help-13

[elfin otter]

Aight

Thanks pals :)

[ancient valley]

no prob :)

[elfin otter]

||I'll be back soon||

.close

[cedar kiln]

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[obsidian umbra]

how do I come up with this polynomial?

i know one of the zeros is (x+4)

but the 1+2i is tripping me up

isn't there like a conjugate your supposed to do for that?

[tropic oxide]

one of the factors is (x+4).

[tropic oxide]

sure is, because the polynomial has real coefficients.

since you know 1 + 2i is a root, you know that what other complex number is a root?

[obsidian umbra]

1-2i

so you do (x-(2+i))*(x+(2-i))

right?

[tropic oxide]

why 2+i and 2-i

[obsidian umbra]

oh

[tropic oxide]

those arent the same numbers that you want as roots

[obsidian umbra]

mt

god

thats the mistake i made

god i can't read

or replicate

lol

so it would be (x-(1+2i)*(x-(1-2i)

?

[spice kraken]

yes

multiply all the (x-root)

[obsidian umbra]

thanks

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[celest heath]

Can someone check both of these?

[dusk finch]

,w 2^24

the second one is correct

[celest heath]

thank you but what about first

[slate lintel]

,calc 240+1080

[wraith dagger]

Result:

1320

[celest heath]

oh

u have to add 1080?

[slate lintel]

well

[celest heath]

why

[slate lintel]

it's any multiple of 360

because 360° is a full circle

[celest heath]

oh

but why not 360 why 1080

[slate lintel]

,calc 240+360

[wraith dagger]

Result:

600

[slate lintel]

,calc 600+360

[wraith dagger]

Result:

960

[slate lintel]

,calc 960+360

[wraith dagger]

Result:

1320

[celest heath]

only one that fits

[slate lintel]

yeah

[celest heath]

oh u have to check all of them

[slate lintel]

I just went right to 1080 because I know that number from skateboarding

[celest heath]

lmao

,w 2^24

[wraith dagger]

Results provided by WolframAlpha

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

Requested by 21savage_

[celest heath]

damn this cool

thank you

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[covert island]

can someone help me on part a please!

im not sure where to go with this

[brave aspen]

Lets start with T1

What does it mean to be in Kernel of T1

I.e. what is true about elements of the kernel

[cedar kiln]

@covert island Has your question been resolved?

[covert island]

sorry! i didnt see this till now. we know that the Kernel of T is the set of all zeros of the transofrmation

right

so

[brave aspen]

So then kernel elements are

Vectors from where

(For t1)

[covert island]

the kernel elements are 0 from r^n -> r^k

??

[brave aspen]

They start as elements of what space

Before being transformed

[covert island]

oh

null space

?

[brave aspen]

They are elements of Rn

The kernel is a set of vectors from the domain space

[covert island]

right

sorry

so before transformation they are all elements of Rn

[brave aspen]

So to define and describe them we start by saying that kernel elements are

v in Rn

But it's not just any v

It's only certain v that adhere to particular conditions

Now this is where you say

Well

They should be v so that

Tv = 0k

[covert island]

im sorry my brain is like stapled to kernel = null space = to 0 vector

cuz i think thats what the professor was getting at

[brave aspen]

Yes they are the same thing

Basically

To define the first

We'd say

{ v in Rn | T1v = 0k }

I.e.

Vectors from Rn such that when T1 is applied to them you get the Zero Vector in Rk

That would be the definiton of the kernel of T1

[covert island]

mhm

let me just process rq lol

so the general equation rn is T(v) = 0 vector

for k

correct?

[brave aspen]

If youre part of the kernel, then that should be true

[covert island]

right ok

[brave aspen]

So that goes on the right side of the set as the restriction

But first you have to identify what kind of objects are even entering T1

Thats the left side of the set

Its vectors v from Rn

Not just any old thing

[covert island]

mhm

[brave aspen]

{ Objects of this type | that follow these conditions }

Thats how you define sets with set notation

You can also sometimes describe it in sentences depending on your class/prof

[covert island]

i think he wants it as a sentence

so then the definition of the kernel of T1 would be T1 * V = 0vector?

[brave aspen]

The definition requires two things:

Mention where the v is coming from

Mention what should be true about the v

So first you must say that

v is a vector from Rn

[covert island]

ohhh i get it i think so then

[brave aspen]

The kernel is:

The set of all v in Rn such that T1v = 0k

[covert island]

can you explain the k please?

[brave aspen]

Thats the sentence version of the definition

Yes

When you transform

Sometimes you go to another space

In this case T1 is defined

Rn -> Rk

Members of the kernel start in Rn

[covert island]

right so

v belongs to N

going to k?

[brave aspen]

But they end up in Rk

Yea T1 is like a vehicle

[covert island]

mhm

[brave aspen]

That carries the Rn vector

To a different location in Rk

So when you say T1v = 0

That 0 is AFTER transformation

So the 0 vector is not the one from Rn

It's the one from Rk

Each space has its own 0 vector

So you specify which one you reach

[covert island]

ok cool

that makes sense

[covert island]

this should be 0 vector k

right?

[brave aspen]

The 0 vector from Rk

Which your prof provided a notation for

I actually always just wrote 0 for zero vector it was never required of me to differetiate which one

[covert island]

ah ok

so then i have ker(t1) is t1 * v = 0k where v is a vector from Rn going to rk

or

no going to rk

is it important that i include that part?

[brave aspen]

Ker t1 is a set of vectors

Its not the equation

Its the vectors that go into it

So ker t1 is

All v in Rn so that T1v = 0k

[covert island]

cool lemme just process

[brave aspen]

Follow that structure

1. Define element

2. Explain restriction

[covert island]

ok so the defining the element part is here

[covert island]

.

and the restriction part is the linear transformation from n -> k

sorry if im a little slow btw, this is really hard for me

;-;

[brave aspen]

Let me rephrase this slightly

There is a 3 tier approach

1. Identify what you are trying to define. In this case I am trying to define Kernel

Kernel is, ultimately, a set of many vectors.

2. Once you know you are defining a set, you follow this procedure:

a. Begin by identifying where elements from this set originate from.

Once you do this, begin by introducing elements of the set.

In your case, Kernel is a collection of vectors. I need to specify what kind of vectors if possible. Because I know kernel includes input vectors, I can start by giving a broad description of elements of Kernel.

• The Kernel is the set of all vectors v in Rn.....

Thats the first half of defining the set

[covert island]

right so we defined kernel to be a set of many vectors within a domain that result in the codomain -> 0

the element of T1 originates in r^n

[brave aspen]

Ok now that i know where to start we proceed...

b. I need to provide some kind of condition or restriction to clarify who is or isnt part of my set.

The restriction in this case is that they must abide by the definition of Kernel.

Therefore, these v must follow this restriction....

.... T1v = 0k

[covert island]

and we can say this

[brave aspen]

You glue the two halves together to define the set

[covert island]

ok so i actually have a question about that

for this to be true

doesnt t1 have to be a matrix

[brave aspen]

All linear transformations have a matrix representation

Basically no not exactly but technically theyre all matrices sort of

Linear transformation can be defined without a matrix in some vector spaces and then you can make a matrix that fits the same transformation

For any possible linear transformation

For that reason the topics of Linear Transformation and Range/Kernel of Transformations is very closely linked to Null Space and Column Space of a Matrix

[covert island]

ok cool that makes sense

lemme write this down really quick 🙂

[brave aspen]

The kernel any linear Transformation is linked to the null space of some equivalent matrix representation of that same Transformation

[covert island]

ok cool

so all of this makes sense now

but

what about the T1 by T2 part

because that would be the dot product of 2 matrixes right

or is it the funciton of t1 with aspects to t2

[brave aspen]

Its basically

Composition

Do T1 and then apply T2

So where do things start in this transformation

Well if T1 is first

Then they came from Rn

So T1v happens

v rides to a new destination in Rk

So now v has become u in Rk

At this point you apply T2

[covert island]

mhm

it t1 first right

or no

[brave aspen]

Yea

T1 first

[covert island]

ok cool so lemme apply it

[brave aspen]

You go from Rn -> Rk -> Rm

In that transformation composition

[covert island]

ok cool lemme try it

so im sorry

[brave aspen]

Its ok no worries

Ask away

[covert island]

wouldnt that just mean its 0vectorN -> 0vector K -> M

0 vector M

**

[brave aspen]

It follows the same structure

Where are the things starting

[covert island]

so we start in vector N

[brave aspen]

Yes

So its

Set of all.vectors in Rn

That follow what restriction?

[covert island]

is it Rn

is the restriction Rn

[brave aspen]

What should be true about things in the kernel

[covert island]

they all = 0

0 vector*

[brave aspen]

Sure they are vectors from Rn

[covert island]

mhm

[brave aspen]

But ones that are in the kernel.of a transformation

So you describe it in terms of the transformation

What should be true about

T2 ° T1(v)

[covert island]

uhhhh im a little mega lost rn

[brave aspen]

So

T2 ° T1

Means do 2 linear transformations back to back

But actuallu

That whole process is itself also a linear transformation

[covert island]

right

[brave aspen]

So if im in kernel of T2 ° T1

[covert island]

so the first transformation would be t1

but what would t1's transformation be

??

[brave aspen]

It means that

T2°T1(v) = 0

It doesnt matter what happens in between

T1 sends everybody to an intermediate vector from Rk

But the final destination is what matters for kernel

The final destination is Rm

So regardless of what T1 did, once I apply T2 you get Zero

So you answer like this

Ker(T2°T1) :=

Set of all v in Rn such that T2°T1(v) = 0m

For a composition youre describing where you start and where you end, just like a regular single transformation

All the in between does not matter for the definition i dont care what happened when T1 was applied

Only what happens after T2 finishes

[covert island]

so no matter what

t1(v) will take us to 0

[brave aspen]

Its T2°T1

Think of that thing as one piece

Another way of writing it ks

T2(T1(v))

[covert island]

right

[brave aspen]

T1(v) may be non zero

[covert island]

but

the 0 vector?

[brave aspen]

But then T2 of that is zero

[covert island]

huh

but we said the def of ker(t1) is t1*v = 0

t1 v = 0 is the transformation right

[brave aspen]

For T1 by itself

When you do T1 followed by T2

[covert island]

right

and its T2(T1(v))

[brave aspen]

This is actually like a completely different transformation

Imagine it's T3

T3 starts in Rn and ends in Rm

[covert island]

true

[brave aspen]

Yes theres stuff in between but i dont care about T1v = 0

[covert island]

wait

u mean t2 right

[brave aspen]

Only the final T3v =0

[covert island]

right

so what im geting is this

T2(T1(v)) = 0

because no matter what the in between steps are

the final step will always be 0m

?

[brave aspen]

Yesssss

It doesnt matter if we hit 0 on the way or at the end

As long as at the end the final answer in Rm is Zero

[covert island]

ok cool

i think i get that

so we have 3 subspaces

rn rk rm

[brave aspen]

Yes

[covert island]

so how do we tell which 2 subspaces belong with eachother

[brave aspen]

Although youd probably just call them vector spaces outright

[covert island]

hmmmmmm

[brave aspen]

Basically

[covert island]

ok cool

[brave aspen]

R1, R2,....

Rn

These are all different universes

They are the "parent" vector space usually

Subspaces would be things inside each of these universes

[covert island]

right

so rn rk and rm are all different universes

[brave aspen]

Yes

[covert island]

ill take that

[brave aspen]

They have different dimensions

[covert island]

true

[brave aspen]

You can create a mini R3 equivalent subspace in R4

But R3 vector has 3 components

And R4 vector has 4

They are not the same class of object

So things from R3 could never be part of R4

[covert island]

so r^n has n components

[brave aspen]

Yes exactly

[covert island]

and things from r4 could never be a part of r3

right

?

[brave aspen]

Yes

[covert island]

so we know that n k and m

could have any amount of components

[brave aspen]

Yes its irrelevant to the question

They could all be equal

We don't know which is greater or less than the others either

[covert island]

right

there is no order to any of them

[brave aspen]

The only order we get is from the way T1 and T2 are defined

You can do T2(T1(v))

But can you do T1(T2(v))?

[covert island]

uhmmmm

i mean

doesnt it matter

on the dimension of the matrix

?

wait

[covert island]

why can we do t2(t1(v)) if we dont have the dimensions

?

[brave aspen]

Where does T2 send you to

Which vector space

[covert island]

t2 sends us to r^m

[brave aspen]

Ok

So if i do T1(T2(v))

Im going to try plugging in an Rm vector into T1

Does T1 accept Rm vectors as inputs?

[covert island]

no

it accepts rn

[brave aspen]

So this is an undefined quantity

[covert island]

WAIT

its Rn and Rk

but t2 doesnt accept Rn

right

or rM

[brave aspen]

If we do the other way

T2 ( T1(v))

We get T1(v) is a vector in Rk

T2 DOES accept Rk vectors

So T2 [ T1(v) ] is defined

Thats why your problem asks you about that one and not T1°T2

I gotta run for a bit im sorry if I have created more questions than I have answered, but hopefully you did get some answers xD @covert island

[covert island]

lol its ok

so then to sum it up

oh i get it now

thanks so much!

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[wintry roost]

Find the indefinite integral

idk how to do this one

[surreal cave]

use $\int cf(x)\dd x=c\int f(x)\dd x$ and $\int x^{n}\dd x=\frac{x^{n+1}}{n+1}+C$ :)

[wraith dagger]

XxMrFancyu2xX

[wintry roost]

okay ty ill give that a shot

i'm confused so 5/(sqrt(x) | dx but then converting it to x^ndx the 5/sqrtx is throwing me off how do i know what x and n are?

@surreal cave

[surreal cave]

so using the first property $\int cf(x)\dd x=c\int f(x)\dd x$ called ``linearity" let's us move the 5 outside the integral so we get $5\int\frac{1}{\sqrt{x}}\dd x$

[wraith dagger]

XxMrFancyu2xX

[wintry roost]

ah

[wintry roost]

so 5 | x^.5 dx + c

oh

5 | x^1.5 / 1.5 + C ?

nvm got it now

5 | x^(.5) / (1/2) + C

which is 5| 2x^(.5) + C

so 10x^.5

@surreal cave is that right

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[wraith warren]

hii

[wraith warren]

i have been on this for 30 mins

half a page of scarrtch paper

i give up

if you can please show work

[slate lintel]

I would start by dividing everything by 2

[cedar kiln]

@wraith warren Has your question been resolved?

[tranquil plank]

also there's always cramer's rule

[slate lintel]

there is no need to drag Cramer into this

substitution or elimination will both work fine

[wraith warren]

hm

[crimson sedge]

you can use https://www.wolframalpha.com and i'll show the work

the step-by-step should be free as long as you keep pressing next step

[cedar kiln]

@wraith warren Has your question been resolved?

[cedar kiln]

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[fervent pike]

Im so stuck

[fervent pike]

my work:

bigger screenshots, I basically did the same analysis on f'(x) and f''(x)

please help me I'm getting no where 😦

[cedar kiln]

@fervent pike Has your question been resolved?

[cedar kiln]

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[dim tiger]

hello everyone

[dim tiger]

shouldnt it become $\sum_{n=0}^{\infty}{n+1}{c_(n+1)}{x^(n+1)}$

[wraith dagger]

calculus is fun

[crimson delta]

it helps to write out the first ~5 terms of the series explicitly

[dim tiger]

i always try to use LaTeX when asking or answering someone but i always fail in that XD

[crimson delta]

$\sum_{n=0}^\infty (n+1)c_{n+1} x^{n+1}$

[wraith dagger]

Denascite

[dim tiger]

tysm

[dim tiger]

shouldnt it be like this

[crimson delta]

no

[dim tiger]

why is that

[crimson delta]

try that

[dim tiger]

i tried first 3 terms and they are all different

[crimson delta]

what are the first few terms

[dim tiger]

for the summation on the LHS the first few terms are $c_1,2c_2 x,3c_3 x^2$..., the first few terms of the summation on the RHS are $0,c_1 x^2,2c_2 x^3.$..

[crimson delta]

yes so right hand side is exactly x^2 * left hand side

[wraith dagger]

calculus is fun

[dim tiger]

wait a sec i totally forgot the x^2 on the LHS

tysm i wasted your time by such a silly question

[crimson delta]

happens

dw

[dim tiger]

have nice day/night depending on your location

.close

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[random kelp]

for ii, is it even possible to get 3 points that satisfy the conditions

because if you take the derivative of y, you get only 2 solutions, 1 of which is (2,3)

[short blade]

L1 is not normal to C

L2 and L3 are normal to C and parallel to L1

[lunar lynx]

Yes, each solution for each line L2 and L3. What's the issue ?

[random kelp]

the issue is that I get x=0 & x=2

[short blade]

show your work

[lunar lynx]

yeah.

[lunar lynx]

This is wrong.

L2 and L3 are parallel to L1. Therefore slope of L2 and L3 will be same as ... ?

@random kelp

[random kelp]

the slope will be the same as L1

[lunar lynx]

Good.

[random kelp]

thats what ive done though

[lunar lynx]

Now, Since L2 and L3 are perpendicular to C, why are you equating them to derivative at those points.

Derivatives give slope of tangents. Not of perpendiculars.

[random kelp]

because thats how you calculate the slope at a given point

[short blade]

you don’t want the slope at those points

you want the slope of the normal at those points

[lunar lynx]

Yes. But that slope is of tangent at the point. Not of perpendiculars.

[random kelp]

they are parallel to L1

so how can they be normals to C

[junior dome]

do you know m1m2 = -1?

[random kelp]

I do..

[short blade]

what you’ve done is set y’ = m

this says

let the tangent line to y have slope m

but that’s not what you want

[lunar lynx]

Why can they not be ?

[short blade]

do you recognize that mcmeny?

[random kelp]

I understand

[lunar lynx]

,w graph |x|

[wraith dagger]

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[lunar lynx]

here tangent at x=1 is parallel to line y = x+1, which is normal to same curve at x=-0.5.

[random kelp]

i see

so I need to locate the points on the graph that are equivalent to the slope -1/3 cuz thats the perpendicular point

[lunar lynx]

Yes

[lunar lynx]

Slope at those points is -1/3 basically.

[random kelp]

so I gotta make 6x^2-12x+3=-1/3

[lunar lynx]

Yes

[random kelp]

why didnt i think of that

[cedar kiln]

@random kelp Has your question been resolved?

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[fair mortar]

$\int_0 ^{\frac{\pi}{2}} \frac{\sin 2x}{\cos^4 x + \sin^4 x} dx$

[wraith dagger]

Bettim

[fair mortar]

I should get pi\4 but idk where I did my mistake

[mighty shuttle]

the answer is pi/2

[fair mortar]

,w integrate (sin2x)/(cos^4x + sin^4x) from 0 to pi/2

[livid hound]

missing /

[wraith dagger]

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[fair mortar]

Oh okay

It's my teachers mistake then

Thank youu

.close

[cedar kiln]

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[fair mortar]

$\int_0 ^a \arcsin \sqrt{\frac{x}{a+x}} dx$

[wraith dagger]

Bettim

[mighty shuttle]

whe are theta and dx in the same expression?

[fair mortar]

I was too lazy to replace, although I didn't integrate so it doesn't make any mistak?

[mighty shuttle]

let me check, just a moment

[upbeat forge]

have you tried using polynomial substitution like t=tan => sin = 2t/1+t^2 cos = 1-t^2/1+t^2 ?

[mighty shuttle]

seems about right, though you techically have to change the limits first

[fair mortar]

I am asked to use x = atan²theta

[fair mortar]

Yes but now I'm struck

How to proceed

[mighty shuttle]

IBP

[fair mortar]

Ohhhh

Can I take the entire remaining trigs to be dv

U is thera

[upbeat forge]

I see but no one stops you to do second substitution

[fair mortar]

Right

[mighty shuttle]

yea

[fair mortar]

I have trouble finding limits

Can you help me

[mighty shuttle]

sure

[mighty shuttle]

upper limit:- $a=atan^2(\theta) $ so $\theta$ is $\frac{\pi}{4}$ unless I'm mistaken

[wraith dagger]

physicsrocks

[fair mortar]

That's what I used

,w integrate arcsin(sqrt(x/(x+a))) from 0 to a

[wraith dagger]

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[fair mortar]

Lol where answer

[mighty shuttle]

,w integrate (xtan(x)sec^2(x)) from 0 to 1

[wraith dagger]

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Upgrade to WolframAlpha Pro!

Requested by physicsrocks

[fair mortar]

,w integrate (xtan(x)sec^2(x))

[wraith dagger]

Results provided by WolframAlpha

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

Requested by bettimsucks

[fair mortar]

Wtf

[mighty shuttle]

yeah, when you IBP that's what you get $x\int(tan(x)sec^2(x)dx)-\int \frac{tan^2(x)}{2}dx$

[wraith dagger]

physicsrocks

[fair mortar]

Howw

[mighty shuttle]

wdym

[fair mortar]

I got xtan²x/2 - tanx+x

[mighty shuttle]

how

can you show your working?

you've sent two pages, send the one before that

[fair mortar]

There nothing before that

[mighty shuttle]

try again

[mighty shuttle]

,rotate

[mighty shuttle]

seems right, but I'm still unable to see how you got the $-\theta$

[wraith dagger]

physicsrocks

[fair mortar]

Tan²x is sec²x - 1 no,?

[mighty shuttle]

yeah, my bad. Let me try integrating this manually. I'll get back to you in a minute

[fair mortar]

Okayy

It's right i got it YAAAAY

Thanks 🙏

[mighty shuttle]

wdym, your original answer was right, how did you geta different answer?

[fair mortar]

My original answer wasn't right

I forgot the /2 for the second 2 terms inside the Paranthesis

[mighty shuttle]

nah, MSE says that your answer to ∫xtan(x)sec^2(x) is right

[fair mortar]

What is msw

Mse

[mighty shuttle]

Math satck exchnage

[fair mortar]

From this pic , see last step, I'm wrong

[mighty shuttle]

ah, got it.

[fair mortar]

Ye

Thanks anyway

.close

[cedar kiln]

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[tidal wasp]

.close

[cedar kiln]

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[crystal raptor]

Maybe try it another time

[dawn junco]

stop ddosing the texit bot they'll find you

[wanton sail]

No spamming

Oh this isn't even your help channel

@tidal wasp Sorry about that.

[steel seal]

can you crash the bot doing that?

[wanton sail]

Feel free to post your question in a new help channel.

[wanton sail]

No it automatically stops responding to you

[cedar kiln]

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[sand cradle]

Determine all $x \in \mathbb R$ s.t. $(x-3)(x+5)(x+9)(x+15) > 3$.

[sand cradle]

How would you approach this?

(By the way, there isn't really a pattern to the numbers here, I picked them randomly and would like to see how one would solve something like this generally)

[tropic oxide]

By the way, there isn't really a pattern to the numbers here,

there's your problem

there's very little hope that this quartic you've got factors nicely

[crimson sedge]

and I'd like to add to Ann and say that there's nothing that can be generalized here

[tropic oxide]

no, one can extract from this the general idea that kepe wants to learn how to solve polynomial inequalities.

[sand cradle]

Oh. So for $(x - a_1)(x - a_2)\dots(x-a_n) > k$, you can't generalize some procedure?

To determine the set that x can be in

[tropic oxide]

not unless $(x-a_1) \dots (x-a_n) - k$ has a nice factorization you can't.

[wraith dagger]

Ann

[sand cradle]

Alright, thanks

[crimson sedge]

if the roots are arbitrarily chosen as Kepe did, then there isn't a way to solve this generally, is what I meant

[sand cradle]

Hm, though if we have something like (x + 1)(x + 3)(x + 5) > 3, we know that the LHS -> infty for x -> infinity and to -infty for x -> -infty, so we just need to find the smallest x that satisfies the equality ... = 3 and our solution set will be (x, infty)

[tropic oxide]

no, that interval will only be a subset of our solution set

[sand cradle]

x, infty, sry

[tropic oxide]

unless you can produce proof that (x+1)(x+3)(x+5) = 3 has only one root

which i am not sure it does

[sand cradle]

oh

Thanks

.close

[cedar kiln]

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[harsh sparrow]

Hey guys I have a question relating to probability and static.
If I was to have a book with 500 pages, how would I count how many letter of "A" are inside the book?

[harsh sparrow]

I had a rough idea of how it would be such as dividing the pages from "a", doing 500 divide by 1/26, would this be correct?

[livid hound]

no

what's your reasoning for doing that

[harsh sparrow]

Cause there was 26 letters in the alphabet I thought it would be 1/26 or something

And use it to divide to 500

How would you approach this?

Or is it simple as just 500/26?

[livid hound]

no

firstly the letters aren't evenly distributed

some letters are more common than others

[harsh sparrow]

Oh okay

Gotcha

[livid hound]

as for

500/26
are you implying that on average a appears once every 26 pages?

[harsh sparrow]

Nope

[livid hound]

because that's essentially what that calculates

[harsh sparrow]

Hmmm

[livid hound]

a basic approach would be to count the number of a in one page

and multiply that by 500
and that'll give a decent estimate

[harsh sparrow]

Hmm okay

CAuse I don't want to use a sample

Since it was incorrect way of doing it according to my senior

[livid hound]

well then you'll need access to some external data

[harsh sparrow]

I was told to do what the question is requesting for

[livid hound]

about how often a is used compared to other letters

[harsh sparrow]

No clue

[livid hound]

size of the page / number of words that could fit would also be relevant

[harsh sparrow]

Would I be able to find that?

Or would it be unknown?

[livid hound]

depends on the exact wording of the question and what is expected

are you expected to reach a numerical amount or
just the approach to how to obtain a reasonable estimate

[harsh sparrow]

I think just the chances of it

So a reasonable estimate

[livid hound]

you'll still need a bit more info

[harsh sparrow]

Hmmm

OKay I see

[livid hound]

some sort of sampling would be required

[harsh sparrow]

Hmmm

Yeah it's tough

I am suppose to do it without sampling

[livid hound]

i have a cabinet with several draws filled with fruit

how would you expect to know how many oranges i have knowing nothing about the cabinet

[harsh sparrow]

HMmmm

Yeah that's good point

[livid hound]

like there's

[harsh sparrow]

But would it be accurate if I just whip out any book to do it

[harsh sparrow]

?

[livid hound]

but that's based off another sample

[harsh sparrow]

Yeah I have found this

Hmmm

This is rough

[livid hound]

some books have bigger pages than others

[harsh sparrow]

and probably more letters etc?

[livid hound]

i suppose you could estimate the number of words on a page

by multiplying words/line by number of lines

[harsh sparrow]

Yeah that makes sense

And then divide it by 8% (a)

?

[livid hound]

multiply

[harsh sparrow]

You don't divide at all?

[livid hound]

~~8% of those words have a

of is multiply

[harsh sparrow]

OHHH

Right

So let's say in those 500 pages there is roughly idk 2500 letters in each page

So it would be like 500 x 2500 x 8%?

[livid hound]

you'd want words/page

the data is % of words the letter appears in

[harsh sparrow]

Oh okay

Would you please show me how you would do it?

Was I on the right track for it or?

[livid hound]

its pretty much all been outlined

you just need to put it together and not make typos

[harsh sparrow]

Ohhh okay

Understood

Thanks so much

I got the gist of it

[harsh sparrow]

So it would be 500 x 8% = 40

[livid hound]

wdym by it

[harsh sparrow]

Oh

I think I am confusing myself now

[livid hound]

you'd want words/page
i suppose you could estimate the number of words on a page
by multiplying words/line by number of lines

500pages/book * number of words/page
will give
number of words/book

[harsh sparrow]

So it would be 500 * 2500?

[livid hound]

2500 is a bit high for number of words on a page

[harsh sparrow]

But just an example it would be something like that?

[livid hound]

if there happened to be a book that has 2500 words on a page, sure

but to my knowledge there aren't any that come close to that

[harsh sparrow]

Ohhh okay

I don't have to multiply by 8%?

[livid hound]

you will

[harsh sparrow]

Oh okay cool

I will be on the right track now right?

[livid hound]

but currently the focus is on having a more realistic number for words on a page

[harsh sparrow]

Oh okay

Gotcha

[livid hound]

i suppose you could estimate the number of words on a page
by multiplying words/line by number of lines

[harsh sparrow]

Thank you so much!

I know what I can do to get an rough answer for this

.close

[cedar kiln]

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[drowsy hare]

does anyone know how this question would acutally work?

[drowsy hare]

I have gotten G(w) and I know b is pi

so i have subbed in G(w-b) but do i have to multiply it out after this?'

[cedar kiln]

@drowsy hare Has your question been resolved?

[cedar kiln]

@drowsy hare Has your question been resolved?

[flint plinth]

have you tried using $\sin(t) = \frac{e^{it} - e^{-it}}{2i}$?

[wraith dagger]

Bungo

[drowsy hare]

yeah i have but I can't tell if I am making a stupid mistake as the calcualtion is so long tho

[flint plinth]

are you able to paste the work you've done? screenshot or photo or whatever? i can take a look

(although i have to leave for an appointment in 7 minutes, so it might have to wait until I get back)

[drowsy hare]

yeah sure but its long hahaha

This is everything with all the things I have tried along the way

[flint plinth]

ah yea, this is more than i have time to look at right now, but i'll check when i get back, and if you don't already have an answer i'll take a look then

[drowsy hare]

yeah don't worry, its been hard so itll probably still be here later

[cedar kiln]

@drowsy hare Has your question been resolved?

[cedar kiln]

@drowsy hare Has your question been resolved?

[cedar kiln]

@drowsy hare Has your question been resolved?

[cedar kiln]

@drowsy hare Has your question been resolved?

[drowsy hare]

.close

[cedar kiln]

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[subtle horizon]

To solve this

I first did trial and error

and tested values of n => 0 and <0

and got 0 1 -1 and -2

when i added them

i got -2

but the answer is 12

however

if we look at the absolute value part

you might think that you need to kind the modulus of each integer then add

but the last part of the questions says nothing about modulus

How should I apply the absolute value thing to solve the question?

[spice kraken]

just solve the inequality

[south tundra]

First pretend like you are solving for |N - 3| in the inequality rather than for N

Meaning you get it to the form a > |N - 3| > b for some numbers a and b

Then you simply consider the cases N >= 3 or N < 3

[subtle horizon]

whhy

[south tundra]

Well if you had an equation like e^(2x^2 + 8) - 1 = 0, you would start with isolating e^(x^2) because you then can use logarithms, similar thing here

[south tundra]

(Hint: ||As the first step I would use the property x1 > x2 > x3 -> 1/x1 < 1/x2 < 1/x3 for positive x1, x2 and x3||)

[subtle horizon]

What

[south tundra]

15 > 30/|N - 3| > 6
So
1/15 < |N - 3|/30 < 1/6

Now you multiply each side by 30 and get 2 < |N - 3| < 5

Let us now consider the cases

The first case being N >= 3

What's |N - 3| equal to in that scenario?

[subtle horizon]

Why not N >= 0

[upper ruin]

What's the definition of |x|?

[subtle horizon]

x

[upper ruin]

Nooo, otherwise we wouldn't waste time to write the bars 😅

|x| = x if x ≥ 0, it is - x if x < 0

[subtle horizon]

Pleasee what do i do with this information

[south tundra]

So |N - 3| = N - 3 if N - 3 >= 0

The latter can be rewritten as N >= 3

That's why we are considering that case

[subtle horizon]

ok

[subtle horizon]

Whata next

[upper ruin]

Consider the two cases, N > 3 and N < 3

[topaz shore]

Can someone help with this one

[upper ruin]

And solve the inequalities given

[south tundra]

If N >= 3, this becomes 2 < N - 3 < 5 and 5 < N < 8

[upper ruin]

Post it on another available room

[topaz shore]

sorry

[upper ruin]

Don't worry

[south tundra]

Same thing for the other case, try considering N < 3 on your own now

[subtle horizon]

=0

[south tundra]

1. |N - 3| >= 0 for all N anyway
2. That's not an equation

[upper ruin]

It's not the answer. However by definition abs values are ≥ 0 always

[subtle horizon]

ok

[south tundra]

If N - 3 >= 0, |N - 3| = ?

[subtle horizon]

3

[upper ruin]

I hope you are able to prove it, directly from the definition

[south tundra]

Why do you think it's 3?

[subtle horizon]

n-3=0

n=3

[upper ruin]

No! Re-read the definition I gave you (and replace x with N-3)

[south tundra]

I never said N - 3 = 0

[subtle horizon]

n-3 ≥ 0

n ≥ 3

[upper ruin]

Yes, so what's the answer to this question?

[subtle horizon]

why is |n-3| =n-3

as n-3 >= 0

[south tundra]

The definition says that if x >= 0, then |x| = x, right?

[subtle horizon]

yes

[south tundra]

Now try writing this exact sentence but replace x with N - 3

[subtle horizon]

why are we finding the modulus of n-3

[south tundra]

Because we have modulus of n - 3 in our problem and we need to get rid of it

[subtle horizon]

ok

So we need to change it to a variba

variable

[south tundra]

You can, but it is not really necessary

[subtle horizon]

|n-3 |= n-3

[south tundra]

Anyway, if you were to use this rule on N - 3, you would get
"If N - 3 >= 0, then |N - 3| = N - 3", right?

[south tundra]

Yes

[subtle horizon]

Ok

Sor

[upper ruin]

Always?

[subtle horizon]

if N-3 >=0

S0

n must be more than n>=3

[south tundra]

So, if N - 3 >= 0, this inequality becomes 2 < N - 3 < 5

[south tundra]

That's the first case that we are considering, yes

[south tundra]

And from here you just add 3 to each side and get 5 < N < 8

Okay, so the solution in the first case is (5, 8)

[subtle horizon]

Why

[south tundra]

If N - 3 >= 0, then |N - 3| is the same as N - 3. Simiarly, the inequality 2 < |N - 3| < 5 is then the same as 2 < N - 3 < 5

[subtle horizon]

ok

so

5< n<8

[south tundra]

Yes, that's the solution to the first case. Now we need to move onto the second case. Namely, we need to consider N - 3 < 0

What's |N - 3| equal to if N - 3 < 0?

[subtle horizon]

that is -(n-3)

-n+3

[south tundra]

Correct

[subtle horizon]

1 < N <- 2

[south tundra]

Yes

Let's now look at the integer solutions

What integer solutions does 5 < N < 8 have?

[subtle horizon]

?

What

[south tundra]

Can you think of any integers that are placed between 5 and 8?

[subtle horizon]

6 7

[south tundra]

Right

And what are the integer solutions to 1 > N > -2 then?

[subtle horizon]

0 -1

[south tundra]

Yup

So, what do we get when we sum up 6, 7, 0 and -1?

[subtle horizon]

12

Wait

So first

Posting for reference

So we look at the inequality

And because we are looking for n-3 without the modulus sign

we need to apply the rules

that | n-3 | = n-3

When n-3 is more than zero

[south tundra]

Yes

[subtle horizon]

Then we imagine situation when

n-3 is positive

and remove modulus signs

then solve

we get the range

then we do that for when it is less than zero

then get range

We then get the integer values of n and sum them

[south tundra]

That is right

[subtle horizon]

But can you please expllain why they began the question like that

[south tundra]

Like what exactly?

[subtle horizon]

no worry

.close

[cedar kiln]

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[late finch]

Hello- I've been stuck on this for quite some time.

this is the full problem

possibly relevant defn

[spice kraken]

!status

[cedar kiln]

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

[late finch]

1

I mean

I don't know where to begin

I was thinking of opening it up and doing any many integrals

but that sounds like a headache >_>

[spice kraken]

you finished part a right?

[late finch]

oh yeah yes

by counter example though

[cedar kiln]

@late finch Has your question been resolved?

[spice kraken]

counter example?

the generalization should be true though

[late finch]

The generalization is an inequality

Direct equality is not possible

[spice kraken]

do you mind showing me your proof?

we use the equality in part a to show the inequality in part b

[cedar kiln]

@late finch Has your question been resolved?

[late finch]

(X,Y)->(e^x cosy, e^x sin y)

Jacobian determinant is always non zero therefore it's invertible or kernel is just the zero vector

But the map is periodic

So LHS 0 RHS non zero

Possible

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[late finch]

.reopen

[cedar kiln]

✅

[late finch]

There might not be a neat solution

[cedar kiln]

@late finch Has your question been resolved?

[cedar kiln]

@late finch Has your question been resolved?

[dire geode]

So?

That's the first thing you should try

This problem is just an application of the definition of the norm

[late finch]

what do you mean?

[dire geode]

Do this

[snow lily]

Question please

[late finch]

part b

[late finch]

So I want to confirm

[dire geode]

?

[late finch]

I'm doing
|| f(x) - f(y) || = || integral ( Jacobian . (b-a)dt) t || being parameter for line

then moving the norm inside and taking inequalities

is that right 0-o?

[late finch]

.solved

.close

[cedar kiln]

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[round raven]

hey there, i actually posted the question before but then i had to go so i couldnt get an answer. I am solving an old exam to prepare for my exam soon and there is a question for a multivariable function f(x,y) = (x^2)y + ((y+1)^2)(y+2). The sub-task I couldnt figure out is the part where it asks for me to find all singular points on the curve 0 = f(x,y) and classify them with the help of the hessian matrix as double point, cusp and hermit. I of course did some research on it but i couldnt find something and the script my prof provided doesnt explain it well. Thanks in advance.

[snow lily]

first singular point are defined with Df(x,y)=0 so df/dx=0 and df/y=0

[round raven]

oh i already know that but thanks, i probably should've been more clear. I am more confused about the classifying part

[snow lily]

for Hessian Matrix you calculate 3 terms d^2(f)/dx^2 (you already calculated df/dx), d^2(f)/dy^2 (you already calculated df/dy) and df/dxdy than you replace the 3 expressions with the coordinates of singular points and you write the matrix : 1st line (d^2(f)/dx^2 df/dxdy)

2nd line ( df/dxdy d^2(f)/dy^2 )

This is so helpful in optimization problems because you can use this matrix to find the convexity of f in this certain point

[round raven]

i see, after that how do i know if a singular point is one of the three categories?

[snow lily]

lets consider H=1stline(r s) 2nd line(s t)

you calculate the determinant D

if D<0 we can't conclude

if D>0 we test r

r>0 so f is convex

if r<0 f is concave

[round raven]

oh i thought this method was being used to find local min/max or saddle points

i mean of course you can also figure out if its convex or concave but i didnt think we'd use it for the classification of singular points

[snow lily]

if it is convex or concave in a point so it's a local min or max

[round raven]

my english is advanced but still not my first language and im not sure what hermit means in this context, I translated it from the file which is in german, which is my 2nd foreign language so could you shortly explain what that is exactly. Idk why doesnt any dictionary provide something for this context.

like is it an isolated-point or something idk

.close

[cedar kiln]

Channel closed

Closed by @round raven

Use .reopen if this was a mistake.

[cedar kiln]

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[boreal trail]

why is it 0.52 - 0.24

[boreal trail]

and not 5*10^8 - the other

.close

[cedar kiln]

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Use .reopen if this was a mistake.

[boreal trail]

nvm

[cedar kiln]

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[potent fiber]

is the 1 cm a third of the whole length or not?

[lunar lynx]

Why should 1cm be a third of whole length?

[potent fiber]

does octagon work like this?

[leaden otter]

Not really

[lunar lynx]

Nope

[potent fiber]

uhh

then where should i start?

[leaden otter]

Do you see the trapeziums

What are the lengths of it's parallel sides?

[potent fiber]

uhh

the smaller side is 1?

[kindred storm]

You can slightly alter that. The middle part is fine if you want to say the sides of the octagon are length 1, but then you have, with the triangle in the upper left, a hypotenuse that's the same width as the top side of the octagon.

[potent fiber]

i think

[leaden otter]

Hmm, not really

[kindred storm]

Then the two legs of the upper left triangle are the same length because of symmetry.

[potent fiber]

legs?

[kindred storm]

Yes, do you see the upper left triangle?

[potent fiber]

these?

[kindred storm]

Well, that's the upper right, but yes.

Those are the same size because a regular octagon is symmetrical.

So, use a variable for the length of one of the legs.

Then write it next to both legs.

[potent fiber]

then it tells you its 1

[kindred storm]

Then write 1 next to the hypotenuse.

[leaden otter]

This might help, same colour lengths are equal

[kindred storm]

No, the legs aren't 1. The hypotenuse, one of the sides of the octagon, is 1.

[slate lintel]

what

the strip is 1cm wide

[potent fiber]

should it be like this?

[kindred storm]

Ahh, yes, that's right.

And then the other leg is 1.