help-23

[woven hull]

the y point at x = 5 is 13.5

so try m >= 13.5?

[safe radish]

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[hallow anchor]

I'm not sure how to solve this type of thing

[pulsar pecan]

First consider $ \lim_{x \to \frac{\pi}2} \cot (2x) $

Thanks

$$ \lim_{x \to \frac{\pi}2} \cot (2x) $$

[flat frigate]

rysrobrgldvoelr👻ep>vneae=u

[pulsar pecan]

I meant this

Your limit exists if this limit exists

Because the exponential function is continuous, so you would be able to move the limit inside the exponent if it exists

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[hallow anchor]

How would I solve this problem

[empty gyro]

Lhopital is a good choice

[hallow anchor]

How do you find the derivative of sin4x?

[sly field]

what is derivative of f(g(x))

[hallow anchor]

Ah i see

What this part then

I'm unsure how I'm supposed to simplify the right button part

.close

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[twin prawn]

in piecewise functions, how do boundary inequalities (ie. 0 < x < 1) dictate which type of endpoint circle to use?

[dry timber]

endpoint circles are based on whether the value is inclusive or not

if the value is included (so having a less than or equal to sign for example) you have a filled in circle

if the value is not excluded (like the number 5 in the set x<5) then you use a non filled in one

[twin prawn]

ohh

can endpoint circles dictate which is part of graph

like for example, the shaded endpoint of slope 1 and the open endpoint circle of slope 2

my classmate said you can use vertical test and horizontal test to check if they are parallel (i think)

[dry timber]

yes. if these points are different for an input value you have what is called a discontinuous function

[dry timber]

can you show me a photo of what you mean i think i might be confusing it haha

[twin prawn]

something in similar to this

[dry timber]

yeah so here you can see

the function is defined for all of the values of x less than equal to 1 and more than equal to 2

but isn’t defined in betweeb

[twin prawn]

so that means

[dry timber]

functions can only really be parallel as long as you’re describing a linear function because they have constant gradients

[dry timber]

here both graphs are parallel because you can see the coefficient of the x term is the same (0 for both cases)

[twin prawn]

yea

wait huh

the x term for function 1 is 1; and same goes for function 2

[dry timber]

if you look at the right hand side, there is no x value in it

if i had a function as say 5x + 4

[twin prawn]

ahhh that makes sense

[dry timber]

that would have an x term with coefficient 5

here there is no x term so the coefficient is 0

[twin prawn]

yeah

also, how do i find the domain and range for this? i have the function expressions already, idk how to do the domain range

[safe radish]

@twin prawn Has your question been resolved?

[dry timber]

so look at the function here

what are the range of values that you can see along the x and y axis

along x you’ll see it is fully defined from -7 to 7 with only a hole at the-1

and for y it’s defined at 6, from 5 to -1 (including -1 but not including 5)

[twin prawn]

a hole at -1 means its not included?

[dry timber]

yes exactly

[twin prawn]

so therefore,

[dry timber]

as you can see both the piece wise functions that go close to x =-1 have blank circles

[twin prawn]

theres going to be unions

[dry timber]

which means they don’t

[dry timber]

yes exactly

you can express it as [-7,-1) U (-1,7]

if you’re using set

notation

[twin prawn]

ohh

question

idk how to word this

but if shaded circle dictates inclusion, whilst unshaded dictates exclusion, if the shaded and unshaded become parallel to each other, does the unshaded circle become part of the line/slope/shape with the shaded circle?

cause our teacher taught us that shaded and unshaded connect each other

[safe radish]

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[fallen thunder]

https://cdn.discordapp.com/attachments/1081837589780234241/1170981134012780554/image.png?ex=655b042a&is=65488f2a&hm=1c53943454b80f34c64cf9254899cb19ab894c1e1e5ac6ee322ac201534517f3&

[fallen thunder]

Hi guys, probably basic question but

https://cdn.discordapp.com/attachments/1081837589780234241/1170981902153416734/image.png?ex=655b04e1&is=65488fe1&hm=14c0c7ec782d343a415e4a13d1f4b440831d2c38188a8b14bb69b6237cbbde1c&

I don't see my answer in any of this lol

average rate of change is literally the secant line connecting day 1 and day 14 point right

,, A(x) = \frac{f(b) - f(a)}{b-a}

[flat frigate]

斗地姿

[fallen thunder]

it looks like (21-11)/(14-1) to me

[flat frigate]

Result:

0.76923076923077

[fallen thunder]

What am i missing?

[buoyant shadow]

nothing?

[fallen thunder]

but where's my answer lol

[buoyant shadow]

0.65 obviously

[fallen thunder]

0.76 to 0.65

that's a wild approximation

or "closest"

[fallen thunder]

don't you think?

i've never come across a question where the "closest" part is that far off

[buoyant shadow]

it doesn't seem strange, you should come across this in the future

[fallen thunder]

okay fair enough

one more thing

or a couple maybe 😭

https://cdn.discordapp.com/attachments/1081837589780234241/1170809336117854368/image.png?ex=655a642a&is=6547ef2a&hm=fdac901605150b0d6707fa2a630ff6d21d8db51db3f06f9718991a5527e11032&

I solved this question by finding the slope through implicit differentiation ofc

which is b/a

so it's between D and E and i just plugged in x = (b+a) and looked at which of them returned (b-a) so the answer is E ig

had a bit of trouble actually making that formula though

,w equation of line passing through (b+a,b-a) with slope b/a

[flat frigate]

Results provided by WolframAlpha

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Requested by kanna22

[maiden jetty]

through (b+a,b-a),slope is b/a then the line is
$y = \frac{b}{a} (x-(b+a))+(b-a)$

[flat frigate]

秋水

[fallen thunder]

oh yeah the y - y1 = m(x-x1) formula right?

idk why i lack the most obvious observations

[maiden jetty]

yes

[fallen thunder]

thanks

[safe radish]

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[lean otter]

Can you use the well-ordering property to prove the statement: “Every positive integer can be described using no more than fifteen English words”?

[buoyant shadow]

that makes no sense

i mean the statement, there's a finite amount of 15 word sentences

[lean otter]

Yeah. You can assume the words come from a particular dictionary of English

i dont see whats the problem with that

[buoyant shadow]

the problem is you can't pointat at every number uniquely

[lean otter]

then idk is my textbook wrong?

.close

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[timid scroll]

Hia

[timid scroll]

I have one possible position vector, being

(9,3)
How can there be one more position????

[shadow glade]

that's it

[timid scroll]

Apparently there is 2 answers

Any of thoes 3 can be the answer

I dont get it

[shadow glade]

hm ok i wasnt seeing it either but plotting the points it does work: (blue are the given ones and red are the solutions)

[timid scroll]

OHHHHH YEAH

They do all make a parallelogram

Man math sucks

[shadow glade]

seems like the red points form a triangle with the blue points at the midpoints of the segments of the triangle, kinda cool

[timid scroll]

yeah

[shadow glade]

anyway what you would do is pick different pairs of the original 3 points and then build the 4th point similar to how you did

[timid scroll]

Maybe you can fit 4 blue triangles in the red triangle

[shadow glade]

youll get different solutions if you pick the different pairs and also extend in different directions

[shadow glade]

yes, that is accurate in fact : )

[timid scroll]

Well well well

Thank u

[shadow glade]

welcome

[shadow glade]

ok badly drawn but

its just similar triangles really, the small ones are similar to the overall large one (same angles, measures divided by 2)

[timid scroll]

Oh thats how it works?

So the tip of large one eg, 40deg = 20deg tip of small one?

[shadow glade]

no, same angles

measures of the sides i mean

[timid scroll]

Oh

[shadow glade]

just stretched out but exact same angles, thats what similar triangles are

[timid scroll]

Oh yeah, right to scale is same angles

But lengths differ

Jajaajaj

Cool

Thanks for pointing that out

[shadow glade]

and it makes sense that 4 small ones fill the large one since

Area of large one is: 1/2bh (one half base times height, just regular area of a triangle right?)

[timid scroll]

Yes

[shadow glade]

since small ones sides measure half the area would be:

1/2(1/2b)(1/2h)

=1/8bh

so 1/4 of the large one

[timid scroll]

Imma write all that down

Good to know for future

[shadow glade]

:p

kinda every problem in geometry you can just draw triangles everywhere and see all this random stuff, its all very interconnected haha

[timid scroll]

[shadow glade]

but seriously, just start drawing random triangles between points when you are confused on how to solve a geometry problem

starts with a circle and some points? great, draw triangles

a rhombus and some points? great, draw triangles

[timid scroll]

Oh acc ok

ill make a note of that too

[shadow glade]

lol ok good luck : )

[timid scroll]

doenst look that fun

tho

I hate visualising stuff

but that might just be me :>

[shadow glade]

well you dont have ti visualize it, you can draw it on paper and actually look at it

😄

or screen

[timid scroll]

Yeah true

Well thank you once again

:>>>>> So much info

Usefull

teheh

[shadow glade]

!done

[safe radish]

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

[timid scroll]

.close

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[lean otter]

if i have a matrix

[lean otter]

5x4

with rank 3

[glacial cairn]

Me too

[lean otter]

does the colums then span R^3

[glacial cairn]

By definition yes

The rank of a matrix is the dimension of the space generated by the columns

Whether it's 5x4 or 78x43

[lean otter]

how about the rows

[glacial cairn]

Identical

[lean otter]

the colums would be linearly independent?

the first 3

[glacial cairn]

Not necessarily the first 3, no

Impossible to say without looking at the matrix

[lean otter]

what about in total?

[glacial cairn]

I believe 3 columns would be linearly independent, yes

[shadow glade]

no, they span a 3 dimensional subspace of R5 since the columns are a subset of R5 not R3

[lean otter]

yes ok

that i what i thought

i have some question where i am less certain tho

[shadow glade]

i mean you can have an isomorphism to R3 but still : )

[shadow glade]

[lean otter]

but the entire matrix

or ratherthe colums are they linearly idependent

[shadow glade]

of a rank 3 matrix? only 3 of the columns

thats what definition of rank is, like Nel said

rank 3 means both the column space and row space have a dimension of 3, thats a result proven in LA

[glacial cairn]

All the columns can't, since there are more than 3

[lean otter]

so if we have a 4x5 matrix with rank 3

then you can find 3 rows or colums that are linearly independent, and the rest of the rows and colums can be expressed as a linear combination of those 3

therefore the colums are not linearly independent

[glacial cairn]

Yes

[lean otter]

or?

[shadow glade]

ya

[glacial cairn]

A matrix being linearly independent doesn't make sense though

Vectors are

[lean otter]

is that better?

[glacial cairn]

Unless you're talking about two matrices maybe, idk

Sure

[shadow glade]

if you are talking about a square matrix n x n

[lean otter]

question 4 and 5, what are they even asking

[shadow glade]

then its rank determines whether its invertible or not, maybe thats what youre thinking of, not linearly independent

any 1 matrix other than the 0 matrix is linearly independent as far as the vector space of m x n matrices it is part of i guess, but thats not really related to what you were talking about with the columns

[glacial cairn]

There are no question numbers in your picture

[lean otter]

from 1-6

[shadow glade]

so Bx = 0 relates to the kernel of the matrix

[lean otter]

it starts with 1, which is colums span r^5.

the last is question 6 (none of the above)

[shadow glade]

the one that would be true is Bx = 0 has infinitely many solutions

since the kernel's dimension > 0 right?

[glacial cairn]

If the rank is less than the number of columns (and rows), then you have more unknowns than equations

So yeah that would be infinitely many solutions

[lean otter]

i havnet heard of kernel yet

[shadow glade]

well think of it this way

[lean otter]

if the rank is 3

dont i have 2 free variables

[glacial cairn]

5 columns, 4 rows, right?

[lean otter]

yh

[glacial cairn]

So you can multiply a vector with 5 elements

[lean otter]

i will have x1,x2,x3,x4,x5

[shadow glade]

5x4 matrix implies 5 rows 4 columns actually

[glacial cairn]

So yes you would have 5-rank = 5-3=2 free variables

[shadow glade]

so its going from R4 -> R5

the kernel and rank i think should add up to the dimension of the domain (4) so dim(kernel) = 1 which is just 1 free variable

isnt it that way?

[lean otter]

isnt this what we have

[glacial cairn]

Yeah that's 4 columns, 5 rows

[shadow glade]

yep so its R4 - > R5 transformation

[glacial cairn]

My bad then, but same thing: you would have 4-rank = 1 free variable

[lean otter]

so still infinitely many solutions

same answer regardless if it is 4x5 or 5x4 i guess

[shadow glade]

so what basically happens is...the domain has 4 dimensions, 3 of those are used to map to a 3 dimensional space in the range...and the leftover dimension maps to 0 (i.e. solutions of Bx = 0). it turns out that the values that map to 0 are also a vector space

dim (kernel) + dim (rank) = dim (domain)
1 + 3 = 4

[lean otter]

ok so

5x4 means

[shadow glade]

something like this, im actually a little bit confused by the non square matrices and rusty but i think its the dimension of the domain they add up to? 🤔

[lean otter]

that we have 5 four dimensional vectors

[shadow glade]

5x4 means you can multiply it by a 4x1 column vector and get a 5x1 column vector so that just should make you think of it as a R4 -> R5 transformation

[glacial cairn]

The kernel is the subspace of the domain mapped to zero

[glacial cairn]

Sounds right to me

[shadow glade]

so if your dim(kernel) = 0 then Bx = 0 will have a single solution...the zero vector

and everything else doesn't give 0

but the statement Bx = 0 has 0 solutions will never be true (5th statement in your list) because there will always at least be the trivial solution of x = 0 vector

even for a matrix with maximum rank

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[turbid light]

how do i start ? without using lhopital

[viral venture]

Do you know the limit as x goes to 0 of sinx/x

[turbid light]

1

[viral venture]

How about x/sinx

[turbid light]

(... 1?) not that sure

[viral venture]

1

What about sin x² / x²

[turbid light]

logically 1

oh i see now

[viral venture]

Yea

You got it now

[turbid light]

thanks !

.close

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[gray badge]

"An urn contains 3 numbered balls (1, 2, and 3). First, a ball is drawn from the urn. If ball "k" is drawn, "k" pieces are selected with replacement from a batch containing 60% defective pieces. Let X be the number of defective pieces in the sample."

[gray badge]

Find the set of possible values of X and its probability function.

I don't know how to start solving this exercise

[shadow glade]

well we're going to end up drawing either 1, 2, or 3 pieces from the batch right?

so I guess there is potential also for drawing either 1, 2, or 3 defective pieces

[gray badge]

If ball 1 is drawn, 1 piece is selected. Then in this case we have P(X = 0) = 40% and P(X = 1) = 60%? Or should I multiply 1/3* 40% and 1/3* 60%?

[gray badge]

yep

[shadow glade]

🤔 1/340%? where are you getting that? what you said in the first part of that sounds good

[gray badge]

1/3*40%

[shadow glade]

oh lol

well, let's just think of each individual case we'll worry about the 1/3 at the end i'd say

so what about if you pull ball 2, then there's potential for X = 0, X = 1 or X = 2

[gray badge]

right

then X = 0 would be equal to 40%^2?

[shadow glade]

yep so 0.4^2

=0.16

[gray badge]

how about X = 1? 0.6*0.4?

[shadow glade]

so we can get 1 in two ways for that case, either by pulling a defective piece first and then a good one, or the other way around

so it would be 0.6 * 0.4 + 0.4 * 0.6, or just twice what you said

[gray badge]

got it, then the X = 2 would be just 0.6^2

[shadow glade]

yup

[gray badge]

X = 2**

[shadow glade]

we can double check our work for ball 2 case makes sense by making sure all the numbers add up to 1

so 0.16 + 0.48 + 0.36 = 1 so it looks good : )

[gray badge]

got it, now i can do it with ball 3

[shadow glade]

yap

[gray badge]

but how about them 1/3?

1/3 of each ball i mean

[shadow glade]

sorry, what?

[gray badge]

Should I take into account the probability of drawing each ball? There are 3 balls, ball 1, ball 2 and ball 3. Each one has a 1/3 probability of being drawn

[shadow glade]

we can handle all that at the end by just multiplying the total by 1/3. let's keep doing the same pattern as the other cases without jumping the gun, for ball 3 we'll have possibilities X = 0, X = 1, X = 2, X = 3 right? so let's start with X = 0 like the other ones

[gray badge]

sure, X = 0 would be 0.4^3

[shadow glade]

yeah

[gray badge]

X = 1 = (0.6 * (0.4)^2)*3

[shadow glade]

sounds right

[gray badge]

X = 2 = (0.6^2 * 0.4)*3

and X = 3 would be (0.6)^3

[shadow glade]

yeah, so maybe multiply those out real quick, write them out here and double check they add up to 1

ball 1: X=0: 0.4, x=1: 0.6
ball 2: X=0 0.16, X=1: 0.48, X=2:0.36
ball 3: X=0: 0.216 X=1: 0.064 X=2: 0.432 X=3: 0.288

[gray badge]

0.216 + 0.064 + 0.432 + 0.288 = 1 (ball 3)

ball 1 and ball 2 also added up to 1

[shadow glade]

alright so now overall we can add up each X = 0, X = 1 etc. cases and multiply by 1/3 the total like you said (saves the work of multiplying every number by 1/3, we can just multiply the added totals)

so for X = 0 we have: P(X=0) = ( 0.4 + 0.16 + 0.216 ) / 3

etc. and that should be about it

does that make sense?

[gray badge]

oh got it now

it is easier than i thought

thanks @shadow glade, i will finish it and maybe come back later with other questions

thx

.close

[safe radish]

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[jaunty finch]

find m knowing that the function is injective

m is from R

[halcyon carbon]

Think about the shape of the function

You want it to be 1-1 for x<=0

[jaunty finch]

wdym 1-1

[halcyon carbon]

What is injective

[jaunty finch]

i am not sure but i found 2 answers and i dont know wich one is correct

one that m<=0

and the other one m>=0

somehow if m>=0 the function isnt injective

[halcyon carbon]

Injective means if we have two different inputs we’d have two different outputs

Your function is a parabola

Go on Desmos and play with slider for m

That might give you the intuition

[jaunty finch]

i already have the intuition but i can work in two different ways and get different answers

[halcyon carbon]

But this just means it’s injective in a given interval if the function is one to one

Show me your work

[jaunty finch]

so

x1,x2 is from R

with f(x1)=f(x2)

(x1)^2+mx1-1=(x2)^2+mx2-1

fast forward and i get

(x1-x2)(m+x1+x2)=0

x1-x2=0

=> x1=x2=0

=> injective

m+x1+x2=0

=>

m>=0

[safe radish]

@jaunty finch Has your question been resolved?

[jaunty finch]

.close

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[slender nexus]

how do i get help

[plucky elk]

this channel is yours, so you just ask your question

[slender nexus]

i have 5 questions is that ok

[plucky elk]

usually just ask one at a time

[slender nexus]

this the first one

[plucky elk]

then when you're done you can either close and open a new one or if it's short enough, use the same channel

[slender nexus]

options are positive negative undefinded and zero

[plucky elk]

the question implies there are choices

slope = rise / run

[slender nexus]

so how do you know if they r any of these

[simple tapir]

Positive=Going up

Negative=Going down

[slender nexus]

its negative

[simple tapir]

zero=no change

[slender nexus]

yeah?

[simple tapir]

yeah

[slender nexus]

nice im so smart

[potent bay]

Take two points on the graph.
(X1,Y1) and (X2,Y2)
Slope = (Y2 - Y1)/(X2-X1)

i wanted to say this for the first but seems like i was fashionably late

[slender nexus]

i dont get that

the slope thing

[plucky elk]

why not

(X1, Y1) = (-7, 2). what's (X2, Y2) = ?

[slender nexus]

yes i dont even know what im looking at

[plucky elk]

X1 is a number

Y1 is also a number

(X1, Y1) is a pair of numbers

Called a coordinate pair

[slender nexus]

ahh

[potent bay]

here is an example:

[slender nexus]

that makes sense

[potent bay]

you didnt learn this formula and the teacher gives these problems?

[slender nexus]

No i just forget everyting

[slender nexus]

can u help me on this

[delicate sphinx]

Use the formula you were given

You have two points

Plug it into the formula

[potent bay]

write it down multiple times

[slender nexus]

4 6

now what i do

[delicate sphinx]

Use the formula

[delicate sphinx]

And coordinates are (x, y), meaning you need the parentheses

[potent bay]

repeat the steps here

[delicate sphinx]

You have two coordinates, label what x1, x2, y1, and y2 are then plug them into the formula

[slender nexus]

x1 is -7

y1 is 2

x2 is 4

y2 is 6

yeah?

[delicate sphinx]

Yes

[delicate sphinx]

Use the formula in this image, plug in those values you just labeled

[slender nexus]

like this

[delicate sphinx]

No that's not the formula

[slender nexus]

fuck

[delicate sphinx]

The $m = \frac{y_2 - y_1}{x_2 - x_1}$ part

[flat frigate]

CaptainNova22

[delicate sphinx]

You labeled what x1, x2, y1, and y2 are

You just need to plug them into that formula

[slender nexus]

yeah?

[delicate sphinx]

Yes

Now do the math

[slender nexus]

lego

4/11?

[delicate sphinx]

Yes

[slender nexus]

no negative?

right

[delicate sphinx]

Did you get a negative?

[slender nexus]

no 4/11

4 - -7 =11

yes?

[delicate sphinx]

Then why are you questioning about a negative if you didn't get one

[slender nexus]

because im not sure

lol

[delicate sphinx]

If you didn't get a negative, then why would you question about it?

Are you just going to throw on a negative?

[slender nexus]

No

im just not smart chill

[delicate sphinx]

You should be confident in your math skills

[slender nexus]

i dont do math

like its not my best subject

[delicate sphinx]

If 6 - 2 = 4 and 4 - -7 = 11, and neither values have a negative, why would you ask if it's negative

As asked, are you just going to randomly make one of those numbers negative?

[slender nexus]

No

[delicate sphinx]

Then it's not negative

[safe radish]

@slender nexus Has your question been resolved?

[slender nexus]

actually im done

[safe radish]

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[slender nexus]

.close

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[kindred nacelle]

Hi, so im currently dealing with the prove of the existence of a fixed point on a monotonically increasing function f:[0,1]→[0,1]
What i dont get here is:
how can f(s) be greater than s, if s is the supremum?
how does this lead to f(x) <x ?

[peak estuary]

well thats the question, can f(s) > s?

turns out, no

because then there is some x between them

s < x < f(s). but because x > s we then need f(x) < x

so f(x) < x < f(s), so f(x) < f(s)

[kindred nacelle]

alright, that makes sense to me now, tyvm!

.close

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[inland salmon]

what does the big theta symbol mean

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[inland salmon]

ok more context i guess

[safe radish]

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[robust mesa]

Need to solve when t is approaching 0. Every app I’ve used said the answer is 1 but I can’t see how they got there. Help please 🙏

[junior smelt]

,rccw

[robust mesa]

I need to get rid of that t in the denominator and the t^2 in the numerator

[junior smelt]

Well also bear in mind that as t^2 + t = t(t + 1), to common denominator them, you only need to multiply 1/t by (t+1)/(t+1)

Also what happened here?

[robust mesa]

oh my I missed that

okay so should I have factored it to t(t+1) at the beginning?

[junior smelt]

That would have made your life much easier

[robust mesa]

I see can I try that and get back to you or should I close this channel?

[junior smelt]

But if you didn't, you have here that the last line is $\lim_{t\to 0 }\frac{t^2}{t^2(t + 1)}$ when you factor the bottom

[flat frigate]

@junior smelt

[junior smelt]

Sure thing, you can keep this open and @ me when you get it

[robust mesa]

okay so thats what the final prodcut should look like ^?

give me one sec 🙂

[junior smelt]

,rccw

[robust mesa]

,rccw

[robust mesa]

I think it looks good and also should I be using the notation "Lim" notation after every step?

[junior smelt]

Also the previous hint was that you'd have that
\begin{align*}
\frac{1}{t} - \frac{1}{t^2 + t} &= \frac{1}{t} {\color{orange} \frac{(t + 1)}{(t + 1)} } - \frac{1}{t(t + 1)} \
&= \frac{t + 1}{t(t + 1)} - \frac{1}{t(t + 1)} \
&= \frac{t + 1 - 1}{t(t + 1)}
\end{align*}
then that's easier to simplify \holothink

[flat frigate]

@junior smelt

[robust mesa]

Okay I see

So when im finding a common denominator I dont have to multiply both sides

that makes my life a whole lot easier

yea this has definitely opened my eyes thank you so much @junior smelt

[junior smelt]

Yep, you just need to "make up" the terms to get the denominantors all being the same really!

[junior smelt]

And a pleasure

[robust mesa]

AS expected from DW

.close

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[night pawn]

Hi, idk how to solve 15, if you can help me, thanks a lot.
Statement : In this part, we look for functions f defined and derivable on I (with I an open interval of R containing 0) that verify: f(x) + f(y) = f((x+y)/(1-xy))
Let f be a function that satisfies these hypotheses.
14.calculate f(0) (I found f(0)=0)
15. let, a∈I with a≠0
determine λ>0 as for all h∈[-λ;λ], the equation
a+h = (a+y)/(1-ay), (of unknown y) admits exactly one solution. This solution is denoted y(h). Then show that h->y(h) is derivable on [-λ;λ]. Calculate y(0) and y'(0)

I have several leads but nothing comes to fruition.
I first thought about solving this equation :
(a+h)(1-ay)=(a+y) which brought me back to a 2nd-degree equation, but I don't think that was very helpful (or idk how to deal with it)
Then, as the lim of (a+y)/(1-ay) is -1/a
So I thought it was sufficient to choose lambda small enough so that [a - lambda, a + lambda] does not contain -1/a. But what lambda?
Or I thought as [-λ;λ] is '' symmetrical '' it contains 0 so we can choose h=0 but with this method I can't conclude about the possible value of lambda...
I'm stuck here, so if someone could help me out 🙂
(the goal of the problem is to find the arctan function so lambda should be equal to pi/2??)
Thanks

[safe radish]

@night pawn Has your question been resolved?

[night pawn]

No

[safe radish]

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[orchid agate]

Hi is there anyone that know how to work out this exercise

[junior smelt]

What does tgh represent?

[main mural]

i assume tanh

[junior smelt]

If that’s the case, what are you allowed to use to evaluate limits OP?

[orchid agate]

we do not use L'Hôpital's Rule yet

we use some algebraic rules I believe

[tiny wraith]

Have you written tgh in terms of powers of e?

[orchid agate]

no how can I do it

[tiny wraith]

Do you know any limits involving tgh(x)?

[tiny wraith]

there is no need for writing it that way btw

It comes down to proving that the limit of tanh(x)/x as x goes to 0 is 1

@orchid agate

[orchid agate]

but I do not understand how can we use tanh(x)/x as x goes to 0 is 1 when original limit goes to infinite

[tiny wraith]

It would mean that (x + 2) * tanh(1/(x + 2)) would go to 1 as x goes to infinity

And with that you can multiply the top and bottom of the original limit with (x + 2)

the denominator could be simplified then

[orchid agate]

but why?

I do not understand

Is there any site that may help me

[safe radish]

@orchid agate Has your question been resolved?

[safe radish]

@orchid agate Has your question been resolved?

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[slow sluice]

so the answer is correct but not in correct form since i have to rationalize the root in the denominator , but i dont have any idea on how the example did it

[latent mountain]

basically you have the form of (a+b) in the denominator, then you can square both a and b by multiplying it by (a-b) because a^2 - b^2 = (a-b)(a+b)

so you multiply the fraction by (8 -15sqrt(3))/(8 -15sqrt(3)) since it doesnt change the value of the fraction

[slow sluice]

makes sense 👍

tyty

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[latent tendon]

can you check the calculations and si units

[mortal sandal]

,w 2*.071591/(997.04799.806.008)

[flat frigate]

Results provided by WolframAlpha

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

Requested by a_mog

[latent tendon]

Thank you!

@mortal sandal is my si units correct?

ping me when you respond

<@&286206848099549185>

[safe radish]

@latent tendon Has your question been resolved?

[junior smelt]

Yep seems so (though I think you meant m/s^2 here)

[lapis wagon]

Can some one help me

[safe radish]

@latent tendon Has your question been resolved?

[latent tendon]

yes!

can you check this

i guess im just concern why my calculations are correct is bc my graph looks odd

and my data differs from my friend

[latent tendon]

<@&286206848099549185>

[safe radish]

@latent tendon Has your question been resolved?

[safe radish]

@latent tendon Has your question been resolved?

[safe radish]

@latent tendon Has your question been resolved?

[safe radish]

@latent tendon Has your question been resolved?

[latent tendon]

.close

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[molten hawk]

from the integral we have ysqrt3<x<sqrt(1-y^2)

why do we use x= ysqrt3 instead of the upper bound?

[glass carbon]

wdym by upper bound

[molten hawk]

x= sqrt(1-y^2)

[glass carbon]

you mean when solving for theta or...?

[molten hawk]

yeah

we would get like r^2cos^2theta+r^2sin^2theta = 1

[glass carbon]

that's equivalent to r = 1

which is true for the unit circle

like you can't solve for theta

[molten hawk]

oh ok so it would be useless

thank you

[glass carbon]

because sin^2x + cos^2x = 1

it's an identity

[molten hawk]

yeah I was wondering if I could do anything for that identity

ty

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[ashen swan]

Help

[ashen swan]

I don’t get this part

please explain

where did x(x-100) come from and why

[wide plaza]

can u see u private

[ashen swan]

what-

[wide plaza]

just see it

private chat

[ashen swan]

I’m not a math expert

idk

.close

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[fossil shell]

what

[safe radish]

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[undone lake]

desperately need help on this homework

[torpid fable]

Solve Ax=0

[undone lake]

I've got no idea how even to do that

[torpid fable]

Make an augmented matrix. Do you know how to do that?

[undone lake]

no unfortunately.. Ive been using apps to help the whole semester and they arent working anymore 😭

math is not my strong side

[torpid fable]

😬

You should look up a video on how to solve linear systems

[safe radish]

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[cinder violet]

Hi there, attached is a picture from a problem sheet im working on. It is asking to obtain the power series solution the ODE shown. I have found X=0 to be a regular singular point and proved such with XP(x) and X^Q(x) . but when It comes to finding the solution by frobenius I am stuck as all I get is A_n being = to 0. I will attach some work also

X^Q(x) . but when It comes to finding the solution by frobenius I am stuck as all I get is A_n being = to 0. I will attach some work also
I would appreciate any help in finding a solution!

[cinder violet]

Some of my workings

In any other example problem I have completed, the powers of X in each summation were different requiring me to change them thus casuing a change in A_n, due to the coefficients here making all the powers the same Im not sure how I can solve

[safe radish]

@cinder violet Has your question been resolved?

[cinder violet]

<@&286206848099549185>

Anyone with any input would be a great help 🙂

<@&286206848099549185>

[safe radish]

@cinder violet Has your question been resolved?

[cinder violet]

<@&286206848099549185>

[cinder violet]

<@&286206848099549185> anyone any input or start for this ?

Even a way to start after differentiating ?

[rough storm]

https://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx

[cinder violet]

This issue im having is that once I multiply back the coefficients back in I get the same powers of X thus not needing to change the index to get a match leaving all A_n the same

Thank you for the resource

@rough storm is there anything I should be doing with the factional power at X^n+1/3?

[rough storm]

Can you show your work so far?

[cinder violet]

This was from earlier, see how my X^n+1/3 are common, but also i have no need to change any index's

based on general solution from Frobenius as X=0 is a regular singular point

@rough storm its really from here im stuck as im not sure what is the best way to approach setting coefficients to 0 as A_n is common and so is the X^n+1/3

It may be a mess up in my indexing but im not sure

[rough storm]

why are you adding a power of a 1/3?

[cinder violet]

Due to us wanting the series solution around X=0 and X=0 being a regular singular point we use Frobenius’ Method

the lower case sigmas here are part of the indicial equation to find which will be the general solution for us to use

as sigma 1 -sigma 2 is not an elemet of Z we can use that case

@rough storm From here

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[tawny vessel]

need help with this

i have no idea where to start

[fleet condor]

this is a calc class right? how do you find a tangent line?

[safe radish]

@tawny vessel Has your question been resolved?

[tawny vessel]

yeah its calc

well you find derivative and then find the line

[fleet condor]

okay, did you do that

[tawny vessel]

but p is not even on the graph so idk

[fleet condor]

you want a line that satisfies two properties:
a) is tangent to f
b) goes through that point

you need to find all lines tangent to f, then solve for the y intercept using that point

[tawny vessel]

sorry i don't get what you mean can't p go fit on multiple tangent lines of the graph?

ok so first i would find the derivative

and then i just have no idea what to do

[fleet condor]

okay, whats the derivative

[tawny vessel]

(3/(x+1^2))

and now i have no idea what to do lol

[fleet condor]

your derivative is written wrong

so how you have a line with slope $\frac3{(k+1)^2}$ that goes through some point $(k, \frac{3}{k+1})$

are you familiar with point slope form?

[tawny vessel]

I think yeah

[fleet condor]

if not, you should look it up, but you get

$y - (\frac{-3}{k+1}) = \frac3{(k+1)^2}(x-k)$

and you want it to go through the point (x,y) = (-2, 8/3)

you should be able to solve all solutions of k from this

and remember that this k is the x value of where the line intersects the original function

from that you should be able to find the lines easily

this seems to not have real solutions

[tawny vessel]

huh

[fleet condor]

oh i forgot a negative somewhere i think

[safe radish]

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[tawny vessel]

.reopen

[safe radish]

✅

[fleet condor]

okay this should have two normal solutions, but I'm missing my own mistake somewhere

[tawny vessel]

btw im looking at what you said earlier im trying to understand it

[tawny vessel]

ok this makes sense to me since it has to pass through that line since its the tangent

[fleet condor]

hmm I dont know what went wrong here, all the steps seem correct

[tawny vessel]

i dont understand what you mean by this

[fleet condor]

im using k as a variable that means (x at a point)

like im using k as the x of the original function

but regardless my point slope form has no solutions when you plug in the x and y, which is wrong

[tawny vessel]

huh i see

[fleet condor]

oh nvm im silly

okay i got it

do you agree up to the point slope form so far?

[tawny vessel]

Yeah okay I got this

[tawny vessel]

yeah

[fleet condor]

okay great

so now we plug in our (x,y) into the point slope form to solve for k

you can verify this by hand, but you get $k = -\frac52,-\frac74$

[fleet condor]

now just plug your two different k's into point slope form and you get your two lines!

(the more roundabout way to do this is using point intercept form where you use k as x in mx+b and solve for b using the (x,y) point of the original function, but this is not needed)

heres the fact checking for the -5/2 solution

obviously you would simplify the green equation to not be a monstrosity, but it is the correct line

purple too, just for fun

I gotta step away for a sec, ping me if you have followup questions or dont understand something

[tawny vessel]

@fleet condor Thanks
So what we did is we first found the derivative
Then we used the derivative in point slope form and then chose a point to find all the possible different slopes that would result in lines passing through that point
Then we solved to find which k values would result in it and then replugged into the derivative to find the slope, and then we solve for b and find the line?

[fleet condor]

note that in the point slope form we both used the derivative and a generic point on the original function

<@&268886789983436800>

[fleet condor]

the last step can also be simplified down to "we solved for k with the point given, then we plugged it back in to find the general form"

https://www.desmos.com/calculator/bk3m6oohpx

sharing that calc because im closing it and wont feel like typing it out again

[safe radish]

@tawny vessel Has your question been resolved?

[tawny vessel]

@fleet condor Thank you I understand most of it I just have a question about the point slope formula itself doesn't the point slope form only consider the slope and not the value of y when x is 0

[fleet condor]

Im not sure what this means

the point slope formula in general is just $y-y_0 = m(x-x_0)$, where (x, y) and (x_0, y_0) are points

[flat frigate]

Compile Error! Click the reaction for more information.
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[pine vale]

10cm paper is equal to how many m^2 of paper?

[lean thorn]

how many cm in a meter?

[pine vale]

ik its .1 m so would m^2 be .01?

[lean thorn]

yep 👍

[pine vale]

ok thx

[lean thorn]

0.01m^2 in particular

np

[plucky elk]

No

,w meters in 1 centimeter

[flat frigate]

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[lean thorn]

If it was 10 cm to begin with, then it'd be 10/100 = 1/10, which squared is 1/100, right?

[plucky elk]

Oh my bad. I thought he was answering your question

[lean thorn]

no worries!

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[lean otter]

is there any textbook where i can learn about lines, planes and distances in 3D?

[lean otter]

like distance between line and point in 3D, distance between plane and point, planes, lines, skew lines etc

its geometry

but is there any book on it??

[plucky elk]

#book-recommendations

[lean otter]

where do i look in there @plucky elk

[plucky elk]

Just ask the same question

[lean thorn]

you can search or ask

[lean otter]

ok

.close

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[lean otter]

so say i have 3 parametric equations

[lean otter]

for x y z

this is for a line in 3 space

now say i have a point (a,b,c)

how do i find the distance between the point (a,b,c) and my parametric equations?

so for example say my parametric equations are

x = -1+ 4t, y = -2+7t, z = -0.5-5t

how do i find the distance between (a,b,c) (whatever it may be) and these parametric equations

do i remove the Ts from the equation, and then find the equatio n ax + by + cz = 0??

what should i do

[plucky elk]

The distance from a point to a line has a known formula

[lean otter]

yes

its

$\dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$

[flat frigate]

Derivative

[lean otter]

but i have a parametric equation with x,y,z

[plucky elk]

Do you know distance between two points?

[lean otter]

no

[plucky elk]

Treat each x,y,z equation as a point

[lean otter]

yes

[plucky elk]

There's a known formula for this too

[lean otter]

so i find the distance between each point

[plucky elk]

sqrt((x-a)^2+...)

No

[lean otter]

i treat the parametric equation as a point

[plucky elk]

Yes

[lean otter]

so three seperate parametric points

x,y,z

[plucky elk]

"the distance" is just the minimum along the line

[lean otter]

so i have point a,b,c

[plucky elk]

Start with the distance from (x,y,z) to (a,b,c). It's a known formula

[plucky elk]

Begins like this

[lean otter]

ok

[plucky elk]

You fill in the rest

[lean otter]

and that's the answer

but isn;t there going to be a t involved?

[plucky elk]

That gives you an answer in terms of t yes

[plucky elk]

.

[lean otter]

derivative?

[plucky elk]

Right

[lean otter]

ahhh

its my name

hahahhaha

[plucky elk]

[lean otter]

so that's the calculus way

what about the linear algebra way

[plucky elk]

Probably a known way yes

[lean otter]

yeah. the calculus way is probably better. but since there are parametric equations in linear algebra, say i have 3 points a(1,0,2) b(3,1,0) and c(2,2,2). Find the distance between point a and the line passing through b and c

so here what i would do is find the parametric equation of b and c

so i find vector BC

then i do the parametric equation stuff

so after i have my 3 parametric equations, i do what we did before (distance formula then find the derivative)

but, this is the calculus way

how do i approach this problem the linear algebra way

cuz i dont think my teacher wants me to find dervatives in a linear class hahah

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[lost ice]

for question d

[lost ice]

this is the question

i shouldve gotten -30 in decimal but instead if i convert that binary number ill get -62

where did i calculate it wrongly

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[ruby edge]

Why can’t logs have negative bases? It should work out theoretically for example
(-2)^3 = -8
log-2 -8 = 3

[split ether]

Well what about $\log_{-2}(-7)$?

[flat frigate]

A Lonely Bean

[split ether]

It is even a real number

Exponentiating with negative base can get really messy

So it's conventional to stick to positive bases

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[frosty mist]

Does anyone have any values that help me with this equation: a^2-b^2 = sqrt c

[frosty mist]

but c is a whole number

[compact rampart]

=>(a-b)(a+b) = c

[simple field]

(= sqrt c)

[compact rampart]

Does the question say anything about the values of a and b ?

[frosty mist]

tbh i dont think i phrased it right lol

let me just use examples

so the problem has this equation: x^2 + 8^2 = 17^2, x = 15

when they square the value it turns out to be a whole number

but im looking for a number combination that gives me a whole number as well

cause i did x^2+8^2 = 12^2 and got x = sqrt 63

[quick crater]

you did the subtraction wrong, also x wouldnt be a real number

[frosty mist]

oh yeah i just realized but why wouldnt x be a real number

[quick crater]

sorry i mean x would be a whole number

[frosty mist]

oh i know that i was trying to find values of b and c that gave me a whole number for x

but luckily i found some !

x^2 + 5^2 = 13^2

i just needed it cause my teacher wanted us to make word problems that related to triangles

like this

Sorry for any confusion

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[fading bridge]

(p -> q) -> (q V (not)p)
Can we say that this is the same as (p V (not)p) given that we can cancel out the -> q’s

[fading bridge]

I know both of then are tautologies

But still can we claim that they are the same expression

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[tall coral]

Justine is paid $10 an hour for x hours. During a particular week, she earns $180.
Write an equation involving x to describe hours worked in a week to earn $180 and solve the equation by
inspection to find the value of x. (3 marks)

[shadow glade]

!status

[safe radish]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[tall coral]

1

1. I don't know where to begin.

[shadow glade]

so if she earn $10/h, how much would she earn working 1 hour?

[tall coral]

$10

[shadow glade]

and 2 hours?

[tall coral]

20

[shadow glade]

what did you do to get those answers?

[tall coral]

so 18 hours

[shadow glade]

so if she earn $10/h, how much would she earn working x hours?

[tall coral]

10x

[shadow glade]

yes

and now set that equal to 180

since thats how much she ends up earning in the problem

[tall coral]

so 10x=180

[shadow glade]

yep, so there's your equation, what next?

[tall coral]

so x=18? right

[shadow glade]

that's it and don't forget to include units with your answer, 18 hours

[tall coral]

ok also can i say another question

[shadow glade]

sure

[shadow glade]

and what is the goal, to solve for x?

how would you start?

[tall coral]

times 3 so than it will be 2x+1=3

so 2x=2

x=1?

[shadow glade]

so if you multiply both sides by 3 you end up with what...?

2x+1 = 3 is not quite right, you completely forgot to multiply one side of the equation by 3

[tall coral]

3 by 3?

[shadow glade]

$$3=\frac{2x+1}{3}$$
$$3(3) = 3\frac{2x+1}{3}$$
$$9 = 2x+1$$

[tall coral]

yep so 2x=8

x=4

[flat frigate]

Soosh

[shadow glade]

ya

there you go

!done

[safe radish]

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[tall coral]

another one please

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[tall coral]

.close

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[lean otter]

what happens after u sub z in

[lean otter]

and why do u sub e^itheta wouldn't u be subbing in the euler's formula of sintheta

do u convert it into cis??

.close

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[teal belfry]

what is the difference between plotting and graphing

[teal belfry]

and thx if you have the answer

[dapper venture]

plotting means marking points

[peak estuary]

I wouldnt make a distinction between those two words

[shadow glade]

agree with Denascite, mostly they are used interchangeably but yeah when referring to drawing invididual points, you usually hear "plot the points" rather than graph the points. But you can also say "plot the graph of the parabola" just as well as "graph the parabola." so pretty much they can be used interchangeably but one might be used over the other more often for specific contexts

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[fringe pivot]

Greeting

I would like to know if my steps are correct

Thank you

[mortal thicket]

thers a much simpler way to do it

when the power of the denominator is greater than the power of denimator

[fringe pivot]

It can be zero instantly

Am my prediction correct?

[mortal thicket]

u just divide the top and bottom by the largest power of x

[fringe pivot]

Is

[mortal thicket]

nopt quite

what was ur final asnwer

i cant quite read it

[fringe pivot]

0

[mortal thicket]

right

u got it right

but two things

u need to specify if its coming fomr below or above the x axid

so in this case its $0^{-}$

[flat frigate]

Big Chicken

[mortal thicket]

how u get it is this

x-2 / x^2 +2x + 1 = {(1/x)-(2/x^2)} / {1 + (2/x) + (1/x)}

now u sub in -inf