help-19

[uncut remnant]

what.

R+ is the positive reals yes?

[valid marsh]

it is R-{1}

[uncut remnant]

huh

[valid marsh]

yes

[uncut remnant]

Im confused

exp : R -> R
exp(x) = e^x

yes?

exp(R) = R+

???

the image of the exponential function is the positives.

[valid marsh]

no.

[uncut remnant]

[valid marsh]

it is non zero positives

[uncut remnant]

ok, maybe this is a language issue

positive in english

means > 0

[valid marsh]

yes sorry

[uncut remnant]

,,\bR^+\coloneqq{x\in\bR : x > 0}

[valid marsh]

in french 0 is positive lol

[uncut remnant]

this is the definition I am familiar with.

[valid marsh]

yes yes you're right

I was told about that

but I used to take french definition

[uncut remnant]

if you say positif, I wouldve understood

I'd advise restating your original question

so what you mean in symbols is clear.

,,\bR^+_0\coloneqq{x\in\bR : x \geq 0}

[uncut remnant]

I assume you want this

,,f : \bR \to \bR\
f(\bR) = {x\in\bR : x \geq 0}

[valid marsh]

let f : R->R.
if f(R)=[0,+oo[, then f is not injective

[valid marsh]

yes

[uncut remnant]

Right, ok, and its still false

[valid marsh]

then f not inj

[uncut remnant]

This is a hint

You utilize the same idea (which is hilbert's hotel)

[valid marsh]

but that's not R

f must be defined in R

[uncut remnant]

Yes but its a hint

[valid marsh]

not N

[uncut remnant]

You use precisely the same idea

Is what we're saying.

If you had to answer the same question for f: N -> N

and f(N) = {n + 1 : n in N}

how would you do this

thats the hint

[valid marsh]

thanks

for now I don't see

[uncut remnant]

You dont see this one?

[valid marsh]

that's still N

[uncut remnant]

huh

{n + 1 : n in N} is definitely not N

thats the set without the 1st element of N

Im asking if you know how to construct an injective function from N into this set

[valid marsh]

x->[x]+x in R+
and x->-[x]-x in R-

that's injective

and positive

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[deep agate]

1037/375 (2056) +7*0

[swift rock]

use google or a calculator

[raw gale]

,calc 1037/375 (2056) +7*0

[clever fjord]

Result:

5685.5253333333

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[glacial gyro]

hey, why do we have the two cases for 2x-1 here?

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[native ocean]

There's a problem where you essentially have to solve the equation (10,000/x) + 500x = 2000 graphically, which is straight-forward and gives two answers, but I tried to convert the problem to 10,000 + 500x^2 =2000x and it has no solutions

[quasi sparrow]

!original

[odd edge]

Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.

If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[native ocean]

I do not believe that the original problem is neccessary here, what I am asking for is why you can't solve the equation $\frac{10 000}{x} + 500x = 2000$ by converting it to $ 10,000 + 500x^2 =2000x$

The intial equation gives the answers x = 0.59 and 3.48. The second one is unsolvable when using the quadratic formula

[clever fjord]

Seed

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[vague sonnet]

From lambda = 1 we get [1, 2, 2]

[lean willow]

Amazing

[vague sonnet]

Picture didn't get uploaded

Don't blame me

.close

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[drowsy swan]

how is the answer 0.8/c?,
i found the intersection, which should be (0.2/a+0.4/b+0.8/c+0/d)

so the lower bound should be 0 i guess?

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[languid comet]

how can I solve this?

[wheat apex]

just plug in 2.0000000001

😎

kidding

first thing, the a can come outside of the limit

actually i think you do plug in 2.00000000000001

because its asking for only the right

which does exist and you can just plug that in

[languid comet]

do u know if this is valid?

[wheat apex]

oh

should be x-2^2

at the top

and+x-2

but you dont need to combine the graction

just evaluate everything seperately

it is just asking you to plug in 2.00000000001 tho

[languid comet]

but i end up with this

[languid comet]

i am solving a bigger thing this is a part i need to use this method

[wheat apex]

then fine, lets combine the fractions

[languid comet]

it should be infinity according to this graph

[wheat apex]

is what you get after combining the fractions

you didnt combine the fractions properly initially

[languid comet]

what did i get wrong

[wheat apex]

here

if you're going to combine the fractions, you might as well also combine a

a=a/1

its not that its wrong, infinity-infinity is still infinity

just that you get a more clear answer

[languid comet]

and what if the question didn't have 'a' then it would be infinity-infinity

[languid comet]

ooh

[wheat apex]

no because you'll just combine the two fractions

suppose it was just 2/b-2 + 1/b+2^2

[languid comet]

ty for the help

[wheat apex]

to me that looks like 1/0^+

[languid comet]

yes

[wheat apex]

also suppsoed to be b-2^2

[languid comet]

is this right?

i think the minus before the first fraction is what got me wrong

[wheat apex]

make a= a(x-2)^2/(x-2)^2

[languid comet]

yes the same

[odd edge]

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[rich rampart]

cosecx-8cosx = 0

[rich rampart]

0 < x < pi

[upbeat drum]

!status

[odd edge]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[rich rampart]

3

[upbeat drum]

what'd you do

[rich rampart]

1 - 8cosxsinx

1-4sin2x = o

sinx = 1/4

[upbeat drum]

that's sin2x=1/4

[rich rampart]

yh my bad typo

then i got my solutions of 0.13 pi and 2.89 pi

not sure if im right

or is second solution 2.35 pi?

[upbeat drum]

i don't know how you got those, can you show your working out

[rich rampart]

i cant as im on desktop

[upbeat drum]

from sin2x=1/4,

,w arcsin(1/4)

[clever fjord]

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Requested by chlamydia3844

[rich rampart]

im using radians

[upbeat drum]

yeah

and then x is half that

[rich rampart]

yh thats 0.13 pi

[upbeat drum]

oh wait my calc might've been in deg lmao

[rich rampart]

relatable

[upbeat drum]

,w 0.13pi

[clever fjord]

Results provided by WolframAlpha

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Requested by chlamydia3844

[upbeat drum]

?

[rich rampart]

is it just 0.13 then?

[upbeat drum]

yes 0.126 radians

no pi

and then for 2.89, that should be for 2x, not x

[rich rampart]

and 1.4 is the other?

[upbeat drum]

yes

[rich rampart]

okay sound

thanks alot

cheers for help

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[forest shale]

Stats.

I understand that $\beta = \frac{Cov(X,Y)}{Var(X)}$

[clever fjord]

AlexanderJ

[forest shale]

Which can be rearranged like: $\beta = Var(X)^{-1}Cov(X,Y)$

[clever fjord]

AlexanderJ

[forest shale]

And I can also find beta using a matrix equation:

$\beta = ({X}'X)^{-1}{X}'y$

[clever fjord]

AlexanderJ

[forest shale]

Clearly, the $({X}'X)^{-1}$ in the matrix equation is analogous to the $Var(X)$ in the OLS equation, and the ${X}'y$ in the matrix equation is analogous to the $Cov(X,Y)$ in the OLS equation

[clever fjord]

AlexanderJ

[forest shale]

But calculating the covariance and variance involves calculating the squared distances of observations from their means, whereas the matrix equation doesn't involve the means at all; it just uses the raw data.

And, I don't understand how we can do that. How does the matrix equation "work" if we're not really calculating the covariance between the independent x and dependent y variables, because we never calculate the means with which we could calculate variance and covariance?

I understand the derivation of the matrix formula, by differentiating ${e}'e$ and setting it to 0. Derivation isn't what I'm looking for. I'm looking for conceptual understanding.

[clever fjord]

AlexanderJ

[odd edge]

@forest shale Has your question been resolved?

[hybrid carbon]

The same information is being used in the matrix formulation of OLS. It's just hidden by notation a little more.

You should make a small example X and y and calculate everything by hand from both definitions. You'll quickly see how they are equivalent

[forest shale]

So, I've done, that, and I'm not seeing the equivalence. 😦

[hybrid carbon]

Did you get different answers?

[forest shale]

Not at all. I just don't understand how ${X}'y$ in the matrix equation is analogous to the $Cov(X,Y)$ in the OLS equation.

[clever fjord]

AlexanderJ

[forest shale]

For example, let's say I have a vector of x-inputs: [ 1, 2, 3, 5, 6 ], and a vector of y-outputs, [ 4, 3, 2, 1, 2 ].

So I can create a design matrix X: $\begin{pmatrix}
1 & 1\
1 & 2\
1 & 3\
1 & 5\
1 & 6
\end{pmatrix}$

[clever fjord]

AlexanderJ

So I can create a design matrix _X_: $\begin{pmatrix}
1 & 1\\ 
1 & 2\\ 
1 & 3\\ 
1 & 5\\ 
1 & 6
\end{pmatrix}$
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.57 So I can create a design matrix _
                                      X_: $\begin{pmatrix}
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info:    Calculating math sizes for size <14> on input line 57.
LaTeX Font Info:    Trying to load font information for U+msa on input line 57.```

[forest shale]

And then when I calculate ${X}'y$ it returns $\begin{pmatrix}
12 \
33
\end{pmatrix}$

[clever fjord]

AlexanderJ

[forest shale]

But I don't really see how that's analogous to the covariance of y and x, which I'd calculate thus:

$\bar{x} = \frac{1}{5}(1+2+3+5+6) = \frac{17}{5}$

$\bar{y} = \frac{1}{5}(4+3+2+1+2) = \frac{12}{5}$

Cov:$\frac{1}{5}((4 - \frac{12}{5})(1 - \frac{17}{5}) + (3 - \frac{12}{5})(2 - \frac{17}{5}) + (2 - \frac{12}{5})(3 - \frac{17}{5}) + (1 - \frac{12}{5})(5 - \frac{17}{5}) + (2 - \frac{12}{5})(6 - \frac{17}{5}))$

[clever fjord]

AlexanderJ

[forest shale]

= –7.8

= –1.56

Edit: sorry, forgot to multiply by 1/5

[hybrid carbon]

So clearly you've made some assumption that doesn't hold

[forest shale]

I don't really see how –1.56 is captured in the vector:
$\begin{pmatrix}
12 \
33
\end{pmatrix}$

[forest shale]

OK, excited to hear where I've gone wrong

[clever fjord]

AlexanderJ

[hybrid carbon]

Does this help?

$\beta$ is a vector

[clever fjord]

Xenophon

[forest shale]

Will stare for 3 mins

I recognize straight away that the last vector there on the right-hand side is ${X}'y$

[clever fjord]

AlexanderJ

[forest shale]

And I recognize that the middle element, the 2-by-2 matrix, on the right-hand side is ${X}'X$

[clever fjord]

AlexanderJ

[hybrid carbon]

Not quite

[forest shale]

Uh oh

[hybrid carbon]

The inverse

[forest shale]

Oh, I was trying to see that maybe the application of the 1/n*[complex sum] was the inverter

[hybrid carbon]

That's the determinant

well, 1/det

[forest shale]

Ah, you mean because the formula for the inverse of a matrix is to flip the matrix like a mirrored transpose, turn the right diagonal negative, and then multiply the matrix by 1 over its determinant

[hybrid carbon]

Yes

[forest shale]

Gotcha, that makes sense. I was wondering why n was on the bottom-right of that matrix for a moment

[hybrid carbon]

The point I'm trying to make is that if your calculation of $\beta$ is giving you a scalar, then you aren't even looking at the same thing as the matrix formulation.

[clever fjord]

Xenophon

[hybrid carbon]

You're just looking at the second element of the vector

[forest shale]

Oh, right, of course, yes

🤦

[hybrid carbon]

Linear algebra is so slick that you sometimes forget how many details are being hidden behind the notation

I have had that problem before too

[forest shale]

So, if I look at the matrix notation, is there a way to see how the Cov(x,y)/Var(x) is captured in the beta-vector's second element?

I'm asking because I really want an intuitive understanding of this, and I definitely have an intuitive understanding of the SLR case where the scalar beta is Cov(x,y)/Var (x).

Like, figuring out the covariance of x and y, then dividing the covariance by the S.D.s of x and y to "scale it" to the x and y variables, returning the correlation, then scaling the correlation by multiplying it by the S.D. of y divided by the S.D. of x...

...that makes intuitive sense to me.

All I'm doing is figuring out how one variable is far away from its mean in relation to how another variable is far away from its mean, and then doing some scaling a few times using the average distance from the mean (or that figure, squared) to give me a number that measures on average how one variable responds when another changes. I get that.

But when I look at the matrix notation, I'm really grasping for the intuition of how it captures that relationship!

OK, I have to go to bed. I hope that by the time I wake up, a magically intuitive explanation, or even just a hint of which direction to look in, will be here. Thank you in advance to anyone who adds something.

[odd edge]

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[silent heart]

can anyone give me a video that will help me solve this?

[hybrid carbon]

https://www.youtube.com/watch?v=cz6UmZLWgEw

[silent heart]

how do I deal with the fraction

that's what I'm stuck on

[hybrid carbon]

A fraction works just the same

[silent heart]

so make the denominator the same and cancel out?

I tired looking for a tutorial no luck

[hybrid carbon]

Yep.

[silent heart]

I'll try ty

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.close

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[little yacht]

Hi could someone help me solve this telescoping series. I'm writing out the partial sum but I think I'm making some silly mistake because the terms arent canceling out

[dark kraken]

Well first you can write that as 7/(2^(n+3))

[royal herald]

Know that

[lime root]

write it as $\frac{7}{8 \cdot 2^n}$

[clever fjord]

nalin

[little yacht]

could you explain how you get it to that form im not seeing it

[royal herald]

$x^{a+b} = x^a \cdot x^b$

[clever fjord]

Stephen

[dark kraken]

2^3=8

[royal herald]

Apply that to the 2^{n+3}

[little yacht]

okay ill try that out

thank you

.close

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[halcyon pond]

proofs

[halcyon pond]

<@&286206848099549185>

.close

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[cursive ingot]

is
2i sqrt3

7
equal to
2sqrt3
------- i
7

[cursive ingot]

for complex numbers in math

$\frac{2i}{7}$

[clever fjord]

A_CheeseFox

[cursive ingot]

how to

oh

is that the same as

$\frac{2}{7}i$

[clever fjord]

A_CheeseFox

[cursive ingot]

is trhis

the same as

[cursive ingot]

this

like does it matter

is one "wrong"

.close

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[slender radish]

Idek where to start

[vapid quartz]

try to move around the numbers to get y on its own

[slender radish]

Alr

[sand night]

right now its in standard form

get in y intercept form

[slender radish]

Y= -3/2+ 4?

[vapid quartz]

yes

dont forget the x

[slender radish]

Ohh

Thank you

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[hollow elbow]

Was both the solution correct?

[vapid quartz]

i cant answer the problem but i appreciate the cursive

[hollow elbow]

Can you see the image@vapid quartz

[vapid quartz]

i can now

i zoomed in

[hollow elbow]

Thanks

[vapid quartz]

i dont know phsyics/engineering

you'll have to find someone else

[hollow elbow]

<@&286206848099549185>

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@hollow elbow Has your question been resolved?

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@hollow elbow Has your question been resolved?

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[frank sapphire]

Hello, please help me with this task, I am using a translator so my answers cannot be quick.

[ivory raven]

use sine rule to find BD

then BCD and PCQ are similar triangles

so you can use ratio of BP/PC to find PQ from BD

[frank sapphire]

Ok thanks

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[azure hull]

@lilac acorn if you're doing induction, you are already assuming n is a positive integer

@lilac acorn your answer is basically done, you may take away stuff to meet your goal

you mean the things I wrote in the last step or..?

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[queen moon]

Help idk how to do this

[desert marlin]

Show a picture of the full screen

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.close

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[elder vault]

Original:

!status

[odd edge]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[elder vault]

4

[elder vault]

Im not sure if the answer is correct

[stoic torrent]

So

F can not be a function of t if you are going to be using t as your "dummy variable" in the integral

Then it all looks correct

Id avoid the arrows though, it gets pretty messy and its hard to follow where youre going

[odd edge]

@elder vault Has your question been resolved?

[tired hollow]

d

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[elder vault]

Given linear regression y'=rx'. Prove that the r would be equivalent to the Correlation coefficient

[elder vault]

why the slope of an linear regression would happen to be the correlation coefficient?

[odd edge]

@elder vault Has your question been resolved?

[elder vault]

<@&286206848099549185>

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@elder vault Has your question been resolved?

[odd edge]

@elder vault Has your question been resolved?

[elder vault]

.CLOSE

.close

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[wispy minnow]

help

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[wispy minnow]

.reopen

[odd edge]

✅

[wispy minnow]

<@&286206848099549185>

[odd edge]

@wispy minnow Has your question been resolved?

[wispy minnow]

<@&286206848099549185> no?

[obtuse moss]

yo

lemme try

[wispy minnow]

ok

[obtuse moss]

ok i think i dont got it

uhh

my brain is not braining

[wispy minnow]

hmm

interesting

[chilly moss]

interesting

maybe maybe

$\frac{1}{d} = 1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$

[clever fjord]

A Note

[chilly moss]

at this point,

start with a

positive integers are 1,2,3,...

because d cannot be negative,

start with a = 2

$\frac{1}{d} = \frac{1}{2}-\frac{1}{b}-\frac{1}{c}$

and the smallest value that b can have

[clever fjord]

A Note

[chilly moss]

is 3

because 1/2 - 1/2 - 1/c is negative

so b = 3

$\frac{1}{d} = \frac{1}{6} - \frac{1}{c}$

[clever fjord]

A Note

[chilly moss]

and the smallest value that c can have

is 7

because when c<6, it's negative

when c = 6, it's dividing by zero

so

$\frac{1}{d} = \frac{1}{42}$

[clever fjord]

A Note

[chilly moss]

solving this equation,

d can have 42

the idea is to substract the biggest value possible that is smaller than 1 using $-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$

[clever fjord]

A Note

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[languid drum]

help 😭

i think i know that its p^2 = 32 x q^2

bc all of the bases are x

dang it

[nimble blaze]

you're not apply product law properly

[languid drum]

is it p^2 = 32q^2?

[nimble blaze]

no

that's pretty much no different to what you had before

[languid drum]

yea…

[nimble blaze]

which I said was wrong

[languid drum]

p^2 = 32 + q^2

[nimble blaze]

yes

[languid drum]

slay

and i coykd set 32 + q^2 equal to 2-q^2 rigtht

and solve for q….?

[nimble blaze]

no

why are you setting those two things equal

[languid drum]

bc p + q = 2

and so p=2-q

and so p^2 = (2-q)^2

[nimble blaze]

that's not what you typed

[languid drum]

oop i forgott he parenthesees

[nimble blaze]

you could do that, but there's a more efficient method

[languid drum]

pls tell

[nimble blaze]

from

p^2 = 32 + q^2
subtract q^2 from both sides
then factor the difference of two squares

[languid drum]

32 is a square..?

[nimble blaze]

no

what do you have after subtracting q^2 from both sides?

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[eager rivet]

How do I solve it using a calculator

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@eager rivet Has your question been resolved?

[viscid flint]

note it doesn't say what the exact shape of the box is

so just pick any shape (that is a rectangular prism aka a box) that gets you a surface area of 8m²

[odd edge]

@eager rivet Has your question been resolved?

[gloomy steppe]

if you increase the side length of a rectangle by double, what happens to its area?

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[latent plaza]

can someone help me with this

[latent plaza]

theres no given statement so i dont know how im suppose to do this

[zinc drift]

since its a reflection u know that angle 1 is congruent to angle 3

[latent plaza]

so would this be correct?

[zinc drift]

for the first line, since it's not really given, i guess you can write that a reflection bounces off a surface at the same angle as it hits it

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[summer pecan]

not sure what to do to get t * (n/3)

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[ivory mural]

and

what are the y coordinates of the points where y^3+x^3+30y^2=4x+3 has vertical tangent lines

diff questoin btw

[odd edge]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[ivory mural]

i conjugated but stuck again

[ocean bramble]

you should not need conjugates, you get that limit by directly evaluating it.
However, you need to calculate the lateral limits (which will be different and thus the limit wont exist)

[ivory mural]

what are lateral limits

[ocean bramble]

limits when you approach the value from a single side

[stark coyote]

lateral != vertical

[ivory mural]

okay

[ocean bramble]

i'm answering first question, not second

[ivory mural]

i don’t know how to do this

ok

from left hand limit and right hand

[ocean bramble]

how he's expected to do the second without knowing what a lateral limit is is beyond me

[ivory mural]

sorry just incompetent

[odd edge]

@ivory mural Has your question been resolved?

[ivory mural]

everyone is conjugating and getting an answer somehow for the first one

[odd edge]

@ivory mural Has your question been resolved?

[odd edge]

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[twilit girder]

How come the answer isn’t C?

[cold sage]

-3 and 1 arent the bounds of the region we are looking at

the region is [-2,5]

(c is in that region) [edit: for f]

[twilit girder]

But aren’t we finding inverse of f?

Oh wait

I think I understand

Is the answer A?

[cold sage]

no

oops

wait

yes

[twilit girder]

Oh ok thank you

.close

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[sage bear]

Is this correct?

[prisma cairn]

Yes

[odd edge]

@sage bear Has your question been resolved?

[sage bear]

.close

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[icy plaza]

So basically I need help on this basic geo logic problem where we need to make slogans into conditional statements, the statements are "I'm lovin it" by Mcdonalds, and "Like a good neighbor , Statefarm is there"

[icy plaza]

i need help making this into a conditional it doesnt flow well

[hybrid carbon]

If I exist, I am lovin it? Lol

[cold sage]

i like that

good ring to it

[icy plaza]

would it be if I am eating mcdonalds?

[cold sage]

it would be valid i guess

[icy plaza]

If I am eating mcdonalds, them I am loving the experience?

or would it be just "loving it""

[cold sage]

just If im eating mcdonalds, Im lovin it

is probably fine

[icy plaza]

because we have to do inverse

and contra positive so it sounds weird

If im lovin it, then im eating mcdonalds

is that fine or too vague

[cold sage]

could replace 'it' with 'the food' or 'my meal' or something i guess

[icy plaza]

how about for the statefarm one

[cold sage]

if im in need, like a good neighbor, statefarm is there

[icy plaza]

k thanks

.close

[odd edge]

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[small acorn]

Can I get some help on starting this

[small acorn]

I am kinda stuck

Its an implication proof so $P\Rightarrow Q\equiv \neg P\lor Q$ right?

[clever fjord]

42

[small acorn]

So its contradiction is the negation which is $P\land\neg Q$

[clever fjord]

42

[small acorn]

But how do I start the proof

just need a bit of a hint

[latent scaffold]

What does it mean if f divides (m+n)?

[small acorn]

$(n+m)\div f=c,c\in\mathbb{Z}$

[clever fjord]

42

[small acorn]

pretty much

so $\frac{n+m}{f}=\frac{n}{f}+\frac{m}{f}$

[clever fjord]

42

[latent scaffold]

Yes, but then what can you say about m/f since f doesn't divide m?

[small acorn]

thats the contradiction?

[latent scaffold]

No

From the definition, what can you say about m/f?

[small acorn]

$\frac{m}{f}=c,c\in\mathbb{Z}$

[clever fjord]

42

[latent scaffold]

It's not divisible though

[small acorn]

yeah

[latent scaffold]

So that's not the case

It means that m/f is not an integer

[small acorn]

oh so $\frac{m}{f}\neq c,c\in\mathbb{Z}$

[clever fjord]

42

[latent scaffold]

That's a weird way to write it. Maybe you can say that there is no integer such that m/f is an integer

Or there does not exist c in Z such that m/f = c

[small acorn]

i dont think i can write that

i have to write it in a certain format and reason it

do i start my proof with f | (n+m)?

[latent scaffold]

You can do either.

[small acorn]

hmm

im confused, but i shouldnt be

so f|n means n = fc, where c is an int, and there is no int l that m = fl

[latent scaffold]

Yes

Now add them together

[small acorn]

so like fc + fl?

[latent scaffold]

Hmm wait this presuposes that this l exists

Try and do (n+m) - n using this divisibility equivalent

That should do the trick

[small acorn]

wdym

[latent scaffold]

Like n+m = af and n = bf, so n+m - n = af - bf

[small acorn]

lemme write this all down 1 sec sorry

[latent scaffold]

It's alright

[small acorn]

like this u mean?

[small acorn]

and then do this

[latent scaffold]

Well in this case what you wrote is just wrong. You should write n=fc and n+m = fd for some c and d in Z

[small acorn]

oh wait am i trying to say that f | m proving a contradiction?

[latent scaffold]

Yes

[small acorn]

ok i was confused before

[latent scaffold]

Once you do the subtraction and factor you'll see

[small acorn]

alright lemme write it

So like this

And when i do (n+m) - n = fd - fc

there will be a contradiction?

[small acorn]

and this step is substiting n in right?

[latent scaffold]

Yes you substitute (n+m) and n in

[small acorn]

ok lemme write this down

Here

I think thats done

just need to write explantion

Thank you so much, Idk why I was stuck for so long

Can you check this proof over for me? I think its all good, just wanted another opinion

[latent scaffold]

It looks good

[small acorn]

alrighty

thanks for the help :D

.close

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[hoary marsh]

Hello

[hoary marsh]

$y=\frac{\frac{e^{x}-e^{-x}}{2}}{\frac{e^{x}+e^{-x}}{2}}$

[clever fjord]

LE SSERAFIM

[hoary marsh]

How do I find the inverse of this

$x=\frac{\frac{e^{y}-e^{-y}}{2}}{\frac{e^{y}+e^{-y}}{2}}$

[clever fjord]

LE SSERAFIM

[royal herald]

ewww

[hoary marsh]

What do I do after this

[royal herald]

well u can get rid of the 2 immediately

cancels out in the numerator and denominator

[hoary marsh]

Oh yay

[royal herald]

do u see it?

why u can do that

[hoary marsh]

$x=\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}$

[royal herald]

nono

thats not what i meant

like

here ill show u

[clever fjord]

LE SSERAFIM

[royal herald]

yea

thats the one

do u understand how to get that

?

[hoary marsh]

Yes

[royal herald]

ok cool

[hoary marsh]

Do I use ln now?

[royal herald]

yea try it, see what happens

[hoary marsh]

$\ln\left(x\right)=\frac{y+y}{y-y}$

[clever fjord]

LE SSERAFIM

[hoary marsh]

Like this?

[royal herald]

hmmm

show me how u got that

[hoary marsh]

??

[royal herald]

nah thats not how it works

u gotta ln the entire side

not each individual term

[hoary marsh]

$\ln\left(x\right)=\ln\left(\frac{e^{y}-e^{-y}}{e^{y}-e^{-y}}\right)$

[clever fjord]

LE SSERAFIM

[royal herald]

yep

now try some log properties

||if only i had a spare riemann around with a preamble filled with factoids||

[hoary marsh]

???

[royal herald]

lol that was a joke

use log properties

how could u separate that fraction into separate logs

[hoary marsh]

I know riemann tho lol

[royal herald]

oh lol

[hoary marsh]

$\ln\left(x\right)=\ln\left(\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}\right)=\ln\left(\left(e^{y}-e^{-y}\right)\left(e^{y}+e^{-y}\right)^{-1}\right)$

[clever fjord]

LE SSERAFIM

[hoary marsh]

I take -1 out?

Wait

No need actually

$\ln\left(x\right)=\ln\left(\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}\right)=\ln\left(e^{y}-e^{-y}\right)-\ln\left(e^{y}+e^{-y}\right)$

[clever fjord]

LE SSERAFIM

[royal herald]

gooood

now what do u think u can do

[hoary marsh]

Um

Hint pls

[royal herald]

rewrite e^-y as a fraction, then make common denominators

then see what happens from there

[hoary marsh]

$\ln\left(x\right)=\ln\left(\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}\right)=\ln\left(e^{y}-e^{-y}\right)-\ln\left(e^{y}+e^{-y}\right)=\ln\left(\frac{e^{2y}}{e^{y}}-\frac{1}{e^{y}}\right)-\ln\left(\frac{e^{2y}}{e^{y}}+\frac{1}{e^{y}}\right)$

[clever fjord]

LE SSERAFIM

[hoary marsh]

Like this?

[royal herald]

yep

now add fractions

then do log properties again xd

[hoary marsh]

How?

[royal herald]

first combine the fractions

then youll see

[hoary marsh]

$\ln\left(x\right)=\ln\left(\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}\right)=\ln\left(e^{y}-e^{-y}\right)-\ln\left(e^{y}+e^{-y}\right)=\ln\left(\frac{e^{2y}}{e^{y}}-\frac{1}{e^{y}}\right)-\ln\left(\frac{e^{2y}}{e^{y}}+\frac{1}{e^{y}}\right)=\ln\left(\frac{e^{2y}-1}{e^{y}}\right)-\ln\left(\frac{e^{2y}+1}{e^{y}}\right)=\ln\left(\frac{e^{2y}-1}{e^{2y}+1}\right)$

[clever fjord]

LE SSERAFIM

[hoary marsh]

This?

[royal herald]

hmm

lemme see

howd u do that last step?

[hoary marsh]

I made an impossible step

Wait

ln(a) - ln(b) = ln(a/b)

No?

[royal herald]

ah ok yea thats what u did i see

but that doesnt get u far does it

[hoary marsh]

What then?

[royal herald]

use log division rule for each individual ln in the second to last step

lemme see if i can work a couple steps ahead to make sure im not leading u on a wild goosechase

[manic sleet]

Consider: $x=\frac{e^{2y}-1}{e^{2y}+1}\implies x(e^{2y}+1)=e^{2y}-1$

[clever fjord]

PajamaMamaLlama

[royal herald]

Aha!

thanks @manic sleet

@hoary marsh

we know da wae

[manic sleet]

[manic sleet]

another great emoji

[hoary marsh]

Thank you for da wae

[royal herald]

u know how to go from ur last step to the answer?

[royal herald]

ur last step in this is good

[hoary marsh]

$$\ln\left(x\right)=\ln\left(\frac{e^{2y}-1}{e^{2y}+1}\right)$$
Remove ln
$$x\left(e^{2y}+1\right)=e^{2y}-1$$

[clever fjord]

LE SSERAFIM

[royal herald]

yeppp

now isolate all terms with a 'y'

[hoary marsh]

$$x\left(e^{2y}+1\right)=e^{2y}-1$$
$$xe^{2y}+x=e^{2y}-1$$
$$xe^{2y}-e^{2y}=-1-x$$
$$e^{2y}=\frac{-1-x}{x-1}$$

[clever fjord]

LE SSERAFIM

[hoary marsh]

e

z

[royal herald]

ye

multiply top and bottom by -1 to make it less ugly

other than that u shd be good

[hoary marsh]

$y=\frac{\ln\left(\frac{-1-x}{x-1}\right)}{2}$

[clever fjord]

LE SSERAFIM

[hoary marsh]

Like this?

[royal herald]

yea thats correct, the fraction just looks a little ugly

[hoary marsh]

$y=\frac{\ln\left(-\frac{x+1}{x+1}\right)}{2}$

[clever fjord]

LE SSERAFIM

[hoary marsh]

Like this?

Wait

Something wrong

[royal herald]

whats (-1-x) * -1

and whats (1-x) * -1

[hoary marsh]

$y=\frac{\ln\left(\frac{x+1}{-x+1}\right)}{2}$

[clever fjord]

LE SSERAFIM

[hoary marsh]

Here

[royal herald]

yep

and thats nice and correct and all

but to make it look amazing

id do

$y = \ln {\sqrt {\frac {x+1}{1-x}}}$

[clever fjord]

Stephen

[hoary marsh]

Where did the root come from?

[royal herald]

the 1/2

thats a log rule

[hoary marsh]

Oh

Thanks so much 🙂

.close

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[robust inlet]

Why do they use 1.645, shouldn’t it be a one tailed test since its only seeing whether they like the exam or not?

[forest sky]

it's a confidence interval, not a z-test

[robust inlet]

O thanks

Just been a while since i did these questions and when i googled it it said u could have one tail

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[blissful jolt]

uhh

[blissful jolt]

is this right

expanded to Logx^1/2 - (1 + 1/2y + 1/2z)

[nimble blaze]

no

[blissful jolt]

why is what I did wrong

I just top minus bottom

keeping the square root on all of them

[nimble blaze]

logs disspeared from the y and z

[blissful jolt]

1/2Logx - (1 + 1/2y + 1/2z)

the first part is correct now right

and y and z are missing logs

but otherwise is good?

[nimble blaze]

well seems that you ignored the root on the 10 as well

its a whole mess

[blissful jolt]

fu

[nimble blaze]

you should be writing clear steps

[blissful jolt]

Log10(10^1/2)

[nimble blaze]

feels like you tried to skip ahead in your head

also unless you were told that x,y,z were positive, this messes up the domain

[blissful jolt]

how??

1/2Logx - (1/2 + 1/2logy + 1/2logz)

whats wrong with this

[nimble blaze]

the domain for that will be x,y,z > 0
whereas in the original, a pair of them could be negative

[blissful jolt]

but this is ok?

[odd edge]

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[odd edge]

Please don't occupy multiple help channels.

[eager vine]

Am I wrong?

[sand night]

i can totally read that

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[sand hill]

3.a) proving by mathematical induction

[sand hill]

It’s messy but here’s what I’ve got so far

I don’t know what to do next

[sharp oak]

I wouldn't set the "= 6p". It's not helpful to have p, which is a variable with no constraints

You're assuming 7^k - 1 is divisible by 6.

Prove 7^(k+1) - 1 is divisible by 6

Obvious step, pull the 7 out

It would be nice if we could pull something with a 6 out of that

[sand hill]

What do you mean by pull the 7 out

As in factorising?

[sharp oak]

7^(k+1) = 7(7^k)

[sand hill]

Oh wow it’s crazy somehow I’ve never thought of doing that

I think I almost got it by using ‘p’ and ‘q’ but I don’t really get how you could do it without them

I’m very new to induction by the way

[odd edge]

@sand hill Has your question been resolved?

[sand hill]

I’m still not sure what to do

[sharp oak]

@sand hill
7^(k+1) - 1

= 7(7^k) - 1

= 6(7^k) + 7^k - 1

The first term is a multiple of 6. The other terms are divisible by 6 because of our inductive assumption

[odd edge]

@sand hill Has your question been resolved?

[sand hill]

Ah I see! Thanks for your help

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[hushed schooner]

If I have this and square both sides, does it still keep the +-

[mossy python]

Make the 20-x^2 in (a-b)^2 form

[hushed schooner]

ahhh so then it just becomes a positive and the - of +- goes away?

[mossy python]

I don't think they go away, I think you still have the +- there because (a-b)^2 = (b-a)^2

So you gotta take both into consideration

Hence the +-

[hushed schooner]

Hmm. Maybe I'll need to give full context then.

I have a system:
A: y=sqrt x
B: y = +- sqrt (20-x^2)

After everything has been plugged in and solved, I'm supposed to have (4,2) as the solution but this +- part is confusing me.

[mossy python]

Sqrt +-... or +- sqrt...?

[hushed schooner]

+- sqrt

sorry

[mossy python]

As, I told you, when you convert '20-x^2' to (a-b)^2 form, the ^2 cancels out with the sqrt, so you have +-(a-b) = sqrt x

They did the +- part for you. I've sometimes been left to figure that out myself when I was given such problems

[hushed schooner]

Okay I see now. Thank you!

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[wooden python]

this is an inefficient way to go about doing things. and i think you also messed up somewhere...

do you know de moivre's theorem?

...do you know the concepts of modulus and argument at least?

the angle a complex number makes with the positive real axis?

ok let's try doing it differently. do the words "cube root of unity" ring any bells to you?

well then

-1/2 + sqrt(3)i/2 is one of those

raising it to the 3rd power yields 1

which helps you a lot

esp. considering 2010 is divisible by 3

i would've said that it has modulus 1 and argument 120° but that'll be lost on you.

on a separate piece of paper work out its 2nd and then its 3rd power.

[wooden python]

you're not doing what i told you, and also you've made a sign error on line 3

$-\frac{1}{2} + \frac{\sqrt{3}i}{2}$

[clever fjord]

Ann

[wooden python]

this is what im asking you to raise to the third power

maybe work that out as $\frac{(-1 + \sqrt{3}i)^3}{8}$ unless you are a masochist

[clever fjord]

Ann

[wooden python]

no, you dragged the full 2010 power along with it.

you can do calculations that are tangentially related to the problem and then tie them back to it later on.

maybe your teacher strictly forbids it

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[round loom]

$$\frac{1}{8}=\frac{1}{4c}$$

[clever fjord]

Lex1729

[round loom]

I know that c = 2, however why do I get c = 0.5 when I rearrange?

[wooden python]

show work

you're probably fucking up when you "rearrange"

[swift rock]

u dont get c = 4/8 , u get c = 8/4

[round loom]

Oh true, my bad

Also one more question

How do I integrate natural log functions like this?

$$\ln\left(3x+1\right)$$

[clever fjord]

Lex1729

[wooden python]

integration by parts helps a lot

[round loom]

Is it possible to do it using u-substitution?

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[wary venture]

Given a pyramid S.ABCD, base is a trapezoid, large side is AB. I, J, K are 3 points on SA, AB, BC respectively
Find find the intersection point of IK and (SBD)
Find the intersection of (IJK) with SD and SC

[wary venture]

,rotate

[wary venture]

So yeah I uh solved the first problem but now stuck on the 2nd one

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[sand hill]

I need help with a)
I’ve honestly just forgotten how to work with permutations, combinations and factorials.
Not sure if what I have is right

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[mystic saffron]

How would I do this?

I don’t know where to start

[viscid flint]

ooh this is a conceptual question, although you can use tools from calculus to solve it

ummm it might help to draw a sketch of some function f(x) and mark points 0, a, and 2a

remember that integrals are area under the curve

[lime root]

$\int_0^{2a} f(x)$ $dx = \int_0^a f(x)$ $dx + \int_a^{2a} f(x)$ $dx$

[clever fjord]

nalin

[viscid flint]

like I said you can also use tools from calculus like combining and splitting integrals (as above), doing substitutions, and renaming variables

if you want to do it analytically, i would go through each of them and try to use those tools to rewrite the integrals into the original form, or argue why they're not equivalent

you could also define F(x) to be a notional antiderivative of f, and "evaluate" each integral

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[torn bolt]

how do i solve for n

[ebon harness]

Multiply both sides by 4

[naive pebble]

same base firstly

[naive pebble]

why?

[wooden python]

unhelpful.

[crisp wadi]

Yeah that's kinda longer

[long tinsel]

write 1/4 as a power of 2

[crisp wadi]

Start with the simple, rewrite 4 as a power of 2

[naive pebble]

ya just turn 1/4 to a base of 2

[ebon harness]

OR use the fact that $$\frac{1}{a} = a^{-1} $$ for $ a \neq 0$

[naive pebble]

💀

oh

[clever fjord]

Cyrenux

[naive pebble]

ok ya but its easier to turn 1/4 to power of 2

[torn bolt]

uhm

so i turn

[naive pebble]

just turn 1/4 to power of 2

and ur good from there

[torn bolt]

4 ^ -1

[wooden python]

4^-1 sure but you need to recall your times tables and recognize 4 as 2 * 2

or 2^2

[torn bolt]

yea

and then

[naive pebble]

2^-2?

[torn bolt]

how do i make 4 ^ -1 into a power of 2

[wooden python]

4^-1 = (2^2)^-1 = ?

[torn bolt]

hmm

so it becomes 2 ^ -2?

[naive pebble]

ya

then solve for n

[torn bolt]

how tho

[naive pebble]

since its same base

it becomes

-2=n

[torn bolt]

no i mean how does it become 2 ^ -2

[naive pebble]

oh

well

u dont want a fraction

[torn bolt]

yea

[naive pebble]

so turning 1/4 to 4^-1

right

[torn bolt]

ye

[naive pebble]

and

4^-1 = 2^-2

[torn bolt]

how

[naive pebble]

wdym how

4 is 2*2

lol

like idk how to explain it

[torn bolt]

uh

okay

[naive pebble]

..

[torn bolt]

i get it now yea

[naive pebble]

alr

[torn bolt]

.close

[odd edge]

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[mellow axle]

Am i crazy?

Why am i getting math error?