help-28

[stuck island]

nah

[astral ermine]

Is there some else method you're looking for?

[willow sedge]

you can factor

[stuck island]

yea factoring

but how would u do that

[willow sedge]

$2x^{2}\left(2x-1\right)+2\left(2x-1\right)=0$

[glossy valve]

B-eard

[stuck island]

why would u factor them seperatly

not as a whole

[willow sedge]

$\left(2x^{2}+2\right)\left(2x-1\right)$

[glossy valve]

B-eard

[astral ermine]

You do get answer in the next step

That's why

[stuck island]

?

[willow sedge]

$x=±i,\frac{1}{2}$

[stuck island]

wth

?

?

?

[astral ermine]

He's writing the solutions

[stuck island]

wait

idk how he factored that

[willow sedge]

bro calm down

[willow sedge]

by taking 2x-1 as common

[stuck island]

but

u seperated them

into two factors

[glossy valve]

B-eard

[willow sedge]

yeah so as to factor

[stuck island]

how do u get 1/2

[willow sedge]

since 2x-1=0

[stuck island]

oh nvm

i got it

but how do u get i

[astral ermine]

Via second factor

[willow sedge]

since 2x^2+2=0

[stuck island]

oh

[willow sedge]

divide by 2

[astral ermine]

2x²+2=0

[stuck island]

ok

[willow sedge]

so x^2+1=0

[stuck island]

can u give me another equation

and see if i can factor it

[willow sedge]

for practice?

[stuck island]

yea

wait

how did u know 2x-1 was a common

[willow sedge]

$15x^{3}-90x^{2}+16x-96=0$

[astral ermine]

He looked out for something common

[glossy valve]

B-eard

[willow sedge]

practice

[stuck island]

no

but

i remember there was a way to tell it

nvm i got it

lol

[willow sedge]

find hcf of two pairs separately (long process)

[stuck island]

ok ima factor that

15x^2(x-6)+16(x-6)

(15x^2+16)(x-6)=0

[willow sedge]

yes

[stuck island]

x=6

and

• or - sqrt 16/15

right?

[astral ermine]

+- √-(16/15)

[stuck island]

yea

im doing it in my head lol

thanks

for the help

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[worn flare]

integrate x*e^(x^3)

[worn flare]

so let's say we have a substitution of u=e^(x^3)

then what we can do is this

also whoever reads this i just need you to check to see if i'm doing it right

[twilit leaf]

This is not a u sub problem

[worn flare]

it si

is

[viral jasper]

there is no analytic solution

[worn flare]

then integrate by parts

well

my teachers have given me this question to do as a "hard" question

so idk

anyways

[viral jasper]

don't know what to tell you

There is no analytic solution

[worn flare]

lemme just show you my working out first

then you can determine if it's wrong

[twilit leaf]

Its literally impossible with just u sub

[worn flare]

and you integrate by parts

it's not

i don't think

[twilit leaf]

Yeah SWR is right

[viral jasper]

Unless you use the incomplete gamma function, you can't solve it

[twilit leaf]

I love when teachers give "hard" or "challenge" problems that are outside the scope of capability

So encouraging of them

[worn flare]

i have an answer so idk what to tell you

it was from the past exam that i just did

[twilit leaf]

Does your answer use the incomplete gamma function?

[worn flare]

no

[twilit leaf]

Ill bite

!show

[full forum]

Show your work, and if possible, explain where you are stuck.

[worn flare]

gimme a second then

[viral jasper]

⏱️

(jk)

[worn flare]

i'm re writing so you will understand

[twilit leaf]

Is it in a different language?

[worn flare]

no

messy

[twilit leaf]

Ah ok

[worn flare]

okay like

basically

my friend did a maths exam and asked me about this question he said he was sure that the question was what i just said

now as i'm doing it

it's seeming not likely

unless

anyone knows how to integrate

1/3ln(x)^-1/3

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[viral jasper]

Nope

Sorry it didn't work out for you, @worn flare

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[spark sleet]

(1, 1) is a generator of z2 x z3 because (1, 1) generates all the elements of z2 x z3. Is (1, 2) a generator of z2 x z3?

[spark sleet]

It generates all the elements and has the same order, but when i search for the generators of z2 x z3 only (1, 1) comes up. Is there some criterion i'm missing for the classification of a generator?

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[feral arrow]

How can we prove,

[feral arrow]

$\forall A \in \mathbb{R}^n,;det(A \otimes k)=k^n$

[glossy valve]

Awesome_Ruler_007

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[mossy anchor]

can you give us more context pls?

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[dim compass]

If determinant of a square matrix is non zero, then must it have a inverse?

[dim compass]

Then why is inverse of identity matrix not there..?

[spice orchid]

Not where?

[dim compass]

Like for inverse, A*A^(-1)=I, if I substitute A by I, then there doesn't seem to be inverse of I

[spice orchid]

It is its own inverse

[dim compass]

Ohh

Ty

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[sullen crown]

which way am i supposed to do this?(or is both ways wrong xd) (Q3)

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@sullen crown Has your question been resolved?

[sullen crown]

[terse karma]

What have you done in both ways

None is right

[sullen crown]

oh

[terse karma]

Editing the first one a bit will give you the right answer

[sullen crown]

do i single out y’ ?

[terse karma]

That's not the problem

e^(xy) . y' is

[sullen crown]

supposed to be e^(xy) . (xy’ +y)?

[terse karma]

Good

[sullen crown]

what do i do after that?

[terse karma]

And -e will be gone

Just put x=1 and y=1

And then get y'(1)

[sullen crown]

why

oh wait nvm i got it

is the final answer y’= -2/(1+e)

@terse karma

[terse karma]

No idea

I'm no expert

[sullen crown]

XD

[terse karma]

XD

[sullen crown]

well thanks anyway

i’ll check with my teacher tmr

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[exotic garden]

Use synthetic division to divide a(x) = 4x^4+5x^2+7x+2 by b(x) = 16x^3+10x+7

[exotic garden]

The actual quotient is x/4

But I am getting 4x

Please point my mistakes 🙏

[rapid rain]

you're putting 4x*(16x^3) in 4x^4, when there's only room for x/4[16(x^3)]

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@exotic garden Has your question been resolved?

[exotic garden]

I understand

But how could I express it in terms of synthetic division

At what step can I derive the quotient to be x/4

I am doing synthetic division for the first time

Sorry if my query is dumb

[rapid rain]

there are no dumb questions

so, we want to get rid of the highest degree term

we want to get rid of 4x^4

and we have 16x^3 + (...)

how many times can I put 16x^3 into 4x^4 ?

(this is like regular division)

[exotic garden]

Can you do it in paper if possible?

[rapid rain]

sorry 😭

[exotic garden]

So my leading number in synthetic division should be 1/4 ?

[rapid rain]

yes!

which you multiply by x to get the x^4

[exotic garden]

Bro till now I thought we should just copy the leading coefficient of the dividend polynomial 😭

Got it

Thx for da help boss

[rapid rain]

no problem

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[modest sail]

Hello

[modest sail]

When converting polar coordinates

To rectangular coordinates

Actually

The other way around

When converting rectangular coordinates

To polar coordinates

[keen spruce]

yes

[modest sail]

Do you have to

Write both

Positive

And negative

Coordinates

Is it required?

For example

[keen spruce]

um i dont think so

[modest sail]

-4 vs +4

But different angles

[keen spruce]

as long it lands on the same pointy

[modest sail]

(-4, 30deg)(4, 210deg)

Like this

[kind jay]

why do u need to send a message for each word xd

[modest sail]

Can you only write one

Instead of those two

[keen spruce]

they're grinding for the active role

[kind jay]

fr

[keen spruce]

it depends on the situation

[spring heart]

something's missing

Angle should be the same

[keen spruce]

like if your coordinate gives that position and its correct then you're fine

but if you need both

[modest sail]

Wait what

?

[keen spruce]

you're asking

[modest sail]

Why

[modest sail]

So

[spring heart]

That's what I remember

[modest sail]

Unless stated to pwrite both

You can only write one?

[keen spruce]

as long as its the same point

it depends on the situation

do you need one coordinate

or does the problem you're asking for require multiple

i dont see why you would need more than one though

[modest sail]

It does not say that

It just gives me

Rentacgular coordinates

And asks me to

Convert

To

[keen spruce]

okay

then

convert

to polar

[modest sail]

Polar coordinates

[keen spruce]

coordiantes

and then

find one correct answer

and you

should be

good

[modest sail]

Ok got it👍

Thank you

It might be a good practice to write both tho

[keen spruce]

eh

[modest sail]

Since in some problems it might help you

[keen spruce]

just make sure that you know the +r -r

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[brazen saddle]

How would I go about solving this equation?

[rapid rain]

Ok, so first of all, I suggest getting everything clearer so we can see what we're facing

[brazen saddle]

Actually let me take a step back first and explain what I'm trying to achieve in the first place

because there might be an easier way

I am trying to find t for a given area A under this parabola

so I tried integrating the function and got this:

[brazen saddle]

then I tried substituting t for v, and that's how I got this

[rapid rain]

And you want t = t(A) for t between the zeros of the function v

[brazen saddle]

Eventually I want v, but that can be calculated with t

Of the two values t I want the lower one

[rapid rain]

You want v(V) ?

[brazen saddle]

I'm not sure what you mean by that

[rapid rain]

You want the value of v in function of the value of the area, A?

[brazen saddle]

Yes

I have a given area A and I want to know what v corresponds to that

I think I might be able to solve it by substituting t^2 by u and using the quadratic formula, let me try that

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@brazen saddle Has your question been resolved?

[brazen saddle]

Is this correct? When calculating 1/4Cb^2+4Cax I get a negative number under the first square root

[rapid rain]

This is not correct, you have u = t², so t³ = u^(3/2) and not u²

You will need to solve a cubic equation Ca/3×t³ + Cb/2×t² - x = 0

[brazen saddle]

ah yes

[brazen saddle]

Honestly I'm not sure how to do that. I've read I could try factoring it like (t+p)(at^2+bt-x)=0 but I am not sure how I would find p.

Wikipedia gave me this formula to calculate the discriminant:

Would I then be able to solve it like usual with this formula?

[rapid rain]

Don't worry, these types of equations only ask you to find one solution

First step is reducing this equation to a cubic equation with no term in t²

To do this, we write t³ + 3Cb/(2Ca)t² - 3x/Ca = 0

Let's now take y = t + Cb/(2Ca)

Notice that y³ = t³ + 3Cb/(2Ca)t² + 3Cb²/(4Ca²)t + Cb³/(8Ca³)

We'll rewrite y³ = t³ + 3Cb/(2Ca)t² + 3Cb²/(4Ca²)(y-Cb/(2Ca)) + Cb³/(8Ca³)

So we can get y³ - 3Cb²/(4Ca²)y + Cb³/(4Ca³) - 3x/Ca = 0

Now, let's name p = -3Cb²/(4Ca²) and q = 3x/Ca - Cb³/(4Ca³)

We now have y³ + py = q

We now apply Cardan's formula

To get y = this monstrosity

To finally get t = y - Cb/(2Ca) = ...

@brazen saddle I'm sure there was a better way but here we are

[brazen saddle]

I'm not sure how you go to this from what you had before

[rapid rain]

We use both this @brazen saddle

[rapid rain]

And this

So we use the fact that t³ + 3Cb/(2Ca)t² = 3x/Ca

[rapid rain]

And we replace it in this equation

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[spring shard]

Can someone please help me with this one?

the points A(2,4) B(t,1) and C(-2,s) makes a triangle. Use vectors to decide the values of s and t so that AB vector is twice as long as AC vector and angle BAC is 60degrees

[safe trench]

It would be helpful to draw the vectors

[spring shard]

https://gyazo.com/0cf74b3e488671f6723a419cb36715fc

this is what I did

is this wrong?

[spring shard]

.

[safe trench]

Sorry I didn't get what you did in the third part. Could you explain it?

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[spring shard]

I used the scalar sentence

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[loud hull]

,w derivative e^{x}+x^{3} -xe^{x}

[glossy valve]

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[loud hull]

[e^{x}+x^{3} -xe^{x}]
[e^{x} + 3x^2 - xe^x]

[glossy valve]

dopediscorduser

[loud hull]

Where am I going wrong?

[glacial pasture]

you need to use the product rule on (xe^x)

[loud hull]

Ah

Right, thank you

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✅

[loud hull]

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[earnest knot]

I am unsure how I got this wrong. Do i not take the derivative of the given equation, then for the part (b) find B'(1)?

[silk flame]

How did you get that derivative

[earnest knot]

just did .25 times (2pi/4.6) and put it in front of cos

d/dx of 4.6 is 0

[silk flame]

,w 0.25 * 2pi/4.6

[glossy valve]

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[earnest knot]

🤔

[silk flame]

Calculator mishap?

[earnest knot]

maybe i did a misinput

bad news though that's still wrong

[silk flame]

It probably wants it in terms of pi

[earnest knot]

okay one sec i'll try that

[silk flame]

,w 0.25 * 2/4.6

[glossy valve]

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[silk flame]

Dumb thing I want it as a fraction…

5/46

[earnest knot]

yeah it does

wish it would've told me

we good

appreciate it

[silk flame]

👌

[light sonnet]

,w 0.25 * 2/4.6 as fraction

[glossy valve]

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[earnest knot]

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[light sonnet]

More words but it works

[silk flame]

Thanks

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[limpid ocean]

Wouldnt one of the crit points also be 0

[limpid ocean]

because u factor out 2x^-3 to get x^4 - 1

[light sonnet]

Why do you think 0 is one?

[limpid ocean]

because wouldnt it be like 2x^-3(x^4 -1)

And i set 2x^-3 to 0

[torn jolt]

evaluate the original function at 0

[limpid ocean]

O

restriction?

[torn jolt]

it's undefined

[limpid ocean]

icic

ty

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[nova island]

is this an acceptable proof?

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[limpid ocean]

Can someone tell me what happens in the second step of finding the first derivative here?

[limpid ocean]

Where did they get -2x^2

nvm i got it

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[hollow moth]

in order to find LCM using prime factorization, you take the product of the uncommon factors and the greatest power for each factor

[hollow moth]

so for example

lcm (2^100, 5^4*2^4)

that would be 2^100 * 5^4, right?

epiccccccccc

thank you

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[dim compass]

If |A|^4= I then will |A|=+-I or only |A|=I?

[dim compass]

I is identity matrix of same order as A

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[cyan vigil]

I don’t understand why it isn’t 4,000,000(0.06)^t

[devout valley]

Growing at 6%, so you'd be 106% of that population at the end of the year

[cyan vigil]

Why 106%

[cloud moon]

mia

[cyan vigil]

Lance

[cloud moon]

the formula you typed

gives 6% of the total population

after you find it you need to add it on top

[cyan vigil]

Oh

[devout valley]

you are 6% greater than you were last year

[cloud moon]

cause population grows with 6% of the value it originally started

[cyan vigil]

I’m dying percentages confuse me badly

Thank you

[cloud moon]

so a way to keep 4M and add it's 0.06

wpuld be

4M(1+0.06)^t

or as charbit immediately answered 106%

@cyan vigil use .close

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[elder cape]

Quick calculus question.

[full forum]

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.

[elder cape]

This is my question:

<@&286206848099549185>

[spark terrace]

So let's start iwth the first case.

[elder cape]

ok

[spark terrace]

If f(0)=0, and f(5)=16, then on average, f'(x) will equal 16/5.

It's just the slope of the line

But since f'(x) is always greater than 5 on [0,5] that's impossible

[elder cape]

hmm

So it's none of the above?

seems wierd

[spark terrace]

Well, the fiest option is wrong

You have to continue with the rest of the answers

[elder cape]

Well u can just see in 2 seconds that the rest r even worse

If f(0)=0, and f(5)=7, then on average, f'(x) will equal 7/5.

If f(0)=0, and f(6)=-5, then on average, f'(x) will equal -5/6.

If f(0)=0, and f(10)=37, then on average, f'(x) will equal 37/10.

They all fail?

[spark terrace]

Give me a second

Yeah

Yeah that's all right.

[elder cape]

hmm

what an odd question lmao

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[devout hamlet]

Me next

[devout hamlet]

I know this is simple but I'm dumb

Where does the (⁴√2+1)(⁴√4+1) come from? <@&286206848099549185>

[spark terrace]

It's just a factorization, root2 * root 4 = root 8, root4 * 1 = root 4, root2 * 1 = root 2, and 1*1 = 1

[fluid monolith]

If you multiply the brackets, you will get the same ans

[devout hamlet]

Yes but how are you uhh

Ong my English is not good sorry

Wait

How do u turn them into that

[fluid monolith]

⁴√8 = ⁴√2 * ⁴√4

[devout hamlet]

And then

[fluid monolith]

And then you take common

Sry if my English is bad

[devout hamlet]

Dw my English too 😔

[fluid monolith]

After you take common you get the brackets

[devout hamlet]

Like how do turn the whole thing. Not just ⁴√8

Eh nvm

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[wind needle]

hey

[wind needle]

need help with the following question

trying to prove or disprove

[torpid stag]

For $d = 1$, this would be $\floor{n + 1} = \ceil{n}$. Can you find any integer $n$ where this is true?

[glossy valve]

Mikkel

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[wind needle]

Thank you

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[bronze magnet]

Is there a good website or image that tells me everything relating to things being equal?

For example x^1/2 = sqrt. (?)

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[bronze magnet]

.reopen

Wow

Rude

[onyx glen]

you deleted your original message, this means you have to abandon this channel and claim a new one.

you cannot reopen channels closed this way.

[bronze magnet]

I see

Ty

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[tame rune]

In a statistical analysis concerning the adoption of minors abroad, the wives of the married couples who applied were divided by age group. The corresponding histogram of the frequencies, referred to a certain year is given in the figure. The examination of this histogram shows that the average age of the wife is between:

[full forum]

@tame rune Has your question been resolved?

[clear abyss]

i dont understand mapping and function

[wispy sage]

I think you're supposed to take a guess at the value of each data point, multiply that by the average age in that age group (assume the lower is 27-30 maybe and upper is 50-55 ish). Then add these up and divide by the sum of the data points. The answer will then be whichever group this number is in.

[wispy sage]

I think they are almost the same. They both take an input and give an output. The only difference is that a function takes in numbers and outputs numbers, but a mapping can take in anything and output anything

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[wispy sage]

for example a function could be f(x) = 2x + 5, but a mapping could take in a shape and output the shape rotated 90 degrees

[clear abyss]

pls explain futher

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[dark escarp]

Why it isn't solved for limit =1

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@dark escarp Has your question been resolved?

[wispy sage]

Here is an example of a mapping of shapes to colours. A map assigns every object in the left group to an object in the right group. The objects in the groups can be absolutely anything. A function is a special case of this where the groups have to be numbers

[dark escarp]

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[edgy owl]

im so confused why this is the uniqueness theorem

[glossy valve]

lball

[topaz valley]

the coefficients are unique because the theorem literally tells you the formula for them in terms of the function

you can't exactly get two different sets of coefficients if you calculate them from the same formula

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[modern wagon]

Help with number 2 please

[hallow walrus]

first get them to the same base

then you can compare the exponents

[modern wagon]

How do I do dat

[hallow walrus]

remember that $$(a^b)^c=a^{b\cdot c}$$

[glossy valve]

kheerii

[modern wagon]

Is x c?

[thin flint]

what is 81?

[modern wagon]

Huh

[torn jolt]

by changing the base of 81 to 9

[thin flint]

81 = 9^2

[torn jolt]

^ and then use this formula

then compare the exponents

[modern wagon]

This?

[kind jay]

no

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[slow sleet]

am i right with B here

[vital ravine]

what is that

what shape is this

[slow sleet]

cylinder

[vital ravine]

but look at the bottom

[modern wagon]

.close

[vital ravine]

https://tenor.com/view/scream-gif-24397066

NOT YOUR CHANNEL

[slow sleet]

its always my channels bro

[modern wagon]

It was my channel at first it got timed out I think because Noone responded to me

[vital ravine]

FINDERS KEEPERS LOSERS SWEEPERS

[pulsar gale]

chill bro

[vital ravine]

sorry i wanted to say that

it was kinda funny

cuz

[slow sleet]

kefir getting mad yall better chill

[modern wagon]

I tried to close it but ok💀

[vital ravine]

lol

[pulsar gale]

im not mad just a bit annoyed

keep it civil

[slow sleet]

oh yea?

oh fasho

[vital ravine]

K

get it cuz your pfp is K

[slow sleet]

am i right with B

[vital ravine]

im still confused on what im looking at tho

[quaint prawn]

Why open a new channel for the same question you posted in the old channel?

[slow sleet]

cuz it got lost in yo nonsense

[teal aurora]

its a cylinder from top to 8.5ft and then a cone after that

ngl its drawn really weirdly

[vital ravine]

ohhhhh

now i see it

[slender onyx]

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[hidden prism]

A vehicle covers a distance of 48.3 km in 2.3 litre of
petrol. How much distance will it cover in 1 litre of
petrol?

[pulsar gale]

use fractions

or ratios

[hidden prism]

kk

😭🫶🏾

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[hidden prism]

yeh.

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[full forum]

Channel closed

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[viral jasper]

Open a new channel

[white flicker]

oh ok

[viral jasper]

If you delete the original message, the channel will close

[cloud shore]

.reopen

[viral jasper]

Doesn't work if original message is deleted

[cloud shore]

I can see that

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[sour peak]

The length of a child's square-based jewellery box is 5cm more than its height. The box has a capacity of 500cm^3. Determine the dimensions of the box. *I have tried this question so many ways and still cannot get he anser on the sheet [ 5cm by 10cm by 10cm] I dont know how I am supposed to format the question.

[brittle steeple]

.reopen

ah right maybe just repost in a new channel

[sour peak]

oh okay one sec

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[clear abyss]

thanks

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[glacial pasture]

they are orthogonal, is the software saying no?

your vector is fine, maybe try putting a * between 15 and i, it looks like you may have accidentally put a space between them if not

,w calc (4+3i)(30+15i)+25(-3-6i)

[glossy valve]

Results provided by WolframAlpha

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

Requested by AℤØ#4979

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[torn jolt]

I need help with a number theory problem.

[torn jolt]

How would I go about solving:

/tex

/close

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[muted pewter]

Can someone please help me in this question? (The picture includes all my approaches and the question itself)
Link: https://math.stackexchange.com/questions/4670117/to-find-the-sum-of-all-possible-perimeters-of-a-triangle

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[tame rune]

is the ans option E ?

[earnest perch]

yes i think so

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[tame rune]

you sure @earnest perch

?

[tame rune]

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[wanton heath]

hi

[wanton heath]

how do I estimate this error

no 2

<@&286206848099549185>

[delicate hornet]

you can write out the series as a finite number of terms plus a "remainder" integral

[wanton heath]

i tried : (

[delicate hornet]

hold on i will try to write it out

[wanton heath]

alr tysm

[delicate hornet]

im new here is there a LaTeX bot?

[wanton heath]

yea

there is

$5x+1$

[glossy valve]

🐬Ryan🐬

[delicate hornet]

cool ty

[wanton heath]

like that

did u figure it out?

[delicate hornet]

typing rn

[wanton heath]

ah ok

[delicate hornet]

$f\left(x\right)=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{1}{4!}\int_{0}^{x}t^{4}e^{\left(x-t\right)}dt$

[glossy valve]

ElectroSpeeder

[delicate hornet]

this monster

[wanton heath]

wait a second

thats not right

thats e^x

[delicate hornet]

o shite

[wanton heath]

its fine

[delicate hornet]

i read it wrong

[wanton heath]

i figured that part out

how do I do legrange for the question tho

[delicate hornet]

my bad sorry man

[wanton heath]

all g

its just 1-x+x^2/2!+x^3/3!-x^4/4!+x^5/5!

i just dont know how to get the answer

do u know what to do?

[delicate hornet]

i think the same method works

one second

[wanton heath]

okay it does

but what method?

[delicate hornet]

$1-x+\frac{x^{2}}{2}-\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{1}{4!}\int_{0}^{x}t^{4}\left(-e^{\left(t-x\right)}\right)dt$

[glossy valve]

ElectroSpeeder

[delicate hornet]

basically

the idea is

if i am reading this correctly

to find an upper bound for the integral at the end when evaluated at x=2

[wanton heath]

yes

but idk how

[delicate hornet]

to evaluate the integral at x=2?

[wanton heath]

well you use legrange

to find an upper bound

idk how to do that

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[frigid wadi]

I just really need help knowing where to start. I want to be able to solve it on my own, but I dont remmeber learning anything to find an integral with the same value as a sum

[frigid wadi]

it looks like it would diverge too

[rapid rain]

Do you know about about riemann sums?

[verbal mantle]

Well, to me it is obvious that it converges though

[frigid wadi]

We learned riemann sums at the very start, I am blanking. We have done series for so long, I was trying to take that route

[frigid wadi]

Oh... yeah. it would go to 0

wouldnt it

[verbal mantle]

Indeed

There isnt a single n on top, so must be

[rapid rain]

What converges to 0?

[verbal mantle]

The sum

[rapid rain]

No

Not the whole sum

[verbal mantle]

It converges to k, not 0

[rapid rain]

k?

[verbal mantle]

But still, it converges, not diverges

Constant

[rapid rain]

Yes, it will converge, but not to 0

[verbal mantle]

Yes

[rapid rain]

For example

[topaz valley]

k

[verbal mantle]

Using k to define a number

[topaz valley]

it converges to L

[verbal mantle]

Lmao ye

I usually go with k

[frigid wadi]

6/n would be delta x. the other one is f(xi), I assume I just do that and convert to integral

[rapid rain]

"It says L is the limit so it must converge"

[verbal mantle]

Well

[frigid wadi]

Theres so many series and integral problems, im having trouble knowing what to do when I look at a problem

[rapid rain]

Don't worry

[rapid rain]

Try to find f(x) such that this is a Riemann sum of f

And from there, writing it out should help you find the result

[frigid wadi]

thank you for your help

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[runic spruce]

Can anyone explain part b

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[chrome raft]

Good Evening!
Does someone have a good explanation on how we work with geometric series etc. when they dont start at k=0?

[fast peak]

either reindex so that they start at 0

or try to factor something out

[chrome raft]

Because I am stuck with this problem and cant figure it out.

1. Am I allowed to combine the approach with the geometric series + e?
2. Are my subtractions correct for the missing k= 0?

[fast peak]

this is not a geometric series

you cant just treat it as two separate series and sum them individually

also write = instead of =>

[chrome raft]

How else can I solve it?

[fast peak]

do you know the exponential series?

[chrome raft]

Yes

[fast peak]

can you write this as sum z^k/k! for some z?

where for a second we dont care about where the sum starts?

[chrome raft]

I tried that approach

[fast peak]

so what is z

[chrome raft]

The 1 on the right hand side?

[fast peak]

no

[chrome raft]

But I cant just multiply the -(1/2)^k to it right?

[fast peak]

I want z^k/k!. you have what you wrote down there

what can you set as z so that they are the same

[chrome raft]

No idea

Sorry

[fast peak]

yes

[chrome raft]

Because that is just wrong no?

[fast peak]

oh wait

the minus is wrong

$\frac{\left(-\frac12\right)^k}{k!}$

[glossy valve]

Denascite

[chrome raft]

But why are we allowed to do that?

I dont get it

[fast peak]

why not

we arent changing anything

[chrome raft]

Because that is not how you multiply fractions?

[fast peak]

how is it not?

[chrome raft]

Because you take numerator by numerator....

[fast peak]

a^k/b^k = (a/b)^k

[chrome raft]

And denominator by denominator

Still

I dont see it

We are splitting up the terms and then change the multiplication

That makes 0 sense to me

[fast peak]

where are we changing the multiplication

$\left(-\frac12\right)^k = \frac{\left(-\frac12\right)^k}{1}$

[glossy valve]

Denascite

[fast peak]

better like that?

[chrome raft]

...thats just so dumb lol

So why would we ever multiply numerator by numerator etc.?

When you can always just divide by 1?

[fast peak]

well I mean depending on situation various ways to write it down could be more or less convenient

[chrome raft]

Hm...I dont know

My math is just not mathing at the moment

[fast peak]

but really, $a\cdot \frac b c = \frac{ab}{c}$ should not be new for you

[glossy valve]

Denascite

[chrome raft]

I mean I am not that dumb. yes

[fast peak]

maybe take a break. sometimes our brain just stops working

[chrome raft]

Its this part for me

Changing it from a standard fraction multiplication

Into dividing it again by 1 and then do the fraction multiplication

[fast peak]

well I want the form z^k/k!

does it help if instead of 1/2 you write it as 0.5 ?

[chrome raft]

I actually thought about that

[fast peak]

(-0.5)^k/k! ?

[chrome raft]

And wondered by that would work but not the other way

[fast peak]

we want to consider -1/2 as one thing

[chrome raft]

Funny enough

[fast peak]

not as a fraction with numerator and denominator. it's a single unit

its just z

[chrome raft]

It just one of the things that you kind of need to know

[fast peak]

I guess

[chrome raft]

And if you have not dont stuff like that it just always feels so bad to struggle with it

[fast peak]

it comes from experience

[chrome raft]

Yeah...my math experience is so long ago

Appreciate the help though!

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[chrome raft]

.reopen

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✅

[chrome raft]

Ah wait, how do we then consider the start at k=1?

[fast peak]

ah yes

so, what is $\sum_{k=0}^\infty \frac{\left(-\frac12\right)^k}{k!}$ ?

[glossy valve]

Denascite

[chrome raft]

So should be e^-1/2 right?

[fast peak]

yes

and the series we currently have is $\sum_{k=1}^\infty \frac{\left(-\frac12\right)^k}{k!}$

[glossy valve]

Denascite

[fast peak]

what is the term for k=0

[chrome raft]

So we plug in k = 0 and get 1

So we need to subtract 1 from the solution?

[fast peak]

yes

[chrome raft]

And in case of k=2

[fast peak]

$e^{-\frac12} = \sum_{k=0}^\infty \frac{\left(-\frac12\right)^k}{k!} = 1 + \sum_{k=1}^\infty \frac{\left(-\frac12\right)^k}{k!}$

[chrome raft]

We also put k=1 etc.

[glossy valve]

Denascite

[chrome raft]

And subtract that

[fast peak]

yes exactly

we write the first few terms of the series extra

and then subtract

[chrome raft]

But should be minus 1 no?

[fast peak]

I generally write it like I did up there to make sure to not mess up the signs

well in the next step you will subtract the 1

because you are solving for the value on the right

e^(-1/2) = 1 + (value we want)

[chrome raft]

Ah I see what you did

Sure

Thanks again!

[fast peak]

youre welcome

[chrome raft]

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[meager ether]

Give a general expression for t, in terms of ω, π and n (where n is any integer), for the times at which −sin(ωt)=0
Give a general expression, in terms of ω, π and n (where n is any integer), for the times at which −sin(ωt)=1

[meager ether]

i got nπ/ω for the first one

but im not sure abt the second

i tried (2n-1)*π/(2ω)

[ember shadow]

well when does -sin(x) = 1

[meager ether]

pi/2

so if i rearragne ωt=pi/2

[ember shadow]

don't forget the negative sign...

[meager ether]

so would it be -(2n-1)*π/(2ω)

[ember shadow]

I meant in front of the sin(wt)

[meager ether]

oh

so its -pi/2

[ember shadow]

yes, or 3pi/2, as well as others

[meager ether]

yeah

since n has to be odd it can be rewritten as 2n+1 right

[ember shadow]

it says n can be any integer, though

[meager ether]

what i mean is it cant just be a multiple of pi/2

[ember shadow]

right

[meager ether]

like if n=2 and u do n*pi/2 u just get pi

[ember shadow]

yes so it can't just be n*pi/2. but there is an expression that works for any integer n

write out the first few values for which sin(x) = -1, you should see a pattern and think of what expression works

[meager ether]

-pi/2, -5pi/2, -9pi/2

so (4n+1)pi/2

?

[ember shadow]

why not start with 3pi/2, but you are on the right track

it won't be (4n+1) because when n=1, 4n+1 = 5. and sin(5pi/2) = 1, not -1

[meager ether]

x = 3pi/2, 7pi/2, 11pi/2

[ember shadow]

yes

[meager ether]

oh so

(4n+3)pi/2

[ember shadow]

yes

[meager ether]

ahhhh

so omega*t = (4n+3)pi/2

t= (4n+3)pi/(2*omega)

[ember shadow]

yes

[meager ether]

okok tysm!

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[fervent shadow]

Hey outage

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@fervent shadow Has your question been resolved?

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[rustic wagon]

can someone explain why it switched

[rocky moat]

The reciprocal means u turn the fraction upside down

And then u multiply

[atomic blade]

$\frac ab \divisionsymbol \frac cd = \frac ab \cdot \frac dc$

[glossy valve]

Umbraleviathan

[atomic blade]

You've done fraction multiplication/division before right?

[gray pendant]

basically just do

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[cyan surge]

Hello

[cyan surge]

so I'm determining if a series converges or diverges, and

the sum is

5/n(log(4n))^4

my initial thoughts were;

Pull the 5 out

5 * sum 1/n(log(4n))^4

and part of me things it's a p-series but i also think integral test would work?

i'm not quite sure which to use

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[cyan surge]

<@&286206848099549185>

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[cyan surge]

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[cursive void]

can someone explain how covariance comes into play here?

solution is here

although i dont understand how htey came to the conclusion

<@&286206848099549185>

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[bright zenith]

why is the server messed up

[normal tree]

it comes into play when calculating Var(U)

[cursive void]

I though variant was E[X^2] + E[X]^2

[normal tree]

it is

well

E[X^2] - E[X]^2

not +

[cursive void]

Yes ur right

Thank you pointing that out

But back to my og question

Why is variance coming into play?

And this equation of variance is different from E[X^2] - E[X]^2

[normal tree]

well because if you computed E[U^2] - E[U^2], you'd find that it involves terms that can be computed from things like Cov(X, Y)

[cursive void]

I see

So for my test I'm allowed a formula sheet

So what generic version should I put to remember this?

[normal tree]

well, it depends on what you've learnt and what you remember best

personally, I don't remember the formula for the variance of the sum of 3 correlated random variables

you can derive it fairly easily from remembering that Var(X+Y) = Var(X) + Var(Y) + 2Cov(X, Y)

[cursive void]

I should be a +?

[normal tree]

yeah my bad lol

I keep writing - because I always used Var(X-Y) lol

[cursive void]

Ty

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[torn jolt]

Does anyone know any good Geometry probability learning website 😭 im suffering rn trying to understand it

[cloud moon]

try and ask for a specific problem

[torn jolt]

d

question d i dont understand

[cloud moon]

send the image as a screenshot not a document

[torn jolt]

alr