help-0

[ocean seal]

.doc

[barren portal]

$\sum_{k=0}^ {k} a_k= a_0+a_1 + a_2 + a_3 + \ldots + a_k$

[ocean seal]

.doc

[clear tulip]

when isolating S_k here

do i treat S_k as a normal variable

[barren portal]

s_k, is just the sum upto (k+1) (counting a_0) terms right?

just like 7 or 8

[clear tulip]

oh yeah, so i can treat it as a variable right? it would make sense

[barren portal]

yes

It’s really what a variable should be

plug k, you get sum upto (k+1)

i have to go now,

Regarding your question on integral as a sum

recall, integration can be seen as sum of rectangles

this means, it can be expressed as a series

i would suggest you to look into Reimann integral for this

[clear tulip]

yes no problem thank you so much great great great help

[barren portal]

Have a great day

Good luck with your studies

[clear tulip]

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[eager cedar]

Say u have 2 types of criteria, getting at least 1 spade and getting at least 1 face card

the user then draws 3 cards from a standard deck of cards

if x represents the number of criteria that is succesfuly met

then how would i caclulate the probability of 0, 1, or all 2 criterions being met

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[cobalt eagle]

can someone walk me through this one?

[cobalt eagle]

does this seem right?

[waxen flame]

The approximations look good to me.

[cobalt eagle]

how bout the equation of the line

[waxen flame]

Looks good, but I would write y = -6x+7.

[cobalt eagle]

kk thx

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[spare dagger]

Statistics: not sure how to do it

[spare dagger]

thanks in advance guys

is it just 0.012 or 0.164?

<@&286206848099549185> sorry guys

[marsh fractal]

0.164

[spare dagger]

On what occasion would the answer be 0.012 then?

wait nvm sorry

I'm stupid

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✅

[spare dagger]

Sorry, but isn't this a one sided test? So shouldn't I divide the P-value by two?

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[cinder sundial]

That is a diagram and scenario of a satellite rotates around Earth

I just assume the mass of them as m and M respectively

[zinc haven]

rotating*
the Earth*

[summer dirge]

minor mistake tbh

[zinc haven]

he asked to correct so i did /shrug

anyway

[frigid hatch]

Looks correct

[cinder sundial]

My question is, if there’s some external force imposed on the satellite and makes the velocity of it to increase

[frigid hatch]

no there isn't

[cinder sundial]

I haven’t finished yet

[summer dirge]

you’re using a uniform circular motion model

[summer dirge]

okay continue

[cinder sundial]

Would the distance between the satellite and Earth increase or decrease

Let just assume the distance between the satellite and Earth increase or decrease as r

[summer dirge]

the tangential velocity, I’m assuming?

[cinder sundial]

Yes

[summer dirge]

then the answer would be yes

increase if the tangential velocity increases, decrease if the tangential velocity decreases

[cinder sundial]

How would you justify your answer

[summer dirge]

hm

[fervent kestrel]

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

[ocean seal]

dragonbreath

[fervent kestrel]

you can solve for v here

which gives orbital velocity at distance r

(technically it's the required velocity for an object to stay in motion around a center body without either (a) crashing into it due to gravity or (b) shoots off into space due a too large velocity)

[frigid hatch]

and that too large velocity is given by this relation

[cinder sundial]

The results seems like if the satellite’s velocity increases then the r would be smaller

[fervent kestrel]

you could also solve for r

$r = \frac{GM}{v^2}$

[ocean seal]

dragonbreath

[fervent kestrel]

larger v = smaller r

[cinder sundial]

Sure

[fervent kestrel]

which is kind of awkward

[outer lark]

[cinder sundial]

That’s quite counterintuitive isn’t it

[fervent kestrel]

yes

[fervent kestrel]

Do you have any clue?

[cinder sundial]

For most of ppl think the r would increase if v increased

[summer dirge]

smth seems wrong to me...

[outer lark]

use kinematics its already a satellite...

[paper mango]

Currently you've solved Fg = Frot to r = GM/v² and under the assumption that the gravitational force remains constant, then yes, if you want your satellite to be closer to earth

[summer dirge]

me thinks these are two different kinds of velocities

[paper mango]

then it has to move faster
If you move the satellite further from earth then it has to move slower to maintain its orbit

[fervent kestrel]

I mean that is what would happen, if somethings velocity would increase it does "escape" the gravitational pull of the central body

[frigid hatch]

if r is too big then it won't be dragged by its gravitational force

and the first relation won't hold

[fervent kestrel]

OH, that's right something does have to move faster to remain in orbit if it is closer

[paper mango]

Sure the pull is stronger

all above calculations were done with the assumption that the gravitational force remains constant

and only r & v are allowed to change

[summer dirge]

but we're talking about an external force are we not?

[cinder sundial]

Yes

[summer dirge]

suppose the rocket suddenly turns on its boosters

think about swinging a ball tied to a rope around your head

if you spin faster and faster, but keep the radius the same

the rope will snap

[frigid hatch]

then if the velocity is greater than sqrt(2GM/r) it will leave the planet's orbit

[summer dirge]

so the ball has been trying to move outwards, as the spinning gets faster

that's why I thought the rocket would move out too

as it's tangential velocity increased

[cinder sundial]

What if gravitational force isn’t constant? Like I impose a external force to the satellite and makes the tangential speed of it to increase, then would the r become smaller?

[summer dirge]

I'm gonna hold firm to my position and say no

r should become larger

but

I am willing to be shown wrong

afaik, the GMm/r^2 = mv^2/r assume uniform circular motion

[paper mango]

That force must lie in the center of the earth.

[summer dirge]

but turning on the thrusters is not uniform

[paper mango]

otherwise you lose the circular orbit

which means your relation Fg = Frot with simple circle orbit isn't valid anymore

you can either alter the force of the earth (which would be your added external force)

or determine the properties of the satellite in a different system of forces

because if you for instance had another force outside the earth

The satellite won't follow a circle orbit anymore, but instead an ellipse, or an arbitrary connection of two ellipsis or it collapses

(well it'd only collapse if we're given starting conditions instead of the assumption of a stable orbit)

[cinder sundial]

My mind blown

The fact that the track might become a ellipse is totally out of my expectation

[paper mango]

Sure, 2-body-system

[cinder sundial]

Ah, I remember it

[paper mango]

except you only regard the satellite

The Fg = Frot works because you have Frot

[cinder sundial]

What is “Frot” by any chance? I

[paper mango]

Frot = rotational centripetal force

which means you have a force from a single point

[cinder sundial]

I see

[paper mango]

not multiple

Note that the centrifugal force

is a, uh how do you say that in english

"apparent force"?

wrong force?

inexistent

idk

[fervent kestrel]

normal force?

[paper mango]

No it doesn't exist, there is no force pushing the satellite outwards

[cinder sundial]

Fictitious Force?

[paper mango]

Yep

You only have one force, which is your centripetal force

in this case gravity

If you look at atoms instead

It'd be a different force

But it's still centripetal

meaning towards the center

and what allows the circular movement is that the body which orbits already has an initial movement

so like in a carousel, if there was no force you'd simply move towards and fly off the carousel

Not directly outward, but in the direction you were moving

I'm not on my setup otherwise I'd draw it

the carousel is the typical example to exemplify that there is no centrifugal force

in any gravitational field

@cinder sundial ok but let's briefly look at what you mentioned before, changing the force

eg let's say we increase the gravitational force of the earth by a factor of 2

so now
2 Fg = Frot

Because the centripetal force is now twice the initial gravitational force

2 GMm/r² = mv²/r

which also solves to

r = 2GM/v²

Note that we now also have the factor 2 in this relationship

which means we can scale the radius of the satellite orbit by a factor of 2 with the velocity unchanged

to be stable in this new gravitational field again

or solving for v:

v = √(2GM/r)

we'd need √2 times the initial velocity to be stable with an unchanged radius

We could also change both radius and velocity, only the equation must be satisfied @cinder sundial

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[slow oasis]

is 0^-1 illegal

[inner haven]

Yes

Cause if you rewrite that that basically becomes 1/0

[slow oasis]

ok thx

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[pulsar spire]

How can I get X alone ?

okay

how can i factor out x here?

ok i got x now

thank you so much

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[autumn spade]

How do i solve this question? cant find their age

[coral flower]

5 years (hence?) , the age of jacobs .....

well first write the question as a equation

[autumn spade]

Age of jacob = x, age of son = y.

[coral flower]

what does 5 years hence even mean?

[autumn spade]

so 5 years later x + 5 = 3(y+5)?

[coral flower]

yeah

[autumn spade]

5 years later

[coral flower]

oh ok

[coral flower]

no

actually nvm its correct

do the same with the other sentence

[autumn spade]

so then 5 years before x-5=7(y-5)

[coral flower]

y-5

but yeah

[autumn spade]

oh yeah

then what

[coral flower]

simplify both equations

and use substitution method to find x and y

[autumn spade]

ok

gimme a min

ight cool thanks man i got the answer

ur the goat

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[scarlet drum]

If X1, X2,..., 1991 are strictly positive numbers and x1 + x2 + .... + X 1991 = 1, show that 2(√x1(x2 + x3 + ... + X1991) + √ x2(x1+x3 + ... + x1991) + ... + √x1991 (x1 + x2 + ... + X1990)) <1991

[scarlet drum]

Still couldnt solve

Jensens doesnt seem to work

[somber blade]

The square roots can be simplified to sqrt(xn(1-xn)). We know that 0<xn<1.
If you could prove that sqrt(xn)-xn is always smaller than 0.5, you would have the answer.

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@scarlet drum Has your question been resolved?

[scarlet drum]

Right, great tip

<@&286206848099549185>

[south verge]

@scarlet drum

I solved it

Imma send the solution as pics

Sry some photos are flipped. And also in the beginning I simplified the expression into a sum of square roots.

[scarlet drum]

What does Df mean?

[south verge]

The domain of the function f

Or the values of x that f can accept in R(real numbers)

[scarlet drum]

I see.

Thank you!

[south verge]

You're welcome

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[copper fiber]

Greetings.

Cash prize for 1st place in URFU math Olympiad is encrypted in this equation. I need help with solving it

[copper fiber]

If you have some free time, u can try to solve it

[modern sedge]

💀

[zinc haven]

ok

[modern sedge]

well so m! will be even integer for sufficiently large m

there is also pi

n....

well

hmm

it's sus

[zinc haven]

in all seriousness though, help channel is for asking, not giving question to helpers

but.....

this looks interesting

[modern sedge]

maybe the cos is jut meant to be 1

[copper fiber]

Ok, so I need help with solving this equation

[modern sedge]

then it's 10^1 which is 10

[zinc haven]

thats the spirit

what have you tried

[modern sedge]

it's just cursed

absolutely cursed

I'd try inputting it in wolfram alpha tbh

[copper fiber]

I've tried to Convert radical expression in the numerator

[modern sedge]

holy fuck the product

okay so

ln(e^sink) is just sink

so its sin^2(tgk) + cos^2(tgk)

that's 1

so it's just square root of 5

wait no

2^kth root of 5

5^1/2^k

5^(1/2)^k

[copper fiber]

I think it will be 1

[modern sedge]

feels more like 5

the whole product will be 5

i believe

[copper fiber]

П is multiplication, right?

[modern sedge]

yep

[copper fiber]

5^0,5 * 5^0,25 etc.

[modern sedge]

mhm

[copper fiber]

What a thing

[modern sedge]

It's simple

5^(1/2 + 1/4 + 1/8 + 1/16...)

5^1

5

[copper fiber]

Aha

I understood

[modern sedge]

alright so that's 1 part

now this damn limit

pi * m! / n

well

both m and n approach infinity

m! will be some large even integer

[copper fiber]

have we found n?

[modern sedge]

n is in limit

its approaching inifnit

[copper fiber]

Aha

[modern sedge]

I feel like the limit isnt even well defined

but my honest guess is that it's pi * even integer

cosine of that would be 1

1^n is just 1

10 ^ 1 is 10

so the whole limit would be 10 i guess

[copper fiber]

Nice

[modern sedge]

*10, not n

meaning the top is 10^5

That is, 100,000

[copper fiber]

Is that answer?

[modern sedge]

now this thing

[modern sedge]

not yet

[modern sedge]

We need this

[copper fiber]

Fr

[modern sedge]

so lets start with this

the antiderivative

would be

umm

y^4/4

so it would be like

something large af

(xnk)^4/2

i think

[copper fiber]

Is it going to be 0?

[modern sedge]

now this scares me

wait yes

you are right

0

y^3 is odd function

and the bounds are symmetric

the integral is zero

therefore whole the sum is zero

so this is just

[copper fiber]

So x^0 is just 1

Not bad

[modern sedge]

integral of cosx/mn + pi^-1

[modern sedge]

indeed

cosx/mn should be quite insignficant as m approaches infinity

so we can just ignore it ig?

[copper fiber]

Idk

[modern sedge]

well if we did ignore it

which i think is right

then we would be left with integral of pi^-1

which is constant

pi^-1(n+pi) - pi^-1(n-pi)

thats

pi^-1*pi

*2

that's 2

so whole that bullshit becomes 2

[copper fiber]

So, the answer will be 50000

And that's pretty bad

[modern sedge]

10^5/2 is 50,000

[modern sedge]

thats also the amount of braincells I lost solving this

[copper fiber]

Because the prize is in rubles(((((((

[modern sedge]

ye, it is bad

[copper fiber]

That's around 500$ for a 1st place

Nah

[modern sedge]

yeo

yep

[copper fiber]

You did nice job solving this

Thx

[modern sedge]

yw

[copper fiber]

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[minor anchor]

If I want to approximate the value of cos(32) using differentials, the function y = cos(x) and its derivative dy = -sin(x) dx, what is the value of dx? would it be 2 if we were using cos(30)?

[modern sedge]

The problem here is degrees/radians

[minor anchor]

radians

[modern sedge]

wait so you are working with cos(30 radians)?

not cos(30°)?

[minor anchor]

yeah

sorry its in degrees

its cos(30°)

[modern sedge]

alright

well

then the derivative would be -pi/180 sin(x)

and dx would be 2

[minor anchor]

so dx = pi/180?

[modern sedge]

if you wish to you can convert it all to radians and work with standard derivatives

2° is pi/90 radians

so dx would be pi/90

and the derivative just -sin(x)

[minor anchor]

ahh okay okay

thank you man

[modern sedge]

yw

[minor anchor]

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[wet minnow]

hi

[wet minnow]

i calculated that there is a 1/6 chance amy doesnt win both games

and so 21/6

so amy loses both on 3.5 fridays

but thats wrong cause 3.5 fridays impossible

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@wet minnow Has your question been resolved?

[nimble fern]

hello there @wet minnow
Since Amy win both games on 21 fridays and she only had 1/4 chance to win both games on 1 friday
the estimated number of Fridays she played is not 21
so it's not 21/6

[wet minnow]

so

[nimble fern]

so the answer isn't 3.5

[wet minnow]

how do i calculate

how many fridays in total

[nimble fern]

Since Amy win both games on 21 fridays and she only had 1/4 chance to win both games on 1 friday
we need to find the number of Fridays with 1/4 and 21, any clue?

[wet minnow]

so 21 times 4?

21/4?

alr

thxc]

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[nimble fern]

21÷(1/4)=21 times 4 =84

Cheers!

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[neat dirge]

Examine the discontinuity of the function given below. If the discontinuity is removable, redefine the function to make it continuous.

f(x)= 2x²-x-3/x+1

I get everything I just need clarification on the piecewise function if I did it right this time:

Here's my piecewise
f(x)={2x-3, if x≠ -1
-4, if x= -1

Feel free to correct me though

[neat dirge]

<@&286206848099549185>

<@&286206848099549185>

[tulip bone]

Mate

Thats 2 pings

!15m

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[olive yarrow]

you hvae the first part correct

but when x=-1

y should be the limit as x approaches -1 of f(x)

you might have made an arithmetic mistake

@neat dirge

[neat dirge]

Ah so -3?

[olive yarrow]

not quite

[weak ruin]

?

Take the limit when x->-1

of f(x)

[olive yarrow]

have you done limits yet?

if not just evaluate 2x-3 at x=-1 and use that

[neat dirge]

Oh it's -5

[olive yarrow]

yeah there you go

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[plain flame]

How to do number 7

[plain flame]

I did attempt like half of it

Then idk what to do

[plain flame]

,rccw

[plain flame]

,rccw

[plain flame]

,rccw

[plain flame]

This much

[vapid kestrel]

hm

[plain flame]

Umm

[vapid kestrel]

nvm this is beyond me

[plain flame]

<@&286206848099549185>

[plain flame]

Number 7

[plain flame]

I attempted this much and got stuck

<@&286206848099549185>

[coral flower]

i think if u expand 2n! and n! u get ur answer

@plain flame

$$2n! = 2n(2n-1)(2n-2)(2n-3) . . . . 3 . 2 . 1 $$

[ocean seal]

JustToPro

[coral flower]

now if we rearrange this a bit and get all the odd values seperate (cuz we want those to stay on the numerator)

$$2n! = 2n(2n-1)(2n-2)(2n-3) . . . . 3 . 2 . 1 = [(2n-1)(2n-3) . . . . . 3 . 1][(2n)(2n-2) . . . . 4 . 2]$$

[ocean seal]

JustToPro

[coral flower]

now we see that in the second term (the even term) its has a common factor of 2 , we can rewrite that as

$$[(2 \cdot n)(2\cdot(n-1)) . . . . 2 \cdot 2 . 2\cdot 1]$$

[ocean seal]

JustToPro

[coral flower]

we can factor all of those 2 out , cuz 2 is multiplied constantly with all the "n" number of terms

we get $$2^n (n(n-1)(n-2) . . . . . . 3 . 2 . 1)$$ which is basically $n!$

[ocean seal]

JustToPro

[coral flower]

so we can rewrite what u have as

[plain flame]

Okay

[coral flower]

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

cancel those n! and we are left with

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

[ocean seal]

JustToPro

JustToPro

[coral flower]

using exponential laws / law of indices we can combine the 2 and -1

cuz they are multiplying and have the same power

[plain flame]

Thanks I now understand

Yes

Thanks

[coral flower]

if something still confusing tell me

ill explain it

[plain flame]

I understood it very well

[coral flower]

good

[plain flame]

Thanks for well explained steps

[coral flower]

np

[plain flame]

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[dense ore]

<@&286206848099549185> 🥹

[dry oyster]

Wdym by putting it back in the denominator?

[rare jolt]

If I understand corrently, you need to derivate that function in dx?

[vocal birch]

What are you solving for?

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@dense ore Has your question been resolved?

[dense ore]

thanks fam i got it

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.close

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[agile sinew]

hi

[agile sinew]

i need help to find the angles

[dry oyster]

Angles on the bottom (U, B, A, P) would be equal since its an isosceles trapezoid. So those four would be 70 degrees like angle A.

The sum of angles of quadrilaterals is 360. So 360 - 140 = 220, and the angles at the top would be equal to 110

[agile sinew]

thxx

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[lunar smelt]

2nd part

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[valid veldt]

Tennis players A and B are going to play a match against each other. Determine the probability distribution for the number of sets in the match if (regardless of the outcome of previous sets) the probability that A wins a set is 0.6, when a player has won three sets.

[valid veldt]

I don't understand where the 3 is coming from.

[lone heart]

@valid veldt Has your question been resolved?

[valid veldt]

<@&286206848099549185>

[valid veldt]

Anyone? 🥺

[wraith stratus]

the left is .6 * .6 * .4 * .6

and the right one is .4 * .4 * .6 * .4

oh you meant the coefficient

[valid veldt]

@wraith stratus Yes

[wraith stratus]

is a match a best out of 4?

[valid veldt]

3 by the tex

text

[wraith stratus]

uh

best out of 3

or first one to 3

[valid veldt]

First one to 3 i guess

[wraith stratus]

so if you've played 4 sets already

and one of the players has won 3, while the other player has won 1

how can you arrange that

[valid veldt]

For me its 0.6^3 * 0.4 + 0.4^3 * 0.6 But that is wrong somehow

[wraith stratus]

answer this

[valid veldt]

They snuck in 3, don't see where it's coming from.

[regal goblet]

What do you want?

[wraith stratus]

ok

you've played 4 matches

player A has won 3

player B has won 1

how can you arrange the order of these wins/losses

[valid veldt]

Hmm, dont know.

[wraith stratus]

one way is

A A A B

do you see any other way

[valid veldt]

Yes' that I get, so AAAB + BBBA

[wraith stratus]

ok

how else can it happen

[valid veldt]

Uh you mean I can combinded AAAB in 3 ways and BBBA in 3 ways?

[wraith stratus]

no

[valid veldt]

Oh ok

[wraith stratus]

you can just have

AAAB
AABA
ABAA
BAAA
when A wins 3 times

[valid veldt]

Oh I see, so thats where the multi with 3 comes from?

[wraith stratus]

no

one of the cases isn't possible

[valid veldt]

First one

[wraith stratus]

why

[valid veldt]

Because the match is over

[wraith stratus]

yes

so only AABA, ABAA, and BAAA are valid

each of those cases has probability .6^3 * .4

so it's 3 * .6^3 * .4

[valid veldt]

Okey that makes sense

Thank you so much, been having a melt down about it 🙂

[wraith stratus]

np

[valid veldt]

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[upper brook]

Hello, can you help me to solve this logic exercise please?

[upper brook]

Ex.13) Write in symbolic form, after defining a dictionary and a universal set, and analyze the validity of the following categorical reasoning, proving by rules of inference or justifying correctly:
d) All matrices having two equal rows are not invertible. Invertible matrices have nonzero determinant. The determinant of the matrix "A" is zero. Therefore, matrix "A" has two equal rows.

How can I approach it?

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@upper brook Has your question been resolved?

[upper brook]

<@&286206848099549185>

[broken charm]

is it necessary that this reasoning is true?

[upper brook]

no

[broken charm]

alr because it seems like it isnt. if matrix A was invertible it would have a nonzero determinant. but maxtrix a has a determinant of 0 . Therefore it is not invertible. However, that doesnt mean that it has 2 equal rows. All matrices having 2 equal rows are not invertible isnt logically equivalent to All not invertible matrixes have 2 equal rows. Now i am not sure how this can be written down probably but thats my thinking

[upper brook]

the problem is that we have not yet seen any matrices! the invalidity would have to be proven by logic

[broken charm]

wdym seen any matrices?

[upper brook]

sorry, I am using the translator, I meant that we have not yet studied matrices.

[broken charm]

yeah i know i have no idea what the text says

i am just concluding logically that matrix A is not invertible.

[lone heart]

@upper brook Has your question been resolved?

[upper brook]

I did this, I don't know if it is correct

at the end I put that it is not valid

PU is universal particularization and MT modus tollens

These would be the universal set and the propositional functions
U = {x / x is a matrix}
p(x): 'x has two equal rows'.
q(x): 'x is invertible'.
r(x): 'x has a non-zero determinant'

[upper brook]

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[gusty fog]

#6 please

<@&286206848099549185>

[willow raft]

what are the values after?

like to the right of that matrix

[gusty fog]

For #6?

[willow raft]

ye

cuz I think there is a theorem that states like general solution is equal to basic solution + particular solution

what's rref of that matrix?

[gusty fog]

Ye I didn't learn that lol

It's an augmented matrix

[willow raft]

what's tthe rref of it tho

[gusty fog]

Row reduced?

[willow raft]

ye

did u try row reducing it?

[gusty fog]

Here

What c and d be free variables?

[willow raft]

I have no clue what the question is asking since there are 5 variables so why are they giving us 4 variables?

also the solutions only three variables

[gusty fog]

Oh

The last row is the solutions of the equation

For example x+y+x=4

So 4 would be the rightmost column

The first row is: a + 0b -2c -2d = 1

[willow raft]

oh

ohhhh

usually there's a bar

ok

that makes more sense

ye c and d are free variables

and u would solve for tha

t

and then u would find the basic solutions

[gusty fog]

Ye the bar is gone

[willow raft]

and then add teh basic solutions to the specific solutino found

and then you would get the general solution

[gusty fog]

Ok how do we know d is free

[willow raft]

cuz its not a leading variable

there isn't a leading 1 for

the d slot

[gusty fog]

Oh okkk

That makes sense

Thank you

[willow raft]

np

[gusty fog]

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[jade gale]

I need to write sqrt(3 + sqrt(5)) as a sum of two numbers

[cinder tundra]

What have you tried?

[jade gale]

nothing tbh, no clue how to approach it

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[brittle osprey]

I've been taking a couple of shots at this problem - I just don't understand why it resolves in the way that it does. I immediately think to take the limit & use L'Hopital's theorem, but that ends up looping over and over again. It's a geometric series (from what the problem is asking, I would not have immediately known that). I'm not sure where to go.

[brittle osprey]

Does something like this work?

Even if so, I'm not sure how I would apply the a/(1-r) way of solving geometric series to it.

[merry depot]

split your fraction, it's the sum of two geometric series.

[brittle osprey]

Something like this, then?

which would then be something more like...

Because N is going to infinity, that (n-1) is insignifcant, right?

It's still just effectively n?

[merry depot]

it's important for the geometric series formula, that depends on the starting value of n and the exponent.

you can use that formula on each part, since theyre both geometric

[brittle osprey]

I end up getting this if I apply the a/(1-r) formula.

This is the thought process behind it - unfortunately, though, says I'm incorrect.

[merry depot]

a should be 7/8 and 3/8

[brittle osprey]

Is that because of the parenthesis?

[merry depot]

because they're all to the power of n
you need n-1
if you just factor out the 7 you still have 8^n in the denominator

and the same with the 3

the series starts at n = 1

[weak ruin]

my bad

$\sum_{n=1}^\infty \left(\frac{7}{8}\right)^n = \frac{\frac{7}{8}}{1 - \frac{7}{8}} = \frac{\frac{7}{8}}{\frac{1}{8}} = 7$

[ocean seal]

Mercurial

[weak ruin]

same for 3

[brittle osprey]

7/8 is acting both as a and r here?

How does that work? I thought that it was one or the other - if A is 7/8, then r is 1, and if R is 1, then A is 7/8.

[merry depot]

series formula is $\sum_{n=1}^\infty a\cdot r^{n-1} = \frac{a}{1-r}$\
You have $\sum_{n=1}^\infty (\frac78)^n = \sum_{n=1}^\infty\frac78(\frac78)^{n-1}$

[ocean seal]

Zybikron

[merry depot]

so both a and r are 7/8

[brittle osprey]

I'm still not sure I understand where the 7/8 is coming from, like -

[merry depot]

$r^n = r\cdot r^{n-1}$

[ocean seal]

Zybikron

[brittle osprey]

OH

OK. To make it a geometric series, though, don't I have to get rid of that r^n immediately? I'm not starting with

I'm starting with this.

[lone heart]

@brittle osprey Has your question been resolved?

[ocean seal]

SollyPolly

[simple imp]

hmm i've also just looked at actual origonal question

so forget that

$\sum_{n=1}^{\infty}\left(\frac{7}{8}\right)^{n}+\sum_{n=1}^{\infty}\left(\frac{3}{8}\right)^{n}$ if you're trying to find this you can use the geometric sequence solution right off that bat

[ocean seal]

SollyPolly

[simple imp]

$\sum_{n=1}^{\infty}\left(\frac{7}{8}\right)^{n} = \frac{\frac{7}{8}}{1-\frac{7}{8}}$

[ocean seal]

SollyPolly

[simple imp]

no need to m-1 it

[brittle osprey]

I thought geometric sequences had to start at zero.

Like - with n being anything except 1, doesn't it also stop being a geometric sequence?

[simple imp]

i actually cant remember convention

but it does give the same value ill have a check

[brittle osprey]

It makes sense to me if in converting the starting n to zero, it becomes (7/8)^(n-1), and in doing so I get the problem I had earlier - I can't decide exactly what a vs r is. In the step of converting (7/8)n to (7/8)x(7/8)^(n-1), don't I still technically have the original function? I wouldn't have changed the starting n.

[simple imp]

i had written this before if this helps, but i thought it was too far off from the origonal question

[brittle osprey]

oh that actually works really well in terms of getting rid of that -1 that was messing with me

Just not sure how I would decide a vs r here. I doubt myteacher expects me to use the method you're laying out, the whole separation thing.

7 can't be A because of the parenthesis, right?

[simple imp]

well a is the first term of the sequence

and if we start at n=0 (7/8)^0 = 1

[brittle osprey]

In the notes I was given, it was split up like this - this is probably what's -

oh yeah this is 100% what's confusing me

[simple imp]

yeah but you see here, there is only a k in the denominator

so they have factored out the 5

and now they have a = 5 and (1/4) = r because when k = 0 this whole thing is 5. as (1/4)^0 = 1 and 5*1 = 5

[brittle osprey]

I see. That makes a lot of sense. I'm not very good with exponents - my best bet for getting rid of the (n-1) (or n minus whatever else) in future problems is to split it up like you showed, right?

[simple imp]

yes

[brittle osprey]

OK, thank you. Do you mind if I take a screenshot of this conversation so I can reference it in the future?

[simple imp]

yeah sure, its a bit of a messy convo

[brittle osprey]

👍

OH-

OOOOOOOHHH

Yeah. This makes a lot more sense now.

Thanks a ton.

was the "a's the first term" comment that made it all make sense.

[simple imp]

:D

[brittle osprey]

.close

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[unique summit]

how do i do 15 and the last part of 18 i been stuck on it for a while

[unique summit]

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[covert dune]

i got 3.34 from doing

tan^-1(500-20sqrt2/20sqrt2)

got 20sqrt2 from doing 40 * sin and cos pi/4

which is sqrt2/2

[covert dune]

like more context

[covert dune]

this spat out 86.56

so i subracted it by 90

gave me the answer u see above which aint correct

[crimson field]

Why is it 500- and not 500+

[covert dune]

isnt y sin in quadrant 4

negative

southeast is quadrant 4

[crimson field]

If it’s blowing feom@the south it’s helping

[covert dune]

wa

[wanton granite]

THis is a draw a pretty picture kind of problem

(They were prettier before I drew 8000 of these during flight school)

You're traveling north, the wind is pushing you from behind

It's blowing from the southwest, not to

[covert dune]

ohhh i see

[wanton granite]

The acronyms stand for head and crosswind component but it's actually a tailwind (TWC) it's just 2am and I'm sleep deprived lmao

Either way it's 500+your tailwind, and then just arctan that with your crosswind

[covert dune]

alr

im kinda burnt out doing out this ill try to see if i get the 3.06 answer

[wanton granite]

It's a pretty soul destroying problem after doing a bunch of them you're not wrong

It does work out as 3.06 though I did check

Counter point to the entire Q though, a) my autopilot knob has increments of 1 whole degree and b) if I'm not on autopilot lord only knows I'm getting nowhere near that accuracy, so if someone told me mid flight that I was drifting by 3.064682504082 degrees I'd die of laughter; so technically you answer rounds to what would actually be flown XD

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[gaunt elm]

I need find whether I include the endpoints

but it seems like I am failing to simplify the expression before using a test

[gaunt elm]

<@&286206848099549185>

[modern orchid]

(13/12)^n tends to infinity faster than n i believe

[gaunt elm]

i haven't learned tayler yet

[modern orchid]

Yeah actually its probs wrong

but uh theres a way

[alpine nacelle]

if (13/12)^n >= n², (13/12)^(n+1) >= 13/12 n² >= n²+2n+1 = (n+1)² for big enough n

[gaunt elm]

why do both of you raise 13/12 to power of n

[alpine nacelle]

idk I was just reading what he was saying and avoiding taylor series just to compare an exp and a linear

if you want to check endpoints just replace directly

[gaunt elm]

thats what i dis

did

it becomes this

[alpine nacelle]

but where's the -4

[gaunt elm]

-1/3 - 4 is -13/4

[alpine nacelle]

(x-4)^(n+1) /(x-4)^n = x-4

not x

[gaunt elm]

yup

your right

[alpine nacelle]

that's why your x wasn't in the good interval so checking endpoints didn't make sense

now you can just replace x with the correct endpoints

in the og expression

I don't have paper but from head, seems like it's the harmonic serie at endpoints

so doesn't cv

wait no exponent error

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

no prob then

another way to see it is to set y = (x-4)/3
then you're summing (-1)^n y^n/n, which converges as long as y^n/n decreases by alternating series (when y positive), if I don't make any mistake, so when -1 < y <= 1
1 < x <= 7

[lone heart]

@gaunt elm Has your question been resolved?

[gaunt elm]

where did the - sign go in your simplification, and do you just not include the n in denominator?

[alpine nacelle]

about the - sign, that's where the endpoint matters, it's not the same for both

for one endpoint, you get -/- = +

and for the other it alternates

[gaunt elm]

but the - sign is inside of the the exponent

[alpine nacelle]

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

[gaunt elm]

yes

[alpine nacelle]

(-1/3)^n = ...

[gaunt elm]

(-1)^n is alternatinc

[alpine nacelle]

you get (-1)^n/(-1)^n = 1 for one of the endpoints

[gaunt elm]

ooh

mb i am slow i guess

[alpine nacelle]

i feel like there's something weird with your endpoints or I did something weird

[alpine nacelle]

yeah that one is obvious just compare to 3^-2n

its sum converges

so 3^-2n /n which is smaller

same conclusion

[gaunt elm]

yes\

[alpine nacelle]

and the other is the same but alternating so converges too

[gaunt elm]

so what is the weird part

[alpine nacelle]

so I did have a mistake in my other point of view earlier

let's say you take x = 7, you get the sum of (-1)^n/n

which converges

7 should be in your interval too

[gaunt elm]

it says wrong so

[alpine nacelle]

yeah

for x = 7 it still converges

I think I = ]-1, 7] by my point above

but what in your method didn't work is an other problem, I still don't see why

https://cdn.discordapp.com/attachments/490557019623915520/1229230830325338172/image.png?ex=662eed6b&is=661c786b&hm=23923752f20b0b9f0499e940d547bafd7bb387d37791185756c800368782eb54&

(-3)^n/(-3)^(n+1) = -1/3

not -3

[gaunt elm]

well it is still wrong

[alpine nacelle]

I excluded the -1

(you didn't)

[gaunt elm]

still nuh uh

[lone heart]

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[gaunt elm]

yes your right ill do it again

.reopen

[lone heart]

✅

[alpine nacelle]

you should find 1 < x < 7 too tho, and 7 should be included

[gaunt elm]

yes i found -1<X<7

trying endpoints now

[alpine nacelle]

when x = 7 it's the alternating harmonic, which converges

and when x = 1, it's the harmonic, which doesn't

wait no sign error (corrected)

oh yeah it was the sign error

I wrote -1 instead of 1

which was the beginning of all confusion

[gaunt elm]

yes lol

[alpine nacelle]

it was the correct sign at this point in time and then I wrote something I cannot explain lmao

[gaunt elm]

i guess this is where we part our ways

it is the end of the question

[alpine nacelle]

yeah unless you need something

[gaunt elm]

I mean I don't

[alpine nacelle]

have a good day

[gaunt elm]

but if i have it later on can i dm

[alpine nacelle]

ping in a help chan, I think I deac dm

[gaunt elm]

I appreciate the help and time. Have a good day!

.close

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[tawdry urchin]

A parametric graph is given by
\begin{align*}
x &= \cos t + \frac{t}{2}, \
y &= \sin t.
\end{align*}How many times does the graph intersect itself between $x = 1$ and $x = 40$?

[ocean seal]

Dork9399

[tawdry urchin]

I know a few points

I know it looks like a sort of spiral graph

but I don't know how to calculate the intersections

<@&286206848099549185>

[vapid steppe]

find the period of each

[tawdry urchin]

well y has period 2pi

but how would x have a period

[vapid steppe]

is calculater allowed

[tawdry urchin]

no

[vapid steppe]

so have u learnt calculus

[tawdry urchin]

no

ik some basics

but barely

[vapid steppe]

well cos is bound between -1 and 1

[tawdry urchin]

yes

[vapid steppe]

and t/2 has no bounds

[tawdry urchin]

yes

[vapid steppe]

wait

[tawdry urchin]

idts

it will form a sort of spiral

like a bunch of cursive l's strung together

[vapid steppe]

ye

[tawdry urchin]

so how do we measure the length of each l

@vapid steppe

<@&286206848099549185>

[vapid steppe]

wai

t

isn’t the equation just a circle

[tawdry urchin]

no

x = cost, y = sint is a circle

[plain vortex]

Find the period of the spirals

Or intersections

however you wanna call it

[tawdry urchin]

how would i do that

find the period

wait is it just 2pi

[plain vortex]

...maybe

but where's the working

[tawdry urchin]

because the period of sin is 2pi

and each loop will be one period of y

[plain vortex]

Wrong*, also how about the start

[tawdry urchin]

what about the start

[plain vortex]

anyway erm I'm pretty sure u can tell the intersections don't occur every 2pi

[tawdry urchin]

yea

[plain vortex]

it's pi, but why

[tawdry urchin]

because itll intersect every half period of sin?

idrk

[plain vortex]

Here's a hint

[tawdry urchin]

wait what

I still don't understand

[plain vortex]

cosx and sinx intersect at its intersection

[tawdry urchin]

yes

[plain vortex]

look at the function

[tawdry urchin]

still not getting it

ohh

is it because cos and sin intersect twice every 2pi

@plain vortex

but even then, how would I determine that the crossing happens when cosx = sinx

without desmos

@plain vortex

<@&286206848099549185>

.close

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[old bronze]

can someone explain how to do b. ii.

[old bronze]

.close

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[quaint sleet]

When do we complete the square?

[charred flint]

when you want to find the vertex of a parabola or solve for x in an equation with x^2's and x's

it's the way to combine the two when trying to rearrange equations

[quaint sleet]

oh okay but how would I know to use that instead of all the other factorising methods? Infact, how would I know what to use when factorising?

[lone heart]

@quaint sleet Has your question been resolved?

[charred flint]

umm factoring is a huge topic but for quadratics completing the square is literally just the quadratic formula written a different way

[quaint sleet]

Oh ok, could you please assist me with completing the square for 2x^2 + 7x + 3 = 0

I've looked at so many tutorials but they all do a different way and it confuses me

[charred flint]

the main thing is that (x+c)^2=x^2+2cx+c^2, notice the 2cx

so since I like having 1x^2 in front I'd divide by 2 for x^2+7/2x+3/2=0, and now we try to complete the square on x^2+7/2x

complete the square means the x part looks like 2cx, so 7/2=2c and c=7/4, so it goes like
x^2+7/2x+3=0
(x+7/4)^2-(7/4)^2+3=0

then you'd just solve normally if that makes sense

[karmic saddle]

so first step is to divide everything by the coefficient of the first number? and in this case it's 2

That doesn't seem right

[karmic saddle]

Yea theres a whole variety of ways to do it

[karmic saddle]

https://www.mathsisfun.com/algebra/completing-square.html
Look through this and go through your question, come back if you're stuck on something

[lone heart]

@quaint sleet Has your question been resolved?

[lone heart]

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[wintry sluice]

im lost

idk where to start

ping me when someone's here to help

[crimson field]

Honestly I think it’s just 10,10,0,0,0,0,0…

Don’t think you’re gonna get any better than that

@wintry sluice

[wintry sluice]

what?

[crimson field]

For ai

[wintry sluice]

100,200,300,400 are the options 💀

[crimson field]

Yeah so what would S be here

I mean idk maybe it’s possible to get higher

I just really don’t see how

[wintry sluice]

how did you get this tho

the answer is 100

but i dont know how

[crimson field]

Hm actually

Wait lemme cook

[wintry sluice]

https://tenor.com/osz9KoDIxof.gif

[crimson field]

Nah I think 100 the max

I can’t even think of a way to get 105