help-49

[echo wharf]

Any idea what you can do to solve this?

[last slate]

i drew a figure and got two cases

[echo wharf]

can you show them

[last slate]

<@&286206848099549185>

hi

[midnight plank]

@last slate Has your question been resolved?

[novel lion]

write conditions for a and check for triangles

[midnight plank]

@last slate Has your question been resolved?

[viral hamlet]

hi

[keen saddle]

@last slate I would say 2

To see how just draw the larger side

And from one end a circle of radius equal to shorter side

[keen saddle]

(depends on the angle and which side it is opposite to but mostly 2)

[last slate]

hi

i am back

[last slate]

Yes

two intersections

OK

.close

[midnight plank]

Channel closed

Closed by @humble meteor

Use .reopen if this was a mistake.

[last slate]

we may form a condition for which the ASS holds true

[midnight plank]

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[smoky meadow]

hello

[smoky meadow]

can anybody help me

[keen saddle]

!da2a

[midnight plank]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[keen saddle]

Sure

[smoky meadow]

there number 2

[keen saddle]

The image hasn't loaded for me

Lol

[smoky meadow]

wait

idk how to write in standard notation

[viral dagger]

dude why do you keep opening new channels

[lethal path]

so the 2nd one for example, 12 is the number of digits after the decimal point
100 = 1 * 10^2, 2 zeroes after the 1, for example

so after you write down 0543, that's 4 digits

you need 8 more zeroes

[smoky meadow]

what

[lethal path]

which part didn't you understand?

[last slate]

2.0543 * 10^12 = 2.0543 * 10^4 * 10^8 = 20543 * 10^8 = 2054300000000

[smoky meadow]

1.2345x 10^4

[lethal path]

using the same logic, 2345 has 4 digits
but the 4 in 10^4 means we have 4 digits after the decimal point

so just 12345

we've used up all our digits

[last slate]

Just shift decimal places dude

[pulsar peak]

shift the dot to right n times if it's 10^n

[lethal path]

yeah

[pulsar peak]

add 0 if theres no other digits

[smoky meadow]

if it positive i understand

[last slate]

and left if it's 10^(-n)

[smoky meadow]

but it negetive

[pulsar peak]

then shift it to left

[lethal path]

ah so when it's negative, you shift the decimal point left

if you can't shift the decimal point any more, just write down zeroes

[smoky meadow]

can u give me an exaple

[lethal path]

4.5 * 10^(-10) on your sheet for example

shift decimal point once to the left, so 0.45 * 10^(-9)

[pulsar peak]

0.12
check how many digits there are after the dot, 2 in this case
you can write it down as 12 * 10^-2

[lethal path]

then we need to write down 9 more zeroes to the left of the 4

[smoky meadow]

so it

[lethal path]

they need to convert scientific form to a regular number

[smoky meadow]

0.00000000045

[lethal path]

yep

[pulsar peak]

yeah well just reverse the process

[lethal path]

that's what they're having trouble with

it helps if you count out loud actually

[smoky meadow]

so the 1.2345 the negetive one

[lethal path]

every time you move the decimal point, or write down a zero, you add 1 to your total

[smoky meadow]

will be

[lethal path]

try it yourself!

[pulsar peak]

12 * 10^-2
add a 0 and a dot
0.
then check the n of 10^-n (in this case 10^-2 so n is 2)
and make it so there are 2 digits after the zero, if the number is not enough to create 2 digits just put zeros before it

[smoky meadow]

0.00004234

[lethal path]

yep

one 4, then four more 0s

which matches the 5 in 10^(-5)

[lethal path]

or you could check by shifting the decimal point 5 to the right and see if you get back 4.234

[smoky meadow]

is this correct

[pulsar peak]

314159 * 10^-9
if i wrote 0.314159 it would be 6 digits after the zero so im just putting zeros before it for it to be 9 digits after the zero
0.000314159

[pulsar peak]

for negatives you can count the digit count after the dot

2 * 10^-15 should have 15 digits after the "0."

[lethal path]

for 2 * 10^(-15) you have an extra zero

there should be 14 zeroes then a 2

to make 15 digits in total

(excluding the starting 0)

[smoky meadow]

what

[lethal path]

so 0, then 14 0s, then a 2

[smoky meadow]

but if i cahnge the o to 2 ill be only 14

[lethal path]

yes and the last 2 makes the 15th digit

[pulsar peak]

count the digits that come after "0."

[lethal path]

0.1 is 10^(-1) for example

0.01 is 10^(-2)

it matches up with the digits after the starting 0

[smoky meadow]

so what the anser to ur exp

[smoky meadow]

is there anything else wrong

[lethal path]

otherwise the rest look all correct

[smoky meadow]

ok thank you

[weary pendant]

the one where u answered $1.32\times10^{11}$ is not quite right

[grand pond]

Duh Hello

[lethal path]

yeah I didn't count wrongly I'm pretty sure

[lethal path]

it's 1.32 * 10^15

[smoky meadow]

what

it 23

[weary pendant]

it was in a previous picture

[smoky meadow]

not 32

[lethal path]

oh jesus

[weary pendant]

here

[lethal path]

no that's the same one

[smoky meadow]

yea it 1.23

[lethal path]

it's actually 1.23 * 10^15

oh right yes

[lethal path]

wait oh

[smoky meadow]

oh

[lethal path]

that's so confusing

ah yeah the answer for that should be 1.32 * 10^13

[smoky meadow]

oh yea

anything else

[pulsar peak]

converting scientific to standard requires attention so make sure to not miss anything when counting

[smoky meadow]

oh ok

[weary pendant]

counting bundles of 3 makes it easier imo

[smoky meadow]

yea

what do they mean

i gotta plus them together or waht

[pulsar peak]

any positive integer (except 1) can be written as a product of prime numbers

[smoky meadow]

example

[weary pendant]

what uve done is right

[pulsar peak]

10 = 2*5

2 is prime 5 is prime

[lethal path]

yeah what you did basically

[smoky meadow]

so i do the same as ex 3

in the picture

[pulsar peak]

yeah

[smoky meadow]

do i repeat number 2

[pulsar peak]

repeat until its not divisible by 2

[smoky meadow]

like id 360 has 2 2 2 3 3 5

do i cahnge it to 360 have 2 3 5

[pulsar peak]

then move to the next prime number

example if its 230

230 / 2
115 / 2 -> cannot be divided with 2 so you move to the next prime number that it can be divided with
1 + 1 + 5 = 7 not divisible by 3 so you move to the next prime which is 5

[pulsar peak]

360 is 2 * 2 * 2 * 3 * 3 * 5, which makes it have prime factors 2, 3 and 5

[weary pendant]

fastest way is to find the product closest to the sqrt. for example 10*12 for 120 instead of doing 2*60, but doesnt really matter much. ull get the correct answer either way and going up the list of smallest primes is the most systematic

[pulsar peak]

true but these don't really look like high school questions so i think it's better to follow just what teacher taught in this case

[smoky meadow]

im solving

guys

37 divided by what

that is a prime number

heh

so 37 dive by it self?

[weary pendant]

treat 37 exactly the same way u do with any other prime

[smoky meadow]

like this

correct or wrong

[weary pendant]

u could fine a nicer way to write it

[smoky meadow]

the thing is that the niceest

[weary pendant]

well i mean that instead of circling everything you just write

a) 120=2*2*3*2*5
b) 640=...
c) 2520=...
d) ...

below the question

[smoky meadow]

yea

[weary pendant]

what u have is right but there is some value in making things look readable for your teacher. but anyways its a small point, but worth giving a thought before you write down your answers

i gotta get on to doing my own stuff tho. gl with the rest

[smoky meadow]

ok

[pulsar peak]

you need to write the greatest one

also its not 2

[smoky meadow]

ok

[pulsar peak]

product of shared prime factors gives you gcf

[mental tiger]

Hello! Just want to point out that your answer for #4 letter d is not correct ‘cause I just double checked it by multiplying the factors you wrote and they’re not equal to 3,330

[smoky meadow]

waht

then idk tbh

that my anser and i cannot find another way

here

my caculations

[pulsar peak]

37.5.3.3.2

3330

its correct

[pulsar peak]

did you see 37 as 3 and 7?

[mental tiger]

Yes lol😭😭
Sorry 😭

[smoky meadow]

guys

i dont understand

[pulsar peak]

it's correct don't worry

[smoky meadow]

Not that

google says it six

[pulsar peak]

see

3 is also

shared

[smoky meadow]

ok

[pulsar peak]

you need to multiply the shared ones

2 * 3 = 6

[smoky meadow]

oh

[mental tiger]

Yes, 6 should be your answer for that first question

[pulsar peak]

also to check if the answer you found is correct, you can divide both of the numbers with the gcf you found

resulting numbers will be relatively prime if the answer you found is correct

[pulsar peak]

72/18 = 4
90/18 = 5
in math consecutive positive integers are always relatively prime => 4 and 5 are relatively prime

[smoky meadow]

are they correct

[pulsar peak]

yes

basically dividing them by the answer you found should result in relatively prime numbers

the concept of relatively prime numbers is important for GCF

[smoky meadow]

ok

Ex 5 correct or wrong

@pulsar peak @mental tiger

[mental tiger]

For question 5a , they are only asking for common factors of those pair of numbers , not all of the factors !

What you wrote is right but it’s only the first step to find the answers. those are not the final answers

the answers for question 5a should only be : 1, 2, 3, 4, 6, and 12

[smoky meadow]

oh

[mental tiger]

From the lists of factors you made for 24 and 36, you just have to look at the factors that are on both lists

[smoky meadow]

ok

that all

thank you

.close

[midnight plank]

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[rare sluice]

Question: There are 5 red balls, 8 black balls and 4 blue balls in a box. If two balls
are drawn randomly, what is the probability that the first ball is not red?

My approach (supposedly wrong):

Case 1: None of the balls are red, so I have to pick two balls from the non-red balls (12)
P = 12C2/17C2

Case 2: The second ball is red, hence the first ball can be either black or blue
P' = (8C1 * 5C1 + 4C1 * 5C1)/17C2

Hence, the desired probability = P + P' = 63/68

Can anyone please tell me what went wrong here and what is the correct approach to
solve this problem?

[lethal path]

isn't it just 12/17 * 16/16 = 12/17

,w (12 choose 2)/(17 choose 2) + (40 + 20)/(17 choose 2)

[grand pond]

Results provided by WolframAlpha

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

Requested by sooouth.

[lethal path]

your approach looks fine though

[rare sluice]

is the answer 12/17 or 63/68?

mine looks kinda off ;-;

63/68 is a bit too large ig

[lethal path]

yeah seems like you need to divide (8C1 * 5C1 + 4C1 * 5C1) by 2

[rare sluice]

thanks a lot

[lethal path]

no worries!

cause if you compare with (8 * 5 + 4 * 5)/(17 * 16), that's a factor of 1/2 smaller

yeah cause in 17C2 you are assuming the order doesn't matter

when clearly it does: the first ball can't be red

[midnight plank]

@rare sluice Has your question been resolved?

[midnight plank]

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[last slate]

Can anyone check this?

[viral dagger]

this is incorrect

the rest is fine i tbink

[last slate]

Oh

How is it wrong?

Whoops mb

Thanks anywayss

.

[midnight plank]

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[midnight plank]

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[ruby moat]

gusy can anyone help me with this question, im stuck on solving

[atomic magnet]

what have you tried thus far

[rugged pumice]

is that sqrt(2) * x or sqrt(2x)

[atomic magnet]

i think its sqrt(2) * x, not sqrt(2x), similarly with sqrt(3)*x

it looks like your x value is correct tbh, the question might be wrong

[ruby moat]

so the question is wrong?

[rugged pumice]

bad typesetting imo

[rare sluice]

yeah I get it now

[midnight plank]

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[midnight plank]

Please don't occupy multiple help channels.

[keen slate]

.close

[midnight plank]

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[twilit field]

This is equivalent to stating if for all integers more than 1 if $n \nmid p \forall n \in [2, \sqrt{p}]$ then p is prime

[grand pond]

A dense set

[twilit field]

The contrapositive would be if $p$ is composite, then there exists n in $[2, \sqrt{p}]$ st $n \mid p$

[grand pond]

A dense set

[twilit field]

right so far?

[runic hamlet]

yes

[twilit field]

We attempt to prove the contrapositive, which is if $p$ is composite, then there exists n in $[2, \sqrt{p}]$ st $n \mid p$
\
As $p$ is composite, it has at least 3 factors. Let the factors be $a_1,a_2,\dots p$.
\
It also follows that $4 \leq n^2 \leq p$

[grand pond]

A dense set

[runic hamlet]

you dont have n yet

you need to show n exists

[twilit field]

I have to show such an exists such that $ \exists n \in [ 2, \sqrt{p}] \implies n \mid p$

right

given that p is composite

Let there not exist an $n$ in $[2,\sqrt{p}]$ such that this is satisfied. Then $p= nq+ r ; 0< r<n$

[grand pond]

A dense set

[twilit field]

I'm lost

[runic hamlet]

you are overthinking

p being composite means that p=ab for some 1<a,b<n

can they both be bigger than sqrtp?

[twilit field]

no

ah

That;s it?

[runic hamlet]

why not

[runic hamlet]

yes

[twilit field]

then their product would be greater than p

[runic hamlet]

yes

[twilit field]

Cool

thanks

one question

Formally, this is if $p \in \Z \land p>1 \land n \nmid p \forall n \in [2 , \sqrt{p}$ then $p$ is prime

[grand pond]

A dense set

[twilit field]

so the contrapsoitive would be if $p$ is composite , then $ p \in \Z \lor p\leq 1 \lor \exists n \in [2, \sqrt{p}] st n\mid p$

[grand pond]

A dense set

[runic hamlet]

p in Z, p>1 is the given setup, you dont include it in the part to form the contrapositive of

the statement is $\forall n\in [2,\sqrt p] n\nmid p \implies p \text{ prime}$

[grand pond]

Denascite

[runic hamlet]

the contrapositive is $p \text{ composite} \implies \exists n\in[2,\sqrt p]: n\mid p$

[grand pond]

Denascite

[twilit field]

Got it

Thanks

. close

.close

[midnight plank]

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[alpine gyro]

$\lim_{x\to0} \frac{e^{-1/x^2}}{x^3}$

[grand pond]

Good

[alpine gyro]

Tried e ln, but didn’t get 0

[novel lion]

do you know lhopital

[alpine gyro]

Oh, lopital for 1/(ex^5)?

[novel lion]

flip it first

x^(-3)/e^(1/x^2)

then apply lhopital

[alpine gyro]

Didn’t work, tried it earlier.

[novel lion]

how

[alpine gyro]

Will this work?

[terse cradle]

No, you can’t combine the exponents like that

[alpine gyro]

After applying, (-3/x^4)/(-2/x^3 * e^-1/x^2)

[alpine gyro]

Using exponent x^2

So you get rid of -1/x^2

[novel lion]

keep the exponents in their place

[terse cradle]

don’t flip it and do lhopital. if u flip it, the derivative x^(-3) won’t go to a constant number

[alpine gyro]

My bad, forgot denominator.

[novel lion]

-3x^-{4}

dont flip the x part, simplify, then apply lhopital again

also that should be e^(1/x^2)

[terse cradle]

$\frac{e^{-1/x^2}}{x^3} = \frac{1}{e^{1/x^2} x^3}$

[grand pond]

Nacho Boi

[terse cradle]

do u see why those powers can’t be combined?

[novel lion]

oh they combined them, i was wondering how they got x^5

[terse cradle]

yes

[alpine gyro]

Wait, is $D[a^{f(x)}]=f’(x)a^xlna$?

[grand pond]

Good

[terse cradle]

it would be $a^{f(x)} \ln a f’(x)$

[grand pond]

Nacho Boi

[terse cradle]

bc of chain rule

let u = f(x), then $D[a^u, x] = a^u \ln a \frac{\mathrm du}{\mathrm dx}$

[alpine gyro]

Yeah, after first time using lopital (1/x)/(e^(1/x^2))

[grand pond]

Nacho Boi

[alpine gyro]

Forget constants

[terse cradle]

it’ll be easier if you don’t flip the fraction because you want at least one of the individual f(x) to go away ie. approach a constant. x^-3 won’t

x^3 will

[alpine gyro]

So lopital this?

[terse cradle]

L’Hopital this so that there is a function on top and on bottom

[alpine gyro]

[(2/x^3)e^(-1/x^2)]/(3x^2)

[terse cradle]

yes

now you’ll still have an indeterminate form 0/0, so u have to apply it again

[alpine gyro]

Then combine the x?

[terse cradle]

you could do that yes

but it will make the lhopital never end in a sense

bc then you’ll have a x^5 on the bottom which is further from a constant than x^2

[alpine gyro]

4e^(-1/x^2) (1-x^2)/(9x)

Do it once more?

[terse cradle]

double check that derivative

but then yes u will do it once more

remember also that $D[f(x)g(x)] = f’(x)g(x) + g’(x)f(x)$

[grand pond]

Nacho Boi

[terse cradle]

for the numerator

[alpine gyro]

Isn’t it using the rule for fraction?

[alpine gyro]

Numerator is a fraction.

[terse cradle]

the quotient rule? since it’s e^(-x^2) you could do quotient rule instead here but product rule will also work

[alpine gyro]

I got this using quotient rule

[terse cradle]

$4e^{(-1/x^2)} (1-x^2)/(9x)$

[grand pond]

Nacho Boi

[alpine gyro]

$4e^{(-1/x^2)} \frac{1-x^2}{9x}$

[grand pond]

Good

[terse cradle]

this is still slightly off, also the derivative of the denominator is wrong, d/dx 3x^2 ≠ 9x

[alpine gyro]

Right, forgot that.

And you’ll get the answer by applying lopital once more time?

[terse cradle]

yes

[alpine gyro]

Alright, thanks for the help:)

.close

[midnight plank]

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[rough breach]

hello am i right on the answer being b here

[placid spoke]

yep 👍

[rough breach]

thank you

[mental tiger]

Yes
B is the correct answer

[midnight plank]

@rough breach Has your question been resolved?

[mental tiger]

This is how you solve the given formula for x :

[rough breach]

yepp thank you and this would be d right?

[midnight plank]

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[rough breach]

.reopen

[midnight plank]

✅

[worthy wing]

Yes it is correct

[rough breach]

thanks

[midnight plank]

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[obtuse totem]

can you use this factoring method for complex, like how you would for reals

or can you only use the quadratic formula

[subtle zinc]

ur using it with real numbers anyways

a and b are real

factoring and quadratic formula are equivalent

[obtuse totem]

ok

[nova yoke]

how did you go from a^2 to (a^2)^2

this line is very unclear

[obtuse totem]

multiplied everything by a^2

so I get a=-2, 2, or i

[nova yoke]

or -i

[obtuse totem]

so how do you determine the right solution

because the answer only shows 2+i

[nova yoke]

well a is supposed to be real, right?

so it can't be i or -i

[nova yoke]

certainly this is not the only answer

[nova yoke]

if w is a square root of z, then so is -w

[nova yoke]

well that's wrong

[visual tiger]

Does sqrt here mean principal root?

[obtuse totem]

yes

[visual tiger]

Meaning argument < pi?

[obtuse totem]

I think sqrtroot has principal of -pi/2<sqrt(argz)<=pi/2?

[visual tiger]

So real part > 0

Or non-negatively imaginary number

[obtuse totem]

wait no the imag could be negative

if its -pi/2

[visual tiger]

-pi/2 < ...

But I don’t think that's the case

[obtuse totem]

so I think img could also be neg?

[midnight plank]

@obtuse totem Has your question been resolved?

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

this was the question I had yesterday

Im not sure how to compare its polar form

ill redo and send a picture

I don’t get it

honestly

im just gonna move on

how is this equation false?

[upper crag]

they explain it right there

[obtuse totem]

but 2e^(ipi) is -2?

pi is ok because it’s -pi < arg(z) <= pi

@upper crag

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

<@&286206848099549185>

I dont understand how my solution is wrong, because 2e^(ipi) is -2, and pi is ok because it’s -pi < arg(z) <= pi which satisfies principal branch

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

.close

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[calm quest]

can someone explain how from part ii) they got this:

[lethal path]

the expression is cyclic

so if you replace $a \mapsto b, b \mapsto c, c \mapsto a$, you get the same thing

[grand pond]

south's secret twin brother

[lethal path]

so replacing a + b - c now
you do indeed get b + c - a

then c + a - b

[calm quest]

ohhhh so its like pretending they are the same thing?

[lethal path]

yeah it's the principle of substitution

[calm quest]

rightttt

[lethal path]

you can let a, b, c be whatever since they are variables

[calm quest]

makes sense!

thank you for your help

[lethal path]

for example, you can substitute $y \mapsto -y$ to get an expression for $\cos(x - y)$ if you know $\cos(x + y)$

[grand pond]

south's secret twin brother

[lethal path]

no worries!

[calm quest]

👍

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[twilit field]

i f $f(x)=x^3$, find $f'(x^2)$

[grand pond]

A dense set

[twilit field]

Now I do know that I can simply derive the chain rule

but that involves treating the derivative as a fraction, which I don't particularly like

[humble torrent]

U don’t need chain rule for this?

[twilit field]

I know

I know it's just f'(x^6)

so 6x^5

[humble torrent]

[twilit field]

Is that wrong?

[humble torrent]

I mean what’s f’(x) first off?

[twilit field]

3x^2

[humble torrent]

Okay so what’s f’(x^2) in that case?

[twilit field]

$2xf'(x^2)$

[grand pond]

A dense set

[humble torrent]

?

[twilit field]

So $6x^5$

[grand pond]

A dense set

[humble torrent]

No

[lofty girder]

x becomes x^2

[humble torrent]

f’ is a function

[twilit field]

I know

[humble torrent]

And you’re plugging in x^2

So you have a formula for f’(x)

Now plug in x^2

[twilit field]

I know

[sharp coral]

the notation is a bit confusing, but i would interpret that as [ \left. \odv fx \right|_{x^2} ] rather than [ \odv f{(x^2)} ]

[grand pond]

cloud

[twilit field]

so just $3x^4$

[grand pond]

A dense set

[humble torrent]

Yeah

[lofty girder]

Yes

[twilit field]

That feels wrong

[humble torrent]

f’ by convention is f’(x)

[twilit field]

pretty sure they're saying if g(x)=f(x^2), find g'(x)

[humble torrent]

No that’s different

That would be written as (f(x^2))’

[lofty girder]

f'(x) is just a function like f(x)

[humble torrent]

Or with the notation cloud used

[sharp coral]

if we were to write out the chain rule you could say
(f(2x))' = f'(2x) * (2x)'
so this comes out to an intermediate step in the chain rule

[humble torrent]

Inherently I think the prime notation is actually less confusing, as x here is just a dummy variable and should not matter

[twilit field]

Lol, just read the footnote, spivak expected readers to make exactly this mistake

[humble torrent]

This becomes more evident once you deal with partial derivatives imo, as the fraction notation can be severely misleading

[humble torrent]

Which mistake?

If you’re careful about how to think about functions this is not ambiguous at all

[twilit field]

treating f'(x^2) as (f(x^2))'

[humble torrent]

Oh right

Yeah so one way to see why they’re different is to write it as a composition

[twilit field]

I see why they're different now

[humble torrent]

Nice, in essence the difference is this

f’(x^2) is just f’(x) composed with x^2

while (f(x^2))’ is (f o g)’(x) where g(x) = x^2

[twilit field]

Got it

Now I'd like to prove the followinf

$g(x) = f(x+c) \implies g'(x)=f'(x+c)$

[grand pond]

A dense set

[twilit field]

We start by finding the derivative of $g(x)$, which is defined to be $\lim_{ h \to 0} \frac{g(x+h) - g(x)}{h}$. We also notice that the derivative of $f(x+c)$is defined as follows : $\lim_{h\to 0} \frac{. f(x+c+h) - f(x+c)}{h}$.
\
If the derivatives are indeed equal , it follows that $\lim_{h\to 0}( \frac{g(x+h) - g(x)}{h}- \frac{. f(x+c+h) - f(x+c)}{h})=0$
\
We now ask the reader to notice that, by definition, $g(x+h)=f(x+c+h)$ and $g(x)=f(x+c)$
\
it thus follows that we have $\lim_{h\to 0} \frac{0}{h}=0$
\
This concludes our proof

[grand pond]

A dense set

[twilit field]

Is this fine?

[humble torrent]

That substitution you speak of at the end could of been done in the very first step or two

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

Hmm, okay

[twilit field]

Do I also have to prove that if g(x) is differentiable, then f(x+c) is differentiable

[humble torrent]

I mean by assumption they’re equal to each other for all x, so if either is differentiable the other must be too

You don’t really need to bother thinking about they’re if differentiable or not. Their limit will be the same if it exits

And it’s just by substituting the thing you said at the end

[twilit field]

hmm

okay

Now I would like to prove if $g(x)=f(cx) \implies g'(x)=cf'(cx)$

[grand pond]

A dense set

[twilit field]

I've done this before, but I'd like to do it again

By definition $g'(x)= \lim_{h\to 0} \frac{g(x+h)-g(x)}{h}$.
\
We also have $f'(cx) = \lim_{h\to 0} \frac{f(c(x+ \frac{h}{c})) - f(cx)}{c\frac{h}{c}}$

What am I doing wrong here

[grand pond]

A dense set

[twilit field]

ah

.close

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[sinful trout]

i need help finding the value of the following expression

[grand pond]

ren

[sinful trout]

A is a symmetric matrix (A^T = A)

note that this is essentially just the higher-dimensional analogue of the generalized gaussian integral

[humble torrent]

I mean it doesn’t need to converge as a start

[sinful trout]

OH mb

i forgot the minus sign

i'm stupid, aren't i?

[humble torrent]

Even in that case

[sinful trout]

why not?

[humble torrent]

Pick A = 0

[sinful trout]

yeah i was gonna say apart from that

;-;

[humble torrent]

Then pick b so that it doesn’t converge in a trivial way

[sinful trout]

such that A is a symmetric matrix not equal to 0I or -bI

[humble torrent]

What’s the XY to this problem

[sinful trout]

i made it up lmao

most of the problems i post here are of my own accord

[humble torrent]

Well I mean again this need not converge

[sinful trout]

what now

[runic hamlet]

diagonalize, split it up into several integrals. then you can also see whether those converge which will tell you whether the original one does

[dawn dagger]

Couldn't you factor x and then have x(Ax+b) = (Ax+b)x ?

[sinful trout]

that
doesn't help much

[humble torrent]

Just pick A and b suitable so that it’s positive

But yeah

[sinful trout]

yeah but i wanna figure it out for a general integral

[sinful trout]

could you explain if it's not too tiresome

[humble torrent]

Right I mean yeah

[sinful trout]

RDR^T = A, where R^T = R^-1

is what i remember

[runic hamlet]

explain what part of it

[sinful trout]

i was gonna say all of it, but imma try it myself and let you know if i get stuck

[runic hamlet]

yes, then sup y=R^Tx

[sinful trout]

hm

ok

[runic hamlet]

then write out how the exponent looks in terms of the y_i

[sinful trout]

uhhhhh

okay

(LA is not my strong suit)

[runic hamlet]

D is diagonal, so that is chill

[sinful trout]

(why is it chill?)

[runic hamlet]

you will see

[sinful trout]

uhm

[runic hamlet]

for convenience it wouldnt hurt to just look at the x^TAx term for now and ignore the rest

[sinful trout]

hang on i really need to revise my LA

[sinful trout]

??

[runic hamlet]

well just so you dont have to write down the rest when that doesnt change for now

[humble torrent]

Maybe alt. it’s possibly useful to write A as the corresponding polynomials of projections

[sinful trout]

wut

explain

hmmmm

[humble torrent]

Like from the spectral theorem you can write A as a sum of projections with its eightvalues as coefficient, but maybe that’s not useful

[sinful trout]

the what now

i don't do LA my guy

[humble torrent]

Then nvm

[runic hamlet]

really just a restatement of the fact that A is diagonalizable

[humble torrent]

Yeah

[sinful trout]

also you mean eigenvalues?

[runic hamlet]

sub y=R^Tx

[humble torrent]

Oops yea lol

[sinful trout]

now what? doesn't seem much better

ahk

dc lagging so hard

[grand pond]

ren

[sinful trout]

we get exp(-y * Dy - b * Ry)

[runic hamlet]

get rid of that x in b*x

[sinful trout]

so

do i just

forget it

for now

[runic hamlet]

if y=R^Tx, then x=... ?

[sinful trout]

Ry

[sinful trout]

and now this is just a quadratic in y, awesome

unless i'm tripping

never mind, i'm tripping

[runic hamlet]

it was quadratic in x before and we did a linear sub, so not terribly surprising

[sinful trout]

*facepalm*

[runic hamlet]

now write out y^T D y

[sinful trout]

?

[runic hamlet]

in terms of y_i

[sinful trout]

okay

$\sum_{i=1}^{n} y_i^2\lambda_i$

lemme think this thru

nope.

uhmmm

[runic hamlet]

no

it needs to be a number

[sinful trout]

yeah but i don't have the columns of D

[runic hamlet]

wdym you dont have them

how does D look

[sinful trout]

eigenvalues of A, ik

[runic hamlet]

open paint, draw D

[sinful trout]

`brb

[runic hamlet]

starts at i=1 but yes

[sinful trout]

o typo

[grand pond]

ren

[sinful trout]

exp( -(that sum) - b*Ry)

[runic hamlet]

yes

[sinful trout]

so we can deal with the sum later

now for b*Ry

i gtg somewhere rn, thanks for all the help denascite! :D

really appreciate it

.close

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[solid pilot]

i need to check if im doing this part of my review right

[last slate]

hi

[solid pilot]

hi

and this one

[raven abyss]

This one is 180 degrees about the origin

This would be good if ur reflecting across y axis

[raven abyss]

Almost there

[solid pilot]

oh? what do i need to change

[solid pilot]

whats wrong with it? (explain pls)

[raven abyss]

One of the points is wrong, when you’re reflecting across the y axis, each point becomes (-x,y)

[solid pilot]

is it point (8,2)?

[raven abyss]

What is (8,2)

[solid pilot]

oops sorry i meant (8,-2)

[signal seal]

I think you are looking at the wrong image.
What MoonCaik said seems to be in regards to the first image (reflection across the y axis), not the second image

[solid pilot]

ohhhh

is it point )6,4)?

[raven abyss]

I’m not sure what you’re talking about 😭

[signal seal]

There is no point at (6, 4)

[solid pilot]

dude

sorry i keep hitting the wrong keys, theyre right next to eachother

i meant (5,3) 😭

this one

[signal seal]

(5, 3) is in the correct location.
It is the reflection of the point (-5, 3)

So the wrong one is one of the other two points.

If you reflect points across the y-axis, the y value stays the same and the x value gets inverted.
So a point (x, y) reflected across the y-axis would be at (-x, y).
A point (-5, 3) reflected across the y-axis would be at (5, 3)

With that in mind, look at the points you have been given and see if the points you put in are at the correct locations.

[solid pilot]

so would it be (3,-2),(3,-8),(5,3)?

sorry if im not picking this up right, math isnt my strongest subject

[signal seal]

No worries.
The given triangle is the Grey one, right?
Try just writing down which points make up that triangle first.

[solid pilot]

(-5,3), (3,-2), (-2,-8) <- grey triangle

so would it then be (5,3), (-3,2), (2,8)?

[signal seal]

Almost.
You took (-x, -y) for the last two points instead of (-x, y)

The y value stays the same for that

[solid pilot]

so just 2,-8

and 3,-2

[signal seal]

Yes, correct!

Those are the 3 points of the new triangle.

[signal seal]

That's correct!

[solid pilot]

yay!!

hb the other one?

was it right or do i need to swtich it up

[signal seal]

For the other one you actually reflected it across the y-axis
If that was the task, it would have been correct, but the task says to "rotate 180° around the origin"
The origin is the point (0, 0)
What you need to do is imagine you put a pin into that point and rotate the entire graph like that.

But because it is 180°, you can do something easier.

Because it is rotated 180°, the new points are basically (-x, -y)

[solid pilot]

(-3,8),(-8,-2),(0,-6) so those would turn into (-3,-8),(-8,-2),(0,-6)?

(i found writing out the points given helpful lol

[signal seal]

Not really.
You just put a - in front of every number.
But if you have a negative number to begin with, it would turn positive

[solid pilot]

ohh so it switches

[signal seal]

It's (-x, -y), so if you insert the values x = -3 and y = 8 for example it would be
(- (-3), - (8))
. - (-3) is the same as positive 3 and - (8) is negative 8, so the point (-3, 8) would be (3, -8)

[signal seal]

That looks correct to me

[solid pilot]

ahh

alright, thats all though

thanks for the help. I understand it better now

WAIT

i have one more if thats ok

[signal seal]

I'm glad to be able to help

[signal seal]

Sure

[solid pilot]

i sent it

[signal seal]

That also looks correct

[solid pilot]

thank you! thats all

should i close this ticket?

[signal seal]

If that's all, you can close it

[solid pilot]

.close

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[glossy compass]

find he intervals over which the graph of f is concave upward or concave downward

[glossy compass]

i dont understand how its increasing from (0, infinity)

when the critical point is -2

and not 0?

[sage helm]

your function is only defined for x >= 0

[glossy compass]

huh 😭

how do we know that tho

[visual tiger]

finding the domain of f

[oak slate]

try calculating $(-2)^{5/4}$

[grand pond]

kaue

[visual tiger]

for which values of x does f(x) make sense

in particular you'll find that x^(1/4), the fourth root of x, is pretty difficult to obtain with a negative number x...

[glossy compass]

i know its undefined if f is less than 0 but

shouldn't we able to find it analytically??

[visual tiger]

not if f is less than 0, if x is less than 0

[glossy compass]

oops yea

[visual tiger]

find what analytically?

the conditions for f to be defined is "everything that appears is well defined"

so x^(1/4) has to be defined

condition for even root to be well defined: x >= 0

same for x^(5/4) = (x^5)^(1/4)

x^5 >= 0

or x >= 0

[glossy compass]

soo

everytime theres like a fraction as the exponent

x has to be greater or equal to 0 ??

[visual tiger]

if the denominator of the fractional exponent is even

x^(1/3) = cube root of x has no limitations in domain

(-1)^(1/3) = -1 is defined

etc...

[glossy compass]

ohh

ty

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[twin kernel]

What I do

Divergence test?

[subtle zinc]

what's that?

examine the convergence of what's inside the cos

[paper inlet]

haha this is cool

[twin kernel]

Yeah how

[twin kernel]

Like if it’s not =0 it immediately diverges

Idk

[subtle zinc]

what

what's the limit of n+1 as n goes to infty

do you know what a limit is

[twin kernel]

Inf

[twin kernel]

Yeah

[subtle zinc]

limit n/(n+1)

?

[twin kernel]

1

[subtle zinc]

limit n*pi/(n+1)

[twin kernel]

Pi/n / 1+1/n

0/1

=0

@subtle zinc

[subtle zinc]

no

[subtle zinc]

it's just pi times this

so the limit is pi

so the whole thing is cos(pi)

[twin kernel]

Oh

But what did I do wrong in mine

[grand pond]

Steel

[midnight plank]

@twin kernel Has your question been resolved?

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[faint nova]

I want to find the surface area of a circle of equation:

[faint nova]

<@&286206848099549185>

[tight pagoda]

you guys are too noob, this is litteraly very ez

[faint nova]

OK

then what, ik that this equation contains a

all real numbers

no imaginary part

then I can't, move on

@tight pagoda then help

[tight pagoda]

the answer of module of z at power 2 + module of z + 8 + (z + z ) +zz =8

is

z=0

dummy

[faint nova]

it's8 × (z+z )

dummy

@tight pagoda you speechless

<@&286206848099549185>

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[inner kestrel]

@mossy glade u there

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[meager ore]

Prove that ( f : \mathbb{R} \to \mathbb{R} ) is continuous if and only if for every ( X \subset \mathbb{R} ), it holds that ( f(\overline{X}) \subset \overline{f(X)} ), where ( \overline{X} ) is the closure of ( X ).

[grand pond]

Halex

[meager ore]

Let ( y \in f(\overline{X}) ), then there exists ( a \in \overline{X} ) such that ( f(a) = y ). Since ( a \in \overline{X} ), we have that there exists a sequence ( x_n \in X ) such that ( \lim_{n \to \infty} x_n = a ); this and the fact that (f) is continuous implie that ( \lim_{n \to \infty} f(x_n) = f(a) = y ) and thus ( y \in \overline{f(X)} ).

[grand pond]

Halex

[meager ore]

Is this right

[midnight plank]

@meager ore Has your question been resolved?

[midnight plank]

@meager ore Has your question been resolved?

[fallow scarab]

looks right for one direction

[meager ore]

thx, kinda stuck in the other direction

[fallow scarab]

might be easier to use the topological definition of continuous

[meager ore]

what is that definition?

[fallow scarab]

Oh if you didn't learn it then don't use it

[midnight plank]

@meager ore Has your question been resolved?

[midnight plank]

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[autumn fern]

I know how to do long division but i keep getting the wrong answer with like triple fractions

[gusty birch]

try showing your work so far, maybe theres a small calculations mistake

[autumn fern]

ok ill write it out on paint rq

yk what i think i figured it out

i forgot minus negative is addition

maybe i should use paint to double check my work

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[gusty birch]

okay ahahah

[midnight plank]

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[alpine gyro]

Show that $f(x)=\frac{1}{x^3} e^{-1/x^2}$ is differentiable at x = 0

[grand pond]

Good

[alpine gyro]

So I'm just suppose to take limit of $\lim_{x\to0}f'(x)$?

[grand pond]

Good

[balmy lintel]

Not exactly

What are the conditions that say a function is continuous at a point?

There are 2

[alpine gyro]

I just came up with continuity.

[balmy lintel]

? Sorry, could you repeat that?

What do you mean by "came up with continuity"?

Oh wait, my definition is stupid

yes sorry

I meant

[alpine gyro]

Oh, I thought you asked what conditionss for differentiable.

[balmy lintel]

Yeah that's what I meant to say

mb

What are the conditions that say a function is **differentiable **at a point?

[alpine gyro]

That the functions is continuous and monoton?

[balmy lintel]

Monoton?

[alpine gyro]

Strictly increasing/decreasing.

[balmy lintel]

Not particulary... at least, not how I 'm interpreting it

From what I know, it needs to be continous and smooth

pardon my spelling

[alpine gyro]

Alright, how will this help?

[balmy lintel]

So we know to to get continuity

and that's by saying the limit as x approaches 0

[ocean hazel]

wait how is it differentiable at 0

at 0 its undefined

[balmy lintel]

is it undefined at 0?

[ocean hazel]

there is 1/0

[balmy lintel]

Oh it is

😭

Wait a minute

huh...

I don't think it's differentiable then

[alpine gyro]

This is the function, where first expression is defined $x \neq 0$ and second is defined $x=0$.

[grand pond]

Good

[alpine gyro]

Will that help?

[balmy lintel]

Okay...

yes

We need to make sure that the limit at x -> 0 is the same as f(0)

for continuity

[alpine gyro]

Yes, I did lim f(0), it was 0.

[balmy lintel]

okay, that's good

so we know it's continuous

now we need to prove that the slope is smooth

do you have an idea on how we might approach this?

[alpine gyro]

Not at all.

[balmy lintel]

hint: we have derivatives

okay

Well, what if teh derivative going towards is the same as the derivative after?

would that make it smooth?

[alpine gyro]

Yeah, I tried it, but after first time lhopital, it became a big fraction.

[balmy lintel]

okay

We shouldn't need lhopital here

but let's derive the $x^{-3}\cdot e^{-x^{-2}}$

[grand pond]

Oreo

[balmy lintel]

I made it into exponents so we don't have to do like quotient rule and that fun jazz

[alpine gyro]

Yeah, I got the derivate as $e^{-1/x^2} \frac{2-3x^2}{x^6}$

[grand pond]

Good

[balmy lintel]

uhhhhh

let me derive it for a sec

one min

So in a blob

I got

$-3x^{-4}\cdot e^{-x^{-2}}+x^{-3}\cdot e^{-x^{-2}}\cdot-2x^{-3}$

[grand pond]

Oreo

[balmy lintel]

So let me simplify that

Yeah...

i don't think I got hte same thing as you

could you show your work?

[alpine gyro]

$$\frac{\frac{2}{x^3}e^{-1/x^2} x^3-e^{-1/x^2}3x^2}{x^6}$$ using quotient rule

[grand pond]

Good

[balmy lintel]

Oh god

[alpine gyro]

Did I made a mistake?

[balmy lintel]

Let me look over it

So

from what I can see

we derive the e^{-1/x^2} first

and multiply that by x^3

so does $\dfrac{d}{dx}[e^{-1/x^2}]=\dfrac{2}{x^3}\cdot e^{-1/x^2}$?

[grand pond]

Oreo

[alpine gyro]

Yeah, isn't it?

[balmy lintel]

uhhhhh

oh it is

huh

wait

no...?

what happened to the negative in the -1?

[alpine gyro]

$D[-1/x^2]=-1\cdot -2 \cdot 1/x^3$

[grand pond]

Good

[balmy lintel]

oh yes

sorry, I'm a bit frazzled

okay

and the second one is right

huh

do we need lhopital?

[alpine gyro]

I got the answer.

It was many lhoptial.

[balmy lintel]

oh no

if it is...

that's a lot of work 😭

[alpine gyro]

Dunno, I guess I did it 3 or 4 times.

Well, thanks for the participation:)

[balmy lintel]

good luck man

that's a lot of stuff to do

[alpine gyro]

Yeah no, I did it already, and the answer would be $lim_{u\to0} \frac{12}{e^{1/u}}$ where $u=x^2$

[grand pond]

Good

[alpine gyro]

.close

[midnight plank]

Channel closed

Closed by @alpine gyro

Use .reopen if this was a mistake.

[midnight plank]

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[obtuse totem]

can you take the slopes of complex cartesian points, and the negative reciprocals will show the two line segments are perpendicular?

[obtuse totem]

cause dot product for sure works

but just wondering if negative reciprocal also works

[rain wasp]

yes, in general, 2 lines are perp if ab = -1, where a and b are their slopes

[blissful totem]

if u can show one is a scalar multiple and rotated by 90 degrees of the other it also works

ie, if a, b are complex numbers

and a = i * k * b

[obtuse totem]

perfect! thanks

[blissful totem]

where k is a scalar that is non zero

then it also works

actually negative reciprocal should work even for complex numbers

[obtuse totem]

lmao cause I just did that on an exam but all of my friends used dot product

[rain wasp]

the fact that its C doesnt really matter, since C is isomorphic to R^2

[obtuse totem]

I just like highschool math🤡

.close

[midnight plank]

Channel closed

Closed by @obtuse totem

Use .reopen if this was a mistake.

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Remember:
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[dim pilot]

Could someone please help me with a differential equation problem?

[lavish venture]

if only it were possible to read

[dim pilot]

Why isn’t it?

Handwriting?

[lavish venture]

because your handwriting is illegible

[dim pilot]

I’m looking for y_1 and y_2

My homogeneous equation is y_h(t)= c_1e^t+c_2te^t

[lavish venture]

<@&268886789983436800>

[dim pilot]

Is y_2 = te^t?

Oh ok. Looks like I’m good then. Thanks!

[lavish venture]

type .close on its own

[midnight plank]

@dim pilot Has your question been resolved?

[night egret]

!done

[midnight plank]

If you are done with this channel, please mark your problem as solved by typing .close

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[upbeat plinth]

One message removed from a suspended account.

[upbeat plinth]

One message removed from a suspended account.