help-0

[loud oriole]

ok in the og pic, the number of terms is n-k, so why is the closed form n*c and not (n-1) * c

[gray isle]

the number of terms isn't n-1

[loud oriole]

oh i see then

[gray isle]

just like how the number of integers from
1 to 100 (inclusive) isn't 100-1 = 99

[loud oriole]

alright, then thx my g

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[fossil mason]

How do I find qtp

[gray isle]

what have you tried?

[fossil mason]

Pythagoras. If I couldn’t find qt or pt

I also tried looking for equilateral triangle

Or like any dif side length

[gray isle]

apply TQS is a sector of a circle with centre T

[fossil mason]

R*theta?

[gray isle]

you're skipping ahead

use what I mentioned to determine the length of QT

[fossil mason]

QT is 15

[gray isle]

yes

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[fossil mason]

What do i do now?

[gray isle]

determine your angle

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[gilded kiln]

How are they rewriting this

doesnt 4/100 = .04 and 25/100 = .25, where are they getting the 5 and 2 squared for a and b

[echo socket]

4/100 = 1/25 = 1/5^2

25/100 = 1/4 = 1/2^2

[gilded kiln]

ohhhh

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[gilded kiln]

$-4\sqrt{2}=81-b^2$

[ocean seal]

bababeeboo

[gilded kiln]

subtracting 81 from both sides gets me -86.66, and then dividing both sides by -1 gets me 86.66, and then sqrting both sides gets me b = 9.31

im finding b to be able to get the co-vertex for this question

but the covertices (9.31), (-9.31) obv arent right since a has to be > b

so im not sure what im doing wrong

[lone heart]

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[gilded kiln]

<@&286206848099549185>

[ocean prism]

O wait

Nvm

Just read what u said

Lemme a sec ill try the question

[gilded kiln]

a calculator said the co-vertices are (7, 0) and (-7, 0) but I have no clue how they got that

<@&286206848099549185> ?

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[alpine sable]

A bit of an abstract question, but does anyone have any idea how I could normalise these results?

[alpine sable]

For example, I would be able to say that if x_0 - x_c < 0 , the equilibrium is "skewed" toward the left, if it's > 0 towards the right and if it's = 0 then it's perfectly balanced

but how could I quantify how much the skew really amounts to?

Say, on a scale from 0 to 1 ?

because x_0 - x_c is unbounded on both sides

does x_0 - x_c / max(x_0, x_c) seem reasonable ?

I'm unsure

yea I think it does

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[shell narwhal]

what does the c underbar math symbol means

[shell narwhal]

this

[timid hazel]

(Source: https://en.m.wikipedia.org/wiki/Glossary_of_mathematical_symbols)

[shell narwhal]

oh thanks

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[tranquil hare]

Can someone elucidate me on what I've done wrong here?

[tranquil hare]

Retraced my steps but can't figure out how I'm screwing up at the signs.

[buoyant kayak]

i'm not sure why you made the 14 negative when you had it positive on your last step

also how'd you get 7 for your denominator

[tranquil hare]

I didn't make it negative, I'm pointing out that that's the answer given as the solution in the end of the book.

[tranquil hare]

Multiplied the complex rational expression by the reciprocal of the denominator.

[buoyant kayak]

are you sure the answer doesn't say -(x-14)

oh wait that entire thing is over 7/(x^2-4)

which shouldn't change the sign of the 14 but makes sense how you get 7 in the denominator lol

[tranquil hare]

I'm gonna sue this publisher for not using parentheses, it's messing me up

You're absolutely right

[buoyant kayak]

#❓how-to-get-help #rules

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[limpid spade]

t=3.4

[naive valley]

is there a question somewhere?

[limpid spade]

.26

[alpine sable]

the refrence number

[naive valley]

thank you for that information

[limpid spade]

9 (2/2 Points) DETAILS PREVIOUS ANSWERS

[alpine sable]

i subtracted 3.4 and got .26

is that bad

[naive valley]

very bad, possibly

[alpine sable]

why it’s hw?

[limpid spade]

[alpine sable]

💀💀

[limpid spade]

we don't know

[alpine sable]

i can show

[limpid spade]

what the referemce

number is

[alpine sable]

see

[alpine sable]

yeah

i saw a youtube video and it said to subtract pi

[naive valley]

extraordinary

[limpid spade]

mam

mam

man

[naive valley]

i would definitely subtract pi in that case

[limpid spade]

sum then all

take the mean

lineair it

find the tangent line

[naive valley]

the pi is a lie

[limpid spade]

last chance

show us

the whole question

part a, b and c

[spark sky]

Lol

That’s weird

He doesn’t wanna @limpid spade

[limpid spade]

why

[spark sky]

I dunno

[naive valley]

no one can answer the question if there is no question

[spark sky]

Yeah for real

[limpid spade]

t=3.4
.26
🧠

[spark sky]

Amazing solution

[naive valley]

❌

[limpid spade]

Need Help?

@alpine sable

[alpine sable]

no sorry i just received it

i posted it twice by accident

[limpid spade]

?

[naive valley]

extraordinary

[limpid spade]

I'll just close it

.close

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[reef trout]

<@&286206848099549185>

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[alpine sable]

how do i find the integral from -3 to 0

using only the graph?

[umbral crest]

The integral is the area under the curve, does that help?

[alpine sable]

I'm aware

but i don't know how to get that from a quarter circle

wait

so that area is 9pi/4, the quarter circle

so... maybe subtract 9 (3*3) from that..? but thats a negative number

ah

9 - that number

kek

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[rich basin]

@bright hedge its because they lie on the unit circle

[last ether]

@rich basin close your session lol

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[viscid dew]

Why is dividing an amount by 0.5 = same as multiplying by 2 and vice versa (multiplying an amount by 0.5 the same as dividing by 2)?

[last ether]

Yeah, good question

So

When you divide by half, you're asking the question "how many halves fit into this?"

For example, when we do 1/0.5, how many halves go into 1?

@viscid dew

[viscid dew]

2 halfs exactly.

Ah, it means the answer for 1/0.5 is 2.

[last ether]

Yeah

So how about this

2/0.5

How many halves go into 2?

[viscid dew]

Four.

[last ether]

Good

How many halves go into 3?

[viscid dew]

6 halfs.

[last ether]

Ye

So you'll notice that

When we divide by 0.5, it results in doubling the dividend

This is also explained by the fraction rule where:

$\frac{a}{\frac{b}{c}} = \frac{ac}{b}$

[ocean seal]

Umbraleviathan

[viscid dew]

What about the opposite?

"Multiplying an amount by 0.5 the same as dividing by 2".

[last ether]

"What is half of this?"

Multiplying by 0.5 is "what is half of this?"

Dividing by 0.5 is "how many halves fit into this?"

Well

Better question is

What is 0.5 of this?

[viscid dew]

I came up with this one: "how much [number] is fit in the sum"?

[last ether]

Well

Ehhhhh

Sum?

[viscid dew]

Because after all, the number could be anything.

[viscid dew]

Isn't sum (summary) = amount?

I'm partly using a translator because I'm not native English-speaker.

[last ether]

Uh

Sum is like

Well

In a colloquial context, I guess

Although mathematically, not really

[last ether]

Regardless, if you're using "sum" colloquially, then it could work

[viscid dew]

Hmm, okay.

Thanks you.

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[ancient fjord]

Hi, can someone help me with span?

I didn't understand from where did the span {1,1} come from

on the first example

[verbal timber]

have you tried solving for x1 and x2 normally?

[ancient fjord]

Yes

[verbal timber]

what did you get

[ancient fjord]

x1=1+x2

x2=x1-1

[verbal timber]

You should try solving it in row echelon form

[ancient fjord]

But that is the result when I solve it with REF

[verbal timber]

multiply the second row by 1/2

then what do you notice about row 2 and row 1

[ancient fjord]

Yes they are identical now

[verbal timber]

So using gaussian elim. the second row is all zeros

[ancient fjord]

yes

[verbal timber]

leaving you with just x1-x2=1

[ancient fjord]

x1 =1

[verbal timber]

x1=1-x2

x2 can be any number

it is not necessarily 0

so x2=k

k is any number

[ancient fjord]

okkk

[verbal timber]

x1=1-k
x2=k

write it in vector form

(x1, x2)=(1, 0)+k(-1 1)

if k can be any number then the solution includes the span of (1, 1)

[ancient fjord]

aaa Okay I see, what if the solution is not any number

The span would be the vectors

that we found

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[alpine sable]

If I pick seven random digits and arrange them as a number, what's the chance the two last digits are equal? So from 0000000 to 9999999 (which has the last two numbers equal), what's the chance?

[alpine sable]

I thought we could isolate the first five digits and think of the last two separately. There are 10 possible combinations for equal digits (00, 11, 22, etc), so I thought maybe one in each (10^5)*(10^2), which is one every 10^7, is that right?

I forgot the properties on exponents, so I'm not sure if we add the exponent or multiply them.

But multiplying is in nested exponents, so I thought my best bet would be adding them.

Because for the 00 ending we have 10^5 possible numbers, for the 11 ending it's the same, all the way up to 99. So it's actually 10*(10^5), leaving me with 10^6 since 10=10^1, right?

One in each 10^6 numbers have the last two digits equal.

[atomic acorn]

you are right in your approach by ignoring the first 5 digits and making the last 2 same

[atomic acorn]

however I didnt understand how you arrived at this conclusion?

the total numbers in your sample space is 0 to 9999999 which is 10^8 if I'm not wrong

and the total numbers with last 2 digits same is 10^6 as calculated by yourself

[alpine sable]

OOH yeah, that's right

My objective is to find the probability that if I pick a random number with 7 digits, the last two are equal, as I've said. So it'd be fair to say that one every 10^6 seven digits numbers would have the last two digits equal, right?

[atomic acorn]

Wait so would 0 be included then?

[alpine sable]

Yep.

0000000

[atomic acorn]

coz that technically isn't 7 digits but ok

[alpine sable]

Just fill it till it's 7

[atomic acorn]

yeah got it

[alpine sable]

12 would be 0000012

[atomic acorn]

yeah but i still don't understand this tho

Your probability would be 10^6/10^8 right?

[alpine sable]

Meaning $\frac{10^{6}}{10^{8}}=10^{-2}$, right?

[ocean seal]

3317

[alpine sable]

Yes

$\frac{1}{100}$ is the probability then?

[ocean seal]

3317

[atomic acorn]

Yea

total cases / sample space right?

[alpine sable]

Heck yeah

Can't thank ye guys enough.

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[fallow pagoda]

can someone help i don’t understand

[atomic acorn]

What's your progress so far?

[alpine sable]

Yeah that's some weird wording you've got there.

[fallow pagoda]

pretty much nowhere

i looked at the answer and k can’t get how they got it

[alpine sable]

You're looking for the domain in which the values increase.

[atomic acorn]

Ok so looking at a graph how can you determine if it's increasing decreasing?

[fallow pagoda]

if it’s going up from to right ?

up to right

[alpine sable]

Generally just up will do.

[fallow pagoda]

okay

[atomic acorn]

with increasing x this is correct

[fallow pagoda]

so

[atomic acorn]

You go from left to right in x

[fallow pagoda]

yep ok

[atomic acorn]

If the graph goes up its increasing

[fallow pagoda]

okay and how do i find the domain doinf that

because the answer doesn’t make sense to me

(0, infinity)

i dont see where the 0 came from

[atomic acorn]

Before zero is the graph increasing or decreasing?

[alpine sable]

Note that in 0 the function doesn't increase nor decrease.

Just in 0.

[fallow pagoda]

on thé x axis??

ohh wait i think i see

so for part ii) is is (0,-infinity)???

because the answer says (-infinity,0)

whys infinity in the x slot

[alpine sable]

Nah, it's an interval, the greater number should be on the right.

[fallow pagoda]

oh so it doesn’t matter about the x and y??

[alpine sable]

Not sure I get what you're asking.

It does matter.

Question is: To which range of x, y increases if x increases?

[fallow pagoda]

yeah no for part ii) i mean sorry

it’s asking about the decreasing

so before x= 0 the graph decreases from forever

y being -infinity

so isn’t it (0,-infinity)

[alpine sable]

It's $(-\infty, 0)$.

Wait, wrong.

[ocean seal]

3317

[alpine sable]

Now it's correct.

[fallow pagoda]

yeah i’m confused why that is sorry

why the infinity is at the front

because that’s the y value is it not??

[alpine sable]

Because it's lesser than 0.

[fallow pagoda]

or is infinity the x value when decreasing

[alpine sable]

Lesser on the left, greater on the right, that's how intervals work.

If you invert the order you just get the null set.

[alpine sable]

No, y approaches negative infinity as x approaches negative infinity.

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[quiet lantern]

Question regarding coefficient of determination within linear regression, does it simply mean how strong the linear regression in predicting future datapoints?

[noble sinew]

not future datapoints, no

[hybrid zenith]

It's more about proportion of change in 1 that can be predicted by the other varibale
As opposed to extrapolation
As far as I remember

[quiet lantern]

Isn't that the slope?

[hybrid zenith]

The slope would surely just be change in y/change in x
This is to do with how much variation in a variable is because of the change in the other

I'm not rlly sure how best to word it sorry

[oblique spire]

Coefficient means the slope of the mx+c linear regression the m there is the coefficient not the accuary of the model

[noble sinew]

no it means the R^2 value

and as I said it doesn't mean strength of predicting future datapoints

[oblique spire]

Did u meant coefficient = r^2 value?

[hybrid zenith]

They said coefficient of determination =r^2

[oblique spire]

Oh

I see

I thought it's the slope

[quiet lantern]

So what does it mean?

[noble sinew]

how much of the variation in respone is explained by the explanatory variables

is one way to view it

[oblique spire]

R^2 does indicate the performance of the model ig

[quiet lantern]

I asked someone about this chart and they said that the R2 is equivalent to a "magnet" to the datapoints on the chart

R2 = 0.314 (31.4%) means that the "magnet" of the blackline is that powerful, if it had 100% then all the datapoints would be on the black line itself and if it had 0% then it would scatter all over the chart, is that accurate explanation?

[oblique spire]

R2 is a measure of the goodness of fit of a model. In regression, the R2 coefficient of determination is a statistical measure of how well the regression predictions approximate the real data points. An R2 of 1 indicates that the regression predictions perfectly fit the data.

It's written directly on wiki ig

[noble sinew]

magnet wtf

[quiet lantern]

yeah its supposedly an analogy to explain it better to me haha

[oblique spire]

Lol like the line is a magnet attraticing the data points

[quiet lantern]

yeah

so like given that its 0.314, thats considered "weak" relatively

but if it was say 0.7 that would be considered 70% powerful

[oblique spire]

From what I learned in machine learning

Ya

R2 is indeed the performance of a model

[quiet lantern]

and 100% would mean that all datapoints is right on the line itself

[hybrid zenith]

Ig the magnet analogy works
Tho u just need to make it clear in the exam that its to do with variables
As opposed to the line being a magnet lol

[oblique spire]

Dude can I just ask if I understand the question correctly did u meant if r^2 is the measure of how good the line of best fit is

Or how well it will perform on unseen data

[quiet lantern]

basically I'm trying to understand what R2 means and how we can benefit from it in future data

Can it be used as a predictor for future datapoints?

Or is it completely useless in that?

[oblique spire]

No?

R2 is a evaluator rather

It tells us

How good the model is

[quiet lantern]

The model being the line itself, yeah?

[oblique spire]

Ya

R2 tells the marks received by the line

[quiet lantern]

And the line itself says "in a perfect case scenario, the datapoint should be on this line"

And that "perfect case scenario" is represented with R2 at 1.0 (or 100%)

[oblique spire]

Ya

The higher r2 the better

Ig

[quiet lantern]

yeah

[oblique spire]

So summing it all up R2 atleast in the field where I worked in is used more of a metric to evaluate the performance of a linear regression model

[quiet lantern]

So I'm asking that because another friend was arguing against the person that explained the analogy... And said that 0.5 R2 means that there is no relation

so I got confused 😂

I'm receiving two opposite explanation and didn't know what to believe

[noble sinew]

you can't just only look at R^2 and say higher R^2 is better

very false

[oblique spire]

I don't think so

[quiet lantern]

about what @oblique spire said? 🤔

[oblique spire]

R2 is a measure of the goodness of fit of a model. In regression, the R2 coefficient of determination is a statistical measure of how well the regression predictions approximate the real data points. An R2 of 1 indicates that the regression predictions perfectly fit the data

It's written on wikipedia ig

Unless u meant something called overfitting

[oblique spire]

Well as I said it's one metric

We still have stuff like RSS

They are different metrics

[noble sinew]

just saying

[oblique spire]

I can't even read the meme

Oof

[quiet lantern]

yeah quality is a bit weak hard to read numbers

[noble sinew]

numbers are useless

just look at the plots

[oblique spire]

Did

[noble sinew]

so clearly R^2 doesn't say anything about how good your model is in itself

[oblique spire]

It is saying in one way

U have to generalize

Between other metrics

LIKE rss

Such that it works well on almost all

It's just one of the many metrics

[quiet lantern]

Yeah I mean I assume there are exceptions to R2 right? but would you say it works in 99% of situations?

[noble sinew]

wdym work?

[quiet lantern]

as in the model is "predicting" the datapoints

[oblique spire]

Well I have seen machine learning engineers use r2 to evaluate linear regression models

All the time

Maybe it's applications are a bit different in normal stats

[noble sinew]

You can fit a linear regression with a R^2 very low just fine

as long as assumptions are met

[oblique spire]

No?

[quiet lantern]

Ah

[noble sinew]

all this means is your confidence intervals will be bigger

yes you can

the assumption that variance is the same at all points and they are independent

only asusmption needed for linear regression

[quiet lantern]

I think what @noble sinew means is that the data has to be:
1- homoskedastic
2- linear
3- independent
4- normality in observations

is that it?

[oblique spire]

I mean a good model can have a bad R^2 but a model with good R^2 will generally be a good model

[noble sinew]

yes

[oblique spire]

That doesn't make my statement

Wrong

Ig

That high r^2 is good

But low r^2 isn't bad either

[noble sinew]

you said the higher the better

[oblique spire]

Oh right mb

[quiet lantern]

In most situations would it still be false?

that the higher is better?

[noble sinew]

step 1 before fitting linear regression is plotting your data (assuming the dimensions actually allow you to look at your data)

but assuming assumptions met then all higher R^2 does is make your length of conf interval shorter

that is all it does

[quiet lantern]

okay one more question, i know that R2 is calculated using different formulas and I've seen many websites calculate R2 by simply multiply correlation by itself...

Was just wondering if R2 can always be calculated by simply multiplying correlation by itself? Because that seems like a very simple formula to calculate something that is supposedly complex with a lot of assumptions in it...

[noble sinew]

if you mean Pearsons correlation coefficient then yes

[quiet lantern]

Isn't R2 coefficient of determination? 🤔

[noble sinew]

which is persons correlation coefficient squared?

[quiet lantern]

That is what I'm asking

Can it always be calculated by multiplying correlation by itself?

[noble sinew]

.

[quiet lantern]

Oh, there are other types of correlation? 😅

(do they give different answers too?)

[noble sinew]

yes there are many (pearsons is the standard one)

[quiet lantern]

So if I want to calculate R2 I can only use pearson

but if I used another correlation?

It would give a wrong R2 ?

I always thought correlation is -1 to 1 and assumed that it is always the same number

[quiet lantern]

@noble sinew

[noble sinew]

Because they are defined differently yes ofc you get another number

[quiet lantern]

Thank you for your help everyone @noble sinew @oblique spire @hybrid zenith

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[strong hornet]

.. don't ping helpers right away

[lone heart]

@thick lynx Has your question been resolved?

[noble sinew]

Depends on the questions. Typically competition math doesn’t have a lot to do with standard syllabus

Depends on the questions you call hard

[velvet pelican]

hard school maths is not really comparable to competition math

[lone heart]

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[delicate widget]

I’ve gotten 48

[delicate widget]

Wrong question

[lone heart]

Channel closed

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[delicate widget]

Help with c

ive gotten 48

but since 48 is in the group 200<w<400 i dont know how to deal with it

its in between the 3rd and 4th group

im getting confused

how to comment on if hes correct

ping me if anyone comes to help me

[lone heart]

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[delicate widget]

d

[delicate widget]

randomlyly closed

[delicate widget]

.

<@&286206848099549185>

noone

sad

.close

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[uneven wing]

The Exercise: ABC a non-isosceles right triangle at A. I is the midpoint of [BC] and [AH] is its height from A. Show that: cos IAH = 2AB*AC/BC²

I didn't understand how to show it , I consider your help thank you

[open folio]

Have you drawn it out?

[uneven wing]

lemme do it real quick

it look like this

hmmm breziboi?

<@&286206848099549185>

[oblique spire]

Cos of I can't read it?

IAH

Ig

[uneven wing]

cos of IAH

can you help me?

[oblique spire]

Lemme try

[lone heart]

@uneven wing Has your question been resolved?

[lone heart]

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[frigid verge]

I just want to check I did it right... in the blue pen

[lone heart]

@frigid verge Has your question been resolved?

[frigid verge]

<@&286206848099549185>

.close

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[alpine sable]

Hello , im having trouble understanding shanon entropy , i need to find out for example how the entropy is for a password consisting of 4 numbers ( 0-9 so 10 posibilities)
So i calculated -sum k=1 to 2 ( 1/10)*log2(1/10) but the answer seems wrong

[alpine sable]

Found the solution the sum was wrong

.close

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[native kite]

so this problem is assigned to me by my teacher

[native kite]

and I answered with this, but I'm not confident with my answer, so I would like to ask anyone if this is correct?

I'm asking because I have very little knowledge on limits and continuities and I just watched some videos about it and tried to answer based on my understanding

[last ether]

Okay so you have the right idea, but not quite

When checking for continuity within [-1, 1]

Check to see if the area to the left of -1 would be equal to the area right of -1

Aka, check a = -1

And a = 1

[native kite]

ohh okay got it

so is 1 and 3 correct?

[lone heart]

@native kite Has your question been resolved?

[native kite]

<@&286206848099549185>

[uncut cobalt]

help

[harsh swallow]

,rotate

[native kite]

so I tried something like this, both a=1 and a=-1 aren't equal

so it's discontinuous?

[harsh swallow]

it's not continuous on the interval [-1,1]

which you have shown

by showing that the function has a jump

from -8 to -3

[native kite]

okay got it

[native kite]

but what about this?

[native kite]

im referring to this

[harsh swallow]

i'm guessing what they mean is at x = 1

if it is continuous at x = 1

[native kite]

yeah this is the problem

[harsh swallow]

meaning the limits from the left and the right are the same and equal to the value at that point

from the left you used the wrong part

you used the first part of the function but you should've used the second part

because we're looking at x = 1

[native kite]

oh I just noticed that

okay thank you very much

[harsh swallow]

also 3 isn't correct if i remember correctly

[native kite]

how do I do that?

[harsh swallow]

the reasoning as to why it is discontinuous is incorrect

[native kite]

okay got it

I'll figure the rest on my own

thank you very much!

.close

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[slender girder]

How do you approach this

[rotund plank]

First multiply both sides by x-1

Then discriminat

[slender girder]

Done

[rotund plank]

Assume x ≠ 1

[slender girder]

I simplified it

lett me show you

where i got to

x^2-2x(m^1/2)+2(m)^1/2((m)^(1/2) + 1)) -3 = 0

since it has no real roots

but how do i do the discriminant from here onforth

whats the a, b , c

[rotund plank]

No

[slender girder]

wait what

[rotund plank]

It’s equal to zero

[slender girder]

it has no real roots

ohh

[rotund plank]

You’ve confused the equation with the discriminant

[slender girder]

the discriminantt is less than 0

my bad

yes yes

[rotund plank]

Just get coefficients of x^2, x, and a constant term

Those are your ‘a’, ‘b’, and ‘c’

[slender girder]

the coefficient of x is 2 ?

and x^2 is 1 ?

[rotund plank]

No

[slender girder]

oof

how

[rotund plank]

What happens to sqrt(m)

You forgot about it

[slender girder]

is it 2sqrtm?

[rotund plank]

Nope

[slender girder]

hmm

[rotund plank]

ax^2 + bx + c = 0

[slender girder]

im a bit lost as you can see

c would be -3 right?

[rotund plank]

There’s a minus in front of that term

No

[slender girder]

what

[rotund plank]

2sqrt(m)(sqrt(m)+1) is a constant

[slender girder]

okay

so thats c

what happened to the -3 then?

[rotund plank]

It’s also part of c

That whole end bit is c

[slender girder]

2sqrt(m)(sqrt(m)+1) - 3 ?

[rotund plank]

Yep

[slender girder]

right

so b would be 2sqrt(m)

[rotund plank]

-2sqrt(m)

[slender girder]

ahh

and a would be 1

[rotund plank]

Yea

[slender girder]

oo

(-2sqrt(m)^2-4(1)(2sqrt(m))(sqrt(m)+1)-3))<0

shouldnt m be greater than or equal to 0

im actually lost

@rotund plank sorry for the tag

could you elaborate on how i'd go about solving it now

[rotund plank]

It’s okay

Yes it should

Just expand it

[lone heart]

@slender girder Has your question been resolved?

[slender girder]

I did

I got -10m -8sqrtm I think

-10 m - 8 sqrt(m) - 3<0

how ?

<@&286206848099549185>

[slender girder]

.close

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[fossil valley]

I don’t even know where to start with these two questions

dP/dt = Kt?

P=Ce^(kt)

For the first one

Is that correct

[fervent timber]

apparently, you do

[fossil valley]

Yeah now I do

Had to remember 💀

Now for the other one

I am stuck

[lone heart]

@fossil valley Has your question been resolved?

[fervent timber]

think about what it means to increase at a constant rate

[lone heart]

@fossil valley Has your question been resolved?

[lone heart]

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[thorn vale]

The speed of sound at sea level is 760 mph (mach 1.0) what is the speed in mph of a bullet traveling at mach 0.7?

I subtracted 0.7 from 1.0 and got a 0.3 difference. Do I then divide 760 by 3?

[marsh rapids]

mach 0.7 means 0.7 times the speed of sound

no need to overcomplicate this

[thorn vale]

Ok

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[slender girder]

I worked out A

How would I approach b and c

[oak chasm]

For b, get the second, fourth, and eighth terms of the arithmetic sequence.

Use those as the second, third and fourth terms of the affine sequence.

Use the formula and fill in values you know.

You should get two equations from that.

And you have two variables to solve for.

@slender girder

[slender girder]

I tried

To do that

so u2 =8

u4 = 18

u8 = 38

in the original arithmetic sequece

they're telling us u2= ru_3 + s and u4= ru_2 + s ?

[oak chasm]

So, in the affine sequence,

u₂ = 8
u₃ = 18
u₄ = 38

u₃ = ru₂ + s

u₄ = ru₃ + s

You got the indexes mixed up.

The larger one is on the left of =.

[slender girder]

oh

so 18=ru_2 + s ?

and 38=ru_3 + s

[oak chasm]

Yes, and fill in u₂ and u₃ there.

[slender girder]

Ohhh

we use our initial values ?

[oak chasm]

18 = 8r + s
38 = 18r + s

[slender girder]

so like u2 is 8

OHHH

im an idiot

[oak chasm]

Now it's just simultaneous equations.

[slender girder]

we can just subtract

and solve for r first

no ?

[oak chasm]

Yeah r in terms of s.

[slender girder]

r=2

[oak chasm]

Oh, I see.

Yes, that works.

[slender girder]

s=2

does that make sense

for both of them to be the same valuie

[oak chasm]

Yes, as long as it works.

Check with the original equations.

[slender girder]

So i found r to be 2

so I did

[oak chasm]

18 = 8r + s
38 = 18r + s

[slender girder]

38=18(2) + s

38-36 = s

s = 2

[oak chasm]

18 = 8(2) + (2)
38 = 18(2) + (2)

If those are truly equal, you got the right answer.

For C, I'd guess you use induction.

On m.

[slender girder]

how

would you use induction

[oak chasm]

Well, you start with m = 1.

[slender girder]

oh wait

they rewrote the q

[oak chasm]

You show that the formula given in part C is equal to the equivalent from just before B.

[slender girder]

lemme send you the reworded q

[oak chasm]

OK.

[slender girder]

Show that the mth term of the affine sequence equals the 2^(m-1)th term of the arithmetic sequence.

[oak chasm]

That's C part ii.

[slender girder]

ii ^

yeah

[oak chasm]

For C part i, induction.

[slender girder]

proof by induction ?

[oak chasm]

Yes.

[slender girder]

so we plug in any two values

of r and s

that are the same ?

[oak chasm]

uₘ₊₁ = ruₘ + s
uₘ₊₁ = rᵐu₁ + s((rᵐ - 1)/(r - 1))

No, you leave r and s as variables.

It's asking you to show it for all r and s.

So, r and s can't take on any particular value when you're proving it, or else you've only proven it for those values.

[slender girder]

ohh

alr

do we use a particular value of u1 etc ?

[oak chasm]

No, but start with m = 1.

[slender girder]

if m=1

itll be u1

[oak chasm]

What will be u₁?

[slender girder]

u_(m+1)

[oak chasm]

No, m = 1 means m + 1 = 2.

[slender girder]

oh itll be u_2

[oak chasm]

But we don't exactly need that.

[slender girder]

okay so u_2=r*u_2 + s

[oak chasm]

uₘ₊₁ = ruₘ + s
uₘ₊₁ = rᵐu₁ + s((rᵐ - 1)/(r - 1))

So,
ruₘ + s = rᵐu₁ + s((rᵐ - 1)/(r - 1))

[slender girder]

okay sure

[oak chasm]

So, prove that for m = 1.

[slender girder]

They're both u_m+1

[oak chasm]

Right, so they're both equal to each other.

[slender girder]

yes

so its like saying

1=1

[oak chasm]

No, it's not.

[slender girder]

why

isnt it

[oak chasm]

The form of the thing on the left and right aren't the same.

[slender girder]

you want me to manipulate it?

[oak chasm]

It's not immediately apparent that they're actually equal.

You need to prove that, unlike with 1 = 1.

[slender girder]

Ohh okay

so should i work out both values at m = 1

[oak chasm]

Yes.

[slender girder]

and show that they're equal?

[oak chasm]

Right.

[slender girder]

I mean

I got to

U_2=r^1 * u_1 + s(r^1-1)/(r-1)

dont they cancel

the r-1/r-1

[oak chasm]

No, m = 1, not 2.

[slender girder]

yes

but its U_m+1

1 + 1 = 2

so it becomes u_2

[oak chasm]

Yes, (r - 1)/(r - 1) cancels.

No.

They give you two formulas for the sequence.

One is known to be correct.

The one above part B.

It's a recursive formula.

One isn't.

It's a more direct formula.

[slender girder]

oh

is the correct one the second

[oak chasm]

No, it's the formula above question B.

[slender girder]

Oh

[oak chasm]

The recursive formula.

In C part i, they give you the closed form version.

Where you don't have to go through each and every previous item in the sequence like with the recursive version.

[slender girder]

right

[oak chasm]

The recursive formula is correct, but the closed formula might not be.

So, you set the formulas equal to each other and then you use induction to show that they are indeed equal for any m.

[slender girder]

the thing is

if we used 1

in the closed formula

we'd get u_2 = r * u1 + s

after cancelling the r-1's

[oak chasm]

Right, but set the formulas equal to each other.

Not equal to uₘ₊₁.

[slender girder]

ohhh

r*u_m + s = r * u_m + s

[oak chasm]

Well, fill in the ms.

[slender girder]

yeah

1

[oak chasm]

So, the formulas are the same for m = 1.

[slender girder]

r*u1 + s = r x u1 + s

yes

[oak chasm]

That's the base case.

Now we do the inductive case.

Let's say it works for a certain m. Does it then definitely work for m + 1?

ruₘ + s = rᵐu₁ + s((rᵐ - 1)/(r - 1))

[slender girder]

I suppose so?

[oak chasm]

We want to prove that.

And the inductive hypothesis is that it worked for the previous m.

ruₘ₋₁ + s = rᵐ⁻¹ u₁ + s((rᵐ⁻¹ - 1)/(r - 1))

The previous m worked, so we know for sure that that equation is correct.

How can we transform that equation into ruₘ + s = rᵐu₁ + s((rᵐ - 1)/(r - 1))?

[slender girder]

m-1 ?

[oak chasm]

Yes, with induction, you have the base case, and we did that with m = 1.

You also have the inductive case, where you assume it worked for the previous number and show it works for this number.

The assumption is called the inductive hypothesis.

[slender girder]

Oh I see

Wait could I ask

how old are you

[oak chasm]

Inductive hypothesis:

ruₘ₋₁ + s = rᵐ⁻¹ u₁ + s((rᵐ⁻¹ - 1)/(r - 1))

Oh, an adult, but I don't want to get more specific than that.

[slender girder]

Ahh I see

No worries

if we set m to 2 ?

[oak chasm]

No.

[slender girder]

it would look the same as the one above

2-1 = 1 ?

[oak chasm]

The base case has the a particular value for m.

The inductive case doesn't.

Here's how induction works.

You show it works for one number.

You show that if it works for a number, it works for the next number.

So, if it works for m = 1, then it will work for m = 2.

Then, if it works for m = 2, it will work for m = 3.

[slender girder]

we want to know if it works when m = 0 now ?

[oak chasm]

And so on.

No.

Do you understand how induction works?

[slender girder]

yes

we prove that a statement holds true

[oak chasm]

How does it work?

[slender girder]

for every natural number

[oak chasm]

No.

With induction, we prove that it works for one number. It could be 53.

Then, we prove that if it works for one number, it works for the next.

[slender girder]

The inductive step in a proof by induction is to show that for any choice of k, if P(k) is true, then P(k+1) is true.

[oak chasm]

Yes, that's what I said.

[slender girder]

So we showed that P(k) is true

[oak chasm]

No.

[slender girder]

bruh

[oak chasm]

That is definitely not it.

We assume P(k) worked.

We don't prove it.

We assume it without proof.

Let's forget the mathematical notation for a bit.

[slender girder]

Could we go onto another question, and get back to this one

in a little while?

a much easier question

I simplified this

as much as I could

Now I'm struggling to find the values of m that satisfy the equation if it has no roots

I knew that the discriminant must be <0

i.e. b^2-4ac <0

[oak chasm]

Please show your work.

[slender girder]

One sec

[oak chasm]

You didn't distribute the 4 to the - 3.

[slender girder]

oh -12

not -3 ?

[oak chasm]

Yes, there are other errors there, too.

[slender girder]

Oh really

[oak chasm]

(-2 sqrt(m))² is not -10 m.

8 sqrt(m)(sqrt(m) + 1) is not 8 sqrt(m).

[slender girder]

its -4m

or

4m

[oak chasm]

Right.

[slender girder]

8m + 8 ?

[slender girder]

.

[oak chasm]

Just distribute.

8 sqrt(m)(sqrt(m) + 1)

8 sqrt(m) sqrt(m) + 8 sqrt(m)

[slender girder]

-8sqrt(m) * sqrt(m) + 1 would be -8m - 8sqrtm

[oak chasm]

Right.

[slender girder]

so

-4m - 8sqrtm - 12 <0

[oak chasm]

How did you get -4m?

[slender girder]

4m - 8m = -4m

oh

double negative ?

[oak chasm]

Ahh, OK.

[slender girder]

so 12m?

[oak chasm]

No, it's 4m for the b²

[slender girder]

yes

so 4m-(-8m-8sqrt(m)-12) < 0 ?

[oak chasm]

So, 4m - 8m - 8 sqrt(m) + 12

[slender girder]

okay

[oak chasm]

-4m - 8 sqrt(m) + 12

[slender girder]

yup

so how would we go about finding m nopw

[oak chasm]

OK, so:

-4m - 8 sqrt(m) + 12 < 0
m + 2 sqrt(m) - 3 > 0

Now, we can take sqrt(m) as the value we find via quadratic formula.

a = 1, b = 2, c = -3 will give us what sqrt(m) can be at the boundary between where -4m - 8 sqrt(m) + 12 goes from greater than or equal to 0 to being less than 0.

Does that make sense?

[slender girder]

yeah

[oak chasm]

You can also do a = -4, b = -8, c = 12.

[slender girder]

now we plug those into the quadratic formula

x=1, x = -3

[oak chasm]

Right, and sqrt(m) can't be negative.

[slender girder]

yeah

so m => 1 ?

[oak chasm]

Now, since the x² coefficient is positive, we know it points up.

The U of the quadratic does.

So, > 0 is on the outside of the roots.

m ≥ 1 is almost correct.

[slender girder]

oh

1/2 ?

[oak chasm]

It's sqrt(m) and >.

> because we don't want when it equals 0.

[slender girder]

ohh

sqrt(m)>1 ?

[oak chasm]

Right.

[slender girder]

so m>1 ?

[oak chasm]

Yes.

Let's check that.

[slender girder]

yeah lets

[oak chasm]

-4m - 8 sqrt(m) + 12

[slender girder]

x^2-1/x-1

[oak chasm]

-4(1) - 8 sqrt(1) + 12 = -4 - 8 + 12 = 0.

So, that's a good sign.

The discriminant is 0 exactly at m = 1.

[slender girder]

that means it has a root though

we want it to be less than 0

[oak chasm]

Sure, but we have m > 1, which doesn't include m = 1.

However, we want to check m = 1 to make sure it's 0.

[slender girder]

Ahh

[oak chasm]

The discriminant is continuous, so when it goes negative, it was just at 0.

[slender girder]

Makes sense

[oak chasm]

Then we try with m = 2.

-4(2) - 8 sqrt(2) + 12 = 8 - 8 sqrt(2) < 0.

So, that's good.

And that's all we really need to check.

[slender girder]

Perfect

Thanks so much!

[oak chasm]

No problem.

[slender girder]

You're really well versed in maths

[oak chasm]

Well, at the calculus and below parts.

[slender girder]

You like calculus ?

Oh lord

[oak chasm]

No, it's not that I like it, it's just about as far as I remember from school.

[slender girder]

Ahh

[oak chasm]

I remember some linear algebra as well and some abstract algebra.

[slender girder]

Do you remember distributions?

[oak chasm]

Not really.

[slender girder]

Like a question like this

[oak chasm]

No, it's been too long.

[slender girder]

Ahh I see

[oak chasm]

Let's go over induction right quick.

[slender girder]

Yeah

Lets

[oak chasm]

You prove it works for a number, like 53.

[slender girder]

which we did

like 1

[oak chasm]

Then you prove that if it works for one number, it works for the next.

[slender girder]

No we didnt

[oak chasm]

You don't prove that it works for that number.

[slender girder]

We assume it does

[oak chasm]

You assume that it works for that number without proof.