help-38

[solid kiln]

Cycadellic

[onyx mist]

This is an open cover of $\mathbb{R}$

[solid kiln]

Cycadellic

[onyx mist]

Can you give me a finite subset of this cover
$$…,(0,2),(1,3),(2,4),…$$
That still covers $\mathbb{R}$?

[warm trench]

no

[solid kiln]

Cycadellic

[onyx mist]

Exactly

So R is not compact

[warm trench]

i see

[onyx mist]

The only sets that are compact are closed and bounded

But we need a metric for boundedness to make sense

[warm trench]

i think i get it

[onyx mist]

Why isnt an open set compact?

[warm trench]

uh

bc its not closed?

[onyx mist]

Lol

So, what does closedness have to do with every open cover having a finite subcover

[warm trench]

nothing

😭

idk man

give me a trivial example that i might be able to process

[onyx mist]

(0,1)

[warm trench]

(0,1) is the set we want to determine if it is compact?

[onyx mist]

Are you familiar with the archimedean property?

[warm trench]

yeah

nx > y

[onyx mist]

Think about 1/1, 1/2, 1/3, …

[warm trench]

converges to 0

[onyx mist]

Make my open covers around it

You cant give me a finite subcover

Has to exist and cover (0,1) by archimedean prop

[warm trench]

{(1/2,2),(-1,1)}

[onyx mist]

A subset of the cover

[warm trench]

ohhhhh

ion think i can

[onyx mist]

You cant

[warm trench]

🤯

[onyx mist]

Once you name like 1/n, ill just say you didnt cover 1/(n+1)

And thats arch prop at the work

[warm trench]

oh damn

[onyx mist]

So, no openness, and it must be bounded

[warm trench]

got it

[onyx mist]

Trickier to show in metrics without archimedean

But we just use open balls

[warm trench]

open balls?

[onyx mist]

The epsilon ball

In R its called an epsilon neighborhood

I digress

[warm trench]

ah

welp, left me with more question lmao

[onyx mist]

Thats good

[warm trench]

thanks for the help

.close

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[umbral onyx]

my teacher showed this weird thing
$X_{n+2} = X_{n+1} - 2X_n \
X_{n+2} - X_{n+1} + 2X_n = 0 \
E^2(X_n) - E(X_n) + 2(X_n) = 0 \
(\lambda^2 - \lambda + 2)(X_n) = 0 \$

[solid kiln]

Nyxzore

[vagrant prism]

what's E? what is X?

context pls

[umbral onyx]

This is a discrete first order homogeneous difference equation

where E is the E operator/ shift operator

$E(X_n) = X_{n+1}$

[solid kiln]

Nyxzore

[umbral onyx]

now the context has scared people away 😭

[arctic silo]

bro ngl

once my teacher said smthing like this

and i never understood

i get it that u mean d(x n+1)/d(x n) = E or smting like dat

[marble wharf]

until the third line you should be fine, yes?

the fourth line is a general trick for linear recurrences like this

[umbral onyx]

It's the lamba nonsense that's weird for me yeah

[marble wharf]

you assume that X_(n+1) = lambda * X_n

that enables you to find particularly simple solutions X_n

[umbral onyx]

Why?

[marble wharf]

it then turns out that the actual solutions will be linear combinations of those simple solutions

[umbral onyx]

Where does this assumption come from

[marble wharf]

someone did it at some point and saw that it worked

[umbral onyx]

💀

[marble wharf]

its a classic example of first assuming a simpler case

then solving that

and then using it to solve the harder original problem

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[soft grove]

anyone's got any idea how to solve this problem (its economics but I can't quite figure out how to get around the math)

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[tough wyvern]

i need some help on the following problem , let $M_3(\mathbb{R})$ be the space of all $3\times3$ real matrices , let $F$ be a subspace of $M_3(\mathbb{R})$ such that $$F = \left{\begin{bmatrix} a & b & 0 \ 0 & a & b \ b & 0 & 2a \end{bmatrix} | a,b \in \mathbb{R} \right}$$ , note that $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{bmatrix}$ form a basis of $F$ , find another subspace $K$ such that $F \oplus K = M_3(\matbb{R})$

[solid kiln]

Sneferu
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[trim lichen]

\matbb at the very end

thats the error

ok so have you made any progress yet

[tough wyvern]

i have just an idea but it just seemed weird that i don't think it's the "intended solution"

[trim lichen]

well share it anyway

[tough wyvern]

the thing we have the canonical basis of $M_3(\mathbb{R})$ and since ${A,B}$ is linearly independent $\dim(F) = 2$ and $\dim(F) + \dim(K) = \dim(M_3(\mathbb{R})) \implies \dim(K) = 9-2 = 7$i thought of using that basis completeness theorem and pick matrices from the canonical basis and add them to the set ${A,B}$ and check whether they are linearly independent or not until i have my 7 matrices and pray that their span will give me $K$

[solid kiln]

Sneferu

[trim lichen]

ok yeah dim(K) = 7 is correct

and actually you can pluck matrices from the canonical basis

thats a perfectly fine solution

[tough wyvern]

is there some other solution you have in mind that doesn't rely on trial and error with the canonical basis ?

i thought it wasn't really the intended solution since the test was like 1 hour and i'm sure this takes some time to check

[trim jolt]

@tough wyvern Has your question been resolved?

[tough wyvern]

@trim lichen

[trim lichen]

ehh

i mean ok like

heres the idea

because the matrices A and B only have 3 nonzero entries each

you can add the canonical-basis matrices with their 1's not overlapping with either A or B, for free

and then for each set of canonical-basis matrices of which A and B themselves are linear combination, you take all but one

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@tough wyvern Has your question been resolved?

[tough wyvern]

thank you so much

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[calm thunder]

Hello so I have been trying my best to do mixed numbers but cannot ever understand how it works. I always get thrown off with the second part with the subtracting or adding portion.

answer= answer given from site

this is one of the problems Im having issues with (subtracting)

I get stuck after attemping to subtract/add them.
There is also usually a last part that has an answer like this:

1 1/2 or 1/4/5 that 1 always in front always throws me off for I dont understand how it always gets there.

[trim lichen]

1/2 + 5/8 = 1/2 + 6/8
how did the 5/8 become 6/8 exactly...? and then it looks like you did the "add top and add bottom" thing, which is a common ish mistake

[worn flume]

Which is the workings for this question?

[trim lichen]

based on the work you've shared i don't think mixed numbers are what's troubling you

i think it's fractions in general that you're struggling with

also yeah none of these like three different exercises look like they have anything to do w/ the problem you shared...

[worn flume]

So we are asked how many more tablespoons of mustard we need than chopped onions

[trim lichen]

you mixed a lot of words up in that

[worn flume]

Ik lol

In short the positive difference between the chopped onions and mustard

[trim lichen]

@calm thunder hey are you still here

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@calm thunder Has your question been resolved?

[worn flume]

,rotate

[worn flume]

@calm thunder

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[calm thunder]

sorry

[calm thunder]

I was

busy

[trim lichen]

https://youtu.be/Vb5c7ay0ziU

[calm thunder]

this is very helpful thanks

[trim lichen]

... was that sarcastic or genuine?

(i can't tell...)

[calm thunder]

geniuine

[trim lichen]

ok

[calm thunder]

I have trouble with fractions and it seems its gonna be a while to study to ackonowledge

[jolly basin]

Look at khan academy for additional practice too!

[trim lichen]

i mean, yeah, shit takes time to get used to/ comfortable with

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[echo rose]

the question is about physics but my problem is math

so i need find Vo

and Vo need to be 3 m/s

i tried use cos

but idk

what to do

if I find Vo the remainder is formula

[trim jolt]

@echo rose Has your question been resolved?

[mellow tendon]

What do you think cos(theta)

[echo rose]

cos(theta) = 5/Vo = 3/5

?

[mellow tendon]

Yes

[echo rose]

but in this case Vo = 25/3

[mellow tendon]

Wait when dealing with vectors

[echo rose]

but I'm using the intensity

no?

[mellow tendon]

Vo = cos(theta) * 5

[echo rose]

yes

[mellow tendon]

Oh yeah

Forget what I said there

[echo rose]

so, cos theta = Vo/5

[mellow tendon]

5 is the hypotenuse

[echo rose]

dont worry

[mellow tendon]

So yeah cos theta= Vo/5

[echo rose]

cos theta top and cos theta bot are equal

[mellow tendon]

Yeah

But here we want to decompose the 5 into its 2 components so it should be the hypotenuse

[echo rose]

how I do this?

[mellow tendon]

Here in that picture I want to decompose 5m/s into 2 vectors

[echo rose]

ohhh

[mellow tendon]

That are perpendicular
So I can move the vectors to be in this shape

So you would be able to see that 5m/s is the hypotenuse vector

Understandable?

[echo rose]

in this case

cos theta = vo/5

and cos = 3/5

Vo = 3

ty

youre smart

[mellow tendon]

Yeah but you understand why right?

[echo rose]

yes

ty

[mellow tendon]

Np

[echo rose]

np?

kk

I will close

.close

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[scenic delta]

$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$\
$f(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}$\
$f'(x) = \sum_{n=0}^\infty (-1)^n x^{2n}$\
$f'(x) = \sum_{n=0}^\infty (-x^2)^n$\
$f'(x) = \frac{1}{1+x^2}$\\
$f(x) = Arctan(x)$\
$Arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ...$\\
$\frac{\pi}{4} = 1 - \frac13 + \frac15 - \frac17 + ...$\\
$\pi = 4 - \frac43 + \frac45 - \frac47 + ...$

[solid kiln]

Oğuzhan

[scenic delta]

Is this a valid way of showing this?

[limpid dawn]

i cant tell what you are aiming to show, but i dont see something particularly wrong so far

[scenic delta]

Our professor proved it this way today, and I've seen it being proven in much more complex ways, is it really just this simple?

[scenic delta]

The famous, simple but slow, approximation of pi

[limpid dawn]

i see, well depends on whats simple for anyone, but yea you can do the things he did

[scenic delta]

I guess that wraps it up

Should I close or wait

[limpid dawn]

if you are done you can close

[sturdy fiber]

If you are done you can .close

LOLLLLL

[scenic delta]

Alright

.close

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[turbid gazelle]

the answers i filled in for e and f are wrong

[turbid gazelle]

but anyways, can someone explain all the parts starting from c

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[obtuse lantern]

\begin{align*}
3g + f + 0c &= -4 \
0g - 2f + c &= -1 \
6g + 0f + c &= -9
\end{align*}

[solid kiln]

Ytterbium

[obtuse lantern]

My calculator shows math error when I try to solve these three

Are they unsolvable

[dusky thunder]

try finding if there is a free variable

[obtuse lantern]

A free variable

I think I get what you're saying

But it's not working

The variables cancel out

[dusky thunder]

take twice the first equation and add it with the second

you will notice a linear combination

which implies...?

[obtuse lantern]

I get $-2f+c=-1$

[solid kiln]

Ytterbium

[obtuse lantern]

Which is the third equation

It's unsolvable

[dusky thunder]

it is solvable

but it has infinite solutions

[obtuse lantern]

Dang

Infinite circles

[dusky thunder]

u can parameterize g=t

and solve the system in terms of that

[obtuse lantern]

Ok thx

.close

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[marsh forum]

I would like someone to walk me through this

[marsh forum]

I have to use green's here btw

so I have $\iint_{D} div(F) \cdot s = \int_{C} F \cdot r$

[solid kiln]

What a wonderful world !

[marsh forum]

have to figure out the boundary

[rigid portal]

u remember what i said about boundaries

if u lived in that 2d plane in the first octant, where would the "end of the world" be

[marsh forum]

at the x,y and z axes

[rigid portal]

uh no

[marsh forum]

wait

[rigid portal]

but maybe u have the right idea but just stated it wrong

[marsh forum]

here it would be a triangle on the xy plane

[rigid portal]

also wrong

[marsh forum]

2x+y=2, x=0, y=0 would be the boundary

[rigid portal]

unfortunately incorrect

[marsh forum]

why

[rigid portal]

say u have a triangle

any triangle

including the inside

what are the boundaries of said triangle

[marsh forum]

the edges

[rigid portal]

right

so what are the edges of the triangular section of that plane?

[marsh forum]

0≤x≤1;0≤y≤2

[marsh forum]

this right

[rigid portal]

no

[marsh forum]

This is part of it

Or is it the bounadry of the trianglar region of the plane

[rigid portal]

well maybe idk i didnt do all the math but the point i was trying to make is that the boundaries are the line segments at the xy yz xz planes

cuz it says first octant

[marsh forum]

I'm still not sure what the boundary would be

The best I can say is0≤x≤1;0≤y≤2;0≤z≤1

[rigid portal]

maybe itd help to draw it out

the triangle ur working with looks like this

its a portion of a plane

that exists in the 1st octant

then the boundary are these 3 lines

findable by substituting x = 0 for the line in the yz plane, y = 0 for the one in the xz plane, and z = 0 for the one in the xy plane

[marsh forum]

Thanks

Ooh

.close

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[marsh forum]

Thanks

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[bronze tinsel]

Hi guys.
Here is an integral that I'm currently trying.

The answer involves Hypergeometric Function.

[bronze tinsel]

I have completed the half of it.

All I need is the other have.

[bronze tinsel]

As you can see here.

Problem is, I'm stuck and I don't know how to proceed to get the 1/8 and 9/8

[trim jolt]

@bronze tinsel Has your question been resolved?

[marble wharf]

something something if you have (1/8)_n and (9/8)_n then they more or less cancel each other out. somewhere along those lines is probably the solution

working backwards probably helps

[bronze tinsel]

My goal is not to prove.
It's to solve.

[bronze tinsel]

Although this might help.

[exotic tundra]

find the ratio between successive terms, then make it look like the definition

[bronze tinsel]

What do you mean by successive term?

I think my main problem lies on 1/(8n+1)

[exotic tundra]

$\frac{\text{n+1th term}}{\text{nth term}}$

[solid kiln]

Sepdron

[bronze tinsel]

Hmm...

Let me see...

Now.

I got

[exotic tundra]

nice!

[bronze tinsel]

1/9, 9/17, 17/25

And then moving on.

[exotic tundra]

oh no, do the general term
not just for a specific n

[bronze tinsel]

Oh.

(8n+1)/(8n+9)

$\frac{\text{8n+1}}{\text{8n+9}}$

[solid kiln]

2KGod

[exotic tundra]

close, what about the $\frac{\qty(\frac12)_n}{n!}$?

[solid kiln]

Sepdron

[exotic tundra]

btw, you don't need the \text if the inside is math

[bronze tinsel]

I straight away copy your stuff.

[bronze tinsel]

What do I have to do with that??

It's fixed.

This is what the form that I want it to be.

[bronze tinsel]

According to here, I have the b and z.
Now I need a and c.

[exotic tundra]

aah, you have that definition
I was following wolfram's
https://mathworld.wolfram.com/HypergeometricFunction.html

[bronze tinsel]

This is generalized one.

[bronze tinsel]

Should have mentioned this earlier...

[exotic tundra]

yeah, so from that, for $2F_1$ you have
$$\frac{c{n+1}}{c_n} = \frac{(k+a_1)(k+a_2)}{(k+b_1)(k+1)}$$

[solid kiln]

Sepdron

[bronze tinsel]

And....

[exotic tundra]

ig the trick is that you can factor an 8 from the numerator and denominator of that

[bronze tinsel]

Hmm....

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@bronze tinsel Has your question been resolved?

[trim jolt]

@bronze tinsel Has your question been resolved?

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@bronze tinsel Has your question been resolved?

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[fair forge]

can i preform the step for whole number z:
z^2 mod 5 = (z mod 5)*(z mod 5)

and what justification do i use?

[vagrant prism]

usually, you have something like
z^2 = ___ (mod 5)

where you have something in place of ___

z^2 mod 5 is a bit cursed

[vagrant prism]

but no that isn't correct: take z=4

[carmine mauve]

yes

[austere cedar]

I know some programming courses like to treat mod as an operator, this is non-standard though

[vagrant prism]

i think this would work if you did
z^2 mod 5 = (z mod 5)*(z mod 5) mod 5

[carmine mauve]

Given $r \equiv z \mod 5$ (where $0 \le r < 5$) means there exists some integer $k$ such that $z = 5k + r$, we have
$$z^2 = (5k + r)^2 = (25k + 5r)k + r^2 \equiv r^2 \mod 5$$
and $z = 5k + r \equiv r \mod 5$

[solid kiln]

Shuba

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[placid radish]

what have you done? its be helpful to sketch a graph

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[stark pebble]

.reopen

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✅

[stark pebble]

i dont think a rough graph will be helpful

[vagrant imp]

its an mcq, you can just assume the function

eg. f(x)=||(1/5)^x||

[stark pebble]

i dont see how's that helping me

🤧

[vagrant imp]

just integrate it....

[stark pebble]

(1/5)^x/log (1/5)

u put the limits u get

(1/5)^10 /- log 5 - 1/-log 5

[vagrant imp]

simplify..

[stark pebble]

1- 5^(-10)/ log 5

[vagrant imp]

alr can you answer the qs now?

[stark pebble]

can you?😭

[vagrant imp]

hint:||f(0)=1||

sol:||0<1-5^(-10)/log5<1/log5<1||

[placid radish]

well you can bound the function by step functions

a graph helps visualize what the steps functions are

the "decreasing function" property is very important

[stark pebble]

its a gp with ratio less then 1

yes its decreasing

[stark pebble]

thats correct, yes

then?

[placid radish]

yeah we can worry abt the ratio later, but the main problem is: how to trasform the integral (which is hard to compute since we dont have any info) to something (perferrably sum) relating to f(0),..., f(10)?

again a step function bound is v helpful

[vagrant imp]

bruh 1=f(0), so 0<integral<f(0)<2f(0)

[placid radish]

i mean this works but it doesn't quite help the op understand the concept

[vagrant imp]

hmmm well yeah but quicker methods are also useful in tests and stuff

[stark pebble]

ohhh i seee 🤩

step function like the graph of [x]

but why not a continuous function

to something (perferrably sum) relating

yes understood

[placid radish]

hint because when we are working with a step function, lets say g(x) is equal to f(0) on the interval [0,1], then $\int_0^1 g(x) \ dx=f(0)$, can you see why?

[solid kiln]

qwertytrewq

[placid radish]

furthermore $g(x)\geq f(x)$ on the interval $[0,1]$ (which is the crucial step, can you justify this?

[solid kiln]

qwertytrewq

[stark pebble]

g(x) is a step function here?

[placid radish]

sure, but lets say one of the "steps" is on the interval [0,1] and g(x). Just to be clear a step function just means that the function is defined as being constant on intervals (you divide the domain into intervals, and on each interval the function is constant)

[stark pebble]

yes got it

how do apply this concept in the qs above

[placid radish]

here is an example

[stark pebble]

yes like the graph of [x]

[vagrant imp]

i believe he means smtn like this

[placid radish]

like that but a little bit more general

if the red line is the function f, then you can see that the constant function (blue line) bounds f from aboce

[stark pebble]

yes yes

make sence

okayyy

[placid radish]

we can improve the bound by dividing into two parts

the blue line still bounds the red line

[stark pebble]

like this right?

[placid radish]

(we need the decreasing condition here for the bounding reason)

[placid radish]

yeah, thats when we refine the bound into 10 parts

can you figure out the function for the bound?

[stark pebble]

btw it can go below the x axis too right?

[placid radish]

yeah

as long as the function is constant on the intervals its fine

its simply partitioning the domain, and then defining the function on each partition

visually it looks like steps

[placid radish]

of different sizes and heights

[stark pebble]

okayyyyy

[stark pebble]

now this has to do smthing with the ratio?

1/5

decreasing

[placid radish]

not yet, but lets work through it

[placid radish]

can you figure out the function for upper bound here?

[stark pebble]

is the partition all the integers?

[placid radish]

not necessarily integers, for example i can partition [0,3] into [0,1.6) and [1.6,3]

[stark pebble]

no for this particular qs... cuz it says the f(0), f 1 , f2.... f10 all are in gp, i want to use this

[placid radish]

if I divide [0,10] into [0,5),[5,10], and define g(x)=f(0) on [0,5), and g(x)=f(5) on [5,10], do you see how g would represent the blue line?

[placid radish]

yeah yeah we are getting there, but ill need to give you sole context (which is done for arbitrary functions)

an intuitive understanding would help

[stark pebble]

yess i think i am getting the full view

[stark pebble]

AHHHH I SEEE

YEAAHH

[placid radish]

This gives us a bound: $$\int_{0}^{10}f(x) \ dx\leq \int _0^{10} g(x) \ dx=5f(0)+5f(10)$$

[solid kiln]

qwertytrewq

[placid radish]

But wait! we can do wayyy better than this!

divide it into 10 parts!

[stark pebble]

okay okayy

[stark pebble]

i got this okay

[placid radish]

ok now try partitioning the interva [0,10] into [0,1),[1,2),...,[9,10]

and defining a step function there

[stark pebble]

i was trying to write a general equation

nvm

[stark pebble]

okay yeah i see how it will be

we put the the first valuve of domain as the upper bound

[placid radish]

its fine to define it one interval by one interval

[stark pebble]

and then make 10 steps

okay

[placid radish]

u mean g(x)=f(0) on [0,1), etc?

[stark pebble]

yes

[placid radish]

yeah precisely! can you now give an upper bound for $\int_0^{10} f(x) \ dx$ using your function?

[solid kiln]

qwertytrewq

[stark pebble]

didnt we just did that?😭

[vagrant imp]

tbh I didn't know about this method , its pretty cool

[vagrant imp]

maybe this diagram would help you answer what he's asking

(sorry for the terrible handwriting)

[stark pebble]

yes i get thatt now how do i proceed?

[vagrant imp]

do you know what upper bound means?

[stark pebble]

yes

[stark pebble]

its the ones with straight black lines ablove the green line

[vagrant imp]

like if you can calculate the area under the graph for this functions

its an upper bound on the area of f(x)

does that make sense?

[stark pebble]

yeah yeah

[vagrant imp]

whats the area of the first rectangle?

[stark pebble]

but doent we get extra area?

[vagrant imp]

thats the point of an upper bound to give us an inequality on the area of f(x) so we can use it to solve the problem

[stark pebble]

hmmmm okayy

so now i find all the areas of rectangle

[vagrant imp]

basically upper bound of the area must have more area than f(x) hence the name 'upper'

[stark pebble]

and then it will be more then $$\int_{0}^{10}f(x)$$

[solid kiln]

Eclipse

[stark pebble]

alright

[stark pebble]

and it will be less then?

[vagrant imp]

the lower bound, but we don't need one right now, just solve this first

if need arises we will check

did you get the upper bound?

[stark pebble]

waittt

[stark pebble]

do i put f(0)= 1

but i cant right?

[vagrant imp]

no leave it as it is

[stark pebble]

i cant get how to put the partitions

[vagrant imp]

whats the width of each partition

[stark pebble]

1

[vagrant imp]

what about the height

say for the first rectangle

[stark pebble]

f(1), f(2)

f3, f(4)

like that

[vagrant imp]

it starts at f(0)

[stark pebble]

yeah sorryy

from f0 to f10

[vagrant imp]

and in the Qs we've been given f(1)=f(0)/5

f(2)=f(0)/25

and so on

[stark pebble]

okayyy

[vagrant imp]

do you know what sum you can apply here

[stark pebble]

f(0)/1-1/5

[vagrant imp]

there isnt inf terms

[stark pebble]

thats the sum of all the height

[vagrant imp]

10

[stark pebble]

OH YEAH

MY BAD

F(0)(1-(1/5)^11)/4/5

[vagrant imp]

im pretty sure there are 10 terms

[stark pebble]

0 to 10?

[vagrant imp]

f(0),f(1),,,f(9)

[vagrant imp]

that would be integrating from 0 to 11

[stark pebble]

yea for the bound

it will stop at 9

to count till 10

right

[vagrant imp]

so power will be 10

[stark pebble]

F(0)(1-(1/5)^10)/4/5

right

[vagrant imp]

alr now you can writ this as 5/4 f(0)(1-5^-10)

which is lesser than 5/4f(0)

but integral<5/4 f(0)(1-5^-10)<5/4f(0)

makes sense?

now check the value of f(0), can it be negative?

[stark pebble]

okay so $$\int_{0}^{10}f(x)$$ is less then 5/4 f(0)

[solid kiln]

Eclipse

[stark pebble]

no it cant be negative

its area

[vagrant imp]

no, its cuz then the gp wouldn't be decreasing

[stark pebble]

that too

okay

great make sencee

[vagrant imp]

alr so f(0)>0

[stark pebble]

yea yea

[vagrant imp]

=>2f(0)>5/4f(0)

[stark pebble]

okayy

[vagrant imp]

from here we get

integral<2f(0)

[stark pebble]

but insted of 2 u can write any integer greater then 2 as well

why u chose 2?

[vagrant imp]

cuz its there in the options

[stark pebble]

cuz its more closer as well

[vagrant imp]

yeah you should also check the lower limit

now we know the area must be more than 0 since f(0)>0 and the function is decreasing, so int>0

[stark pebble]

okayyyy

[vagrant imp]

so the int must also be more than 0

[stark pebble]

yeah hence the ans

option 1

GREAT MAN

THANKS A LOT

[vagrant imp]

well you should thank qwertytrewq

[stark pebble]

thats a nice way to slove

[vagrant imp]

I didn't do much, it was him who introduced it

[stark pebble]

@placid radish thankssss

[placid radish]

[stark pebble]

its a great way to approach this

[placid radish]

np

[stark pebble]

really appricated!!

[stark pebble]

u helped me reach the ans, also thankss!! plus the short way too!!

[placid radish]

you can also get a lowr bound this way

i believe its 1/5 f(0)

[stark pebble]

alrighttt!!

[vagrant imp]

wouldnt it be 5/20

1/4

[placid radish]

or something close to that

[placid radish]

yeah maybe

its a bit less than 1/4 f(0)

cuz we can really add to infinite anymore for the lower bound

[vagrant imp]

you consider these rectangles for the lower bound btw

[placid radish]

2441406/9765625 f(0)

[vagrant imp]

lmao nice

[stark pebble]

hmmmmmmmm

[placid radish]

its the same method, just with a lower approximation instead

these type of "step function bounds" are sort of common when it comes to riemann integral, in fact it is used to define riemann integral

we say riemann integral exists when the highest step function bound, and the lowest step function bound gets arbitrarily close when our partition is small.

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[vagrant imp]

Im not sure how to simplify this

C represents choose or combination or binomial coeff or wtever you call it

[carmine mauve]

what does nCr mean?

by which I mean, what is the formula for it?

[vagrant imp]

n!/(n-r)!r!

[carmine mauve]

yep, maybe use that to simplify?

[vagrant imp]

ok what now

i dont think it helps, besides the options are in terms of A10 B10 and C10, so maybe expanding it isnt a great idea

[austere shard]

split it into two summations, take the constants outside

after that, hint: use nCr = nC(n-r)

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@vagrant imp Has your question been resolved?

[vagrant imp]

(1+x)^10 * (x+1)^10=(x+1)^20
and whenever the powers add up to 10, like say x and x^9, their coefficients will be the same since 10C1=10C9, thus we get sum of Ar^2=coeff of x^10 in (x+1)^20=B10

is that right?

but im not sure how to do sum ArBr

[austere shard]

yes

[austere shard]

do the same thing

[vagrant imp]

oh so its C10

[austere shard]

yup

[vagrant imp]

Wow so ans is 0

[austere shard]

yeah

[vagrant imp]

wth uh is this wrong?

[austere shard]

huh oh wait

oh shoot

left summation is only for 10 terms

so doesnt include all B

[vagrant imp]

yeah but coeff of x^10 wont have stuff like coeffs of x^11 or x^20 from B

or maybe im understanding it wrong

oh bruh 0 is also not included

[austere shard]

ok i got it

yeah 0

even for the right summation

[vagrant imp]

Thanks a lot for your help, really appreciate it!

[austere shard]

np happy to help

[vagrant imp]

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[fluid bloom]

trying to show this series converges to 3

not sure how to do so. is there a way to express Sn as a sequence so i can take the limit?

[trim lichen]

your S_N is incorrect bc you just take the first, second, third and then jump straight to the N'th term

[fluid bloom]

yeah i know

[trim lichen]

yes there is, look up geometric progressions

[fluid bloom]

i meant to add ...

[fluid bloom]

this?

so 2r^n-1

what's r

[trim lichen]

yes but you need the formula for the sum of the first n terms.

[trim lichen]

1/3.

[fluid bloom]

how can i tell it's 1/3?

could i have a scenario in which a = r?

[fluid bloom]

is it because of this?

[trim lichen]

2/3^(n-1) = 2 * 1/3^(n-1) = 2 * (1/3)^(n-1)

[fluid bloom]

where'd 2/3rds come from?

[versed jasper]

$\frac{2}{3^{n-1}} = 2 \cdot \frac{1^{n-1}}{3^{n-1}} = 2(\frac{1}{3})^{n-1}$

[solid kiln]

BuilderDolphin

[fluid bloom]

okay

that makes more sense thanks

alright thanks

cheers

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[valid beacon]

Hey everyone, heres a question from Silverman's friendly intro to number theory

im having a difficult time with demonstrating the first part of the hint

i was thinking about using quadratic residues mod 3 but it doesnt seem to work well enough

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@valid beacon Has your question been resolved?

[zinc ginkgo]

Also try #elementary-number-theory

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@valid beacon Has your question been resolved?

[hot root]

I think I might be able to help

First you can easily say that x will not be negative

Second by further analysing you can say that x has to be perfect square

[valid beacon]

i dont think x has to be a perfect square

[hot root]

No it has to be

[valid beacon]

i might have solved it now but that was one of the cases i checked

it seems that checking the cases of how squarefree x is seems to work

so either x is a square, x is not square but not squarefree, or x is squarefree

[valid beacon]

i mean even x being a square doesnt work

because you get smth like

[valid beacon]

$\begin{aligned}
x = A^2 \
\implies y^2 &= A^2(A^4 + z^4) \
\implies (\frac{y}{A})^2 &= Y^2 = A^4 + z^4
\end{aligned}$

which doesnt work bc thats a stronger statement of FLT at n = 4

[solid kiln]

hiidostuff

[hot root]

This is the equation that forms so x is either equal to
X^2 + Z^4
Or x is perfect square but
If all x,y,z are integers x has to be perfect square

[hot root]

For this

[valid beacon]

i dont think it has to be equal

i mean consider smth like $x = a^2 b$ where b is squarefree

[solid kiln]

hiidostuff

[valid beacon]

then we just get an implication that b divides x^2 + z^4

rather than saying that x is forced to be equal to x^2 + z^4

[hot root]

But if that's the case then y ^2 will not be perfect square

[valid beacon]

why not

x^2 + y^4 could be divisible by b

[hot root]

Which it has to be as y is integer

[valid beacon]

then we get a^2 b^2 times smth else that could very well be square

[hot root]

Got it I forgot bout that possibility

[valid beacon]

luckily im pretty sure the "not squarefree but not square" case can be shown to be impossible due to descent

i didnt use the advice the book gave though

i still wonder what their idea was

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[leaden nebula]

is the empty set a basis for a subspace with dimension 0?

[leaden nebula]

oh yeah it is

and {0} would be such a subspace right?

[royal remnant]

yes

[leaden nebula]

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[lean kraken]

Any shortcut?

[trim lichen]

1/6(A^2+cA+dI)*A = I

write this as a cubic equation in A

then use cayley-hamilton

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@lean kraken Has your question been resolved?

[lean kraken]

But how can I find A^3 by any short way?

[native shuttle]

Idk

[lean kraken]

Thanks

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[stark tapir]

.reopen

[stark tapir]

use characteristic Equation |A-(lambda)I| = 0

Got the ans. loll

[lean kraken]

I have solved out

[lean kraken]

Yes same

Thanks

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[warm sedge]

Part c

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[boreal apex]

Open a new channel

This one is currently closing

[warm sedge]

Yea lol mb

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[vagrant imp]

shouldnt f-1(2) be 0

[vagrant imp]

e^(-0) f(0)=2+int(0,0)sqrt(t^4+1)dt
=>f(0)=2+0=2

so f-1(2)=0

am i wrong or is this question wrong?

they've given the answer as|| B) 1/3||

.

[spiral ocean]

Derivative

Of inverse?

[vagrant imp]

oh bruh missed the dash

ugh

yeah i got 1/3

basically i took f(x) as y, found dx/dy, interchanged x and y, found values of x and y from the function, subtituted and solved

[vagrant imp]

thanks for the help!

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[glad patio]

does that seem right

the area of R being 100/3?

[carmine mauve]

where do you get x = 2 from?

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@glad patio Has your question been resolved?

[woven python]

how are you finding the turning point x=2 when the function (by extension its turning points) are dependant on k?

@glad patio

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[fervent leaf]

what part are u stuck at?

[hushed ferry]

I just dont know how to do it

[dusty sleet]

Do you know the quadratic formula?

[hushed ferry]

yes

b2 -4ac

[dusty sleet]

Then apply it to the equation

[dusty sleet]

This is the discriminant, yeah

[trim lichen]

b**^**2. you should use the symbol ^ to write exponents.

[hushed ferry]

so I get 8 +/- square root of 120/ 4

[dusty sleet]

Exactly

So what's the value of k?

[hushed ferry]

so the answer would be 120

[dusty sleet]

There you go 👍

[hushed ferry]

thank you!!

[dusty sleet]

Yw 🤗

[hushed ferry]

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[glad patio]

i got area of R as 100/3

[empty prairie]

What is your question?

[glad patio]

b)

[grim saffron]

Look at the second and third picture

[split chasm]

can you show your work

[split chasm]

you didn't find the area of the correct region

you've found the area between curve C and the x-axis, from x=0 to 2
i.e. the red part

[glad patio]

thats what i originally thought

then i did this

[split chasm]

yes, that works

[trim jolt]

@glad patio Has your question been resolved?

[trim jolt]

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[blazing geode]

ahhhhhhhhhhh

[trim jolt]

Please don't occupy multiple help channels.

[blazing geode]

I do not

I get tired about an hour of studying

But I'm supposed to solve 20 questions in a 100 minutes test

what should I do

[empty prairie]

https://tenor.com/view/omori-memes-omori-wendy's-sir-this-is-a-wendy’s-mcdonalds-gif-10606255422517932247

[late flume]

Well the best suggestion is to simulate exam conditions a week before the exam

[blazing geode]

should I take a nap, I'm so tired rn

[late flume]

That isn't that bad

[blazing geode]

should I take a nap if I feel tired before exam

[late flume]

We have 75 questions in 180 mijs

[blazing geode]

really

[late flume]

Yes

Check out jee mains

[blazing geode]

.close

[trim jolt]

Channel closed

Closed by @blazing geode

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[trim jolt]

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[trim jolt]

@pearl lark Has your question been resolved?

[trim jolt]

@pearl lark Has your question been resolved?

[trim jolt]

@pearl lark Has your question been resolved?

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[boreal shuttle]

$\lim_{x\to\infty}x\left(\frac{\ln(x+1)}{x}\right)$

[solid kiln]

<rajel />

[boreal shuttle]

im trying to calculate that limit

$\lim_{x \to \infty} x \left( \frac{\ln(x) + \ln\left(1 + \frac{1}{x}\right)}{\ln(x)} \right)$

[solid kiln]

<rajel />

[boreal shuttle]

this is what i ended up wtih

i believe theirs a rule that says for any large value x , ln(1 + 1/x) = 1/x

and then i ended up with $\lim_{x \to \infty} x \left( \frac{\ln(x) + \frac{1}{x}}{\ln(x)} \right)$

[solid kiln]

<rajel />

[trim lichen]

hold up

doesnt this just simplify to ln(x+1)

or did you mess it up at some point writing down the original

[boreal shuttle]

i pasted the wrong equation

its ln(x) not x, sorry

$\lim_{x\to\infty}x\left(\frac{\ln(x+1)}{\ln(x)}\right)$

[solid kiln]

<rajel />

[frozen plover]

this is just infinity

[boreal shuttle]

no its not

[frozen plover]

ln (x + 1) ~ ln (x)

[boreal shuttle]

it should be 1

[frozen plover]

,w limit as x goes to infinity of xln(x + 1)/ln(x)

[trim lichen]

hold on a minute

[solid kiln]

Results provided by WolframAlpha

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

Requested by neonperseus

[frozen plover]

^

[trim lichen]

are you SURE you arent messing up the original eq again

log(x+1)/log(x) -> 1

x*that -> ∞

[boreal shuttle]

yes im messing it again lmao

i need some rest lol

$\lim_{x\to\infty}x\ln(\frac{x+1}{x})$

[frozen plover]

better

[solid kiln]

<rajel />

[eager pendant]

put u= (x+1)/x

[frozen plover]

this would be much easier if you know the e limit

do you?

[boreal shuttle]

nope

[eager pendant]

Then solve for x

[boreal shuttle]

@trim lichen pasted the correct one ^

[eager pendant]

U know how to do that transition that I told you?

[trim lichen]

bruh ok finally

$\lim_{x \to \infty} \frac{\log(1 + \frac{1}{x})}{\frac1x}$

[solid kiln]

Ann

[trim lichen]

this is your limit

[eager pendant]

Why do that

[frozen plover]

LH incoming

[boreal shuttle]

hmm we dont work with log in my school

[eager pendant]

No need for DLH yet

Try what I’m telling you brother

[boreal shuttle]

neither l'hopital

[eager pendant]

No need for that

Put u = (x+1)/x

Solve for x

[boreal shuttle]

i'll try

[eager pendant]

Then find where u tends to

U know how to do that?

[gusty plank]

what do you mean you don't work with log

[trim lichen]

ln

just pretend it is ln

i write it as log without a base out of habit

[trim lichen]

fuck no

tossing a thousand stones at you for that @frozen plover

no LH

just limit definition of the derivative @boreal shuttle

[boreal shuttle]

oh i see

[eager pendant]

U see what I am seeing ? Doesn’t it work also that way , way more simple

[trim lichen]

i think my way is simpler

you just recognize 1/x goes to 0

[eager pendant]

With that?

[boreal shuttle]

@trim lichen @eager pendant you both got cool methods , solved it with both

thx

[eager pendant]

Yo also

[trim lichen]

yes

x goes to infinity so 1/x goes to 0

[eager pendant]

Ln(x+1/x) = ln(x+1) - lnx

[trim lichen]

and you end up with the limit just being ln'(1)

= 1/1

[boreal shuttle]

nice

[boreal shuttle]

and ?

.close

[eager pendant]

Nvm that doesn’t get anywhere

[boreal shuttle]

.close

[eager pendant]

But what I told you in the beginning does

[trim jolt]

Channel closed

Closed by @boreal shuttle

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[trim jolt]

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[signal island]

i need help understanding something

[signal island]

can somebody explain why this would be true

instead of using -1 please use a variable like a or c or something

[grim saffron]

What is the definition of limit?

[signal island]

aproximation

aproximating the value at a given point of a function

[ionic pendant]

the limit $\lim_{x \to c} f(x)$ is completely independent of the value of $f(c)$

[solid kiln]

cloud

[ionic pendant]

because it only quantifies the behavior of f near c

[grim saffron]

Wait, that's statement is not true. What if f(x) and g(x) is not continuous, like Weierstrass function

[signal island]

ohh

so even if it doesnt exist at that point

we can aproximate what it would be if it existed at that point?

[signal island]

whats that

im lockind in for my calc bc exam

[wooden plover]

then both limits don't exist

[signal island]

im self studying and i jsut started

i ahve it on may

[ionic pendant]

we can say that the limit is the point that the function "would" approach if it's continuous

[signal island]

that makes sense

[stable ore]

The weierstrass function is continuous

That’s like the whole point

[grim saffron]

Yeah, not weierstrass

meant for Blumberg function

[stable ore]

I’m not familiar

[grim saffron]

Well, or 0 for rational, 1 for irrational, dirichlet function

[ionic pendant]

the implication here is that the limits are performed over the same set. it would only be untrue if you dropped that assumption

[stable ore]

The indicator function of the rationals is the Dirichlet function

[vast egret]

I could say, f(x) = g(x) at $(-1)^+$ and $(-1)^-$ so f(-1) = g(-1)

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

Let's add limit

[vagrant kraken]

^that is not true

jump discontinuity

[vast egret]

I could say, $\lim f(x) = \lim g(x)$ at $(-1)^+$ and $(-1)^-$
so $$\lim_{x \to -1} f(x) = \lim_{x \to -1} g(x)$$

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

What I know is that if a function is continue to the right of a point and to its left, and left = right. that means it is continuous at that point.

[trim jolt]

@signal island Has your question been resolved?

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[somber siren]

the blue is the correct answer. i don’t understand b)

[somber siren]

i don’t really get what it means to be colinear, but it says it’s on the same line

wait

“but i don’t know how they’d be on the same line”
is it because if you multiply a scalar quantity by the vector, you only increase/decrease the magnitude, basically make it longer or shorter

u can ofc change its direction but we aren’t talking about that rn

so if u multiply by the scalar vector, you’ll be able to find ones that will be on the same line

is this correct?

but still what does it ACTUALLY mean to be collinear though

[supple copper]

You’ve got it

[stuck plover]

Yeah collinear just means they're on the same line

Aka they're scaled versions of each other

[supple copper]

2 vectors v and w are colinear if there exists some scalar k such that v = kw

[trim jolt]

@somber siren Has your question been resolved?

[trim jolt]

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[plain tusk]

can i get some help

i just started the q so no progress

[tidal igloo]

What have you tried mate

[plain tusk]

7c+5t = 480

thats it 💀

this question got me bamboozled

[tidal igloo]

Now what do we know about c and t that relates them?

[plain tusk]

the ratio 5:9

[tidal igloo]

c / t = 5 / 9

Can you use this to substitute

[plain tusk]

whats that slash sign

[plain tusk]

does that mean divide?

[tidal igloo]

Divide

Yes

[plain tusk]

so c/t = 5/9

[tidal igloo]

Thats what ratio means

[plain tusk]

oh

so would u sub in 9 for the T

[tidal igloo]

No

[plain tusk]

oh

[tidal igloo]

c = (5/9)t

Sub this in for c maybe