help-33

[lilac siren]

ofc when writing the number you'd only write "1"

but we can perceive it as analogous

[still temple]

the hint on the hw says to consider how many 1 digit, 2 digit, 3 digit and 4 digit numbers are possible

[lilac siren]

to a sequence of 5 digits

[still temple]

the main thing though i dont really understand the only 1-5 thing

[lilac siren]

that's not the best approach since we allow repetition

[still temple]

and the repetition is allowed makes it even more weird for me

[lilac siren]

well you have an arbitrary 5 digit number that's odd, less than 3000 and only uses the digits 0 to 5

your first digit has 3 options (0,1,2)

[still temple]

yes

[lilac siren]

your second digit has 6 options

[still temple]

second digit is in the hundreds?

[lilac siren]

your third digit has 6 options

your fourth digit has 3 options

=> 3*6*6*3

[still temple]

im starting to understand

but where did you get the options for the second, third, and fourth digist?

[lilac siren]

given the requirements

you can use digits 0 to 5

which are 6 values

[still temple]

oh right

[lilac siren]

so second digit has 6 options

likewise for third

likewise for fourth, except it's odd

so 3

[still temple]

oh and fourth has to be odd because an odd number is defined by its last digit?

[lilac siren]

ys

[still temple]

ok tysm

[lilac siren]

if you want to test your understanding attempt the same but without repetition.

[still temple]

i understand it now, was a lot more confusing when i first saw it but now im realizing how it works out

[still temple]

im unsure how repetition would change anything

since each number is different anyways

[lilac siren]

say we use 1 for the first digit

[still temple]

ok

[lilac siren]

that means we can't use 1 for the other digits anymore

meaning for the others we now strictly have ≤5 options

[still temple]

oh

[lilac siren]

for the second we take another digit

meaning now you only have 4 numbers left

[still temple]

so 3 * 5* 4* 2?

[lilac siren]

and the tricky part comes when considering that the last digit must be odd

[still temple]

yeah

[lilac siren]

nop

because you have different cases

say we took 2 odd and 1 even digits for the first three digits

now you only have 1 option left for the fourth digit

but if instead we only took even digits for the first three digits, then you have 3 options left for the fourth digit

[still temple]

how would we know whether the fourth digit would have 3 options or only one?

[lilac siren]

depending on which digits we took for the first three

of course we don't want to determine every single possible number that can be formed

so we split into cases

[still temple]

i thought there had to be a conclusive answer

[lilac siren]

there is, it's just a bit more nested due to the cases

[still temple]

ok

[lilac siren]

shall we briefly go it through?

[still temple]

ok

[lilac siren]

let's split it into these 4 cases:
For the cases I'll denote them with a number indicating how many odd numbers we choose for the first three digits:
[0]
[1]
[2]
[3]

meaning we can either choose 0, 1, 2 or 3 odd numbers for the first three digits

[violet badger]

odd number = ends with 1 3 5 7 or 9
less than 3000
using digits 0 1 2 3 4 5

how many numbers can you make with 0 1 2 3 4 5, which end in 1 3 or 5

[lilac siren]

But it's already solved

[violet badger]

oh

ok

[lilac siren]

Notice that if we took three odd numbers for the first three digits

[still temple]

i dont think my teacher taught us that

[lilac siren]

eg. "135?"

[still temple]

yeah

[lilac siren]

doesn't need to be taught, it's about the principle to make you more familiar with how to calculate the total number of possibilities

[still temple]

ok

[lilac siren]

Different tasks require different solutions

[lilac siren]

Then we can't choose a number for the fourth

because the fourth has to be odd

and 1,3,5 are already used up

meaning for the case [3] we have 0 options

now let's regard case [0]

so we only take even numbers for the first three digits

first digit has 2 options (0,2)
second digit has 2 options left from among (0,2,4) and third digit has 1 option left

and the fourth digit has 3 options: (1,3,5)

meaning in total we have 2*2*1*3 = 12 options

[0] = 12
[3] = 0

[still temple]

ok

[lilac siren]

ok now onto [1]

now comes the slightly more difficult part

since we now want one of the first three digits to be odd

we have from among three places to choose

if the first digit is odd, then there are 1*3*2*2 = 12 options

if the second digit is odd, then there are 2*3*2*2 = 24 options

if the third digit is odd, then there are 2*2*3*2 = 24 options

it's fine if you don't get to these products that quickly, but it's just going through how many options we have for each digit

[still temple]

alr

[lilac siren]

meaning our total for that case is: [1] = 12+24+24 = 60

three cases solved, one to go: [2]

now we have two odd and one even number among the first three

so we can regard it as "placing" a single even number among the first three digits

if the first digit is even then there are 2*3*2*1 = 12 options

If the second digit is even there are 1*3*2*1 = 6 options

and if the third digit is even there are 6 options too

[2] = 12+6+6 = 24

total options = [0]+[1]+[2]+[3] = 12+60+24+0 = 96

this would be the case-Splitting approach

as you see a lot more steps required

[still temple]

yeah

[lilac siren]

just for no repetitions

but there are concepts which simplify the above calculations

[still temple]

all of my homework which has that sort of idea in the problems all have with repetition

[lilac siren]

you might have also noticed that many numbers repeat, which can be summarized

[lilac siren]

Now you know why :]

It's mainly because your requirements like "the last digit is odd" become more convoluted

since now the number of options depends on what you picked for the other spots

nywys gl with the topic 🐛

[still temple]

ty

but yeah I think I wont be a fan of this chapter

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[marsh citrus]

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[digital saffron]

can someone please help me understand where it says y(0)=-1 what am i supposed to do

Like I don't understand what is going on there

am i supposed to plug 0 in for y and x

because it says C=0

if i plug 0 in for both that wouldn't get me C=0

so I am so lost

[dry prawn]

You plug in x=0 and y=-1

[digital saffron]

;o

ohhhh ok

[dry prawn]

Every y on the left side of that equality (post-integral) is still a function of x, y(x) if you prefer

[digital saffron]

also quick question this work is wrong right it should be 3y+y^2 or am i tripping

so like C should actually be -2

because I think he did the integral wrong

[dry prawn]

It should be 3y shouldn't it

That's awkward

[digital saffron]

yeah

I had a friend write these problems so they probably rushed the solutions out to me

its ok

just making sure I am not dumb and making a mistake

[dry prawn]

Oh it's not the instructor, alright that's not so bad

[digital saffron]

yeah yeah

[marsh citrus]

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[signal rapids]

Hi

[signal rapids]

I'm not sure how I can do this, I'm missing (not)p or r

[marsh citrus]

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[signal rapids]

close

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[gritty mango]

Can I have help with 1.9

[gritty mango]

So far I have

I'm just not sure how to generalize for n. I have up to n=2 is correct. It's just after that

[copper raven]

Maybe you should try posting in #linear-algebra or one of the advanced channels (idk where that would fit)

Not sure what that Lambda^p notation means

[gritty mango]

It's antisymmetric p type tensors

[copper raven]

Ah yeah I prolly saw that notation once, not very familiar with tensors tho

What book do all these questions come from ?

[gritty mango]

There from an introduction to differential manifolds

[copper raven]

Yeah you prolly should try the diff geo channel then

[gritty mango]

That makes sense

[copper raven]

#diff-geo-diff-top

[gritty mango]

I've just put it in adv algebra

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[solar narwhal]

can someone help me with this inner fellowship tangent and
tangent to external alliances, so basically i started with

6² = d² - (R+3)²
36 = d² - (R² + 6R + 9)
36 = d² - R² - 6R - 9
45 = d² - R² - 6R
then im finding the tangent to external alliances,
(4√6)² = d² - (R - 3)²
96 = d² - (R² -6R + 9)
96 = d² - R² + 6R - 9
105 = d² - R² + 6 R

and now im confused what should i next cause im stuck there

[solar narwhal]

GSPD is inner fellowship and GSPL is tangent to external

<@&286206848099549185>

i cant find the r and d?

bruv

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[rigid mountain]

i need help in further solving this aplication of laplace transform :

[rigid mountain]

This is what I have done so far

[marsh citrus]

@rigid mountain Has your question been resolved?

[rigid mountain]

<@&286206848099549185>

help

help me

Laplace transforms, a complex scheme,
Help me decipher, this mathematical dream.
With your expertise, guide me through,
To unravel solutions, tried and true. <@&286206848099549185>

[rigid rivet]

Chat got

[rigid mountain]

?

help me out 🙂

pls

@rigid rivet

[mighty seal]

You have the Laplace transform of the solution to the ODE, to get y you just need to perform the inverse laplace transform

To start write $$\frac{2}{s^{4} - 4s^{2}}$$ as a partial fraction

[elfin berry]

TomB

[mighty seal]

Apologies! $$\frac{2}{s^{4} + 4s^{2}}$$

[elfin berry]

TomB

[rigid mountain]

my final part was L{y}= 2/s^4 + 4s^2 ...... so now i have to do inverse laplace transform?

[mighty seal]

Indeed, once you do the Inverse Laplace transform, you will have the solution to the ODE

[rigid mountain]

ohhhhhhh ok thanks man

[mighty seal]

Happy to help!

[rigid mountain]

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[rigid mountain]

.reopen

[marsh citrus]

✅

[rigid mountain]

@mighty seal do i have to do it on both the sides ?

[mighty seal]

Indeed, but $L^{-1}[ L [y] ] = y$ since you are doing the inverse of the Laplace Transform on a Laplace transform.

[elfin berry]

TomB

[mighty seal]

And you can use the property of the Linearity of the laplace transform to help find the Inverse on the right, hence why you may want to write the RHS as a partial fraction

[rigid mountain]

what formula for inverse of this ^

[mighty seal]

Generally, the Inverse Laplace transform is a hard procedure to do, so there is no formula for the inverse. The reason why you want to write the RHS as a partial fraction is because the 'technique' per-se is to go "Hang on, I recognise that this is just the Laplace transform of this". If that makes sense?

[storm dagger]

yep

[mighty seal]

Write the RHS as a partial fraction and I can guarantee you will recognize a Laplace transform

[rigid mountain]

got it

is the ans : y=sin2t-sin2 sin t ?

[mighty seal]

Not quite, what did you get as your partial fraction?

[rigid mountain]

A/s + B/s^3 + Cs+D/s^2 + 2

[mighty seal]

Not quite, you should get something of the form $\frac{A}{s^2} + \frac{B}{s^2 + 4}$

[elfin berry]

TomB

[rigid mountain]

this was my second last part

[mighty seal]

You are finding the partial fraction of $$\frac{2}{s^{4}+4s^{2}}$$

[elfin berry]

TomB

[mighty seal]

which is what you want to do the inverse laplace trasform of

[rigid mountain]

ok got it

[mighty seal]

$$\frac{2}{s^{4}+4s^{2}} = \frac{2}{s^{2}(s^{2}+4)}$$ makes it a tad clearer

[elfin berry]

TomB

[rigid mountain]

is the ans t sin 2t?

[mighty seal]

t * sin(2t) or t - sin(2t) ?

[rigid mountain]

t * sin(2t)

[mighty seal]

Barring some constants, you are in the right area

Should be the you should get something along the lines of At + Bsin(2t)

[rigid mountain]

yesss!!! thank you soooo much got it

[mighty seal]

Awesome!

[rigid mountain]

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[wary pagoda]

@spark otter hi sorry for ping because my channel got closed

If we had $\left\langle f,g\right\rangle =\int _a^bf\left(x\right)g\left(x\right)dx$ for a set of 3 functions, in terms of axioms, why can't we extend it to all other functions in the space of real continuous functions? (defined for domain R)?

[elfin berry]

quan

[wary pagoda]

because im not sure what you wrote in the other channel, im not that advanced 😔

[devout mauve]

it would help if you linked your other channel

[wary pagoda]

#help-15 message

Sorry

[devout mauve]

continuity implies that the integral exists (over closed intervals), so actually you dont have to worry about what rafilou wrote

if you didnt have an integral over a closed interval, so something like integral from -infty to infty or maybe integral from 0 to 1 of 1/x, then you have to worry about the integral existing at all

[wary pagoda]

yep but because this is continuous for all R

we dont have to worry about integral not existing

so it satisifes the axioms the exact same way as the original set of 3

[marsh citrus]

@wary pagoda Has your question been resolved?

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@wary pagoda Has your question been resolved?

[marsh citrus]

@wary pagoda Has your question been resolved?

[devout mauve]

what is still your question

[marsh citrus]

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[hybrid ibex]

hello

[hybrid ibex]

i have a simple question

find sin(2x) if sin(x) = -5/13 and pi<x<3pi/2

when i applied the formula i got a postive answer

but the correct answer is negative

how come?

[cold horizon]

for the future, send your actual problem in the first message so that it gets pinned

[hybrid ibex]

oh

my bad

[cold horizon]

It seems like the correct answer is positive

what was the answer you got

[hybrid ibex]

120/169

do u think the book is wrong?

also do i have to take into account the interval for x

like 2pi<2x<3pi

sorry if my question seem stupid

<@&286206848099549185>

[marsh citrus]

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[hybrid ibex]

<@&286206848099549185>

.close

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[shy barn]

can someone please direct me to the right direction for proving this?
if a<b and c<d then a/c < b/d
I've already studied this and my brain is just not braining I can't remember how to start proving this, it seems glaringly obvious but I forgot

[broken dome]

are a,b,c,d real number? can they be positive/negative

[shy barn]

yeah forgot to mention,

[leaden monolith]

-10 < 1, -2 < 1, then 5 < 1?

[broken dome]

Does this even hold then? I mean -500<1 and -1<2, but clearly -500/-1 is not < 1/2

[leaden monolith]

can you show the original question in its entirety?

[shy barn]

it's in Hebrew, I have 3 statements in total and I have to either prove or disprove them

this is the first one, so from what I understand I have to disprove it by giving an example of it not being true

[leaden monolith]

you should try your best to translate them literally

[leaden monolith]

yeah but now there's nothing to "teach" because we've already given you counterexamples

[shy barn]

true

[leaden monolith]

it's just a bit better if you translate word for word in terms of your learning

but it's ok

[shy barn]

how should I have known by myself to find those counter examples?

just use basic logic and plug in numbers that I think wouldn't hold to see if I get it?

[leaden monolith]

it doesn't sound true

[broken dome]

it seems to me that the original problem has another limit

maybe it says for any positive a,b,c,d from R

[leaden monolith]

there are a lot of "cases" to think about where the a b c d are positive and negatives

the main thing is whenever you do dividing or multiplying in inequalities

[random ledge]

How do I get help solving math problems on this discord server

[shy barn]

a,b,c,d for all R

[leaden monolith]

they flip when you multiply/divide by negative numbers

[broken dome]

Go to an available channel, this one is occupied

disproving a statement is as easy as finding a counter example

but you could also try to do it systematically

[shy barn]

yeah, that's enough in my case

inequalities + proofs = my weakness lol

[broken dome]

What are the other statement you have to check?

[shy barn]

if a/c < b/d then a<b and also c>d

if a<b and also c<d then (a-b)/(c-d) > 0

[broken dome]

well let's take my example from before

a = -500, b = 1, c = -1, d = 2

does the first statement hold for it?

oh wait

it doesn't work

let's take a = -1, c = 2, b = 30, d = 1

the first condition is fulfilled

the other are too, but this isn't enough to prove it

I'm trying to disprove it by finding a counter example

let's take a = -1, c = 2, b = 30, d = 3

now the last statement is false

I mean the second implication in the first statement

this is a counterexample making the statement "if a/c < b/d then a<b and also c>d" false

[shy barn]

as I remember correctly to prove a statement I'd have to take a statement that's already true and then through induction I think you get to the statement that you have to prove

if that makes sense

[broken dome]

through deduction

not induction

and yes

but it takes only 1 counterexample

to prove a statement false

[shy barn]

right

it's easier to find a counterexample

is there a general rule of thumb kind of thing for finding counter examples?

[broken dome]

no I just try to adjust numbers until I find something that works

Like here I just lowered d until it broke the rule

Let's try to find a counterexample for "if a<b and also c<d then (a-b)/(c-d) > 0"

let's start with a = 1, b = 2, c = 3, d = 4

[shy barn]

I think I got the hang of it

[broken dome]

the solution (1-2)/(3-4) = -1/-1 = 1 > 0

holds

hmmm

now let's try to break

it

when wouldn't it hold?

if the numbers were both negative

or rathr

if top was negative

and bottom positive

hmmmm....

this one might actually be true

I can't think of a counterexample

let's try to prove it them

if a<b then a-b<0 right?

so (a-b) is a negative number

same for c<d making (c-d)< 0 making it a negative number

and two negative numbers divided always give a positive which is > 0

therefore the statement is correct

[shy barn]

I see how the logic works here, hold up I'm gonna try to write a proof for it

@broken dome is it enough to begin proving by stating "if a<b then a-b is a negative number...etc"

[leaden monolith]

1 < 2, 0.1 < 1, 10 < 2 is also not true

you just look at it

and it looks wrong

[broken dome]

hm? regarding which statement?

[leaden monolith]

like the original question

it needn't even use negative numbers

[broken dome]

right

whatcha think of the last statement

[leaden monolith]

which one

[broken dome]

let a,b,c,d be real numbers, let a < b, from this follows that a - b < 0, therefore a - b is a negative number, let c < d, from this follows that c - d < 0, therefore c - d is a negative number, since both a - b and c - d are negative, (a - b) / (c - d) must be positive, therefore (a - b) / (c - d) > 0

as such a < b and c < d implies (a - b) / (c - d) > 0

does this seem reasonable?

[leaden monolith]

sounds fine

[shy barn]

my proof looks something like this: if a<b then a-b is a negative number, same is true for c<d, we'll define x,y are in R+, a-b=-x,c-d=-y and get a-b/c-d = -x/-y, with the property of '-a/-b=a/b' we get -x/-y = x/y which is larger than 0

is that good enough for a proof

[leaden monolith]

you didn't state that x and y are positive

[broken dome]

I'd just add to that x and y are positive from R

as a consequence of a<b and c<d

[shy barn]

I've edited it to include that x and y are positive from R, is that okay now?

[leaden monolith]

"part of" isn't so good

in is better

[shy barn]

oh okay

[marsh citrus]

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[elfin berry]

EbonyRocketship

[marsh citrus]

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[wispy shadow]

i dont know hwo to start to compute this derivative

[plain granite]

mr

[cobalt sentinel]

Well here’s the thing

Notice that in the integral, the x is not the variable of integration

[wispy shadow]

yh, so cant you take x out?

[cobalt sentinel]

t is the variable of integration

Yea

[wispy shadow]

but how would you integrate the remaining one is what im asking, would it be a form of usub

[cobalt sentinel]

Nah I don’t even know, but I don’t think they expect u to integrate it

[wispy shadow]

would i just say using ftc this would be x(F(3)-F(1))

[cobalt sentinel]

Yea look it’s nonelementary

[wispy shadow]

lmao ive been staring at this problem so confused

[cobalt sentinel]

Well we don’t have any function F defined

Why don’t u just leave the answer in terms of the integral

d/dx [x * integral]

[wispy shadow]

oh, then would i the ingerala with bounds 3 to 0 - 1 to 0

i see thankyou for the help

:))

.close

[marsh citrus]

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[cobalt sentinel]

Wait what

[wispy shadow]

.reopen

[marsh citrus]

✅

[cobalt sentinel]

Wdym

[wispy shadow]

just splitingt up the integral first to show more working wasa done then leaving it in simplified form

[cobalt sentinel]

Splitting it up into what? And even so, what’s the point of splitting it up if u can’t solve it

[wispy shadow]

because the question is an assignment question worth 7 marks, and i alreaaday had that solution down but i thought it was too short aand couldnt figure out the integral

so i came here

[cobalt sentinel]

Yea I mean I don’t think u can figure out the integral without some super complex Cleo method or something

So I think the point is just to leave the answer in terms of the original integral

[wispy shadow]

yh, my course is onyl first year introduction to analysis with a bit of computing aswell so i dont know what tf this even was

[wispy shadow]

aight

thanks for the help :))

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[coarse furnace]

I'm trying to create a stochastic matrix for the snakes of ladders game depicted alongside. Does this look correct so far? If so, how would I continue in completing section (b) of this question using this matrix. Would it be following the path of highest probability, i.e., after move 1, it'll be on square 4, then after move 2, it'll be on square 5 as that has the highest probability. Or if I had to square the matrix to correspond with the 2nd move, what value would I be looking at to figure out the most probable square?
Assumptions I made were that we start on square 1 (mainly for completing the stochastic matrix for part (a) as there was no clarification), and you cannot land and 'stay' on squares 2, 6, or 8, hence, I added their probability to the square they lead to, i.e., landing on square 2 is the same as landing on square 4 so square 4 has double the probility etc.

[marsh citrus]

@coarse furnace Has your question been resolved?

[coarse furnace]

<@&286206848099549185>

[crystal lintel]

set x = [1,0,0,...0]?

(for part b)

[limber marsh]

consider this:
The probability of landing on 5 on after two rolls is the probability of landing on 1 from 1 * probability of landing on 9 from 1 + probability of landing on 2 from 1 * probability of landing on 9 from 2 + probability of landing on 3 from 1 + probability of landing on 9 from 3 + ... + probability of landing on 9 from 1 * probability of landing on 9 from 9.

Seems an awful lot like a dot product, doesn't it?

[crystal lintel]

then A^2x gives you the probability distribution for each square and you can pick the highest

[coarse furnace]

Do you mean to set the values of [x1, x2, ..., x9] as [1, 0, 0, ..., 0] as part b states it starts on square 1? So move 0 would be [1, 0, 0, ..., 0], making A a 9x10 matrix?

[crystal lintel]

first question yes, 9x10 matrix no

x is like

your initial state vector

with entries assigned as probabilities we start on each square

the probability we start at square 1 is 1

and the rest are 0

[coarse furnace]

So I would make 2 matrices?, a 9x1 matrix [1, 0, 0, ..., 0] and the original 9x9 matrix A?

[crystal lintel]

yes

[coarse furnace]

So given the two vectors now, what would the approach be in getting the most probable square after the second move

I've calculated matrix A^2 as:

[crystal lintel]

this describes the process pretty completely

[coarse furnace]

Would I multiply this matrix with the initial state vector? So, (A^2)*(initial state vector)

[crystal lintel]

yea

[coarse furnace]

And then I would select the highest probability from that table, and whichever column it corresponds with would be the most probable square it'll be on after the second move?

Wait no, that would make a 9x1

Whichever row it corresponds with, i.e., if row 5 was highest probability then square 5 would be most probable?

[crystal lintel]

yep

[coarse furnace]

Ahhh okay

Thank you for your help!

Sorry for the troubles o:

[crystal lintel]

[coarse furnace]

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[crystal lintel]

what trouble

[marsh citrus]

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[mental bear]

What is the missing number

[jagged forge]

5x6+5x5

[mental bear]

So whats the answer ☠️

I dont get it

[jagged forge]

use the right number to multiply the top and the left, and add them

[mental bear]

So 55?

[jagged forge]

yes

[mental bear]

Ok bet

Thanks

Its for a tapjoy ☠️

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[mental bear]

Does ? Equal.

Y. W
X. Z

[marsh citrus]

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[harsh vine]

Can someone please help me with question c

[marsh citrus]

@harsh vine Has your question been resolved?

[marsh citrus]

@harsh vine Has your question been resolved?

[marsh citrus]

@harsh vine Has your question been resolved?

[raven bloom]

hi

[marsh citrus]

@harsh vine Has your question been resolved?

[harsh vine]

yo

[cursive ermine]

e= mc

[cursive frost]

suree

[harsh vine]

that is insane

[marsh citrus]

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[marsh citrus]

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[velvet badge]

I've got the following problem for my number theory class, and I'm honestly just entirely lost. I have the result that $\left\frac{2}{p}\right)$ is 1 if $p\equiv \pm 1 (\text{mod} p)$ and -1 otherwise, but I don't know how I'd go about applying it (if that's even the right thing to do). Does anyone have tips on the right direction to follow? My class uses Burton's Elementary Number Theory, if that helps.

[velvet badge]

Corrected Latex:
I've got the following problem for my number theory class, and I'm honestly just entirely lost. I have the result that $\left(\frac{2}{p}\right)$ is 1 if $p\equiv \pm 1 (\text{mod} p)$ and -1 otherwise, but I don't know how I'd go about applying it (if that's even the right thing to do). Does anyone have tips on the right direction to follow? My class uses Burton's Elementary Number Theory, if that helps.

[elfin berry]

cat_food_sounds

[normal sky]

have you shown the two equivalent definitions of legendre symbols? if not which way are you defining it

[velvet badge]

The definition we're using is that it's 1 if a is a quadratic residue modulo p, and -1 if a is not.

[normal sky]

ok, have you seen gauss's lemma

[velvet badge]

yes!

I'm having a hard time figuring out how to apply it, though. (I'd assume that's probably the point of this problem tho)

[normal sky]

yeah so

[velvet badge]

I get the general idea of using it for computing legendre symbols at least

[normal sky]

in 2, 4, 6 ,... , p - 1, there are (p-1)/ 2 integers, and let 2k be the smallest integer larger than (p-1)/2, that is 2k is the first number showing up in the second half of the list, so we split the list up
2, 4, 6 ,... , 2(k - 1) = (p-1)/2 is the first half and 2k , 2(k + 1), ... ,p - 1 is the second half

can you see how to apply gausss lemma from here

[velvet badge]

Maybe. I'm gonna take a few to play with it

thx for not immediately telling me the answer, i rly need to get htis down myself so I dont bomb a second test

[normal sky]

*had a small typo, fixed now

[velvet badge]

I think I get it

I'm seeing the process, albeit slowly

[normal sky]

[velvet badge]

Never mind I don't see how to apply Gauss' lemma

[normal sky]

ok so in second half of the list, we need to figure out what happens if theres an even amount or an odd amount right

let $g = ord({2k, 2(k + 1), \ldots, (p-1)/2})$

[elfin berry]

chebyshev's infinite pee norm

[normal sky]

can you find a nice expression for g?

[velvet badge]

Oh shit

Hm

[velvet badge]

The second half is the set we're considering for the sake of n for gauss' lemma, right?

[normal sky]

correct

[velvet badge]

OK

[normal sky]

it might help to rewrite 2k as (p+1)/2

as a hint

[velvet badge]

Hm

[normal sky]

sorry this hsould have been ord({2k, ... , p -1 })

[velvet badge]

Ok

I'm not familiar with what ord is referring to here, sorry

[normal sky]

just the size of the list

[velvet badge]

Oh gotcha

thx

[normal sky]

ord as in order of a set ( how many elements in a set)

[velvet badge]

gotcha

Uhhh

I'm gonna try playing with this more, tysm

[normal sky]

okay, will leave this here if you get stuck

||the last element in the whole list is p -1, the last element in the first half of the list is (p - 1) / 2|| so the size of the list is ||g = 1/2( (p - 1) - (p-1)/2)||

now to apply gauss lemma, ||what conditions are needed to make g even? to make g odd?|| if you're still stuck ||look at what it is you're trying to prove in the first place|| if you're still stuck, consider what happens if ||p = 1,3,5,7 mod 8||

[marsh citrus]

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[zinc gulch]

ok so
why is 1+1=2

[desert dirge]

go read the principia mathematica by Whitehead and Russell, theres a short proof in there 🙂

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[warm dome]

Need help for part b

My answer seems to be going in loops

[zenith python]

the first integration by parts is good

but after that you do the same thing but in reverse

so you will indeed be stuck in a loop

hint: x^4=(x^2-1)(x^2+1)+1

[marsh citrus]

@warm dome Has your question been resolved?

[warm dome]

Just trying the calculation ><

@zenith python so I've split it into 2
How do I deal with the first fraction

I plus 2 in numerator in attempt to make them equal or easier to integrate but I know that's no the right way to go about it TvT

[zenith python]

first fraction is quite hard

let me suggest another way to split it :

[ \int \frac{x^4}{x^2+1} dx = \int \frac{(x^2-1)(x^2+1)+1}{x^2+1} dx = \int (x^2-1) + \frac{1}{x^2+1} dx]

[elfin berry]

LF

[warm dome]

Ohh

That looks more like it :'>

Thank you

[zenith python]

yess just plug that with the rest

youre welcome

[warm dome]

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[viscid horizon]

Hey, just a small question. Is this a valid reasoning when finding the determinant of a matrix?

[copper raven]

wdym by B + lambda though ?

[viscid horizon]

Lambda is any scalar value and B is a matrix

[fervent rampart]

it doesn't make sense to add a scalar to a matrix in the first place

[viscid horizon]

I'm aware of that but we're debating a problem with a friend and he came up with that

Only thing we know is that |B| is 2

he factored it as |A(B+2)| and then split it as |A|*|B+2|

[fervent rampart]

it doesn't make sense to factor it as A(B+2) for the reasons mentioned, but 2A = A(2I) so it does make sense to do 2A + AB = A(2I) + AB = A(2I + B)

[viscid horizon]

I see

[slow root]

Help i literally forgot everything

[viscid horizon]

Not to be rude but you should post this on an available channel

[slow root]

im sorry im new to the server idk how it works

[viscid horizon]

That's alright

[fervent rampart]

#❓how-to-get-help

[slow root]

so do i put it in the help forum?

[fervent rampart]

either in #1021175428326633542 or in any of the help channels under "math help (available), like #help-9 for example

[viscid horizon]

Exactly

So, anyway, was his reasoning correct? I tried thinking of |2A + AB| as a linear combination of A but the determinant changes with 2A so it's not valid anymore

both A and B are 3x3 matrices btw

[fervent rampart]

it's not wrong to write |2A + AB| = |A| |B + 2I| for the reasons discussed, but you should write it as B + 2I from the start

[viscid horizon]

Alright, guess he was right after all then. Thanks again, have a great day/night!

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[soft nacelle]

I need to Find the area of a segment formed by a chord 8" long in a circle with radius of 8".

[onyx flume]

do you know the formula for the segment

[soft nacelle]

no

[onyx flume]

ok

[onyx flume]

to find the area of the segment first we need to find theta

do you know how to do that @soft nacelle

[soft nacelle]

can u remind me

[onyx flume]

ok

[onyx flume]

from here subsitute the known values so you can get theta

[soft nacelle]

what's theta

[onyx flume]

the angle

[soft nacelle]

oh

ok

how do i find the sin

[onyx flume]

the sin?

substitute the known values into AB=2r sin(theta/2)

ok

so 8=2* 8 * sin(theta/2)

then simplify to find theta

is u alright?

[soft nacelle]

idk

um

[onyx flume]

o

[onyx flume]

so this become 8=16*sin(theta/2)

then 1/2=sin(theta/2)

do inverse sin to both sides

30=theta/2

multiply each side by 2

60=theta

make sense?

[soft nacelle]

yeah

[onyx flume]

ok

we have theta, and the radius

so we can find the area of the segment

lets find the area of the segment in radians

theta=60, radius = 8 so now subsitute into this

[soft nacelle]

32 (60 - sin*60)?

[onyx flume]

yeah

i hope it all makes sense

[soft nacelle]

me too
sry if not though im really bad at this

[onyx flume]

thats fine that's why we're here to help

[soft nacelle]

ty

what next

[onyx flume]

that's it

that's the area of the segment

[soft nacelle]

so how do i plug that into this thing

[onyx flume]

not sure sorry

[soft nacelle]

oh ok
thx for the help

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[still temple]

what's the f^-1 (x) = ..guys?

[quick moth]

do you know what an inverse function means?

[still temple]

i mean

yea opposite of function ig

[quick moth]

it the x values of the original function become the y values

[hard gull]

is there nothing else given?

[quick moth]

of the inverse function

[hard gull]

or do u need to find inverse of f(10)?

[quick moth]

this is because

the inverse function is just the origami function reflected over the line y=x

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

New standard notation $Origami(f)\equiv f^{-1}$

[elfin berry]

4C

[proud ice]

!original

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

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[pearl pulsar]

gradient function of a curve is given as
f'(x) = 16/x³ - 4.
given 2y - x - 6 = 0 is a normal to the curve,
find equation of the curve.

[pearl pulsar]

how do i find C after integrating

integral of f'(x) -> -8/x² - 4x + C

[hard gull]

do u really have to integrate?

dy/dx is the slope

[pearl pulsar]

i think i found the answer but i'm not sure if the method is appropriate

[hard gull]

find slope of the given line , and we know that
if they are normal (aka perpendicular) slope 1 * slope 2 = -1

[pearl pulsar]

ok so what i did was solve 16/x³ - 4 = -2

x = 2

[hard gull]

oh wait nvm i read the question wrong

[pearl pulsar]

and i plugged x = 2 into the equation

integral = normal

and got C as 14

14 is correct but not sure how it works

[marsh citrus]

@pearl pulsar Has your question been resolved?

[marsh citrus]

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[devout mist]

Is my answer correct?

[marsh citrus]

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[marsh citrus]

@devout mist Has your question been resolved?

[amber birch]

Yes that's correct

[devout mist]

Is it correct

[amber birch]

Yes that's correct

0.75 (old price) = new price

[marsh citrus]

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[cloud gorge]

Hi! I was wondering if anyone has any idea as to why they have taken f(x) as dy/dx equation and not the y equation?

[late geode]

where, wdym

[cloud gorge]

So as you can see in the question (left side) they've given a y = equation and then its derivative

so for newton raphson i assumed x- (y = (x-5)lnx / 1-5/x + lnx)

subbing in 2.4 for x

[late geode]

they're trying to find the location of the turning point

[cloud gorge]

but they've used the first derivative as f(x) and then got the second derivative for f'(x)

which is what im confused about

[late geode]

so they're interested in when dy/dx = 0

[cloud gorge]

so is that then what you do when its a turning point? f'(x) as f(x) and f''(x) as f'(x) ?

[late geode]

essentially you're trying to approximate when
1 - 5/x + ln(x) = 0
starting from x_0 = 2.4

[cloud gorge]

ah okay so in newtons method we need f'(x) = 0

and since in this case it doesnt and the derivative does we use that instead ?

bcs when finding stat. pnts dy/dx = 0 ?

[late geode]

they used f(x) to represent the function you're trying to find the roots of, not the derivative of the original y

[cloud gorge]

but what they've set as f(x) in the markscheme is what it says is dy/dx in the question tho

[late geode]

yes

[cloud gorge]

so because its a stat point we use the derivative hence them setting it as f(x)?

[late geode]

that's what i'm saying, f(x) isn't y here

[cloud gorge]

yeah okay i get it now

thankyou

have a good day

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[hot cosmos]

really lost for 7 lol

could someone draw the tree diagram out for me please 🙏

[vernal forge]

give me a sec

[still temple]

no thats against the rules

[vernal forge]

lmao

i won't add too much detail

[still temple]

are you sure? it practically is what you need to solve this

[hot cosmos]

lmao my problem is i did the tree diagram wrong, but i cant find a way to change it to make it right to get the answer

[vernal forge]

then show it

[hot cosmos]

this is def wrong but i used diff numbers several times but i keep getting funky numbers

[still temple]

what are those numbers on the top right branch

[hot cosmos]

wait hold on how this is against the rules?? (isnt this server to help with assignments that you cant do or smth)

[hot cosmos]

4/7 (red), 4/7(blue)

[still temple]

you misunderstood. It is to help guide you to solve it yourself. Doing the work for you is what's not allowed, meanwhile

[hot cosmos]

yeah i get that, i need guidance to get the right tree diagram

idc bout to answer i wanna know what i did wrong

[still temple]

Wait why exactly are you getting 7 in the denominator by the way? you are excluding a ball, not two. So why the jump from 9 to 7?

[hot cosmos]

2 further draws are made without replacement

[still temple]

right but its like

you start from 10 and pick a ball - > you end up with 9 so u pick up a ball - > you end up with 8 so u pick up a ball

so all in all 3 times

you overshot it by 1

to put it in a better way, you should label the edges of ur tree diagram with the probabilities and there should be 3 of them in total

something like this

[hot cosmos]

ahhhh thanks man

no wonder

[still temple]

yeah so i think u r fine with just replacing ur denominators with 8

and that should make ur thing hopefully right

[hot cosmos]

yaya its correct after that

thanks a lot man 🙏

[still temple]

aight good luck! sorry if i seemed confrontational earlier

[hot cosmos]

lol its ok, have a good day!

[marsh citrus]

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[dull lodge]

Hello good people I have a problem regarding Sequences.
Sum of all terms of an infinite geometric sequence is equal to 24 The sum of its terms with even numbers is 8. Calculate the first term and the quotient of this sequence.
I have translated that with the help of google so there maybe a misunderstanding but hopefully there isn't. The answers should be a₁ = 12 and q = ½

[umbral hare]

I need help with an electrical engineering assignment, i dont understand the math used in the solution.

[leaden monolith]

#❓how-to-get-help

[ruby mulch]

What is the formula of the geometric sequence?

And what does "sum of its terms with even numbers" mean?

[tranquil mist]

That answer is correct. Are you asking how to work it out?

[dull lodge]

yes

[dull lodge]

1 sec

that's literally the whole assignment

it only says word to word: Infinite sequence (an) is equal to 24, and it's Sum of even numbers is equal to 8

nothing more

After looking around I only found this formula

[tranquil mist]

If you write down the sequence of even terms, you will realise that this is itself a geometric sequence:

ar, ar^3, ar^5 ...

So you can work out its sum the same way as any other geometric sequence.

[dull lodge]

oh by using y^2 = x * z ?

[ruby mulch]

Oh even terms, I thought the terms where the value is even.

[ruby mulch]

Plug in the formula, then.

[dull lodge]

yeah but what will plugging the formula give me when i have no numbers at a1 or q

[dull lodge]

oh wait this one

[ruby mulch]

Yes, and you don't have to plug in a number.

[dull lodge]

to the both of these?

[ruby mulch]

Yep

[dull lodge]

that looks hella weird

oh wait looks like i made a mistake there

idk I've done something but It gave me 1 instead of 1/2

[ruby mulch]

Wait:
$$8=\frac{a_1}{1-q}$$

[elfin berry]

Xwtek

[ruby mulch]

That doesn't sound right.

[dull lodge]

Yeah that was supposed to be the Sum of equal numbers

[ruby mulch]

In the sequence that sums to 8, what is the first element in sequence?

Definitely not $a_1$

[elfin berry]

Xwtek

[dull lodge]

oh right

loool

so i need a2 there

which turn to a1q

that makes sense

[ruby mulch]

And what is the ratio of the adjacent elements in a sequence that sums into 8?

[dull lodge]

q^2

[ruby mulch]

And from now, it's trivial to obtain $q$ and $a_0$

[elfin berry]

Xwtek

[dull lodge]

got it

[inner zenith]

عذراً، لم أفهم ما تقصده.

عذراً، لم أفهم ما تقصده.

عذراً، لم أفهم ما تقصده.

عذراً، لم أفهم ما تقصده.

[ruby mulch]

English please?

[dull lodge]

He said he doesnt understand

idk

thanks anyway m8

still got 21 more to do

I will come back soon I guess lol

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[marsh citrus]

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[marsh citrus]

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[marsh citrus]

@still temple Has your question been resolved?

[quaint elm]

there is a lot of terminology in those images

[elfin berry]

00

[quaint elm]

that seems right to me

i think you'd prefer to use strong induction here

you don't know how big Y is but you know it's strictly smaller than X

it'll work but it means you end up writing a huge paragraph

[marsh citrus]

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[marsh citrus]

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[still temple]

Would multiply both sides by (b+rd)(b+sd) first, since it's clear that this product is positive it won't invert the inequality

The latter

You'll end up with something like
(a+rc)(b+sd)<(a+sc)(b+rd)
(a+sc)(b+rd)-(a+rc)(b+sd)>0
I think some terms cancel out there when you multiply

I subtracted the left side from the right side

I'd say just multiply them, you'll get
(ab + scb + ard + scrd) - (ab + rcb + asd +rcsd) > 0

The first and the last term cancel out

[marsh citrus]

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[still temple]

,rotate

[still temple]

#4, how do I start

[pliant tangle]

First find the radius of the circle

[still temple]

I did that

10

[pliant tangle]

The area of shaded region is the total area - area of rectangle

Ok then find the area of the rectangle

[still temple]

Is the radius of the circle the same as like CD?

[pliant tangle]

No

[still temple]

How would i relate the radius of the circle to the rectangle

[pliant tangle]

First of all, in the diagram, AD is half of the radius

So find AD

[still temple]

Okay so AD is 5

[pliant tangle]

Yes

Now AC is also the radius of the circle

We have AD and AC, by Pythagoras thereom, find CD

[still temple]

pythagorean thereom

yes

[pliant tangle]

Yep

[still temple]

So what is the part of the circle above BC

Nevermind

That is part of the 90 degree sector

Thank you bro

[marsh citrus]

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[still temple]

What am i doing wrong?

[still temple]

$$A = \M[ 8 1 ; -2 6]$$
$$x^2 - 14x + 50 = 0$$
$$(x-7)^2 = -1$$
$$\lambda = 7+- i$$
$$(A- \lambda_+ I) = \M[(1-i) 1; -2 (-1 -i)]$$
$$(A- \lambda_+ I) = \M[(1-i) 1; 1 (1 +i)/2]$$

[elfin berry]

Book Reader

[still temple]

Oh nvm

I did it right

just forgot last step

...

$$x_1 = -\frac{1+i}{2}x_2$$

[elfin berry]

Book Reader

[still temple]

bruh

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

.close

[marsh citrus]

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[marsh citrus]

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[full stream]

Hi

[fathom sun]

hi!

[full stream]

<@&286206848099549185>

My question is just why they put log a why the •a•?

Is it because we can put anything we want in the place of b?

This is the formula can anyone help?

[cyan parcel]

its log base 10

[full stream]

No no we only use letter we prove that the left side is equal to the right side thats the exercise

[cyan parcel]

?

[cyan parcel]

You trying to prove this?

[full stream]

Yes

We have these formulas

Not sure if u can read it but the bottom one states that we use log b/ log b

But they used a instead of b why is this?

[full stream]

.

[trim quest]

Yes, that formula works regardless of the base of the logarithm

[full stream]

But why where does a come from?

[trim quest]

They could have used anything, but by using a, they get log_a(a) in the numerator

Which is equal to 1

[full stream]

Oh right so we can use anything? Alright

[trim quest]

Yeah, the base can be anything in that formula, it just has to be the same in both the numerator and denominator

[trim quest]

Notice on your formula sheet, they started with log_a(x) and they used a generic base b

[full stream]

But do we constantly work out the whole equation like how do we know its gonna equal what we want to?

For like a long exercise how do we tell what is best used in the place of b?

[trim quest]

It probably won't matter most of the time

[full stream]

Wdym is there no way to tell whats best used in that place?

[trim quest]

I mean, there's all different kinds of problems, so I guess it kind of depends on the situation

but like in this case, you didn't have to use a

You could have done something like

[full stream]

This situation lets say? We just got introduced to logs so

[trim quest]

$\log_b(a) = \frac{\log_c(a)}{\log_c(b)} = \frac{1}{\frac{\log_c(b)}{\log_c(a)}} = \frac{1}{\log_a(b)}$

[elfin berry]

tatpoj

[trim quest]

Even with a different base, it's still not too complicated to get the same result

[full stream]

Euler?

[trim quest]

yeah usually I'd just use e, but I just used a generic base c here

The base doesn't actually matter

[full stream]

I am afraid I didn’t learn that formula?

Where u made it into 1/loge/loge

[trim quest]

it's the one from the bottom of your sheet here

[full stream]

Ye the first part bu what about 2nd part?

[trim quest]

That's just

x/y = 1/(y/x)

1/s is the reciprocal of s

[full stream]

Oh get it

But ye that was new to me but alr thnx

[trim quest]

Yeah, it's just the same as (5/6)(6/5) = 1

just rearranged a little bit

[full stream]

Thank u for u help

.close

[marsh citrus]

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[trim quest]

np 👍

[marsh citrus]

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[still temple]

is dis alright

[still temple]

uh x not y

forgot to replace em💀

y = x -1/2

[winged pier]

Uhh this is hard to read 😭

[still temple]

😭😭👍

lemme fix it

💀💀

[winged pier]

Can you send the original problem

[still temple]

it was to find f^-1 (x)

.close

[marsh citrus]

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[still temple]

.reopen

[marsh citrus]

✅

@still temple Has your question been resolved?

[marsh citrus]

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[thorny slate]

Hey all need someone to do a math check on a really large piece of math

[thorny slate]

Just see where / if i went wrong

[marsh citrus]

@thorny slate Has your question been resolved?

[thorny slate]

<@&286206848099549185> c;

[marsh citrus]

@thorny slate Has your question been resolved?

[marsh citrus]

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[still temple]

how does this span

if it has a 0 z component

[broken dome]

it has a z component

it's value is irelevent

it exists, whether it's 0 or 231424

[still temple]

what defines if something spans over R2 or R3

[fervent rampart]

it spans "a set in R^3"

[still temple]

but say you had a set of vectors

what would define if it spans or if it doesn't

[fervent rampart]

every set of vectors spans some vector space, the question is what vector space that is

[still temple]

What I had previously understood it to be was

they can't be scalar multiples

of eachother

and they can't have a 0 vector

this doesn't fulfill the second criteria

[fervent rampart]

that's (part of) the criteria for being a linearly independent set of vectors

the set doesn't have a 0 vector, idk where you got that

[still temple]

is the z value not a zero vector?