help-10

[obtuse pebble]

@queen dragon Has your question been resolved?

[queen dragon]

.close I give up

[obtuse pebble]

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[carmine sable]

can someone teach me how to find the coefficient of x^n in a power series

[carmine sable]

i am trying to solve it for f(x) = 1/(1-x^2)^2

[kind hawk]

in such cases where its just x^2 or some other power plugged into a function, its often easier to first do 1/(1-x)^2

which you can either do by computing the taylor series or by integrating and recognizing something

[carmine sable]

ok got it but what do i do after that?

[kind hawk]

wdym. afterwards you are done

[carmine sable]

no but what im trying to find is the coeffiecent of 1/(1-x^2)^2 not 1/(1-x)^2

[kind hawk]

its the same series

just x^2 plugged in

so instead of (random numbers) 1+3x+7x^2+15x^3 or something its 1+3x^2+7x^4+15x^6 etc

[carmine sable]

ahh okay got it thank you, i thought it would change the coefficent for some reason

[kind hawk]

thankfully not

[carmine sable]

okay thank you 🙂

[obtuse pebble]

@carmine sable Has your question been resolved?

[carmine sable]

but hold on considering this for n's the x^n's become only even so does it mean that when n is odd the coefficent will be just zero?

[kind hawk]

yes

[carmine sable]

so the coefficent is (n/2) +1 when even and just 0 when odd then

[obtuse pebble]

Channel closed

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[carmine sable]

.reopen

[obtuse pebble]

✅

@carmine sable Has your question been resolved?

Channel closed

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[warm depot]

Guys how do I solve exponential equations

[warm depot]

😂

[restive gorge]

your pfp is so cool

[warm depot]

I can drop the image if you want

[restive gorge]

dont mind

[brazen gorge]

show an example

[warm depot]

Stuff like this are in the power point

[wary badger]

just matching bases

and exponents

[wary badger]

^

[warm depot]

All I want to accomplish today is develop an understanding of these equations

Like the ones on the SAT a

Knief i haven’t even started precalculus and i’m trying to learn how to find the derivative in a function

[wary badger]

why..

that isn’t on the SAT

SAT is just algebra

and geometry

[warm depot]

I just want to expand my knowledge if ykwim

Algebra 1-2 and geometry

[wary badger]

yea but learn the fundamentals first

[warm depot]

I am trying to learn that all on a week

in*

[wary badger]

this?

[warm depot]

I can’t find where to find the elementals

[wary badger]

you mean the elements?

idk if i’d recommend that to learn geo

[warm depot]

yeah sorry my autocorrect is annoying

Basically i’m trying to master algebra and geometry before sophomore year and before december

I have the shsat soon

[wary badger]

tbh i didn’t do SAT specific studying so i wouldn’t know

i took some practice tests

that’s it

[wary badger]

oh you’re young

you’ll be chillin

[warm depot]

Yeah but i’m trynna learn all of that right now

[wary badger]

if you want to get unnecessarily good at geo and algebra then pick up the AOPS books

[warm depot]

My “peers” are doing precalculus in their freshmen year

[warm depot]

I never heard of that, AOPS?

[wary badger]

ahh you go to an advanced school

[wary badger]

art of problem solving

they have subject specific books

more for competition math

but still useful

[warm depot]

My friends do, but unfortunately life dealt me the wrong cards and I am in the worst DOE school in all of new york

[wary badger]

i’m in new york

[warm depot]

I am solely depending on the shsat to get me out of this school

[wary badger]

unfortunate

[warm depot]

Cool, i won’t have to unnecessarily explain why I take the regents lol

[wary badger]

you’re in a better spot than most at your age though

[wary badger]

i barely took any regents

only junior and sophomore year

covid for freshman and 8th

[warm depot]

Oh, I guess that’s what happens because of the pandemic

[wary badger]

yep

didn’t have to take living environment

🙏🏻👍🏻🙏🏻

[warm depot]

But feeling smart by DOE standards amongst those who don’t care at all isnt useful to me

Living environment is annoyingly long

[wary badger]

yea living environment was just boring

didn’t give a shit about finches

earth science was wack too

[warm depot]

its more like the idea of biology lol

[wary badger]

i didn’t have to take either

[warm depot]

earth science is 10th grade?

[wary badger]

at my school it’s 9th

LE 8th

chem 10th

then physics was optional

[warm depot]

I took the global history regents a while ago

[wary badger]

but i took ap physics 1 in 11th

then ap physics c mechanics in 12th

[warm depot]

My goal is ALL AP CLASSES

[wary badger]

and self studied for ap physics 2 and physics c e&m because my school didn’t offer it

[wary badger]

nah you don’t want ap latin and all of those

just take all of the science, math, and history

[warm depot]

More appealing for colleges

[wary badger]

and lang

lit

[warm depot]

stem classes are where its at

[wary badger]

fr

i credited out of the first two years essentially

of college because of my ap credits

[warm depot]

Its just my teachers lack the ability to explain ANYTHING

[wary badger]

learn on your own

[warm depot]

regardless of what it is, they don’t care about the students

[warm depot]

I’m trying, kinda directionless for the most bit

[wary badger]

start with khan academy

then whatever subject interests you after, just go on youtube to find books and or videos

mit ocw was always good too

but with what you’re doing khan academy is all you need

[warm depot]

What if i’m still lost 😂

[wary badger]

but i’d supplement with a lot of problems

[wary badger]

do more practice

[warm depot]

I can’t find any practice questions or tests because they are all subscriptions

[wary badger]

there are books online

openstax probably has some algebra

they’re open source

yea i checked

they do

they’re great

[warm depot]

I’ll check it out

[wary badger]

they have everything up to calculus 3 i think

full textbooks with exercises and everything

all for free

[warm depot]

What would you advise me to do

[wary badger]

do more problems

watch videos on khan academy/youtube

and use textbooks online

no need to spend money

[warm depot]

I’ll do my best, thank you

[wary badger]

you’re welcome

[late elbow]

Just sharing my thoughts:

Before i was in this math discord, I thought that I am good at math 🗿 but now that I think of it, I’m not that good compared to most people 😭 I’m only better than my classmates back in junior high (grade 7-9)

I lack knowledge in many math topics. The highest math I’ve studied in highschool is pre-cal

And then I took accounting in college 🫠

[patent crag]

ping?

[wary badger]

there’s always a bigger fish

[obtuse pebble]

@warm depot Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[remote pivot]

This is a practice exam I'm looking at, can someone explain to me whats going on in this problem?

I know that the FTC says that the derivative of an integral is just F(x), so whats going on here?

[fossil crag]

FTC tells you that if F is an antiderivative of $f$, then $\int_a^bf(x)dx = F(b)-F(a)$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

so just let $F$ an antiderivative of $2\ln(t)$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

$g(x) = F(e^x)-F(1)$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

now we differentiate, and we need chain rule

I think you mistook with another version of the FTC that says that $F(x) = \int_a^xf(t)dt$ is an antiderivative of $f$ when $f$ is continuous

[warm shale]

rafilou is not not born in 2003

[fossil crag]

notice that the lower bound is ALWAYS a constant, and the upper bound is ALWAYS the input x

[fossil crag]

which is not what we have here

since the upper bound is e^x, not x

[fossil crag]

so this version of FTC is not applicable

[obtuse pebble]

@remote pivot Has your question been resolved?

[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[glossy pike]

could anyone help me with this?

[glossy pike]

i thought the "trend" should be the other direction, since as t increases, x is approaching 0 while y also goes up

why is the "trend" given in the soltution says its from left to right, not the other way

t=0

,calc e^0

[warm shale]

Result:

1

[glossy pike]

,calc 3*e^0

[warm shale]

Result:

3

[glossy pike]

t=1

,calc e^-1

[warm shale]

Result:

0.36787944117144

[glossy pike]

,calc 3*e^-1

[warm shale]

Result:

1.1036383235143

[arctic bloom]

did you know calc is short for calculator

its just a fun fact i learned back in the days

[glossy pike]

yeah?

[vale pelican]

me taking calculator 3 in college

[arctic bloom]

Managed to find the only flaw in my arguments procks to you

[vale pelican]

I thought it was props to you?

[arctic bloom]

Im spamming random english cliques

wordplay maxxing

[vale pelican]

that's fair. spam, king

[glossy pike]

hmm guys? Are you here to answer my question?

[glossy pike]

^

[vale pelican]

honestly in a vacuum, I like your interpretation of how it should be more, but a variable doesn't have to change in just the positive direction I guess

actually, I don't know what the point of the arrow heads in the first image is. they seem irrelevant

[glossy pike]

i only just started parameter topic, so what what does the "trend" represent here? Just point

[vale pelican]

it's not even the curve on (0, inf)

[glossy pike]

oh, then what does the arrow usually mean

[vale pelican]

it usually does denote how a point on a curve moves as a parameter changes

[glossy pike]

yeah, thats how i interpreted at first.

[glossy pike]

this is the orignal question btw

maybe they just wanted to illustrate its the same graph as y=3\x?

[vale pelican]

I think so

[glossy pike]

except with some restriction

[vale pelican]

but is there another figure? neither of these are the actual curve, they show too much

[glossy pike]

thats all

thats why im confused

[vale pelican]

you're right to be, it should be equivalent to the function f : y = 3/x, 0 < x < 1

[glossy pike]

i thought its supposed to be like this? purple curve

[vale pelican]

starting from the point (1,3), the x-coordinate shrinks dramatically but never reaches 0, and the y-coordinate rises dramatically

as per the given parameterization

[glossy pike]

i guess the one pointing right and approaches y=0 is when t<1?

[vale pelican]

more generally t < 0

you cut out some middle section if you say t > 0 or t < 1

allowing t to be any real number gives the curve of figure 7a, the left one

[glossy pike]

ah yeah

[glossy pike]

so could you agree my interpretation is correct?

[glossy pike]

^

[glossy pike]

*t<0

[vale pelican]

yeah, the purple section should extend arbitrarily leftwards/upwards, but you highlighted what you'd actually see in the window

I already gave it the nonverbal thumbs up lol

[glossy pike]

endlessly and not touching zero right

[vale pelican]

yeah

[glossy pike]

yeah ik i just want one more confirmation

[vale pelican]

no problem

[glossy pike]

thank you so much for answering

[vale pelican]

no problem part 2 lol

[glossy pike]

thank you so much part 2

.close

[obtuse pebble]

Channel closed

Closed by @glossy pike

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[sudden steppe]

Right so I'm studying and I am after completely forgetting how to simplify quadratics, can someone help?

[arctic bloom]

Yoo

[sudden steppe]

hiya!

[arctic bloom]

Thats text and texts

[sudden steppe]

correct

[arctic bloom]

ok factorise

think time

[sudden steppe]

real

so dont i have to like find something that divides into 10

[arctic bloom]

yeah

[sudden steppe]

and those two numbers when added have to also make 11?

[arctic bloom]

yes

well yeah you kinda have to go through factors

1 10
2 5

[sudden steppe]

so 1 and 10

because they make 11

[arctic bloom]

yes

[sudden steppe]

so now what do i do after that?

[arctic bloom]

well actually since the x^2 has a coefficient

thats bigger than 1

you have to do a little more

[sudden steppe]

oh its the other one

sorry could you do a little explaining for me my minds gone blank lol

[arctic bloom]

sure

[sudden steppe]

thank you so much

[arctic bloom]

so you have to also take into account the multiples of 6

which are 3 and 2 and 6 and 1

[sudden steppe]

and how do i know which one to use

[arctic bloom]

trial and error honestly but id start with 3 and 2

remember since the coefficient is 6 we actually have to find numbers that multiplies to -60

[sudden steppe]

so

1 x -60

2 x -30

[arctic bloom]

the general form is something like this

(2x-15)(3x-51)

the numbers are random

[sudden steppe]

oh okay

let me try work it out

[arctic bloom]

👍

[sudden steppe]

3x + 2

2x + 5

i think that should be it

[arctic bloom]

Yup thats correct

now you have to factor the bottom

[sudden steppe]

thats 2x + 5 i can do the bottom its just the top one i cant do

thank you for the help!

[arctic bloom]

nice

[sudden steppe]

.done

[arctic bloom]

haha two people have done this

its .close

[sudden steppe]

oh haha thank youu

.close

[obtuse pebble]

Channel closed

Closed by @sudden steppe

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[obtuse pebble]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[ripe leaf]

this is a physics exercise but i think this is math-related
could someone explain to me how the first equation becomes the second one, please
thank for your help really much

[ripe leaf]

I found an explanation online i understand everything before these lines
shouldn't it be v(t) = e^(-b/m)t + v0 (take e both side) instead of v(t) = v0e^(-b/m)*t like this ?

[obtuse pebble]

@ripe leaf Has your question been resolved?

[glacial verge]

it is taking e on both sides
but e^(a+b) = (e^a)(e^b)
not (e^a) + (e^b)

[ripe leaf]

thank you i got it now

[glacial verge]

np

[obtuse pebble]

@ripe leaf Has your question been resolved?

[obtuse pebble]

Channel closed

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[somber patio]

Hi can someone help me with this question

[sturdy juniper]

Alright so what have you tried so far

[sturdy juniper]

Alright so let’s try writing out the function

If y is the mass of the population, we can write the function y = 3(2)^x

And what we’re asking is

For what value of x is y = 60?

[wary badger]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[keen venture]

._.

[sturdy juniper]

First time as a helper and I’m already getting reported 😭😭😭🙏

[wary badger]

._.

[wary badger]

nah you’re chillin gigagoat

[keen venture]

real

u got this

[wary badger]

we believe in you

you’re my goat

[sturdy juniper]

I think Rav left anyways

I’m talking to a brick wall 💀

[wary badger]

@somber patio

have some respect

gigagoat is helping

[keen venture]

low key, i think Rav just wanted the answers from my goat

[wary badger]

manless universe

[keen venture]

real

[wary badger]

@somber patio

boss

i’m sorry @sturdy juniper

[keen venture]

dw

[wary badger]

you didn’t deserve this

[keen venture]

im consulting him rn

for his loss

[sturdy juniper]

My first student ditched me 😢

[wary badger]

hate it when a helpee leaves

[keen venture]

yeah, i would ditch u too ngl

[wary badger]

the audacity

[keen venture]

real

[wary badger]

damn

[tired mica]

Why isn't this $P(t) = P_0e^{kt}$?

[wary badger]

harsh

[warm shale]

Bean Man

[sturdy juniper]

???

[wary badger]

Bean Man!

[keen venture]

nice try bean man!

[wary badger]

well what do we assume t to be

if we consider time in hours

[keen venture]

wait

[wary badger]

then doubling every hour is consistent with Gigagoats model

[keen venture]

why isn't this chat shut down?

didn't Rav leave?

[sturdy juniper]

I think there’s a command or something

[wary badger]

because it takes a long time

i can close it

[keen venture]

okie

[wary badger]

but i wont

[keen venture]

i see

[sturdy juniper]

I’m new to helping

Is it .close?

[wary badger]

yea

[sturdy juniper]

okay imma close it

[wary badger]

go ahead!

[sturdy juniper]

.close

[keen venture]

oh

[wary badger]

🙏🏻💀💀

[sturdy juniper]

Uhhh

[wary badger]

you can’t

you don’t have the power

[sturdy juniper]

Something isn’t right 💀💀

[keen venture]

i think only Rav can or smth

[sturdy juniper]

Maybe it’ll work the second time

[tired mica]

I see that gigagoats is correct but I'm asking why is what I typed wrong

[sturdy juniper]

.close

[somber patio]

sorry i was afk

[sturdy juniper]

He’s back!!!!

[somber patio]

i had pings off

[sturdy juniper]

Ok Rav

[somber patio]

for some reason

[keen venture]

Rav, we missed you

[sturdy juniper]

My first student!!!

[somber patio]

srry lol

[wary badger]

it isn’t

[wary badger]

we forgive you sir

[wary badger]

now everyone let gigagoat cook

[sturdy juniper]

What have you tried so far Rav?

[somber patio]

so i said let y = 3^t

and then if y = 60

then 60 = 3^t

but idk where to go

from there

[sturdy juniper]

Hmmm

So let’s take a look at your function

You wrote y = 3^t

[somber patio]

oh wait do i just convert it to log

form

[sturdy juniper]

Well, we know that every hour, the population doubles

Does that make sense with your function?

[somber patio]

not really

[sturdy juniper]

So what do you think your error was?

[somber patio]

do i have to do y = 2^t x 3?

[sturdy juniper]

Yes!

Because 3 is the initial population, and because it doubles every hour, we have 2^t

[somber patio]

yes

so then 60 = 3 x 2^t

[sturdy juniper]

So how do we actually solve the equation? What are we trying to silence for?

Yes

[somber patio]

20 = 2 ^t

[sturdy juniper]

Yes keep going

[somber patio]

log2(20) = t

[sturdy juniper]

Yes you’re on the right track

[somber patio]

log10(20)/log10(2)

[sturdy juniper]

Yes

[somber patio]

and whatever that comes out as

is answer i think

[sturdy juniper]

Perfect!

[somber patio]

how do i convert from decimal to minutes and hours tho

[sturdy juniper]

Do you have the image of the original question

[sturdy juniper]

I don’t think it requires you to express it as minutes and hours

It should be fine expressing it as a decimal in hours

[wary badger]

≈

[somber patio]

the answer book has it in minutes and hours

so i wanna know how to convert it

to that

[sturdy juniper]

Ok so take the decimal part of the answer

And multiply by 60

[somber patio]

4.32192809489

259.315685693

[sturdy juniper]

Ok there was a slight error

You multiplied the whole thing by 60

[somber patio]

oh

[sturdy juniper]

So what do you think you should do differently

[somber patio]

multiply 0.32192809489

by 60

[sturdy juniper]

Yes

Because everything else is greater than 1, we can express those in hours

[somber patio]

19.3156856934

[sturdy juniper]

Everything in the tenths digit and below is less than 1, so we only convert those into minutes

Yes

[somber patio]

ah so answeer is 4 hours 19 mins

i see

[sturdy juniper]

Ye

[somber patio]

ty for the help 🙂

[sturdy juniper]

Thank you 🙏 🙂

[wary badger]

Gigagoat the true goat fr

[keen venture]

real

[sturdy juniper]

I did itttttt

[somber patio]

ill close now

[sturdy juniper]

My first student

[somber patio]

.close

[obtuse pebble]

Channel closed

Closed by @somber patio

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[feral flume]

hi

[timid silo]

Did you try row reduction?

Until you got all of it down?

[quartz shuttle]

answer was
a) lambda=2 and lambda=-2
b)lambda = 1
c) labda in R except above values

i understood why

.close

[obtuse pebble]

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[silent acorn]

Can someone help explain how this is E

I understand that inverse proportion is where a=k/b

[tardy epoch]

a is inversely proportional to b yes

Find the k in p=k/sqrt(m)

[silent acorn]

K=p x sqrt(m)

[silent acorn]

If you substitute it doesnt sqrtm cancel out to p=p

Or does that not matter

And we just test it

[tardy epoch]

I have no idea what you're substituting

[tardy epoch]

You should be substituting your 3 pairs of numbers

[obtuse pebble]

@silent acorn Has your question been resolved?

[obtuse pebble]

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[novel totem]

How do I begin this?? Do I use intermediate value theorem??

[zenith raft]

yea that would work

[novel totem]

How do i start exactly?

If f is continuous on (a,b)

Then its also continuous (ri, ri+1)??

[zenith raft]

sure

[novel totem]

Okaayy I got it

How do I do (b)??

And is f(1) = 0??

[civic dune]

change x and y by 1

does it work?

i think so

[alpine raven]

f(xy/y) = f(x/y) + f(y)
f(x) = f(x/y) + f(y)

🤔

[novel totem]

Woooow

Why didnt I think of that?? 😭

Thnx I got the answers now

.close

[obtuse pebble]

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[timid silo]

Lemme claim this

[timid silo]

So.. uhhh can you like explain one by one this question and answer

question: log 10 = 1
answer:
2 - log 5
2 log_10 - log 5
log 10^2 - log 5

log 10^2/5

log 100/5

log 20

[void holly]

um what ?

[timid silo]

What... do you not understand?

[void holly]

$\log_{10} 10=1$ is not really a question

[timid silo]

Well

Given

[warm shale]

convergence

[rose scroll]

You question is a bit incomplete. It should be Calculate/Simplify 2-log_10(5) using log_10(10) = 1

[timid silo]

I can.. even show you..?

Because its literally just that

[willow crescent]

Hello

[timid silo]

This is literally given

and I don't understand how that happened

[rose scroll]

That seems to be the step 1 of the answer, and not the statement of the question

[timid silo]

Given, Question, Step 1

Sure

Let's just

idk anymore

[rose scroll]

Whoever wrote it doesnt seem to have written the question here

[willow crescent]

Red is just a fact

Yellow is the solution

Their is no question here

Just a fact

And a solution

[timid silo]

IDK OKAY It's like that on the board I don't know what to tell you guys 😭

[willow crescent]

To some arbitrary question

[rose scroll]

This or a statement equivalent to it should be the original question for this solution

[void holly]

alright so log_{10} 10 =1 so 2=2log_{10} 10

[rose scroll]

It is important to understand the question before understnading the solution

[rose scroll]

You dont have to tell anything. We are making sense of the stuff in the pic, and letting you know what is and isnt in there.

[timid silo]

There is no question 😭

[rose scroll]

But to the image at hand

[timid silo]

NO WAIT

I FIGURE IT OUT

[rose scroll]

Do you understand why log10 = 1?

[timid silo]

Is that better???

[rose scroll]

Yes... That circled text is the question

[timid silo]

I think the question
2 - log 5 = log 20 can be log 5 x 2 = 20 = log 10 = log 20 to match the right side

Is that correct?

[rose scroll]

No

[timid silo]

oh

How about if I multiply each side by 2?

[rose scroll]

what do you mean log10 = log 20 here?

[timid silo]

Well since the solution (step 1) here is log 10 = 1 I could just multiply 5 x 2 together

[rose scroll]

tbh individually all the equalities that you have written here are wrong in
log 5 x 2 = 20 = log 10 = log 20

[timid silo]

log 5 x 2 = 20
log 10 = log 20

[rose scroll]

That is fine, but that gets you log(5) + log(2) = log(10)

This doesnt mean log(5x2) = 20

Or
log 10 = log 20

[timid silo]

Dang

hmm...

[rose scroll]

you can see 10 is not same as 20

so log 10 cant be equal to log 20

[timid silo]

Divide both sides by 2

?

log 5/2 = 20/2
log 5/2 = log 10

[rose scroll]

that doesnt help us much

[timid silo]

uhhhhh dang this is difficult to come up with a first step

Uhhh

[rose scroll]

log 5/2 = 20/2 is still wrong

[timid silo]

Subtract?

Both sides by 2?

[rose scroll]

Can I try and explain what is being done in the image?

[timid silo]

Yes

[stiff mortar]

Yo

[rose scroll]

Ok

[stiff mortar]

What’s going on here

[rose scroll]

So in the first step here

[timid silo]

🪵

[stiff mortar]

Ah I see

[rose scroll]

They took log 10 = 1

[stiff mortar]

What’s the questions

[timid silo]

How did log 2 - log 5 = log 10 became log 10 = 1?

[rose scroll]

They didnt do that

[timid silo]

Huh?

[rose scroll]

there is no log2 to start with

[timid silo]

???

[rose scroll]

What they did was, they wrote 2 as 2 * 1

[stiff mortar]

@rose scroll what’s the original question

[timid silo]

His trying to explain how did the solution from the photo happen

[rose scroll]

@stiff mortar this is the question

[timid silo]

So... just,,, 2 * 1?

[rose scroll]

Yes

[timid silo]

I don't really get the logic here

How is that important?

[rose scroll]

and they replaced the 1 with log10

because log 10 = 1

[timid silo]

So log will be 0?

[rose scroll]

so 2 becomes 2*log10

[rose scroll]

nonono

[timid silo]

Uhhh

Where did 5 go?

[rose scroll]

Wait, we will get there

[timid silo]

So...

[rose scroll]

do you first of all understand how 2 and 2*log10 are equal?

[timid silo]

2 - log 5 = 20
2*log10 5 = 20?

[timid silo]

No

[stiff mortar]

Pretty sure that’s correct working out

[rose scroll]

It is correct, she wants explaination for steps

[stiff mortar]

Ah ok

[rose scroll]

Uhhh
The steps say
2 - log 5 = log20
2*log10 - log 5 = log20

[stiff mortar]

Wait does she understand that log means log base 10?

That might clear things up

[timid silo]

OHHHH SEPARATE

[rose scroll]

Yes

[timid silo]

Okay okay I got it

[rose scroll]

So how do you go from 2 - log 5 to 2*log10 - log 5?

[timid silo]

Separate 2 - log 5 into 2*log10 - log 5

I have a question

Why 2*****

[rose scroll]

I think the separate doesnt mean what I think you mean.

[stiff mortar]

Because log base 10 of 10 is equal to 1

[rose scroll]

Thats what I meant by 2 = 2*1 = 2*log10

[stiff mortar]

So 2 x log10

[timid silo]

What happened to 2 - log

[rose scroll]

when two things are equal, like log10 = 1, you can replace one of them by other

[rose scroll]

Nothing has happened to it. We are only looking at 2 from the eexpression 2 - log5

log5 is a nice guy, and we have no trouble with it

2 is a bad guy coz he has no log

[timid silo]

So... we have to times it to a log?

[rose scroll]

so we need to give it a makeover so 2 also has a log

yes

anything times 1 is the same number

[timid silo]

Okay

[rose scroll]

and we know log10 is just another way of writing 1

[timid silo]

I got ti

I'm locked in

[rose scroll]

so we can multiply 2 with log 10, and it still remains 2, but now has a free log term

[stiff mortar]

It’s just some manipulation of formulas. Usually you want to make it easier for you to group terms or whatever so replacing something with another that is mathematically equal is a good thing

[rose scroll]

that log term helps us apply rules of logarithms to it

[timid silo]

Sooo What do I do now?

[rose scroll]

Do you get why 2 is same as 2*log10?

[timid silo]

Yes

[rose scroll]

Then you can apply an appropriate rule of logarithms so that 2*log10 becomes log(10^2)

can you name that rule?

[timid silo]

Power of a Quotient Law?

Like log_b x^something = y log_b x

[rose scroll]

This, yes

[timid silo]

okay

Whats the next step

[rose scroll]

So that law gets you 2-log5 = log(10^2) - log5

Coz 2 = log(10^2)

[timid silo]

Okay next step

[rose scroll]

Now on log(10^2) - log5, you choose to apply another law of logarithms

Can you tell what you should use here?

[timid silo]

Quotient Law is the closest when I analyze this

[rose scroll]

Yes

[timid silo]

So.. log 10^2/5?

[rose scroll]

Correct

Also, use that 10^2 = 100

so you get the expression log(10^2) - log5 as log(10^2/5) which becomes log(100/5)

Now can you tell what you do here?

[timid silo]

ye

It becomes log 20

[rose scroll]

There you go, thats the answer

[timid silo]

I guess I wanna practice

Give me an equation similar to that

[rose scroll]

One second, while I make up a problem ok?

[timid silo]

kk

[rose scroll]

Prove that $\log_{10}(4) + 3 - \log_{10}(5) = \log_{10}(800)$

[warm shale]

Bacter10Fr4g is not fr0g

[rose scroll]

@timid silo Heres a new problem, very much similar to the previous problem

[timid silo]

@rose scroll

[rose scroll]

Yes

[timid silo]

Is it correct or...

OH

YIPPPEEEE

[rose scroll]

Its correct

[timid silo]

Lemme get my power bank real quick and picture the next one

[rose scroll]

Do you know all these rules of logarithms?

[rose scroll]

Its just using the product rule and quotient rule on the equation

[rose scroll]

You have the wrong expansion written in 1.

It should be $\log_{b}(x\cdot y) = \log_{b}(x) + \log_{b}(y)$

[warm shale]

Bacter10Fr4g is not fr0g

[timid silo]

I'm confused cause I thought the quotient law goes something like
Log 4y^8/Z^3X^2 = Log 4y^8 - Log Z^3X^2?

[timid silo]

Oh okay

[rose scroll]

Yes, it does. You are right. Thats what is written there too

[timid silo]

No it's not?

NO WAIT IT DOES

[rose scroll]

There is a typo tho

they have used y^5 everywhere

[timid silo]

Oh yeah..

I noticed that too

We might as well think it's 8

[rose scroll]

Yeah, 8 everywhere

Other that that, theres no mistake that I see in there

[timid silo]

Log 4y^8 - Log Z^3X^2?
So I guess since there is z and x I should separate everything like
Log4 + Logy^8 - Logz^3 + Logx^2

Correct?

[rose scroll]

yeah.

thats the quotient and product rules used

[timid silo]

then.. For the exponent I guess put it like
Log4 + Logy^8 - Logz^3 + Logx^2
log4 + 8*
logy - 3*
logz + 2*logx??

Sorry I have to keep separating em cause it looks like this

Me and My boyfriend agreed to like take breaks :p

[rose scroll]

yea

[timid silo]

So it's correct?

[rose scroll]

you should use \ and * together so it looks properly like what you want e.g. 8\*logy

[timid silo]

Okay

Thanks

I'll take a break then come back

Is it okay like while I'm back I could answer a similar question?

[rose scroll]

breaks are good for your health

Sure. But you should close this channel till then

[timid silo]

Okay then thank youuuuu

[rose scroll]

npnp

[timid silo]

.close

[obtuse pebble]

Channel closed

Closed by @solemn wadi

Use .reopen if this was a mistake.

[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[timid silo]

I'm back

[timid silo]

.

What-

This is occupied

[nova cobalt]

whats the question?

[timid silo]

.

[timid silo]

.

What?

[nova cobalt]

that seems correct

[timid silo]

.

[nova cobalt]

i dont understand ;-;

[timid silo]

Can you give me a question similar to the encircled question? I wanna practice.

[nova cobalt]

sure

[timid silo]

My brain is literally dead right now

[nova cobalt]

Try solving it one at a time

Like the largest log

Etc

I'm busy rn tho

[timid silo]

I did

Oh

[nova cobalt]

Give me around half an hour

[timid silo]

<@&286206848099549185>

[timid silo]

Nah it's fine

[timid silo]

Would you like to continue with this (sorry for the ping Kreutzer this is for all helpers who would like to help me at this time)

or

[timid silo]

Give me a question similar to the encircled question?

[rose scroll]

You need change of bases formula for solving this

which you said, you havent been taught

[timid silo]

oh

[rose scroll]

so its fine if you dont get this right now

[timid silo]

We might as well get another question

[timid silo]

So that we can follow this

[rose scroll]

You can solve this:
$\log_{10}(x) + \log_{10}\left(\frac{y^2}{x^3}\right) - 3 = ?$
You need to simplify this expression so it all becomes a single term

[warm shale]

Bacter10Fr4g is not fr0g

[rose scroll]

The answer you should get is $\log_{10}\left(\frac{y^2}{1000x^2}\right)$

[warm shale]

Bacter10Fr4g is not fr0g

[timid silo]

correct?

[rose scroll]

YES

[timid silo]

How about next time without telling me the answer because I kinda figured out very easily on what answer I should go for ;w;

Meaning It's not challenging

Can you find me another question similar to it?

[rose scroll]

Sure

$5\log_{10}(z^2) + 4\log_{10}(15z^2) - 2\log_{10}(25z)$

[warm shale]

Bacter10Fr4g is not fr0g

[rose scroll]

Simplify this

[timid silo]

I'm going to die

Alrght then

[rose scroll]

(You dont have to evaluate the numbers. Just use the laws of logarithms more times)

[timid silo]

https://tenor.com/view/azumanga-daioh-sad-cry-girl-anime-gif-5985908

[rose scroll]

You can start again, by first using the product law everywhere

Your first step was good, but you started messing up near the end

Are you trying @timid silo ?

[rose scroll]

No...
Go step by step

Can you use the product rule on first term?

[timid silo]

Wait I kinda realized it

[rose scroll]

Nooo

Why do you split 5 log_10 (z^2) as log_10 (5) + log_10 z^2 ?

Can you tell me what rule that is?

[timid silo]

uhhh

idk

[rose scroll]

if you dont know then you shouldnt write it like that

cmon, use power rule on 5 log(z^2)

and tell what should be the answer

you know you can do it

[rose scroll]

You have to use the 3rd rule

[timid silo]

I keep failing over and over

Is this never going to end?

[rose scroll]

Alright, I'll help you on steps

5 log(z^2) = 5 * 2log(z) = 10log(z)

ok?

[timid silo]

https://tenor.com/view/anya-spy-x-family-anime-anya-crying-crying-crying-anime-girl-spy-x-family-gif-25998648

[rose scroll]

From power rule

[timid silo]

what

[rose scroll]

you do know the power rule right?

[timid silo]

Yes

[rose scroll]

We use that on the first term

z has power 2, so we take it out of the log

[timid silo]

YOU MEAN TO TELL ME YOU WROTE THE QUESTION WRONG?!

I COULD HAVE DONE THAT

[rose scroll]

No

The question is correct

I am using the rules of logarithm to solve

[timid silo]

Why is z^2 log_z^2 instead of log_10?

[rose scroll]

log_10 is the base of log

[timid silo]

anddd???

[rose scroll]

we are taking log of z^2

to the base 10

[timid silo]

....

[rose scroll]

You know how log and exponent are related right?

[timid silo]

https://tenor.com/view/madea-fadreturns-angry-no-messing-everything-up-gif-24319516

Let me repeat it to a clearer version

[rose scroll]

10 in $\log_{10}(z^2)$ is same as the b in your notes. The rules do not apply to 10.

[warm shale]

Bacter10Fr4g is not fr0g

[void holly]

z for real numbers is absolutely cursed

[rose scroll]

hey, an unknown can be anything ok? Doesnt have to be x or y all the time. You can use z

NO. Leeave the 10 in log_10 alone

do not touch the 10

[void holly]

yes i am aware of that but...

[timid silo]

THIS IS CONFUSING

I GIVE UP

I SURRENDER

https://tenor.com/view/died-of-cringe-anime-cringe-dies-from-cringe-aharen-san-wa-hakarenai-gif-25337087

[void holly]

https://tenor.com/view/galaxy-quest-never-never-give-up-gif-4398729

[timid silo]

https://tenor.com/view/anime-bocchi-bocchi-the-rock-death-dead-gif-12123813093863538966

[rose scroll]

There is only one thing. Not playing with the numbers written in subscript. Leave the 10 in $\log_{10}(z^2)$ alone. No playing with that 10. That means do not multiply it. Do not divide it. Do nothing

[warm shale]

Bacter10Fr4g is not fr0g

[timid silo]

Okay

[void holly]

just keep the fact the base is 10 in the back of your mind for now

[timid silo]

IT'S MY BREAK

https://tenor.com/view/run-away-freak-out-happy-excited-run-gif-5324613

I GIVE UP NAH I AIN'T DOING THIS AAAAAAAAAAAAAA

.close

[obtuse pebble]

Channel closed

Closed by @solemn wadi

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[primal iris]

I am having trouble with how to define this thing so that /0 case never is and I don't know what to do for from end with 2d as said to substituting in an integral for a straight line. I think I can define arbitrarily a set of points all is relative to such that no divided value is zero but that v_2, is that a point or a number? How does it go to end result like this: https://mathproblems123.wordpress.com/2022/09/13/integrating-polynomials-on-polygons/ ? @worn coyote, sorry, you did this, so if you want to weigh in, go ahead. Using Green's Theorem on a planar, straight line, closed, not self intersecting polygon in 3d+ space. I prefer to avoid division as it can be undefined. Sorry for all this, just those "pictures" inside should be enough. Actually, dot product make numbers but v1 is a point so multiplying gives another point so yes v_1 and v_2 are points. I guess all I need is how to go from that Green's Theorem end to substituting into that integral from 1 to 0. X E.

[primal iris]

Kind of sorry to do this, but I can't explain not understanding to a computer like this very well. X E.

[obtuse pebble]

@primal iris Has your question been resolved?

[primal iris]

Would it be like inverse of partial derivative of a function over partial derivative of x?

[obtuse pebble]

@primal iris Has your question been resolved?

[primal iris]

f(x,y)=sqrt(sum per 3d+ dimension (O+Xx+Yy-P)^2). X E.

[primal iris]

Polygons are straight line, planar so like 2d, closed, and not self intersecting. X E.

[primal iris]

Should I just focus on triangles, rectangles, circles, circle parts, lines, and points because I think I can do all those?

[primal iris]

Just realized, need polygons. X E.

[worn coyote]

Hi, sorry for late response. I've just returned home

[primal iris]

Okay. Thanks for eventually coming. X E.

[worn coyote]

Admittedly, yesterday I did a poor job at explaining the idea

Hopefully today I will manage to properly explain everything

Let's start with reducing the problem to 2d. You mentioned that v_2 may encounter division by zero. In reality, that's not the case if p_1 - p_0 and p_2 - p_0 are linearly independent.

Since p_1 - p_0 and p_2 - p_0 are not collinear, they're linearly independent and therefore v_1 and v_2 are well-defined.

[primal iris]

Okay so are we defining p1 relative to p0?

[worn coyote]

Yes, but yesterday you said that p0 = (0, ..., 0) so I started saying p_1, p_2 instead of p_1 - p_0, p_2 - p_0

[primal iris]

Oh, p0 can be 0s but not necessarily, should I define things so p0 is 0s way for this?

Sorry, I was bad at explaining. X E.

p0 is (0,0) in 2d though. X E.

[worn coyote]

Well, we can let p0 be any vector, that doesn't complicate anything

Here's an image

The red vectors are p_1 - p_0 and p_2 - p_0

Then we orthogonalize them and get v_1 and v_2 (I will send an image in a moment)

[primal iris]

Thanks for all this. X E.

[worn coyote]

v1 and v2 are orthogonal to each other and have length 1. That will be convenient later

[primal iris]

So what are v1 and v2 equations?

[worn coyote]

Just like yesterday

[primal iris]

Okay, got that. X E.

[worn coyote]

$v_1 = \frac{p_1}{||p_1||}$, $v_2 = \frac{p_2 - (p_2, v_1) v_1}{||p_2 - (p_2, v_1) v_1||}$

[warm shale]

EQUENOS

[worn coyote]

Oh, wait, we now have arbitrary p_0

$v_1 = \frac{p_1-p_0}{||p_1-p_0||}$, $v_2 = \frac{p_2-p_0 - (p_2-p_0, v_1) v_1}{||p_2-p_0 - (p_2-p_0, v_1) v_1||}$

[warm shale]

EQUENOS

[worn coyote]

Here we go

[primal iris]

Working so far. X E.

[worn coyote]

We now can calculate the coordinates of $p_k$ in terms of that planar basis

[warm shale]

EQUENOS

[worn coyote]

It's simple: let $x_k=(p_k - p_0, v_1)$, $y_k = (p_k - p_0, v_2)$. Then $(x_k, y_k)$ are the coordinates of $p_k$ in our orthonormal 2d basis

[warm shale]

EQUENOS

[primal iris]

Those , are dot product correct?

[worn coyote]

yes

[primal iris]

Try *?

[worn coyote]

I can use this notation, if you want: $x_k = (p_k - p_0)\cdot v_1$

[warm shale]

EQUENOS

[primal iris]

Easier to see it is dot, not , like another coordinate thanks. X E.

[worn coyote]

alright

[primal iris]

I can tell this will be complex. X E.

[worn coyote]

Now let's recall the original goal. Let $\Omega \subset \mathbb{R}^n$ be a planar polygon, and let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be an integrable function. The goal was to compute
$$\int\int_{\Omega} f(p) d\sigma,$$
where $d\sigma$ is the element of the area.

[warm shale]

EQUENOS

[primal iris]

Yes. X E.

[worn coyote]

We now know that each point on the polygon can be expressed as $p = p_0 + x v_1 + y v_2$, where $v_1, v_2$ is an orthonormal basis, so the integral got reduced to
$$\int\int_D f(p_0 + x v_1 + y v_2)dxdy$$

[warm shale]

EQUENOS

[worn coyote]

Here $D$ is a polygon on the xy-plane that we obtained after calculating all $(x_k, y_k)$.

[warm shale]

EQUENOS

[primal iris]

I notice that vectors v1 and v2 are more like of 1 or less any dimension, can be greater than 1 distance. X E.

Or am I wrong?

[worn coyote]

$v_1$ and $v_2$ are vectors of length precisely 1

[warm shale]

EQUENOS

[primal iris]

Okay, sorry. X E.

[worn coyote]

And they're perpendicular to each other

[worn coyote]

That follows from this construction

Anyways, let's continue

For convenience, let $F(x,y)=f(p_0 + x v_1 + y v_2)$.

[warm shale]

EQUENOS

[primal iris]

Next?

[worn coyote]

Due to Green's formula, if $F(x,y) = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$, then we have
$$\int\int_D F(x,y)dxdy = \int_{\partial D} (Pdx + Qdy)$$

[warm shale]

EQUENOS

[worn coyote]

Let's just say that $P(x,y)=0$ and $Q(x,y) = \int_a^x F(s,y)ds$.

[warm shale]

EQUENOS

[worn coyote]

Then, if $(x(t), y(t))$ is the curve that parametrizes the boundary of $D$, our integral is equal to
$$\int_0^T Q(x(t),y(t))\dot{y}(t)dt$$

[warm shale]

EQUENOS

[primal iris]

I don't know what that s does but this seems maybe correct. I can do single integrals. X E.

[worn coyote]

Since $D$ is just a polygon, the curve is a piecewise linear function, where the $k-th$ segment can be paramterized as $(1-t)(x_k,y_k)+t(x_{k+1},y_{k+1})$

[warm shale]

EQUENOS

[worn coyote]

s is just an integration variable

For example, if $F(x,y)=xy$, then $\int_a^x F(s,y)ds = \int_a^x s y ds = \frac{s^2 y}{2} \mid_a^x = \frac{x^2 y}{2} - \frac{a^2 y}{2}$

[warm shale]

EQUENOS

[worn coyote]

Shall we continue from here?

So now we have
$$\int_{\partial D} Q(x)dy = \sum_{k=0}^n \int_0^1 Q((1-t)x_k+tx_{k+1},(1-t)y_k+ty_{k+1})(y_{k+1}-y_k)dt$$

[warm shale]

EQUENOS

[primal iris]

I am back, power outage. X E.

And those "," are dot?

[worn coyote]

In the blog post you shared, the person was integrating monomials. Let's do that too, because why not. Let $F(x,y)=x^p y^q$. In the formula for $Q$ we're free to choose any $a$. Let's choose $a=0$. Then we have
$$Q(x,y) = \int_0^x s^p y^q ds = \frac{s^{p+1}}{p+1}y^q |0^x = \frac{x^{p+1}}{p+1}y^q.$$
After that we obtain
$$Q\left((1-t)x_k + t x{k+1}, (1-t)y_k + t y_{k+1}\right) = \frac{((1-t)x_k + t x_{k+1})^{p+1}}{p+1}((1-t)y_k + t y_{k+1})^q$$
Now we can just substitute that to our sum

[worn coyote]

Inside the integral "," separates two arguments of Q

[primal iris]

Oh then solved already?

[warm shale]

EQUENOS

[worn coyote]

yes

You now have the algorithm for integrating f(p) over polygons in R^n

[primal iris]

I am on a phone, have to wait for internet to be online again, then I will document this on a computer and hopefully have no more questions. X E.

[worn coyote]

Alright

[primal iris]

So partial derivatives went away?

[worn coyote]

Yes, here's how we dealt with them: #help-10 message and #help-10 message

[primal iris]

Okay, there will be a pause but going to document now. X E.

[worn coyote]

okay

[primal iris]

I think I got it, if this does not work then I may call you or someone else again. X E.

That T above is like x in F(x,y) correct?

Anyway, thanks. X E.

.close

[obtuse pebble]

Channel closed

Closed by @primal iris

Use .reopen if this was a mistake.

[worn coyote]

T is some upper integration bound that depends on the curve we chose. I meant that (x(t), y(t)): [0, T] --> R^2

So if we integrate over such curve, the integral is from 0 to T

[primal iris]

.reopen

[obtuse pebble]

✅

[primal iris]

I still am unsure what happened to |partial derivative p/partial derivative x+partial derivative p/partial derivative y|. X E.

As from origin formula. X E.

This looks as solved as it should be. X E.

Great, power out, no internet. X E.

Just so you know, those last two not from now. @worn coyote . X E.

[worn coyote]

oh ok

[worn coyote]

We want to find $P(x,y)$, $Q(x,y)$ such that $F(x,y)=\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$. Note that if we let $P(x,y)=0$ and $Q(x,y) = \int_a^x F(s,y)ds$, we're done. Because in this case $\frac{\partial P}{\partial y}=0$ and $\frac{\partial Q}{\partial x} = F(x,y)$, so we indeed have $F(x,y)=\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$.

[warm shale]

EQUENOS

[primal iris]

Before that, p as in f(p(t)). X E.

[worn coyote]

$F(x,y) = f(p_0 + x v_1 + y v_2)$

[warm shale]

EQUENOS

[worn coyote]

Each point on the original polygon can be represented as $p_0 + x v_1 + y v_2$

[warm shale]

EQUENOS

[worn coyote]

So integrating f over the polygon is equivalent to integrating F(x,y) over D

[primal iris]

Looks like we re-defined to this. X E.

[primal iris]

Also this. X E.

I guess solved. X E.

Any closing words?

No I guess. Thank you. X E.

Maybe write a paper on this one EQUENOS, this is great stuff for graphics. X E.

[worn coyote]

I think the morale is that as long as we have a planar figure, the integral over it can be reduced to a 2d case

[primal iris]

I guess I have permission to use this in a 3d+ graphics system?

[worn coyote]

I think a more meaningful generalisation of the blog post you shared would be to integrate over polyhedrons, but again, all this was invented long ago in far more general cases, such as manifolds of arbitrary dimensions

[worn coyote]

Of course

[primal iris]

Now fun part, how to tell a computer this. X E.

[worn coyote]

I'm gald this problem has applications in computer graphics

[primal iris]

I get to do that. X E.

Even more fun when I need to calculate where plane is in view in 4d+. X E.

Anyway, thank you. X E.

[worn coyote]

It's a painstaking work but it's definitely doable

You're welcome

[primal iris]

.close

[obtuse pebble]

Channel closed

Closed by @primal iris

Use .reopen if this was a mistake.

[obtuse pebble]

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[full elm]

What are the number of 10 digit binary sequences such that no more than two 1's or 0's are placed beside each other

Example :
1001101001 --> ✅
1101011101 ---> ❌

[twin sandal]

should be 2^n - (number of sequences in which there are 3 1s in a row or 3 0s in a row) right?

since if there are 4 in a row, there are 3 in a row for example

so no need to count those

[full elm]

actually can you tell me how we can approach the 3 1s or 0s in a row problem without repetitions?

cuz my approach initially was that we group 3 1's or 0's in a box then we place them in between a series of 7 digits like so

1010101 ----> place -----> 1010[000]101

but then this as well as this are same

1010101 ------> place -----> 101[000]0101

these two cases are equivalent and so will be repeated

[twin sandal]

that's a good point, I think you may have to do it by considering cases in which there are exactly 3 zeros in a row, exactly 4 in a row, etc.

there's probably a simpler way but i can't think of it right now

oh duh, just methodically count the number of sequences with exactly two 0s in a row

same for 1s, it shouldn't be too difficult

[full elm]

but then placing will be tough as we cant place [0,0] near another 0 or [0,0]

[twin sandal]

start with only one 00 and count those, then two 00s, then three 00s, etc

if there is only one 00, then you know there must be 1s on either side of it, and the rest of the digits must be alternating

does that make sense?

[full elm]

yes