help-49

[last slate]

^

[simple field]

which matrix

[last slate]

1 min

i will send the question again

$A^, = A^-1$
This is how i did it
Assuming $B^-1AB = X$
And applying DET on both sides
I would be left with $"IAI=IXI"$

[grand pond]

Cnidarian

[last slate]

so

[simple field]

skew symmetric means A' = -A not A^-1

[last slate]

Yes but my question is about inverse

[simple field]

oh right sorry youre saying A' means the inverse of A?

[last slate]

Can we can A is skew symmetric if X is skew symmetric

[last slate]

yes

[simple field]

ok

[last slate]

^

[simple field]

$\mathrm{det}\left(B^{-1}AB\right) = \mathrm{det}\left(B^{-1}\right) \cdot \mathrm{det}\left(A\right) \cdot \mathrm{det}\left(B\right) = \mathrm{det}\left(B\right)^{-1} \cdot \mathrm{det}\left(A\right) \cdot \mathrm{det}\left(B\right) = \mathrm{det}\left(A\right)$

is true

[grand pond]

Acman

[simple field]

but i dont see how that helps with skew symmetry

[last slate]

for a skew symmetric matrix DET A is 0

[simple field]

ah yeah

ok are you SURE like 100% that A' means A inverse and not A transpose

[last slate]

According to the question A' is transpose

but

say A' means inverse

[simple field]

why

are you saying, ignoring the question and just you want to think about if any of these statements are true if its B^-1 A B instead ?

[last slate]

yes

[simple field]

right ok ok

we can say that B' A B is skew symmetric iff A is skew symmetric

but we cannot say for sure anything about the symmetry of A or B' A B

[last slate]

Okay

so my method will not always work

[simple field]

can you be more specific?

which method and work for what?

[last slate]

This one

[simple field]

what you have written there is not totally correct

oh do you mean for IAI to be |A|

right yes ok

it is always true that det X = det A

[last slate]

okay

what about A's nature when the nature of X is given

[simple field]

this is all you can say

actually sorry no thats not right

1s

you can say that if X is skew symmetric A has determinant 0

and conversely if A is skew symmetric then X has determinant 0

[last slate]

For a matrix having Det 0 doesn't mean its skew symmetric ?

[simple field]

not necessarily

[last slate]

oh

[simple field]

$\begin{vmatrix} 0 & 1 \ 0 & 0 \end{vmatrix} = 0$

[grand pond]

Acman

[cobalt swan]

Determinants of every skew symmetric matrix is zero

[last slate]

Okay,that clears it up

[cobalt swan]

But other way round is false

[last slate]

Okay

Thanks everyone

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[steep quest]

this is one of the final parts of the problem, i need to find k for which modulus will be the lowest, what does that mean?

[steep quest]

In theory lowest modulus is 0 right

why cant i just say k = 0

i got the answer, k = 3/5 but im not sure why

[sage helm]

you need to minimize that expression

whats the minimum value of x^2

[idle phoenix]

try squaring both sides

x^2 is always +ive

[midnight plank]

@steep quest Has your question been resolved?

[midnight plank]

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[twilit field]

so this is the same as

$r^2-16r(sin(\frac{\pi}{3}+ \theta )+15$

[grand pond]

ƒ(Why am. I here)=I don't Know

[sage helm]

why not just convert it to cartesian right away

[twilit field]

how do I do that?

x=rcos(\theta)

and y=rsin(\theta)?

[sage helm]

[r\cos \theta = x \quad r\sin \theta = y ]

yeah

[grand pond]

Bob l'éponge

[sage helm]

so $r^2 = x^2 + y^2$

[grand pond]

Bob l'éponge

[twilit field]

wait, what

$(x^2+y^2)^2+\sqrt{3}x-8y+15=0$

right?

[sage helm]

you need to get rid of r and theta entirely

[grand pond]

ƒ(Why am. I here)=I don't Know

[sage helm]

uh no

that is also not a circle

[twilit field]

yeah, just realised

got it

thanks

x^2+y^2+....

yeah?

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[sage helm]

yep

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[twilit field]

so I was thinking of starting with $z= e^{i \theta}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so I have $e^{-i\theta}-e^{2i \theta}= i(z^{-i\theta}+e^{2i\theta})$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

now I know that multiplying by $i$ rotates the number by pi/2

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so I have

$e^{-i \theta}-2e^{i \theta} = e^{\pi/2 - i \theta)}+ e^{\pi/2 -2i \theta}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

let $e^{i\theta}=u

so I have to find the number of solutions to

$\frac{1}{u}-2u=\frac{e^{\frac{\pi}{2}}}{u}}+\frac{e^{\frac{\pi}{2}}{u^2}$

[grand pond]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

$$ \frac{1}{u} - 2u = \frac{e^{\frac{\pi}{2}}}{u} + \left( \frac{e^{\frac{\pi}{2}}}{u^2} \right) $$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

now what?

[last slate]

wow

[twilit field]

this is from JEE adv 2022

[last slate]

you could solve it by quadratic

[twilit field]

I know other ways to do it

but I thought using rotations would be elegant here

[last slate]

my way would have been to take z= x+iy making the solution really big

[twilit field]

yeah

that's the problem

,w \frac{1}{u} - 2u = \frac{e^{\frac{\pi}{2}}}{u} + \left( \frac{e^{\frac{\pi}{2}}}{u^2} \right)

[grand pond]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
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Requested by physicsrocks

[last slate]

solutions are (e^pi)/2 and 1/2

[twilit field]

huh

desmos says only 1

[twilit field]

how?

[last slate]

mb

nvm

[fresh sparrow]

where did r go?

[twilit field]

I assumed r=1

[hard umbra]

you can prove it

rewrite as
[ \f {1 - \f {z^2} {\conj z}} {1 + \f {z^2} {\conj z}} = \f {1 - w} {1 + w} = i, \quad w = \f {z^2} {\conj z} ]

[hard umbra]

by the calculation we did earlier

[twilit field]

ah

[hard umbra]

(1 - w)/(1 + w) is imaginary iff |w| = 1

[twilit field]

yes

[hard umbra]

with this, you can rewrite the original equation into
[ 1 - z^3 = i(1 + z^3) ]

[hard umbra]

you can make quick work of this

[twilit field]

so that gives me $z^3=-1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[hard umbra]

probably not

[twilit field]

$1-z^3=i+iz^3$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

ah

[hard umbra]

(1 - i) = (1 + i)z^3

,w (1 - i)/(1 + i)

[twilit field]

yes

[grand pond]

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[hard umbra]

anyway its a cubic so its most likely just 3

[twilit field]

I see

thanks!

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[west iron]

snoseph

[hard umbra]

mogged

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[twilit field]

$\frac{\left(a\sin\left(x\right)+\sin\left(2x\right)\right)}{x^3}$, find the value of a that makes this function continuous at $x=0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[carmine sigil]

No value of a does that

The function is undefined no matter what value of a is chosen

[twilit field]

let make sure the question is right

ah yeah

and $x \neq 0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$\frac{a\left(x-\frac{x^3}{6}\right)+2x-\frac{\left(2x\right)^3}{6}}{x^3}$

so a=1 works

no?

[west iron]

That is not the right Taylor series

[twilit field]

,w taylor series of sin(x)

[grand pond]

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[twilit field]

ah

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so a=-2 should work

right?

oh

right

no a

got it

[west iron]

it should, though the question is still pretty poorly defined

[twilit field]

a=-2 will give a 0/0 form though

I mean 0/x^3

[west iron]

If you're allowed to add a point (0,c) wherever you want then it's still continuous

[hard umbra]

@junior flower

[twilit field]

hmm

ok

thanks!

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[twilit field]

thanks

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[twilit field]

I was thinking of using induction

so let's start with $n=1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[fresh sparrow]

start with n=2, cuz you need n-1

[twilit field]

oh yeah

[subtle blaze]

No then you’ll need to start with n=1 and n=2

Oh wait maybe not

[twilit field]

n=1 works

1 digit

[sage helm]

3

[visual tiger]

There is no induction to do here

[subtle blaze]

Ah you do still need the n=1 case

Hey let him try induction

I want to see how it goes

[fresh sparrow]

yeah but what's a_0?

[twilit field]

1, which isn't exactly useful

[visual tiger]

like showing a_2 is divisible by 3a_1 will not help you to show a_3 is divisible by 3a_2

[fresh sparrow]

oh yeah. i'm silly

[visual tiger]

like you could use base cases to find a general proof though

[fresh sparrow]

the derivative of a_3 is 3a_2 /j

[twilit field]

ok, so I'll have to prove 2 things here

firstly that the number is divisble by 3

and secondly that it's divisble by $a_{n-1}$

[visual tiger]

i don't like where this is going

[grand pond]

ƒ(Why am. I here)=I don't Know

[visual tiger]

yep i knew it

3 and a_n-1 are probably not coprime

doesn't work

[west iron]

yeah

prove 111111111 is divisible by 333 without brute force

took me about 6 seconds

[visual tiger]

like there is only one thing to realize

[twilit field]

no spoilers please

[sweet wing]

$a_0=1\a_1=111\a_2=111111111\a_3=111111111111111111111111111\\vdots$

[west iron]

no

[visual tiger]

no

[sweet wing]

no?

[visual tiger]

a_2 has 9 digits

a_3 has 27 digits

[sweet wing]

ah

[dreamy lichen]

3^n isnt 3n

[grand pond]

Flappie

[sweet wing]

there

[visual tiger]

sure

[subtle blaze]

Incredible

[sweet wing]

if you see this its trival 🤓

[sweet wing]

$a_4=111111111111111111111111111111111111111111111111111111111111111111111111111111111$

[grand pond]

Flappie

[subtle blaze]

This is amazing

[twilit field]

what are you doing

[subtle blaze]

What’s the next one

[dreamy lichen]

It's not

off by one digit

[subtle blaze]

Ok but seriously, the first one he sent actually should make sense

[subtle blaze]

This one

[sweet wing]

from here you should start to see where its going

[west iron]

$a_5 = \foreach \n in {1,...,243}{1}$

[twilit field]

yeah, Ive played around with the numbers a bit, getting a rough idea

[grand pond]

thewizardofOU

[sweet wing]

hahaha

[subtle blaze]

@hard umbra u gotta come see this

[visual tiger]

but if you squint enough, I think with this next one you'll really understand what happens

$a_6 = \foreach \n in {1,...,729}{1}$

[dreamy lichen]

:( I cant help without nitro

[grand pond]

rafilou2003

[subtle blaze]

@twilit field the question is claiming that every line is divisible by the last one * 3

[twilit field]

yes

[subtle blaze]

Can you at least see that it’s divisible by the last one?

[twilit field]

I have a conjecture

[subtle blaze]

Ignoring the *3 part

[hard umbra]

,tex<pgffor>
\gdef\n{111}
\foreach ~ in {1, ..., 5} {
$a_~ = \n$ \
\xdef\m{\n\n\n}
\xdef\n{\m}
}
$a_{6} = \n$

[subtle blaze]

POINT AND LAUGH

[dreamy lichen]

Haha snow failed latex

[fresh sparrow]

my favourite resolution

[twilit field]

$a_{n-1} \cdot \Xi =a_n$

[twilit field]

where n is some natural numebr

is my conjecture

[visual tiger]

'some natural number'?

[subtle blaze]

How can you tell

[visual tiger]

um

try 3*a_2

see if it's a_3

[subtle blaze]

I feel like perhaps we shouldn’t use n here

[sweet wing]

its kind of like placing the previous term back to back 3 times

[visual tiger]

if they meant k instead then that's the wildest conjecture ever

[dreamy lichen]

Quite a valid conjecture though

[visual tiger]

like

a_n = something*a_n-1

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

this better?

[visual tiger]

this is a pretty wild assumption

[subtle blaze]

I feel like they meant like it’s some natural number, we haven’t yet figured out which one

But that’s the conjecture

[twilit field]

I've played around with the numbers a bit using WA, seems to work

[dreamy lichen]

Yeah

probably because this is part of what you have to prove

[sweet wing]

try looking at a few terms and seeing what theyre divisible by

[twilit field]

they're all divisible by 3

[sweet wing]

like a_3/a_2

[visual tiger]

"Show that a_n is divisible by 3a_n-1"
"guys I have a conjecture"

[sweet wing]

or a_4/a_3

[dreamy lichen]

This is a definition of an is divisible by an-1

[twilit field]

yeah

I'll have to prove that

the question is how

[visual tiger]

so

have you tried computing a_n/a_n-1 for some values of n?

and see if a pattern arises?

[subtle blaze]

Unironically look at this

And if you can’t see it look at it a bit more

[twilit field]

$a_n=10^{3n}+10^{3n-1}....1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$a_{n-1}=10^{3n-1}+10^{3n-2}....1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

are you sure?

[twilit field]

uh

[dreamy lichen]

with this

[twilit field]

one minute

[dreamy lichen]

the first one looks fine

[twilit field]

$a_{n-1}=10^{3n-3}+10^{3n-4}....1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

How many digits does this have?

[west iron]

3n?

[twilit field]

3n

[dreamy lichen]

oh wait, n is used twice

neither is correct then

i thought n was some newly introduced variable, nvm

[west iron]

it shouldn't

[twilit field]

oh

right

[dreamy lichen]

It should have 3^n digits

[twilit field]

yes, so upto $10^{3n-1}$ would be the place value

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

Why?

[west iron]

3n?

[dreamy lichen]

where are you getting 3n from?

[twilit field]

like take 100

n=1

it's upto the 100th place

[dreamy lichen]

What?

whats "orn"

[sweet wing]

have you looked at $\frac{a_2}{a_1}$,$\frac{a_3}{a_2}$,$\frac{a_4}{a_3}$ yet?

[visual tiger]

It seems like it's a bit all over the place

[grand pond]

Flappie

[twilit field]

not yet

[visual tiger]

yeah do that asap

[sweet wing]

always look at that first

[subtle blaze]

Now would be a good time if you still haven’t figured it out

[visual tiger]

it's like a key to understanding

[dreamy lichen]

try doing the long division algorithm instead of using calculator btw

[twilit field]

oo

100100100

one one , two zeros

followed by one

[dreamy lichen]

That's if you divide by 111 i think

[twilit field]

yes

[dreamy lichen]

others are gonna be bit different

[twilit field]

if I divide by 111111111 I should have 8 zeros between two ones

wait

no

[dreamy lichen]

Why not?

[visual tiger]

you mean 1001001?

[twilit field]

yes

[visual tiger]

there's an extra one

[twilit field]

that's if I divde by 111

oh

right

yeah, it's 8

so using this I can argue $a_n|a_{n-1}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

though not sure how I'd formally write it

[visual tiger]

you mean the opposite

[twilit field]

my bad

yeah

[visual tiger]

well

[west iron]

uh

[twilit field]

$a_{n-1}|a_n$

[grand pond]

ƒ(Why am. I here)=I don't Know

[west iron]

ah there we go

[visual tiger]

it would be nice to have a formula for a_n/a_n-1

can you find one and prove it?

[twilit field]

yeah

how do I put the bracket under a number ?

like to say n times

[sweet wing]

$\underbracket{123445}_{lol}$

[twilit field]

thanks

[grand pond]

Flappie

[sweet wing]

$\underbrace{1...1}_{3^n \text{ times}}$

[twilit field]

thanks!

yeah

[grand pond]

Flappie

[twilit field]

so $\frac{\underbrace{11.......11}}{3^n \text{times}}}{\underbrace{1...1}{3^{n-1} \text{times}}$

[sweet wing]

$\overbrace{1...1}^{3^n\text{ times}}$

[grand pond]

Flappie

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

give me 5 minutes to sort out the jax

sorry

[sweet wing]

$\frac{\overbrace{1...1}^{3^n\text{ times}}}{\underbrace{1...1}_{3^{n-1}\text{ times}}}$

[grand pond]

Flappie

[twilit field]

$\frac{\underbrace{111 \cdots 1}{3^n}}{\underbrace{111 \cdots 1}{3^{n-1}}}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so now

uh<, I think I'll just write it out and send it

[west iron]

spoiler boi ||try long division (unironically)||

[twilit field]

that's what I'm doing

,rotate

[twilit field]

,rotate

[twilit field]

,rotate

[west iron]

not quite with that top part

[twilit field]

that's 3n-2 if it wasn't clear

[west iron]

Yeah it was clear, just wrong

[fresh sparrow]

idk if anyone mentioned it but it simplifies a lot if you do 111...1 (3^n times) = 111...1 {3^(n-1) times} + 111..1 {3^n - 3^(n-1) times} * 10^(3^(n-1))

one sec let me latex it

[twilit field]

should it be n-2 times

[west iron]

No

[fresh sparrow]

$\underbrace{111...1}{3^n} =\underbrace{111...1}{3^n - 3^{n-1}} * 1\underbrace{00...0}{3^{n-1}} + \underbrace{111...1}{3^{n-1}}$

[grand pond]

artemetra

[fresh sparrow]

you get what i am talking about

[subtle blaze]

$111,111,111 = 111,000,000 + 111,000 + 111$

[grand pond]

Frosst

[fresh sparrow]

yeah exactly

[subtle blaze]

This is the crux of what you should be seeing

[twilit field]

yeah, I had that idea, didn't think it would go anywhere, is it actually useful?

[fresh sparrow]

yes

[subtle blaze]

It is the main part of the proof

[subtle blaze]

Write the right side in terms of 111 * something

Show that the something is divisible by 3

[subtle blaze]

That’s this

[twilit field]

there are 3^n 1(s) so it will be divisible by 3

[subtle blaze]

Then you’re done

[twilit field]

wait, what

[west iron]

Well no, this is wrong

[subtle blaze]

Do this

[twilit field]

I have to prove $a_{n-1}$ is a factor too

[grand pond]

ƒ(Why am. I here)=I don't Know

[west iron]

You can't just show a_n is divisible by 3 and a_{n-1} and call it a day

[subtle blaze]

My bad I misread

[subtle blaze]

Do this step

Write it out

[twilit field]

you mean the left side?

[subtle blaze]

I mean the right side

[sweet wing]

you should be able to show that $a_n = a_{n-1}10^{2n}+a_{n-1}10^n+a_{n-1}$

[grand pond]

Flappie

[hard umbra]

idk why but i think LTE would work great here

[twilit field]

LTE?

[subtle blaze]

Lebesgue’s tantalising equaliser

Idk

[twilit field]

3(37000000)+3(3700...)

[hard shard]

lifting the exponent

[west iron]

lifting the exponent

[twilit field]

$3(3710^6)+3(3710^3)...$

[hard shard]

why not show 3a | 9b where a is 3a_n-1 and b is the other

[grand pond]

ƒ(Why am. I here)=I don't Know

[west iron]

If we multiply our string of 1s by 9 we get 10^(3^n)-1

[twilit field]

yes

[subtle blaze]

Can you do this first

[twilit field]

I though I did

oh

I wrote it in terms of 3

my bad

[hard shard]

i dont think thats useful here

[twilit field]

(1009009....)(11)

[subtle blaze]

What

Why is it 11

[twilit field]

something *11

right?

that;s what you wanted

[hard shard]

ok it is, ill brb

[subtle blaze]

It literally says * 111

[twilit field]

oh

misread

sorry

[subtle blaze]

I didn’t pull 111 out of my ass, it’s the last in the sequence to 111 111 111 which is on the left side

[twilit field]

111(10^6)+111(10^3)+111(10^0)

[subtle blaze]

Now factorise

[twilit field]

$1111(10^6+10^3+10^0)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

The inside is 1001001

[twilit field]

yes

[subtle blaze]

This guy has 3 1’s and the rest are 0’s

[twilit field]

which is divisible by 3

[subtle blaze]

The digit sum to 3, it is divisible by 3

We are done

111 111 111 = 111 * something divisible by 3

It should be obvious how this extends to the n case

111111111 111111111 111111111 = 111111111 000000000 000000000 + 111111111 000000000 + 111111111

[twilit field]

it will become $111(10^{3n}+ 10^{3n-3}+ 10^{3n-6}....10^0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

You should be able to extend to this as your general case

[subtle blaze]

No it will not

We want to factorise out the last number in the sequence

[twilit field]

yes

[subtle blaze]

We don’t always care about 111, only in the case of 111111111

[hard shard]

$\frac{\frac{10^{3^n}-1}{9}}{3\frac{10^{3^{n-1}}-1}{9}}$

use diff of cubes

i forgot the -1s, one sec

[grand pond]

the grind

[hard shard]

use diff of cubes

show the other factor is a multiple of 3

[subtle blaze]

lol that’s a more algebraic approach

[hard shard]

i think its simpler

[twilit field]

ooh

ok, yeah, this clicks

sorry

sorry for wasting everyone's time

[hard shard]

dont worry about it

literally

stop worrying about it

[subtle blaze]

I imagine this as solitaire

But you’re cheating, taking a back portion of a stack out

[twilit field]

wait,so $\frac{\frac{10^n-1}{10^{2n}+1}10^n}}{3 * 10^{3n-1}-1}$

[grand pond]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

$\frac{10^n - 1}{10^{3n} - 1}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

wait

I messed up my Tex

$\frac{\left(10^n-1\right)\left(10^{2n}+1+10^n\right)}{3\left(10^{3n-1}-1\right)}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[west iron]

why does 3^n keep turning into 3n over and over and over again

this is like the 6th time you've done that

[twilit field]

I'm using an external Tex editor, sorry

[west iron]

ah

that makes more sense

[west iron]

But this is wrong because you then proceeded like the exponent was 3n instead of 3^n

[twilit field]

yeah, my bad, correcting it now

ok, I'm lost, how do I divide this using difference of cubes, I see no cubes here

[subtle blaze]

._.

[hard shard]

write 3^n as 3^(n-1)*3

[twilit field]

ah, so I have $\frac{\left(10^{3^{n-1}\cdot3}-1\right)}{3\left(10^{3^n-1}\right)}$

?

ok, so that would be

$\frac{\left(10^{3n-1}-1\right)\left(10^{2\left(3n-1\right)}+1-10^{3n-1}\right)}{3\left(10^{3n-1}-1\right)}$

$\frac{\left(10^{3^{n-1}}-1\right)\left(10^{2\left(3^{n-1}\right)}+1-10^{3^{n-1}}\right)}{3\left(10^{3^{n-1}}-1\right)}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is

$\frac{\left(10^{2\left(3^{n-1}\right)}+1-10^{3^{n-1}}\right)}{3}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

is that right?

[twilit field]

.

now further factorising this

[sweet wing]

Still on this? How

[twilit field]

New to proofs

very new

and to number theory too

[sweet wing]

Why not use this?

[twilit field]

$\frac{10^{3^{n-1}}\left(10^{3^{n-1}}-1\right)+1}{3}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I thought an algebric approach would help more

it made more sense

sorry

[sweet wing]

Sure

[twilit field]

could you help from here? have I made a mistake?

[sweet wing]

What are you trying to show with this

[twilit field]

proceeding from here

[west iron]

This is wrong

[hard shard]

you did diff of cubes wrong

[west iron]

You did sum of cubes instead of difference of cubes or something

[twilit field]

this is the formula I have to use, right?

ah

I see my miskatke

$\frac{\left(10^{2\left(3^{n-1}\right)}+1+10^{3^{n-1}}\right)}{3}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

is this right?

[hard shard]

yes

[twilit field]

is this supposed to be an easy problem? Feels like the most mindbending thing i've ever seen

[hard umbra]

i feel like you need someone to wave their hands in front of a blackboard to demonstrate this to you

[twilit field]

I guess now I can just say , the ones add up to a number divisible by 3, and thus this is divisble by 3

[twilit field]

what do you mean

[hard umbra]

the solution is supposed to be quite simple but you need to draw it out and stuff

[twilit field]

ah

[hard shard]

youre familiar with 3|digit sum <=> 3|that number

[twilit field]

yes

[hard shard]

whats the digit sum of three powers of 10

[twilit field]

that's what I've written here

[hard shard]

yeah

experience definitely helps here

[twilit field]

thanks so much everyone!

[twilit field]

any resource suggestions?

funnily enough, the polynomial question I asked a few hour ago, was from the same paper

[hard shard]

alcumus on artofproblemsolving

[hard umbra]

,, \replicate3{\underbrace {111\dots1}} = \underbrace {111\dots1}(1000\dots01000\dots01)

[hard shard]

its a good contest math trainer

[hard shard]

i think thats whay frosst was trying to say

[hard umbra]

that's what everyone was trying to say

[hard shard]

yeah, but frosst was trying to do the whole factor out the 11...11

[hard umbra]

that's also what everyone was trying to do

[twilit field]

sorry, I didn't really understand what was going on

[hard umbra]

can you at least see this

[twilit field]

yes

[hard umbra]

the bracketed thing is divisible by 3 cuz there's 3 1s

[twilit field]

yeah, got it

[hard umbra]

the thing outside the bracket is a_(n-1)

end of proof

[twilit field]

ok, wow

how did you even notice that

[hard umbra]

you stare at 111 111 111 for like 10 seconds

and prove that it's divisible by 333

[sweet wing]

bruh thats literally what i had

[hard umbra]

you gotta be more visual

[twilit field]

I didn't realised, sorry

[sweet wing]

just written differently

[hard umbra]

visual

[hard shard]

❌ visual
✅ algebra

jk, visuals do it justice

[hard umbra]

gotta use LTE

[hard shard]

lose the experience

[twilit field]

Lifting the exponent lemma?

[hard umbra]

v_3(10^(3^n) - 1) = v_3(9) + v_3(3^n)

[hard shard]

ah

[hard umbra]

so 10^(3^n) is 1 mod 3^(n+2)

[twilit field]

can I close this now?

thanks again everyone! sorry for wasting your time

.close

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[normal plover]

how do i differentiate this

[subtle zinc]

product rule

[normal plover]

how to differentiate the 10cos^2x

[hearty rune]

make it loggy ig

a^b=e^{bln(a)}

[midnight plank]

@normal plover Has your question been resolved?

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[wraith oriole]

How do you get the radius

[raw quail]

draw it out
we have the following info:

1. (-4, 5) is the center
2. (-8,8) is a point on the circle

how do we get the distance between two points?

[midnight plank]

@wraith oriole Has your question been resolved?

[midnight plank]

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[dense tulip]

why am i getting 1/4 here? wolfram alpha says 1/2 but im sure i did everything right

[wary trail]

do u=x/2

and then you tell me why

[tribal tartan]

you forgot to compensate for that 1/2

[tribal tartan]

this too

[dense tulip]

wait, cant i just use this formula and replace x with x/2?

[wary trail]

no

remember substitution

or how to integrate by substitution

i think i spelt it right

[dense tulip]

hm ok, i see now, why it is 1/2

[wary trail]

remember when you replace dx

you find the relationship between your differentials dx and du by differentiating

[dense tulip]

i mean, i know how to integrate with subtituion

ig i didnt know how to work with formular in the table 😂

[wary trail]

okay

then show me the steps after the substitution u=x/2

[wary trail]

there's no point telling me you can do it if you're not doing it buddy

[dense tulip]

there you go

[wary trail]

excellent

understood?

[dense tulip]

sorry, flip it now

yep

.close

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[wraith oriole]

Hi dragi

[grizzled gazelle]

Hey, im wondering why you post this. You are so close to the answer. Try to use that y>4.

🙂

[wraith oriole]

Yeah so I’m stuck what to do with that. The answer key plugged 4 into the top equation

But I’m confused because since y is anything above 4, can’t you use 7,8, or any number above 4?

[grizzled gazelle]

Yes you can. But you search for every x s.t. x satisfies the system of equations. for every x > 3 you can find a y so that its true

[fair star]

4x-8>y>4,therefore 4x-8>4

[wraith oriole]

Uhhhh

[fair star]

get the first equation

bring the 8 to the left

and you get that

[wraith oriole]

You used 4

My question is why not any number above 4

[fair star]

4x-8>y

and y>4

therefore 4x-8>4

[wraith oriole]

OHHHHHHHHHHHHHHHHHHHHHH

wait

[fair star]

hm?

[wraith oriole]

Uhhh so

Y is bigger than 4

[fair star]

yes

[wraith oriole]

And 4x-8 is bigger than y

[fair star]

yes

[wraith oriole]

Ah I see

[fair star]

so clearly 4x-8 is always bigger than 4

[wraith oriole]

Right

That makes sense tysm

[fair star]

and then solve for x

[wraith oriole]

Right

X is greater than 3

Yayyy

[fair star]

easy thing

yep

[wraith oriole]

@grizzled gazelle I had another question for u btw

[grizzled gazelle]

sure

[wraith oriole]

Is it true, that in every finite-dimensional C-Vectorspace V, if i have a linear map y with y: V-> V with dim(Ker(y))=0, there exists a base of V with eigenvectors of y?

[fair star]

[grizzled gazelle]

[wraith oriole]

😭😭

[wraith oriole]

He asked that question somewhere lol so I was just messing around but seriously what the heck is that

I don’t understand a word

[fair star]

i know i know

[wraith oriole]

Anyways thanks so much for your help guys I was stuck on this for so long

Have a great rest of your day

Or night

.close

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[runic wraith]

Hello, I'm having a bit of trouble conceptualizing and setting up direct comparison of infinite series and would like help with an example

[runic wraith]

how do I type latec here

[fallen vigil]

$$

[runic wraith]

I just sent a photo

okay, so I can see that this respembles a p-series, specifically the divergent harmonic series correct?

so If I where able to prove that the form lets say.... $$\sum_{n=1}^{\infty}\frac{1}{2n}}$$ is lesser than or equal to my original series then the origial series diverges

it didnt work

nvm

[nimble copper]

\infty

[runic wraith]

ty

[grand pond]

穴熊
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[runic wraith]

Am I taking the correct approach?

[nimble copper]

well i'm not sure that you'd have to prove that this new series is less than or equal to your first one

but showing that this new one diverges is pretty sensible

[runic wraith]

I'm trying to do direct comparison

[nova yoke]

try comparing with 1/n instead of with 1/(2n)
oops wait, that might not work

[runic wraith]

okay

[nova yoke]

hang on, i was doing something in my head, might be wrong, lemme write it out haha

[runic wraith]

sure

[nova yoke]

ah yea sorry, i had something reversed mentally

try 1/(3n)

[runic wraith]

can't I just make a tautolgy with 1/(2n)

[nova yoke]

oh yea 1/(2n) works too, since you have (2n-1) in the denom, not (2n+1)

[runic wraith]

This is what I tried

Excuse my handwriting

[nova yoke]

yea, that works since all those manipulations are reversible

[runic wraith]

but idk if this is correct

[nova yoke]

so 0 >= -1 (true) implies 1/(2n) <= 1/(2n-1)

damn i can't type

fixed now

[nova yoke]

it's correct, your series is termwise larger than those of a divergent series (with all terms positive)

[runic wraith]

so I found a series that is less than my original series, and it is divergent, therefor I can conclude my given series is divergent. Right?

[nova yoke]

yep

[runic wraith]

Okay, I think I'm getting it

.close

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[young spire]

I got an answer but it’s 3 times less than the answer given

I have done 9.8x10^-4=(9.0x10^9xq1q2)/.1^2

How do u put it into a picture so it’s easier to read

[dull yoke]

,calc 0.00098*0.01

[grand pond]

Result:

9.8e-6

[dull yoke]

bruh

$9.8 \cdot10^{-4}=\frac{(9.0 \cdot 10^9 \cdot q_1 q_2)}{0.1^2}$

[young spire]

Times 0.1^2 on both sides

[young spire]

Then divide 9.0x10^9 on both sides

Q1q2 equals 1.1x10^15

[dull yoke]

,calc 9.8/9

[grand pond]

Result:

1.0888888888889

[young spire]

Then I’m not sure what to do after

[dull yoke]

why not?

it tells u to find the charge of each

[young spire]

Does q3 equal 1.1x10^-15

Nvm

[dull yoke]

read line 2 of the problem again

1.0888 * 10^(-15) = q_1 * q_2

[young spire]

Sqrt?

[dull yoke]

read the start of the second sentence of the problem

[dull yoke]

wdym

[young spire]

Equally charged so square root to find 2 equal values to get 1.0888…x10^-15

It is 2

$3.3x10^-8$

[grand pond]

Storm09

[young spire]

Oops

$3.3.*10^-8$

[dull yoke]

,calc sqrt(1.0888 * 10^(-15))

[grand pond]

Result:

3.2996969557825e-8

Storm09

[dull yoke]

yea that looks right

[young spire]

Ye 2 sig fig

I think

Doesn’t say

Anyways ty

.close

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[limber shell]

i think you can use this, y*cos(30°)=12

[uncut topaz]

how

[limber shell]

The cosine of an angle in a right triangle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse.

In this case, the angle you can use to find the value of y is 30°, Cos(30°)=12/y

if you multiply al of this by "y", you get "y*Cos(30°)=12"

[uncut topaz]

y=13.86?

[limber shell]

yes

[uncut topaz]

alr

thanks

u helped me a lot

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[pastel dome]

is this wrong?

shouldnt it move |d| units right when d is less then 0?

[hearty rune]

did they write cos(x-d)?

[pastel dome]

wdy,

wdym

yes

[hearty rune]

then its fine

[pastel dome]

nopthing else should horizontally shift the function

[pastel dome]

the thing the teextbook wrote?

[hearty rune]

yes

[pastel dome]

how>

shouldnt it move |d| units right when d less then 0

[hearty rune]

if d was -5 then its cos(x+5) which is a left shift by 5, ie |-5|

id d was 5 then its cos(x-5) which is a shift right by 5

[pastel dome]

what

????

-5 is negative

[hearty rune]

im fully aware of that, your point being?

[pastel dome]

so its less then 0

[burnt flame]

this is the general case for all functions

[pastel dome]

so it sohuld shift right

[burnt flame]

here is a useful chart

[pastel dome]

yes

I know

c is negativ

I was saying

when c is negative

it sohuld shift right

and when its positive

[burnt flame]

oh right

[pastel dome]

it should shift left

but he keeps saying

[burnt flame]

periodic functions are neato like this

[pastel dome]

if d is 5

then its sin(x-5)

thats not true

-5 makes d equalt to -5

not 5

[pastel dome]

wdym

-90 makes it shift to right

[hearty rune]

but there d is 90, not -90

[pastel dome]

wwhat?

no its -90

[hearty rune]

its cos(x-d) not cos(x+d)

so d is 90

[pastel dome]

hm

[hearty rune]

if it was cos(x+90) then d is -90

[pastel dome]

what?

[hearty rune]

cos(x-(-90))

[burnt flame]

they are defaulting to writing cos(x-d)

so d=90 is a a shift to the right 90 degrees

[pastel dome]

thats implying

there is a negative sign infront

by default

but isnt d negative

since -90

[hearty rune]

then it becomes a +

so left

[pastel dome]

no...

when its -90

shouldnt it shift 90 right

[hearty rune]

x--90 is x+90

(x-90) is a shift right, yes d=90
(x-(-90)) is (x+90) a shift left d=-90

[pastel dome]

is this right then

[hearty rune]

yes

[pastel dome]

is it x-d by default

?

????

[pastel dome]

if k is greater then 0, shift left

see

[hearty rune]

thats not contrary to anything, im not sure what the see is for

x-d is a common form but you can do it however you want as long as you understand it

in the one above theyre doing x+d

[pastel dome]

thanks

do different people ahjve didfferent preferences

[hearty rune]

i most commonly see x-d, but youll likely get a mix

[pastel dome]

alr thanks bro

.close

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[twilit field]

so to start

[burnt flame]

10

[twilit field]

what's 10

[burnt flame]

so it should be leq

geq*

oh wait

does N include zero in your class?

[twilit field]

no

but 100 is valid I think

[burnt flame]

10 is a weird case then...

[hard shard]

i would include 0

[burnt flame]

yea

if your function goes from N to N you must include 0

[twilit field]

$0 \leq g(n) \leq 9^{\gamma}$

[burnt flame]

wait

0 leq

[grand pond]

ƒ(Why am. I here)=I don't Know

[burnt flame]

ye

[twilit field]

I have to show this hold true

[burnt flame]

9^gamma?

[twilit field]

$\gamma$ is the number of digits

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so say 9999...

[burnt flame]

that's not 9^4

[twilit field]

I know

g(n) is the product of the digits

[burnt flame]

oh

yes then its true

using powers of ten definitely... Maybe this would be better with an induction argument...

[hard shard]

the problem with this idea is that 9^gamma is not always less than your number

[twilit field]

you mean more than?

[burnt flame]

9^2<99

but its product

we're trying to bound g(n)

not n

[twilit field]

hmm

yeah

[burnt flame]

so 9^gamma works as a bound

for g(n)

but this might not be the right approach

[twilit field]

right

[twilit field]

so this

[burnt flame]

i think an induction arguement work would better

say for n=2

a,b your digits

[fleet moss]

i mean the lower bound on n is 10^gamma right

[twilit field]

$81<99$

yeah

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

it is?

[fleet moss]

oh sorry gamma - 1

hmm yeah induction might work

[twilit field]

ok, so the inductive argument would be that $g(n) \leq n$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I have to prove that $g(n+1) \leq n+1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[burnt flame]

mmhm

note you can factor out 10 from most of the digits of n provided n is large enough

[twilit field]

let's consider it ccase wise

let n+1=0

then $g(n+1) \leq n+1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

similarly for 1

the problem is proving this when n+1 ends in 2

I don't think induction is the way here

[burnt flame]

well 9^gamma approach might be correct

think about the expansion of n

[twilit field]

eh, but it's wrong for $\gamma =2$

no?

[grand pond]

ƒ(Why am. I here)=I don't Know

[burnt flame]

10^gamma(gammath digit)+10^(gamma-1)(gammath-1 digit)...=n

we're trying to bound g(n)

[twilit field]

yes

[burnt flame]

g(99)<=9^2

g(999)<=9^3

wait

9^gamma<=10^gamma+(other stuff)

ez

wait

[fleet moss]

first digit is x10^(gamma -1) i believe

[burnt flame]

oh right

it is

damn

brain

malfunctioning

[fleet moss]

i made the same mistake above lol

[burnt flame]

so more accurately,

[west iron]

to find the most elegant proof i imagine what i would do if it asked me to prove it for every base

and that made me stop trying to brute force the problem and figure out that you should induct on the number of digits

actually

[fleet moss]

strong induction on all n fixing gamma?

[west iron]

let me think

something like 98765 idk

given that 8*7*6*5 < 8765