance throughout my semester in statistics. I just want guidance if lost because I am not the brightest in math. Im taking regular entry level statistics in community college

[odd edge]

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[tardy horizon]

What is the easiest way to determine which one is the top function

[odd edge]

@tardy horizon Has your question been resolved?

[sand horizon]

See between 0 and 1 whats happening

Check if x^2 - x is negative or positive between 0 and 1

If its negative then y = x is above

If its positive then y=x^2 is above

And the really easiest way is to look at the graph of both function

[odd edge]

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[true hazel]

how to justify with no knowledge of how the graph looks like?

[upper onyx]

draw it, if you need the graph

[true hazel]

yes

its hard to draw the graph

is there another way to justify without knowing how the graph looks like

[mystic saffron]

Hi I am studying for the ASBAV and AFOQT both to join either the Army or Airforce but the math problems on the real test you can't use a calculator which I can do any math with a calculator but honestly I forgot how to do it from scratch with a piece of paper any one can help me like maybe once a week or twice?

[hardy panther]

Use the definition of inverse

[true hazel]

ye

its supposed to give the x by information of y

[hardy panther]

Get a fresh channel

[mystic saffron]

were?

[odd edge]

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).

[true hazel]

the inverse would be this if it werent for that x<0 domain

[mystic saffron]

Thanks man

[true hazel]

how to use that to justify the specific inverse function?

[hardy panther]

You would just show, algebraically, that that formula is the inverse of f(x)

[true hazel]

yes

the inverse would be this

but now they are saying to reject the one that is +sqrt(b^2-4ac)

they are only taking the question with the - sign

theres that + and - sign on the numerator

but due to the domain of f(x), the - sign is only accepted

the question is how to deduce whether to use the minus or plus sign for the inverse function from the domain of f(x)

[mystic saffron]

bcz its specifically tells you for x < 0

[true hazel]

ok that doesnt explain much

ok now

[mystic saffron]

why doesn't it ?

[true hazel]

what if the question ask to find f(x) inverse

how would u know to use the - or + sign?

[mystic saffron]

it would tell you if they want a specific solution

otherwise you pick both

[true hazel]

ok

actually

[mystic saffron]

?

[true hazel]

one of the justification is this

look at the graph of f(x)

the range of f(x) is f<0 because the domain is x<0 in the first place

right?

[mystic saffron]

if you meant f(x) < 0 bcz x < 0

yea

[true hazel]

yes

[mystic saffron]

notice how it also touches the point x = 0

[true hazel]

and now f inverse is this

[mystic saffron]

that means it can be f(x) <=0 for x <=0

[true hazel]

ye

[mystic saffron]

for the case f(x) = 0 , x = 0

[true hazel]

now the f inverse is that

[mystic saffron]

yes

[true hazel]

now an inverse function gives the x value from the y value given into the inverse function right?

it takes a y value then gives the coressponding x value

thats what an inverse function do

[mystic saffron]

yyes

[true hazel]

ok

now since the inverse function gives the x value

and the x value has a restriction of x<0 from the domain

[mystic saffron]

yes

[true hazel]

this means that f inverser cant be more than 0

[mystic saffron]

yea

[true hazel]

now because the inverse function can gives values more than 0

how do we know that the - sign is the solution instead of the + sign

[mystic saffron]

yes it can gives value more than 0 but you cant plug x's more than 0

bcz u have restriction

remember?

f(x) >0 <=> x>0

now if you would choose the + sign solution, for x <0

you would still get f(x) > 0

bcz x² grows faster than x

(in the discrimnant)

[true hazel]

like this?

on the left

[mystic saffron]

correct

[true hazel]

the one on the left says this

the left part means that the (f inverse with + sign) will only give a negative value for x>0

thats what it means right?

and on the right

[mystic saffron]

yes

crrect

correct

[true hazel]

the f inverse function with the minus sign will only give negative values for x values inputted that is <0

[mystic saffron]

yea

[true hazel]

now since we know the range of f is f<0

meaning the domain of the f inverse is x<,0

there was a typo

[mystic saffron]

yea

[true hazel]

so we would choose the equation with x<0

[mystic saffron]

yea

[true hazel]

which is the minus sign one

that justification is alright

buttt

[mystic saffron]

yes

[true hazel]

we only know to choose the minus sign equation because we know the rawnge of f

[mystic saffron]

coreect

[true hazel]

and we know the range of f because we know how the graph looks like

what if we idk how the grpah looks like

how do we know to choose the minus sign one

[mystic saffron]

not neccesarly , we could know the range of f by inputting values.

[true hazel]

yep

[true hazel]

actually thats kinda hard

[mystic saffron]

its not

[true hazel]

look at this graph

[mystic saffron]

u can just sketch the parabola

[true hazel]

if u input values from just x=0,1,2,3

and x=-1,-2,-3

u would assume the range for x<0, is y<0

[mystic saffron]

not neccesarly

[true hazel]

hmm

is inputting values the only way to find the rnage of graphs?

because inputting values is similar to looking at how the graph looks like

because a graph is simply all the x values plotted with the coressponding y values too

like desmos

they know how the graph looks like by also inputting all x values and getting the y values then plotting it on a graph

so in a sense its similar to looking at how a graph looks like

[mystic saffron]

yea

[true hazel]

nvm forget about that question

how about this question

how to find x

[mystic saffron]

wdym

[true hazel]

one of the way is using the lambert w function

[true hazel]

how to solve for x

[mystic saffron]

i cant see the whole equation

[true hazel]

which x value can satisfy that equation

[true hazel]

its just a simple isolating x question

finding an x value that can satisfy that equation

[mystic saffron]

hello

i cannot see the whole equation

[true hazel]

thats the whole equation

[mystic saffron]

just 2^x

aaa

sorry

i see it now

[true hazel]

oh ok

[runic swift]

2 works

[true hazel]

real solutions?

[mystic saffron]

yea

[true hazel]

how to find it?

lambert w function?

but listen to this tho

an eqation is like this right?

7=7

[mystic saffron]

i haven't studied lambert w

[true hazel]

agree?

[mystic saffron]

i studied something else

[true hazel]

its ok

[mystic saffron]

wait

let me finish

[true hazel]

oki

[mystic saffron]

how i would solve this equation, i would first find a solution by eye, e.g. (2), then i would prove that there doesn't exists any other solutions, bcz the function is monotonly increasing for x > 0

[true hazel]

interesting

thats not bad

but what if the x value is some irrational number

but thats for later

but listen to this tho

an equation is like this right?

7=7

[mystic saffron]

that's another disccusion, there are other tools we can use in the complex system.

[true hazel]

whatever on the left hand side is same as the on right hand side right?

[true hazel]

ok

[mystic saffron]

okay

so x = x

[true hazel]

yep

[mystic saffron]

what about it

[true hazel]

so whatever we do on left hand side, also have to be done on the right hand side

for it to remain an equation

[mystic saffron]

correct

[true hazel]

now to the orginal equation

what happens if both sides are differentiated?

it doesnt give the answer 2 tho

but why?

it should

[distant token]

thats not how it works

[mystic saffron]

hah

[distant token]

that would be true if both sides of the equation were equal for all values of x but this equation is only true for one value of x

[true hazel]

ooooooooo

so thats wh differentiating both sides doesnt work

[mystic saffron]

yea

[true hazel]

ty for the help

[mystic saffron]

np

[odd edge]

@true hazel Has your question been resolved?

[odd edge]

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[raven nebula]

how to find the derivative of this?

[pale atlas]

Do you know the formula of d/dx(u/v)

[raven nebula]

i cld only simplify this much

but i prolly think this is wrong

[raven nebula]

yupp

[dapper canyon]

check on https://www.desmos.com/calculator

define f(x) then graph f'(x) and ur answer to see if they match

[alpine hawk]

I'm getting like this idk whether it's correct

[pale atlas]

This is correct Volga

[raven nebula]

but the only prob is

[dapper canyon]

,rotate

[pale atlas]

Yes break the numerator

Or distribute

[alpine hawk]

Wait let me

[pale atlas]

Is?

[alpine hawk]

Ain't gonna do that

[raven nebula]

these r the options available

[raven nebula]

yeaa but how to simplifyy

[wanton bison]

This looks off

[alpine hawk]

Solve this you may get the answer

[raven nebula]

uhmmm but when u apply uv rule thts wht i got

[pale atlas]

1st one

[alpine hawk]

I have been wondering too

[raven nebula]

but how did u get?

[dapper canyon]

looks like the last option

take out x^-2/3 as a factor

[raven nebula]

its showing last option is correct but how

[pale atlas]

Multiply x in numerator and denominator

[raven nebula]

u will get this then ryt?

[dapper canyon]

no

x^-2/3 (1 - 5x^6/3)

then mult top and bottom by x

[alpine hawk]

How did u get this?

[dapper canyon]

how did i not get this

[pale atlas]

@raven nebula try this

You will get

[dapper canyon]

it's correct

[raven nebula]

okays will do it

[raven nebula]

basically we r adding the powers right?

i mean when we simplify tht

[alpine hawk]

Exactly

[dapper canyon]

yes that's how exponents work

[raven nebula]

ohhhhhh

wai

(x^3)(x^2) soo its gonna be x^5 or x^6?

[dapper canyon]

5

[raven nebula]

ohhhhh

[dapper canyon]

(2^3)*(2^2) = 2^5

just numbers

[raven nebula]

now it looks much easier

[dapper canyon]

8*4 = 32

this rule is not made up, it's real

[raven nebula]

ohhh

thnks

.close

[odd edge]

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[tardy horizon]

Why is this

[mystic saffron]

1/3 * x/y

let me try to type it in latex(im bad)

[tardy horizon]

Isn't it supposed to be reciprocal

Cuz 1/3 is at the numerator

It's like 1/3 divided by y

Oh wait

[mystic saffron]

$\frac {1}{3} \cdot \frac {x}{y}$

[tardy horizon]

1/3 divided by 3 = 1/3 * 1/y

[wanton bison]

no space between $ and \frac

[clever fjord]

daniel

[mystic saffron]

there we gooooooo

i guess i am decent

[nocturne brook]

decent daniel

[mystic saffron]

yea

[tardy horizon]

So this yes?

[nocturne brook]

$\frac{\frac13x}y=\frac{\frac13x}y\cdot\frac33=\frac{\frac33x}{3y}=\frac x{3y}$

[clever fjord]

Flip

[odd edge]

@tardy horizon Has your question been resolved?

[odd edge]

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[mystic bone]

Did i do this right?

I had to plot a piecewise linear function(idk if its the right wording)
and then i had to solve g(x)=6

[mystic bone]

so the answer to g(x)=6 is 10, 0 or 4 right?

[nocturne brook]

looks real

[mystic bone]

oo alright ty so to write a conclusion i could write

To sumarize then g(x)=6 will either be 10, 0 or 4.

would it be wrong to say x is either 10, 0 or 4?

[nocturne brook]

it's acceptable but not the best way

you can say something to the effect of "the only values x for which g(x) = 6 are x=10, x=0, or x=4"

or

"g(x) = 6 if and only if x=10, x=0, or x=4"

or with sets,

"g(x) = 6 if and only if x is in the set {0,4,10}"

[mystic bone]

oooo

ty ty

much appreciated C:<

[odd edge]

@mystic bone Has your question been resolved?

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[noble solstice]

I'm reading a book in measure theory and it stated that the Cantor set is perfect and totally disconnected. How is this possible? Here are the given definitions:
$\ E$ is perfect if ${\forall x \in E, \nexists r > 0 : B_r(x) \cap E = {x}}$ where $B_r(x)$ is the open ball with radius $r$ centred at $x$. In other words, There aren't any isolated points in $E$.
$\ E$ is totally disconnected if ${\forall (x < y) \in E, \exists z \notin E : x < z < y }\$
To me it seems impossible that any set can be both perfect and totally disconnected. By definition perfect means no isolated points, and totally disconnected implies that every point is isolated...

[clever fjord]

woomy

[prime basalt]

The answer is in the construction of the cantor set

[noble solstice]

how so?

[native canopy]

set theory?

[fickle silo]

consider {1, 1/2, 1/3, 1/4, ...} U {0}, is the point 0 isolated? It kinda is, but also not, depends on what exactly you mean by an "isolated point"

[prime basalt]

Denote K1 = [0, 1] - (1/3, 2/3)
K2 = K1 without middle intervals of the segments K1 consists of, etc.

Note that if x is a point from the Cantor set, then there exists Kn such that the edge of Kn contains x. We consider a ball B_r(x). Then we remove intervals of decreasing length, until one of those intervals intersects the ball and we get Km. The edge of Km will intersect B_r(x)

[noble solstice]

Its a book on real anaylsis, specifically measure theory, integration and hilbert spaces

[noble solstice]

(im on the first chapter, which is measure theory)

[native canopy]

I'll pass but cool

[noble solstice]

by "edge" do you mean that its on the boundary of the interval?

[prime basalt]

yes

The explanation is a bit vague but it should give the idea

[noble solstice]

So:
Kn = [0,1] - (1/3, 2/3) - ((1/9, 2/9) U (7/9, 8/9)) ...
Take x = 1/3 or 2/3 or 1/9 or 2/9 or 7/9 or 8/9 ...
Consider B_r(x) where r > 0

I'm stuck here

Then we remove intervals of decreasing length, until one of those intervals intersects the ball and we get Km. The edge of Km will intersect B_r(x)

[prime basalt]

We can prove that for any x from the edge of Kn there exists a sequence of intervals of decreasing length, approachign x

It's enough to show that for edge points of K1

Because other cases are just scaled down versions of K1

[noble solstice]

Sorry I dont think i really understand, could you maybe draw out what you mean

ohh wait i think i understood.
So we take: (a, x); (a1, x); (a2, x);
for all r > 0 there's an a_m < r so that (a_m, x) in C

therefore the ball with radius r contains both a_m and x

or am i making a mistake

[prime basalt]

Here white set is K1, red point is x, and purple intervals will be removed in future steps

[noble solstice]

okay

[fickle silo]

ok looking up the definition, a point is isolated, if there exists a neighborhood that contains no other points, so there needs to be some fixed epsilon>0, such that all other points are more than epsilon away from it. This matches the definition of a perfect set and it guarantees there are no isolated points. The definition of a totally disconnected set though does not really imply this. It means we can separate any pair of points, yes, but they may lie arbitrarily close together, as my previous example shows, and so we cant really fix any epsilon

[noble solstice]

So basically just because we can seperate 2 points, this doesnt mean that these 2 points are arbitrarily close, and thus any ball with r > 0 will contain these 2 points?

[prime basalt]

The purple sequence eventually intersects B_r(x) for any r > 0

[fickle silo]

the set of rationals should be another example, being both perfect and totally disconnected

[noble solstice]

ok yes

[noble solstice]

ohhh right

Like i understand

It just seems weird

[fickle silo]

this doesnt mean that these 2 points are really far apart from each other, they may be as close to each other as you want to

[noble solstice]

Lets say they are epsilon > 0 close

taking r = epsilon

wouldnt b_r(x) contain x but not y

(y being the arbitrarily close point)

[fickle silo]

so you have (x < y) but they are separated by x<z<y, but there is no reason why there shouldnt exist some y', such that x<y'<z<y

or x<z<y'<y

ofc for sets like Z, this y' doesnt exist if we consider numbers like 3<4

but in Q, no matter what we choose for x and y, we can always get closer

[noble solstice]

I understand this but then you fall into this iterative behaviour where you can always say "there is a y' s.t. x<z<y'<y" and i'll rebutal with "but there's a z' s.t. y'<z<y" etc...

so which one "wins"

This isnt the first time I came across something that requires this iterative process of continuously finding a number that should contradict the statement, but hey there is another even smaller number which means its correct, but hey there is an even smaller number that contradicts the statement, but he-..... and i have no idea how to not fall in this trap or even how to get out of it hahaha

[fickle silo]

I guess it doesn't really matter who wins, if it's z or y'. In the end, the fact of the point x being not isolated will win, as this iterative process will yield you numbers that are as close to each other as you want to

[noble solstice]

okay i think i get it

so you said that also Q is perfect and totally disconnected by the same reasoning?

[fickle silo]

like, throw me some r, and we just repeat this process until we are inside this r region with y'

[noble solstice]

unlike the cantor set though Q isnt compact

[noble solstice]

yea makes sense

[fickle silo]

yeah should be, as the rationals and irrationals are dense in the reals

[noble solstice]

probably a fun exercise to prove it

that Q is perfect and totally disconnected

ill have a go at it once i finish this 400 page book (aka never) (i have to return it by the end of september. . .)

Thanks alot though @fickle silo @prime basalt :)

ill close the channel now

[fickle silo]

alright

[noble solstice]

.close

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[timber yacht]

u have a questions

[timber yacht]

i have a question

[dapper creek]

hello guys am new here ,wanna kniw how this works

[timber yacht]

what do u need help with?

[low locust]

then ask your question

[low locust]

read #❓how-to-get-help

[timber yacht]

3x + 1

[low locust]

thats not a question. thats an expression

[timber yacht]

ok

what about this one

The equation above shows how temperature F, measured in degrees Fahrenheit, relates to a temperature C, measured in degrees Celsius. Based on the equation, which of the following must be true?

A temperature increase of 1 degree Fahrenheit is equivalent to a temperature increase of 59 degree Celsius.
A temperature increase of 1 degree Celsius is equivalent to a temperature increase of 1.8 degrees Fahrenheit.
A temperature increase of 59 degree Fahrenheit is equivalent to a temperature increase of 1 degree Celsius.
A) I only
B) II only
C) III only
D) I and II only

[low locust]

well there is no equation

[timber yacht]

a,b,c or d?

[upper onyx]

Do no toccupiy more than in channel with the same question. close on by typing .close.

[timber yacht]

.close

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[dapper creek]

Understanding this app

[timber yacht]

ok

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[rustic sail]

Is the vector <a,b,0> orthogonal to the unit vector k for any a,b?

[onyx cloak]

yes

[brave gale]

try taking dot product

[rustic sail]

Okay thanks! I tohught so intuitively (and now that i think of it i couldve taken dot product)

but google didnt answer my quesiton

[warped badger]

I assume by unit vector you mean v=(1,1,1)?

[onyx cloak]

unit vectory k -> vector in z direction with magnitude 1

0i + 0j + 1k

[rustic sail]

Yeah thought about that as I was typing the question lol

[warped badger]

Ok, didn't see the k part

[rustic sail]

Thanks

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[errant timber]

I apologize if this is the wrong place for this but, for converting cm to meters wouldn't using 10^-2 m on the bottom equate to the same thing too?

[uneven osprey]

it seems like that's what they did

[errant timber]

ok

so im not going crazy

[odd edge]

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[sonic beacon]

hello, i'd like to do partial fractions for this before integrating it, how am i supposed to do it?

[turbid compass]

This isn’t a partial fractions integral

Atleast there’s a much easier way to do it

[sonic beacon]

should i simply consider the denominator as u squared?

[turbid compass]

Yeha I would do it with a u-sub of u=8x^3 + 3x^2

It’d be much easier that way

[sonic beacon]

mmm i see, if i in some case had to go through partial fractioning here, how would i do so

just out of curiousity

[turbid compass]

You’d have to completely factor the denominator and then set up your different fractions/equautions

But in this case it would be especially bothersome because it’s a degree 6 polynomial in the denominator

It would be easier to show with a degree two or three

[sonic beacon]

i see, so yes it would make sense to do it your way

thank you brother, i really appreciate it 🙏

.close

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[vivid kelp]

What will be the integration of S = integration of (2Pt/m)^1/2*dt where P , m are constants and t is from 0 to t

[limber glade]

just bring all constants out of the integration symbol

then it beconmes simple enough to apply power rule

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[steel sky]

I need help

[steel sky]

can someone solve the limit

[native canopy]

0 right?

oh wait its a Sum infinity sorry

[autumn bolt]

do you know partial fractions?

[steel sky]

no

[autumn bolt]

thats what were using here

[steel sky]

i know considerable amount of integration and diff

I could understand

[granite ferry]

Would you convert this thing into a Riemann sum?

[autumn bolt]

no

see

basically

$\frac{1}{(g)(g+1)(g+2)} = \frac{1}{2}(\frac{1}{g(g+1)}-\frac{1}{(g+1)(g+2)})$

[autumn bolt]

you can try with simple cases like g = 1, 2

[granite ferry]

This is a telescopic form right?

[autumn bolt]

yea

[granite ferry]

Is this also called partial fractions?

[native canopy]

Just for curiosity what level of math is this?

[granite ferry]

I mean they are partial and fractions really

[steel sky]

the ans is 1/4 then

[granite ferry]

I think it is highschool level

[steel sky]

i dont have access to the ans too

[native canopy]

grade 12 was my guess but i wanted to check

[autumn bolt]

yep

[quasi sparrow]

,w sum n=1 to inf 1/(n(n+1)(n+2))

[clever fjord]

Results provided by WolframAlpha

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

Requested by rie.mann

[autumn bolt]

you got it fast

wow

[steel sky]

thank you so much for the idea @autumn bolt

[quasi sparrow]

This was the hard part

[autumn bolt]

welcome (:

you can also extend this to longer denominators

but you need to multiply by 1/3, 1/4, 1/5 and so on

[steel sky]

can u give some eg

like 4 terms in deno means mul by 1/3?

am i right?

[autumn bolt]

yea 1 min

$\frac{1}{(g)(g+1)(g+2)(g+3)} = \frac{1}{3}\bigg(\frac{1}{g(g+1)(g+2)}-\frac{1}{(g+1)(g+2)(g+3)}\bigg)$

[steel sky]

thanks i got it!

[odd edge]

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[potent jasper]

Can someone point me to a resource that can teach me how to solve equations like this in limits?

[elfin zodiac]

What do you mean equations like this?

[potent jasper]

I’m not really sure what is the desired answer nor do I really know what would be the correct manner to solve this limit. Any resource that help me better understand it fundamentally would be appreciated.

[potent jasper]

This limit contains an unknown variable x and I am not sure how to correctly answer it

[nimble blaze]

could try looking up
limits with conjugates

[potent jasper]

Most of what I can find deals with only a singular known variable. I can’t find anything that talks about a single known variable and a single unknown variable. Do you know of any resources that covers this specific case?

[nimble blaze]

limit definition of derivative if you want a broader search

[potent jasper]

Ok sweet this looks more like what I am looking for

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[plain apex]

"Josh flips a fair coin until he receives an equal amount of heads and tails. For example, HHTHTT would correspond to 6 flips. What is the expected number of flips Josh does?"

[plain apex]

Not even sure how to approach this

[red surge]

if you let X_n be the difference between the number of heads and the number of tails flipped

then X_n defines a simple random walk on Z starting at 0

the question is then asking for the expected number of steps before we hit 0 a 2nd time

[plain apex]

Yeah that's what I did with my markov chain

If it was probability I could easily get the answer

Not sure how to find expected value

[red surge]

ok so i kinda expected this problem to be easier/a standard problem but i just checked my prob lecture notes and we didn't cover this in lectures lol

checked the internet and the numerical answer is apparently ||∞ ||

but i can sorta see why

if you define like x_n to be the expected number of steps required to reach 0 starting at n

you can easily create a recurrence relation with x_n, x_(n-1) and x_(n+1) using the law of total probability

we wanna find out x_1

obviously x_n >= n

playing around with it i found that

x_1 = 1 + 1/2 x_2

x_1 = 2 + 1/3 x_3

so you can probably induct to show that x_1 = n + 1/n (x_n) which implies x_1 >= n for all integers and hence is infinite

hope that helps @plain apex

[odd edge]

@plain apex Has your question been resolved?

[plain apex]

how did you get 2 + 1/3x_3?

[red surge]

x_2 = 1 + 1/2(x_1 + x_3)

then plug that into x_1 = 1 + 1/2 x_2

(with x_0 = 0)

[red surge]

i haven't actually done it properly so don't take my word for what it actually inducts to, but you could probably show x_1 must be unbounded

[plain apex]

hmm got it thanks for all the help

are you a math major rn @red surge

[odd edge]

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[quartz pier]

I was reading about closed subsets in $\mathbb{R}^d$ and there was this one definition that said a set $F \subset \mathbb{R}^d$ is closed if every convergent sequence of points in $F$ has its limit also inside F.

[clever fjord]

Calc III Victim (Pt. 2)

[quartz pier]

this is the example that was talked about in class

So now for any $n$ if the sequence $x_n$ is inside the set $F$ then its closed ?

for example

For this example would something like

$F = (0, 2] \times \mathbb{R} \times \mathbb{R} \times [0, 1]$ work?

I feel like Im doing it correctly but Im not 100% sure

oh wait

I thought the third componenet said n+1/n

[clever fjord]

Calc III Victim (Pt. 2)

[quartz pier]

1/n is inside (0, 2]
2, n+1 are inside R
1/n-1 is inside [0, 1]

oh wait I think Im missing something so I have to check the limit of the sequence and make sure thats inside F too right

so instead of (0, 2], [0, 2] would have worked because when taking the limit as n -> inf we have

L = [0, 2, 1, 0]

ye idk

[mystic saffron]

For closedness, it is necessary that all convergent sequences that take values in F have limits in F. So, in your case, the sequence x_n = [1/n,0,0,0] are in F for all n, but its limit, [0,0,0,0] is not, so F is not closed.

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[marsh hound]

Can someone help me understand the intuition

[marsh hound]

I feel I almost have it

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[marsh hound]

How do I do this

Here’s my workings

I see forgot to add f(x,y) but that’s not the root of the mistake

[odd edge]

@marsh hound Has your question been resolved?

[marsh hound]

Tried to split it into two integrals

Shit is =0 in the top half here

Regardless of fxy

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[steady spoke]

no idea where to even start

[summer river]

have your read the problem

[steady spoke]

yes i have read it and looked over my notes but we didn’t do any problems related to this

[summer river]

yeah no worries I acc TA for this kinda material

what do you know about work

[steady spoke]

oh perfect!

i can do work problems related to springs. i just haven’t done one regarding a water tank

[summer river]

are you given water density

[steady spoke]

1000kg/m^3

[summer river]

great great

let's get an idea about force

F = ma

[steady spoke]

ok. i know F=mg, where m is mass and g is gravitational constant

i think

[summer river]

yes

great great

what is g btw

[steady spoke]

9.8m/s^2

[summer river]

okay okay great great

how can we use water density to find mass

[steady spoke]

mass is equal to density times volume

[summer river]

great great

any ideas about volume

how can represent that

[steady spoke]

umm idk

[summer river]

yeah we need a little geometry for that

can you draw two triangles for me

[steady spoke]

mhm i can!

btw i’m confused. is the image depiction the tank as a triangular prism?

[summer river]

I believe so

[steady spoke]

since that’s just what the end of the tank looks like

[summer river]

this gives me like google form vibes

[steady spoke]

okok

lmaoo yeah

[summer river]

lets goo

[steady spoke]

well it’s not actually i meant i was just agreeing with that ahah. my online program has very poor graphics 😂

[summer river]

yeah haha that makes sense

[steady spoke]

so this is what the tank looks like?

[summer river]

yep!!!

[steady spoke]

okok!

[summer river]

what's your fav letter for a variable

x y z

[steady spoke]

x works

lol

[summer river]

cool cool

think of x to be like the water level

like x is from the

top of this rotated triangle

to the middle

[steady spoke]

oops i just realized the triangles are upside down

[summer river]

can you send me a pic with that labeled?

[summer river]

no that's good

[steady spoke]

like this? idk 😭

[summer river]

give me a sec

[steady spoke]

kk

[summer river]

i can prob draw what im saying

[steady spoke]

ty

[steady spoke]

ohh i gotcha

[summer river]

ye ye

can you draw me another triangle

with the 5 and 4

honestly I could acc do it

[steady spoke]

kinda drew a water level like u said

[summer river]

WOW

ty

do you remember similiar triangles

[steady spoke]

i think so. somewhat

[summer river]

im sorry can you give me 10 mins

[steady spoke]

sure ofc!

i have other problems i can do in the meantime so take your time :)

[summer river]

my parents are calling

hehe

[steady spoke]

gotcha, no worries!

[summer river]

hi hi

[steady spoke]

hello!

[summer river]

how did the other q go

[steady spoke]

good! this the other ones i had were spring problems

and lifting a mass

[summer river]

wanna label the 10 ft long

and draw delta x

[steady spoke]

delta x would be slice of the triangular prism right?

[summer river]

mhm

[steady spoke]

i’m not sure if i drew it right 😅

tried to draw a slice

[summer river]

hmm

it's kinda like a slice

like a rectangular prism slice

[steady spoke]

yeah. is that how it was supposed to be?

[summer river]

do you want me to draw it out

im not the best artist

[summer river]

not exactly but ur close

[steady spoke]

yes please. i’m not either ahah. definitely won’t judge

[summer river]

@steady spoke it's so bad

but like

uhh try and bare

basically it's a rectangular prism piece

I put a formula for the volume

[steady spoke]

that’s very helpful ty!

[summer river]

also @steady spoke I looked back at the problem

and look at all the answer choices

they are in feet pounds

😭

[steady spoke]

oh 😭

[summer river]

yeah so it's not 9.8

[steady spoke]

32.2 ft/s^2

?

[summer river]

ye

[steady spoke]

okk

[summer river]

yeah sorry I'm boiling some water

[steady spoke]

ur good 😊

[summer river]

I have not been so attentive

[steady spoke]

no worries lol

[summer river]

but hopefully the diagrams

were pretty good huh

[steady spoke]

yeah that definitely helped a lot. i think i can probably figure it out from here

[summer river]

are you sure?

if you need help

you can always ask

[steady spoke]

ok! i’ll just leave this open and close it once i figure it out

and i’ll ask if i get stuck

[summer river]

that sounds good!

im a bad physics TA 😭

[steady spoke]

noo u were very helpful

[summer river]

biggest cap I've heard all day

today

[steady spoke]

😂

u got me further than i could get myself. so u definitely helped

[summer river]

is this physics or like a calc subpart

[steady spoke]

calc is part

omg auto correct

[summer river]

ohh yeah then it doesn't matter

[steady spoke]

yeah. this question is just part of a quiz that i can take unlimited times. but the final is made up of all of the exact quiz questions

[summer river]

didnt the sem just start

[steady spoke]

i started my school year in march bc of health issues and i have 9 months to complete it from the start date

so i’m finishing it up

[summer river]

oh wow that sounds really nice

(not the health issues)

[steady spoke]

yess it definitely is. i get a lot more free time than in person classes

caught onto that part 😂

i got it right. tysm for all the help

.close

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[feral igloo]

I need help with this. How do I figure out its correct if I only have one table?

[forest sky]

what information is given about f(x)?

[feral igloo]

wait...

oh...

i know now

bruh i thought they were two separate problems

bro... tf?!

https://tenor.com/view/waiting-gif-4372156584801469333

,rotate

[feral igloo]

what did I do wrong?

<@&286206848099549185>

[quasi sparrow]

You assumed f(x)=h(x) which are none of the choices

[woeful bobcat]

no solution - 1 is -1?

im dumb nvm

[feral igloo]

So. What do i need to do instead for it to get the correct answer?

[quasi sparrow]

Do this for one of the choices. Try f(x+1)

[feral igloo]

So, h(x) = f(x+1)

h(-2) = f(-2+1)

h(-1) = f(-1)

;-;

[quasi sparrow]

Where did h(-1) = f(-1) come from

[quasi sparrow]

that says h(-2)=-1

[quasi sparrow]

Not this

[feral igloo]

Whats the correct numbers to plug in for x then?

[quasi sparrow]

Here x=-2

Plug that into f(x+1)

[quasi sparrow]

See if it equals this

[feral igloo]

f(-2+1)

f(-1)

[quasi sparrow]

What's f(-1) =?

[feral igloo]

2

according to the table

[quasi sparrow]

No

f(x) doesn't equal 2 anywhere

[feral igloo]

;-;

[quasi sparrow]

Did you understand this

[feral igloo]

ok.

huh?

https://youtu.be/-dGA_77fLbw?si=VLgZXqJOY6uizclZ&t=78
I've been solving the way this guy from the provided video guide did it

[feral igloo]

Probably not.

[quasi sparrow]

I'm not watching a video to help you

[feral igloo]

Right. So maybe help me understand where I went wrong for this one, and what I should have done to get the correct answer for it?

I think if I can maybe understand that one first, then I could maybe figure out the confusing h(x) problem

[quasi sparrow]

Do the same thing I said above

[quasi sparrow]

This is exactly the same mistake as before

You're already assuming g(x) = f(x) which is none of the options

find a few values of g(x) and see if it equals f(x+1)

g(-1) = ?

[feral igloo]

3?

[quasi sparrow]

No

Stop guessing

[feral igloo]

So its not g(x) = x
g(-1) = 3
which I was basing off the table

[feral igloo]

then how do I find what g(-1) equals? Or, perhaps, where?

[quasi sparrow]

Look for x =-1

Then go down one row

[feral igloo]

from the table? Wouldnt that be g(x) = 0 ?

Yeah

[quasi sparrow]

That'll be g(-1)

Yes 0

[feral igloo]

ok..

[feral igloo]

plug in that 0 into f(x+1) ?

[quasi sparrow]

No

g(-1)=0

x=-1

[feral igloo]

oh

so the -1 for f(x+1)

[quasi sparrow]

Go back up and read

[feral igloo]

So by this idea, g(0) = -3
x=0
f(x+1) --> f(0+1)

f(1)

[feral igloo]

so the -1 for f(x+1) --> f(-1+1) --> f(0)

[quasi sparrow]

Does g(x) = f(x+1)?

[feral igloo]

yes

[quasi sparrow]

No

g(0) = 3, while f(0+1) = -4

g(-1) = 0 and f(-1+1) = f(0)=3 which again aren't equal

[feral igloo]

Im still confused as hell.

[quasi sparrow]

Say in words what you think the notation f(x) means.

[odd edge]

@feral igloo Has your question been resolved?

[odd edge]

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[eternal ravine]

for the function ((x-4)(x-2))/((x-2)^2(x-1)) would there be a hole or a vertical asymptote

[forest sky]

there is still a factor of (x-2) in the denominator after you cancel

[eternal ravine]

yeah

would it take over? so there’s a vertical asymptote instead of a hole

[forest sky]

yes

[eternal ravine]

alright thank you

.close

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mystic saffron]

Can someone help me with what those symbols represent in these images the U upside U the C and the underlined C - I keep getting these answers wrong- I think I’m confused about they represent

[mystic saffron]

<@&286206848099549185>

[odd edge]

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

[slow yoke]

hi

do you know set theory

basicalyl

first of all

[mystic saffron]

No im so lost tbh

[mint mirage]

$\subseteq$ means subset

[slow yoke]

sets are always represented in curly braces

[clever fjord]

CaptainNova22

[slow yoke]

and with capital letters

like

R = {p/q : p,q are integers and coprime}

this is called set builder form

[mint mirage]

So for example choice 1, is asking if g, m, p, and s are a subset of B

[slow yoke]

thosesymbles represent operations of sets

C is subset

U is union of sets

so A U B = every element in A + every element in B

[mystic saffron]

Okay but idk why I keep getting those wrongs because if it’s subsets right it would be the ones within the circle?

[slow yoke]

se

see

a subset is like

A C B

so

every element of B is also an element of A

but underlined C indicates that they might be exactly the same

[mint mirage]

Not only just within

[mystic saffron]

Oh I understand - how about if it’s A upside u B upside u C

What’s upside U and if it’s all three does it all add up

[mint mirage]

Upside u meaning $\cap$?

[clever fjord]

CaptainNova22

[mystic saffron]

Yes

[mint mirage]

That means intersection or the terms that they have in common

[slow yoke]

yeah

[slow yoke]

how do i get helpful role maan

[mint mirage]

With time

[mystic saffron]

Okay so regular u is everything the have together and upside u is what intersects but if it’s three in a row is it also asking what the first and last intersect as well ?

[slow yoke]

hmm

[slow yoke]

wdym 3 in a row

see

[mystic saffron]

Let me take a picture

[slow yoke]

alr

[mystic saffron]

So is this also asking what a and c intersect

[slow yoke]

ok

so

that just means

what is common in all of them

[mint mirage]

It's asking what term is in common with all 3 sets

[slow yoke]

u intersect the first three

then intersect the last

same thing

its commutative

[mystic saffron]

Ohhh what intersects with all three that makes a whole lot more sense

[slow yoke]

so

[mint mirage]

So is a common in all three, ie does a exist in set A, B, and C

[mystic saffron]

How about this with parenthesis

[slow yoke]

A = {c, w, s} C = {s, m} D = {a, g} B = {f, h, g, a, p, m}

[mint mirage]

Start with the parentheses, first

[slow yoke]

ah yes

theres also a law for that

distributive law of sets of intersection

[mint mirage]

Find $A \cap C$ and $B \cap D$

[clever fjord]

CaptainNova22

[mint mirage]

Then union them

[slow yoke]

yes

[mystic saffron]

I don’t get how to do that

So is it just all that intersects

[mint mirage]

First what would $A \cap C$ be?

[clever fjord]

CaptainNova22

[mystic saffron]

Just s

I think

No s and m

[mint mirage]

Yes

What about $B \cap D$?

[clever fjord]

CaptainNova22

[mystic saffron]

A and g

[mint mirage]

Now union them

[mystic saffron]

s m a g?

Like just all?

[mint mirage]

Yes