help-42

[torn surge]

ohh

i tought about another thing with inequalities

that's actually what i did btw

[oak elm]

Forgot to remove roots when copying my bad

[potent lotus]

AkitoLite

[oak elm]

I mean you just solve them 🤷🏻‍♂️
we know that the leading power has a coefficient of +ve to both of them, so the y value must decrease as x decreases in large values and increase as x increases

since we have 2 critical values (starting with the first one), that means the curve cuts and touches the x axis

$x=0,x=1$ \
$1>0$ therefore we have $x(x-1)^2 \ge 0$ at $x \ge 0$

It is also obvious from pure looking at it that ANY negative value of x will result in a negative value of the expression. Because we have x(f(x))², so even if f(x) is negative, it will become positive, and the x=-ve will cause it to be negative as a whole

@torn surge you can try solving for the other one aswell 🙂

[potent lotus]

AkitoLite

[calm coral]

@torn surge Has your question been resolved?

[calm coral]

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[remote mural]

Can I get help

[versed nest]

yes

!justask

[pure kayak]

sure

[remote mural]

I hope you can help me with acceleration 😭

It's probably not math but it has math

[glad parrot]

!da2a

[calm coral]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[glad parrot]

:)

[remote mural]

I wanna know what I need to write on my paper all the acceleration formula

[drifting seal]

what

[pure kayak]

every single formula for acceleration?

we'd be here a while

[remote mural]

Not every

[latent widget]

Like the equations of motion you mean or what?

[pure kayak]

it usually boils down to F=ma or dv/dt, and the suvat equations

[remote mural]

I don't get what this teacher is teaching me

I just write things down in my paper

But I don't get it 😭

[pure kayak]

theyre doing (change in velocity)/(time for that change to happen)

[remote mural]

Can u explain how to calculate them

[pure kayak]

say i have velocity v1 at t=t1
and velocity v2 at t=t2

my (average) acceleration between those two times would be (v2-v1)/(t2-t1)

[latent widget]

Its similar to how you would find velocity from displacement and time

[remote mural]

I meant like when I get to my exam and I'm question with 50km/h I don't how to calculate that and the teacher told me to use v and t

[latent widget]

You won't be asked to find acceleration from just 1 velocity
Thats not possible

[pure kayak]

they may say 'blah is travelling with velocity u and experiences uniform acceleration a, after t seconds what is blah's velocity, then its just u+at [suvat equations]'

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[torn surge]

I'm studying the 3 dynamics laws but i feel that's not the context to use them

[pure kayak]

ive been summoned

[torn surge]

@oak elm btw thanks

[pure kayak]

I meant it in the sense of finding the acceleration if there is a general force acting, eg in centripetal force

usually youre given either a force, a function for displacement or velocity, or some information to use the suvat equations. As a generalisation for formulae you might have for accelerations

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

No

[swift dragon]

clikc the ❌

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

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[simple musk]

Let $L_1$: $X = \lambda (0,-1,1) + (4,2,1)$, $L_2$: $X = \lambda (0,1,2) + (4,-2,-4)$ and $A = (4,2,1)$.

Find a point $B$ that belongs to $L_1$ and $L_2$, and a point $C \in L_2$ such that the triangle $ABC$ is right-angled at $A$.

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[calm coral]

@simple musk Has your question been resolved?

[calm coral]

@simple musk Has your question been resolved?

[calm coral]

@simple musk Has your question been resolved?

[flat gorge]

What did you get for B, C?

For B, you need to solve (4, 2 - x, 1 + x) = (4, -2 + y, -4 + 2y)
For C, you need to write the parametric: (4, -2 + t, -4 + 2t)
Write vectors BA, CA and make the dot product zero

[calm coral]

@simple musk Has your question been resolved?

[flat gorge]

@simple musk

[calm coral]

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[cyan knot]

hey guys, i need help, to verify whether or not these 25 derivatives are correct.

the first column is the function, and the 2nd is the derivative.

[calm coral]

@cyan knot Has your question been resolved?

[hasty mulch]

Looks all good to me. I'd represent derivative number 18 as -cot(x)csc(x) though.

[cyan knot]

alright, thank you. ill take note of that 🙂

[calm coral]

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[simple musk]

\textbf{2.} Let
$A = \begin{pmatrix} 2 & 1 \ 0 & -6 \ 4 & 5 \end{pmatrix}$
and
$B = \begin{pmatrix} 1 & -1 \ k & -k \ 3 & k+1 \end{pmatrix}$.

Determine all values of $k \in \mathbb{R}$ for which there exist infinitely many matrices $X \in \mathbb{R}^{2 \times 2}$ such that

$AX = BX$.

For each of these values, find all matrices $X \in \mathbb{R}^{2 \times 2}$ that satisfy the equation.

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

AX = BX

AX - BX = 0

(A-B)X = 0

,, \begin{bmatrix} 2 - 1 & 1 + 1 \ -k & -6+k \ 4 - 3 & 5 -k - 1 \end{bmatrix} = A - B

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

,, A - B = \begin{bmatrix} 1 & 2 \ -k & -6+k \ 1 & 4 -k \end{bmatrix}

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

,, (A - B)X = \begin{bmatrix} 1 & 2 \ -k & -6+k \ 1 & 4 -k \end{bmatrix} \begin{bmatrix} x_1 & x_2 \ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \ 0 & 0 \end{bmatrix}

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

,, (A - B)X = \begin{bmatrix} x_1 + 2x_3 & x_2 + 2x_4 \ -kx_1 -6x_3 + kx_3 & -kx_2 -6x_4 +kx_4 \ x_1 + 4x_3 -kx_3 & x_2 + 4x_4 -kx_4 \end{bmatrix}

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

,, \begin{cases} x_1 + 2x_3 &=0 \ x_2 + 2x_4 &=0 \ -kx_1 -6x_3 + kx_3 &=0 \ -kx_2 -6x_4 + kx_4 &= 0 \ x_1 + 4x_3 -kx_3 &=0 \ x_2 + 4x_4 -kx_4 &= 0 \end{cases}

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

1. x1 = -2x3
2. x2 = -2x4
3. -k(-2x3) -6x3 + kx3 = 0
4. -k(-2x4) -6x4 + kx4 = 0
5. -2x3 + 4x3 -kx3 = 0
6. -2x4 + 4x4 -kx4 = 0

3.1) 2kx3 - 6x3 + kx3 = 0
3.2) (2k-6+k)x3 = 0
3.3) 3k - 6 = 0
3.4) 3k = 6
3.5) k = 2

[boreal rose]

looks good

[simple musk]

6.1) (-2+4-k)x4 = 0
6.2) -2+4 = 2
6.3) 2-k=0
6.4) 2 = k
6.5) k = 2

so either k is 2 or k is NOT 2

k = 2 or k ≠ 2

[boreal rose]

u can check this by just substituting something thats not 2 for k

and seeing if its true

[flat gorge]

You need infinitely many solutions for AX = BX... Why would k be neq to 2?

[boreal rose]

^

[simple musk]

I need help and hints

I am trying . . .

[boreal rose]

i don't really undertand where k neq 2 comes from

[simple musk]

idk

[boreal rose]

and also u can save a bit of trouble for urself by checking the minors of A-B

[flat gorge]

You assumed AX = BX here, and solved for a "k" that always satisfies it

That is, k = 2 gives infinitely, many solutions for AX = BX no matter the X

[simple musk]

(A-B)∈R^(2x3)
whats the dimension of the minors? 1x2?

[simple musk]

determinant is only for square matrices doe, wdym

[simple musk]

ye

[boreal rose]

i meant by checking the 2x2 minors of A-B

idk i also started lin alg only recently but i recall that u can check that A-B is rank 1 by inspection for (A-B)X to be true for infinitely amounts of X

which then can be solved by checking the determinants of the 2x2 minors

idrk whats happening geometrically since im a bot but algebraically its just finding when A-B is of rank 1

[simple musk]

wdym bot

[boreal rose]

sorry, its just colloquial slang from where i live that describes a person being bad/poor at something

im just saying i havent learnt enough to see whats happening geometrically

[simple musk]

you from america?

🗽🦅

[boreal rose]

so i can't really give an explanation of why im checking for it other than the fact that i've just been taught to do it as such

[boreal rose]

australia

[simple musk]

A-B being rank 1 after plugging k = 2 and rrefing(?)

🤔

[boreal rose]

ur 2nd column is a multiple of 2 of ur 1st column

[simple musk]

okay

[boreal rose]

u know rank is 0, 1 or 2

since A-B is not trivial 0 matrix

we know its 1 or 2

ur 2nd column must be a multiple of the first to satisfy having infinite solutions

so that they are linearly dependent

[simple musk]

rank = dim(column space)

[simple musk]

its not 2 bec col 1 is lin dep col 2

[boreal rose]

yes

thats the whole point

[simple musk]

. . . so?

[boreal rose]

im just saying u can check urself whether its k=2 or not

beacuse it satisfies rank 1 of A-B

[boreal rose]

U essentially solved k=2 but u aren't sure whether its k=2 or k =/=2

hence i tried to provide a way to self-check

[simple musk]

I need to find X

1. x1 = -2x3
2. x2 = -2x4
3. -k(-2x3) -6x3 + kx3 = 0
4. -k(-2x4) -6x4 + kx4 = 0
5. -2x3 + 4x3 -kx3 = 0
6. -2x4 + 4x4 -kx4 = 0

1. x1 = -2x3
2. x2 = -2x4
3. -2(-2x3) -6x3 + 2x3 = 0
4. -2(-2x4) -6x4 + 2x4 = 0
5. -2x3 + 4x3 -2x3 = 0
6. -2x4 + 4x4 -2x4 = 0

[boreal rose]

ok ok let sjsut go back to Ax=0

do you know when is this true for non-zero A?

like what does it imply about x

[simple musk]

1. x1 = -2x3
2. x2 = -2x4
3. -2(-2x3) -6x3 + 2x3 = 0
   3.1) 4x3 - 6x3 + 2x3 = 0
   3.2) (4-6+2)x3=0
   3.3) -2 + 2 = 0
   3.4) 0x3 = 0
4. -2(-2x4) -6x4 + 2x4 = 0
   4.1) 4x4 - 6x4 + 2x4
   4.2) (4-6+2)x4=0
   4.3) -2 + 2 = 0
   4.4) 0x4 = 0
5. -2x3 + 4x3 -2x3 = 0
6. -2x4 + 4x4 -2x4 = 0

[simple musk]

idk

[simple musk]

no

[boreal rose]

do you know what nullspace is

[simple musk]

ye

[simple musk]

the kernel

[simple musk]

all the vectors that map to 0

[boreal rose]

would you agree any matrix A, multipie by vectors from its nullspace will give 0

[simple musk]

rrefing an matrix augmented by zeros

[simple musk]

A multiplied by vectors in the kernel give 0

[boreal rose]

naturally x is vector from nullspace of A

[simple musk]

so?

[boreal rose]

so we can find x

if we are finding x for (A-B)x=0

we can just find nullspace of A-B

[simple musk]

ye

,, (A - B)X = \begin{bmatrix} 1 & 2 \ -k & -6+k \ 1 & 4 -k \end{bmatrix} \begin{bmatrix} x_1 & x_2 \ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \ 0 & 0 \end{bmatrix} \ \textbf{ k } = 2 \ (A - B)X = \begin{bmatrix} 1 & 2 \ -2 & -6+2 \ 1 & 4 -2 \end{bmatrix} \begin{bmatrix} x_1 & x_2 \ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \ 0 & 0 \end{bmatrix}

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[boreal rose]

yes

the quesiton is asking when k=2, what X satisfies the equation (A-B)X=0

[simple musk]

,w Nullspace[{{1,2},{-2,-6+2},{1,4-2}}]

[potent lotus]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[boreal rose]

naturally ur X is?

[simple musk]

Ker(A-B) = (-2x,x) = x(-2,1)
Ker(A-B) = <(-2,1)>

[simple musk]

X ∈ R^(2x2)

[simple musk]

help

[simple musk]

Ker(A-B) ∈ R^2

[simple musk]

X ∈ R^(2x2)

[hollow lion]

You already had the parametrisations x_1 = -2x_3 and x_2 = -2x_4

So plug those back in to X

[ X = \begin{pmatrix} -2a & -2b \ a & b \end{pmatrix} ]

[boreal rose]

its true as long as the 2nd column is a multiple of the first

[potent lotus]

shsgd

[boreal rose]

they dont all have to be a

ok nvm u changed it

[simple musk]

where did "a" come from

[boreal rose]

arbitrary values

there are infinite Xs

[simple musk]

ye

[hollow lion]

a = x_3 and b = x_4

Renamed because easier

a and b are free variables so you know there are infinite solutions

[hollow lion]

Also notice the columns of X are the same format as the kernel that was given by wolfram here

This makes sense because a and b are free variables, so any choice of a and b means the column will be found in the kernel of A-B

[simple musk]

ye

[hollow lion]

That’s it, you solved another one

[simple musk]

they should AFAIK all be a

because otherwise the dimension of the kernel will be 2

unless I am tripping hard

no, because, a(2,1) is a linear combination of b(2,1)

dimension of the kernel is still 1

[simple musk]

[hollow lion]

3.2) (4-6+2)x3=0
4.2) (4-6+2)x4=0
It doesn’t matter what x_3 or x_4 you choose, it will go to zero

[simple musk]

ye

[hollow lion]

Which means there is no constraint on x_3 = x_4

[simple musk]

ye

I appreciate the help

thanks

.solved

[calm coral]

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[calm coral]

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[twilit gust]

is there a way to solve this type of problem other than just guess and check?

[swift dragon]

i think youre going to take a while with that

what do you need for this to become an integer?

[twilit gust]

70

[fathom solstice]

whenever you have a square number, all of the powers of the prime factorization should be even

[twilit gust]

gotcha, thanks

.close

[calm coral]

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[rare inlet]

$(2x-3)^2 = 6-5x$

[potent lotus]

Simon James B

[rare inlet]

i am doing something wrong idk what

$(2x)^2 - 2 * 2x(-3) + (-3)^2 = 6-5x$

[potent lotus]

Simon James B

[pure kayak]

you have a double -

shouldnt be there

[rare inlet]

i don't see it

[pure kayak]

either - 2 * 2x(3) or +2 * 2x (-3) not both

[eternal shard]

$(2x)^2 \textcolor{red}{-} 2 * 2x(-3) + (-3)^2 = 6-5x$

[potent lotus]

anti-algebraist 𝔸dωn𝓲²s

[rare inlet]

if we have (2x-3)^2 don't we use a^2 - 2ab + b^2?

[pure kayak]

yeah, but b would be 3 then

not -3

[potent lotus]

anti-algebraist 𝔸dωn𝓲²s

[rare inlet]

i never seen this before

[eternal shard]

me too

[rare inlet]

usually i was not changing signs when doing -

[eternal shard]

but it's to explain

[nocturne heron]

you know the formula for $(a-b)^2$ ?

[potent lotus]

ss

[rare inlet]

yea

a^2 -2ab + b^2

[nocturne heron]

yeah

try to identify a and b in your case

[rare inlet]

2x and -3

(2x)^2 -2ab i see it now

[eternal shard]

that would be the case in (a+(-b))²

[rare inlet]

-*- = +

[nocturne heron]

$(a-b)^2 = (2x-3)^2$, whats the a and b ?

[potent lotus]

ss

[rare inlet]

2x is a and b is -3

??

[nocturne heron]

no bro

[pure kayak]

youre saying b is -3 then it would be (2x-(-3))^2

[rare inlet]

well how can it be 3? even +(-3) makes it -3

[pure kayak]

then youd have to use (a+b)^2, here b would be -3

but youre using (a-b)^2 so b would be 3

[rare inlet]

i get that it should be 3 but right now for me it makes no sense lol

[pure kayak]

lets say 2x-3=2x-b

what is b

[rare inlet]

when i had a-b i always did it with - with no issues lol

[nocturne heron]

we have $(a-b)^2 =(2x-3)^2$, so if b=-3, we would have $(2x \textcolor{red}{-}-3)^2$

[potent lotus]

ss

[nocturne heron]

we would have a double $-$

[potent lotus]

ss

[nocturne heron]

you see it ?

[rare inlet]

yea i see it

[nocturne heron]

okay its good so whats the result of $(2x-3)^2$ ?

[potent lotus]

ss

[rare inlet]

$4x^2 + 2* 2x * 3 +9$?

[potent lotus]

Simon James B

$4x^2 + 2* 2x * 3 +9$?

[nocturne heron]

its $a^2 -2ab+b^2$ , $a=2x, b=3$

[potent lotus]

ss

[nocturne heron]

the a^2 and b^2 are good

[rare inlet]

(2x)^2 is 4x^2

ah ok

-2ab = -2 * 2x * 3 wdym. 2ab if our a is 2x and b 3 it is - 2* 2x *3

[nocturne heron]

yeah it’s $-22x3$

[potent lotus]

ss

[eternal shard]

In this formula of (a-b)² = a²-2ab+b² = a²+2a(-b)+b² the minus is already taken into account

[split nymph]

is there somone who understand frensh

[nocturne heron]

oui

[rare inlet]

i don't speak frensh but i speak french

okay ty

.close

[calm coral]

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[split nymph]

est ce que tu comprend integral à parametre

[nocturne heron]

retourne dans le channel d’avant

[calm coral]

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[calm coral]

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[calm coral]

@errant knot Has your question been resolved?

[slate field]

please don’t ping the mods for math help

[calm coral]

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[fluid swallow]

Needa help

[fluid swallow]

Is my integral done correctly

<@&286206848099549185>

[flat gorge]

The answer is correct, and I could not spot any errors :|

[fluid swallow]

🥰🥰🥰🥰🥰🥰

Those calculations are too intimidating in many ways 😭😭

Thanks for looking it for me 🥰

Another question for cal 3 course can I use frechet differentiablity for related problems and discuss it in normed metric spaces (which isn’t taught during the lecture but I feel it would make life easier)

[calm coral]

@fluid swallow Has your question been resolved?

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[exotic cosmos]

for finding the number of solutions in sin(x) = x/100

[exotic cosmos]

how do i know for each period, there's exactly two intersection points and not any more any less

the pattern suggests that is the case

but like uh

i'm guessing it has something to do with sin being periodic but hmm not convinced

[brazen elbow]

there cannot be more than 2 intersection points per period

the only case where there is 1 intersection point per period is when the maximum/minimum of f(x) = sinx - x/100 is also the root

show that this is not possible

[exotic cosmos]

this looks like using ivt or calculus in general

can't use that

i'm reviewing trig and this is a trig exercise so only concepts below/equivalent to trig

[mortal orbit]

complicated to only use trig for that

though you know that for n such that (2n+1)*pi < 100:

sin(2npi)-2npi/100 < 0

sin((2n+1/2)pi) - (2n+ 1/2)pi/100 > 0

and the function is increasing on [2npi, (2n+1/2)pi]

so there is a unique point on that interval where the function is 0 by corollary of IVT

the tricky stuff comes for the interval [(2n+1/2)pi, (2n+1)pi]

[exotic cosmos]

i'm thinking about it 😭

[mortal orbit]

oh actually it's the opposite

[mortal orbit]

the function isn't increasing on that interval

f'(x) = cos(x) - 1/100

the function is increasing on [2npi + arccos(1/100), (2n+1/2)pi]

which isn't a problem when n is big enough

[exotic cosmos]

okay so i'm guessing that calculus is almost required for this problem?

[exotic cosmos]

yeah this part makes sense

[mortal orbit]

well trig and calculus will be needed

IVT corollary is also a need

[exotic cosmos]

did they skip over the calculus part here?

also i guess this solution messed up the part where 30pi <= x <= 31pi

where you get another 2 solutions

2 * floor(50/pi) = 30 + 2 from considering values of 30pi <= x <= 31pi

but since it's a stackexchange answer i'll assume i'm probably wrong somewhere

[exotic cosmos]

😭

[mortal orbit]

there might be a way to consider (2npi,(2n+1)pi) immediately

so f(x) = sin(x) - x/100

or f(x) = 100sin(x) - x

to get rid of the fraction

[exotic cosmos]

are we going to ignore the scaling factor?

[mortal orbit]

something = 0 is the same as 100something = 0

[exotic cosmos]

yeah okay sure

[mortal orbit]

so

[mortal orbit]

taking this function

[exotic cosmos]

sure

[mortal orbit]

on the interval (2npi,(2n+1)pi):

f(2npi) < 0

f((2n+1)pi) < 0

no actually take any n > 0

so

two cases

solving for f'(x) = 0 on that interval

we get x = arccos(1/100) + 2npi

[exotic cosmos]

right

[mortal orbit]

so the max of the function on the interval is f(x) = 100sin(arccos(1/100)) - arccos(1/100) - 2npi

[exotic cosmos]

😭 mhm okay evaluating at the critical point

[mortal orbit]

yes

f and f' all being smooth functions

it won't be hard to prove that

f(x) is indeed a maximum

(for example computing f'' < 0 on that interval can be helpful)

[exotic cosmos]

yeah fair showing a max can be done with first derivative test or second derivative test

bla bla

[mortal orbit]

so

if max > 0

by IVT, splitting the interval into (..., x) and (x,...)

two solutions exactly

now to understand when f(x) = 100sin(arccos(1/100)) - arccos(1/100) - 2npi > 0

[exotic cosmos]

righttt okay f(2npi) < 0 and f(2(n + 1)pi) < 0 and if max > 0 then yeah by ivt there's 2 solutions

okay i think the general thought process makes sense

[mortal orbit]

yes

so now it's all about computing 100sin(arccos(1/100)) - arccos(1/100)

[exotic cosmos]

or just showing it's > 0 for our case?

do we really care about the exact value?

[mortal orbit]

well if the approximation is good enough it's ok

we just need to know the max value of n that works

[exotic cosmos]

i see okay fair enough

but wait so how do the amc folks solve this question

do they take this route too?

[mortal orbit]

those?

[exotic cosmos]

or is there a heuristic approach to this

[exotic cosmos]

i mean the source

"source: AMC" and well i'm guessing they are only expected to know till precalc

[mortal orbit]

graphic

[dusky flax]

in a single sine cycle there is a bump above the x axis and one below it

[mortal orbit]

they do it graphically

basically they ask "when the positive bump of sin becomes below x/100"

[dusky flax]

so in a cycle it wouldn't matter if the positive or negative side of the line intersects

it would intersect twice for a considerable number of cycles

[exotic cosmos]

yeah that's fair but they would have to assume that the "pattern" holds right?

the pattern being that x/100 intersects sin(x)

twice every period

[dusky flax]

which you can approximate

[exotic cosmos]

or do they have a tool to justify this

[dusky flax]

self evident from graph

there is a bump

[exotic cosmos]

yeah but u can't draw the graph for 0 <= x < = 100

[dusky flax]

it's not that hard

not 100

[exotic cosmos]

for a period yes

for several periods too

but then they have to resort to assuming this pattern holds, no?

[dusky flax]

but you can be sure that it intersects twice for some 15 cycles

both sides

[mortal orbit]

well they are 'assuming' it but the reason why is because sin is concave on its positive bump

[dusky flax]

like it's almost a horizontal line passing through a bump

[mortal orbit]

so if it instersects with a line

it intersects exactly twice

[exotic cosmos]

unless it's tangent

hmmmm right

[mortal orbit]

here that would mean there is an n such that 1 = (2n+1/2)pi/100

[exotic cosmos]

and what's the justification for 30pi <= x <= 31pi, for example

do we just say that 0.94ish <= x/100 <= 0.97 and 0 <= sin(x) <= 1 and since sin(x) is concave in said interval then it must intersect the line twice?

[mortal orbit]

integer = multiple of pi

not that possible

[mortal orbit]

sin(x) is concave, and sin(30.5 pi) = 1 > 30.5pi/100

the line has to intersect

and isn't tangent

so intersects twice

[exotic cosmos]

oh so without the calculus thing we did prior

we just see heuristically

that okay yeah a line intersects the concave bumps ( > 0 ) of sin twice each period

[mortal orbit]

ofc there are values of k such that sin(x) = kx will show some tangents

but here picking k = 1/100

being rational

and under 1

it won't happen

so bonus question: what are the values of 0 < k < 1 such that sin(x) = kx has an even amount of positive solutions?

[exotic cosmos]

wait 😭 i'll tend to this

okay just to recap what we did

we wanted to examine the roots of f(x) = x/100 - sin(x)

and for that we considered the derivative

if there's a max in our interval then there must be a root in some closed interval [2npi, 2(n + 1)pi]

and well since there's only one max in the said interval

wait why can't we just stop here?

do we have to mention ivt for f(2npi) and f(2(n+1)pi/2)

[mortal orbit]

ivt is the completely rigorous way

[exotic cosmos]

no i mean like

if we show that there's only one max in the said interval

then there must be two roots, no?

uhhh

[mortal orbit]

uh

no

[exotic cosmos]

okay yeah i can think of a function > 0

over the entire interval

and still having one max

that would have 0 roots

okay i see how ivt ties in with this

ivt guarantees that you must cross the x axis twice

because of f(2npi) and f(2(n+1)pi < 0

okay

and the "heuristic" approach was to just uh say that a line intersects the concave bumps of sin (when the y values are > 0) exactly twice?

[exotic cosmos]

no idea 😭 but just a rough sketch already suggests every value of k >= 0.1 works

but between (0, 0.1) it gets pretty hard

[mortal orbit]

are you sure?

we're not counting x = 0 as a positive solution

[exotic cosmos]

oh whoops

but like just the opposite i guess

[mortal orbit]

not really

some k around 0.13 works

[exotic cosmos]

i got down to eliminate all k between [1/(2pi), 1)

maybe 1/(2.47pi) ish

😭

i think it's time to use calculus to find a line that is tangent to sin(x) between 2pi and 3pi

wait uh u can't have a tangent of the form y = kx so that is out the window too

[exotic cosmos]

is it that you can't have even amount of positive solutions?

it's cuz y = kx can't possibly be tangent to any of the concave bumps of sin

i first got rid of all the cases where kx just intersects sin(x) once

so like around k < 1/(2.5pi)

now if we admit the line passes through one of sin's concave bump then it must again intersect an even amount of times to the next sin bump

it can't be tangent

so yeah, uhhhh it's not possible?

this is really hard 😭

thanks for the help btw

.close

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[subtle gate]

so i did 7x7 x 3.14

got 153.9

and then im lost

[still marlin]

ok that's the area of the base

[subtle gate]

and whats the circumfrens

would i just x 153.9 by 21?

[still marlin]

no, that would give you the volume

[subtle gate]

so what would i do

[still marlin]

you need what's called the "lateral area" now (and since there's two bases, multiply the area of the base by two)

[subtle gate]

307.8

so TheEnthusiast

what now?

[still marlin]

Ok now the lateral surface area is the area to the side of the cylinder

[subtle gate]

ok

[still marlin]

You can basically roll it out like a rectangle https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcRPEH-ncqjxP4sjqsT0L-1-dNq6s9cPHh6Ppw&usqp=CAU

[subtle gate]

what do we do with the 21?

21 x 3.14

307.8

2 x 3.14 x 7

44

307.8 +44 x 21

.close

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[indigo cloak]

Today I've started to learn about the use of cos sin and tan. I learned a trick called 'soh cah toa' which I'm sure many know here.

My question really is what we find out when using for example here, the question is

"Use sinus and cosinus to find A and C."

The book used this trick to find A, but what do we really get when we get to 0.598? Since it doesnt really have any unit, is it relevant to know, or is it just a constant we find in the process?

[dull wagon]

it's the ratio of the relevant sides

[indigo cloak]

My understanding is that these 3 functions have quite a lot of uses for them correct? We can enter degrees, but also constants?

Or is this why we would use sin^-1 instead

[dull wagon]

sin,cos,tan of an angle gives the ratio of sides

applying the inverse to a ratio gives the angle

[indigo cloak]

.close

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[copper notch]

How to solve this limit?

[nocturne heron]

try to factorize by $e^{xln(x)}$ maybe

[potent lotus]

ss

[nocturne heron]

or you can use taylor series

[copper notch]

How should i do it using taylor?

Dunno how to continue

<@&286206848099549185>

Pls help

[vagrant oak]

i think this would work

factorize e^(-xln(x)) from the denominator

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

[potent lotus]

MæthIsAlwaysRight

[vagrant oak]

this is already a decent simplification

limit of e^(xln(x)) is 1

[vagrant oak]

after this, the denominator can be "compared" to a function with similar asymptotic growth

there is a well-known limit (e^x - 1) / x = 1, as x -> 0

2xln(x) tends to 0, so we may use that as well

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

[potent lotus]

MæthIsAlwaysRight

[vagrant oak]

the limit in denominator would now evaluate to 1, so we can cancel it

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

[potent lotus]

MæthIsAlwaysRight

[vagrant oak]

that gives us this

and this is sum of 2 limits, one of which tends to 0

@copper notch you here?

[copper notch]

Nvm i got it

But Nice work

Thank you a Lot

.close

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[sage girder]

hi

i am learning product rule for the first time

is this right or wrong

[delicate charm]

Your solution is correct

[sage girder]

thank you

lasty question

i learnt

first principles product rule

and power rule

what do i learn next

[delicate charm]

Quotient rule, chain rule

You can try also worded problems

[sage girder]

Thank you

where can i find som,e worded problems

[delicate charm]

Try the book of Ron Larson, just search ron larson calculus pdf

[magic tulip]

When you write fg(x), do you mean f(x)g(x) or f(g(x))?

[sage girder]

f(g(x)) im pretty sure

[magic tulip]

yeah then it's kinda off

[mortal orbit]

then you shouldn't use product rule of differentiation

if fg is not a product

[magic tulip]

you need chain rule

[delicate charm]

You're implying chain rule there

[sage girder]

oh is it cos its a function inside of a function

chain rule is like

g'(f(x)) f'(x) isnti t

[mortal orbit]

for g(f(x)) yes

[sage girder]

okok wait

wait

wait

dont tell me anything im gonna try and solve it rn using chain

tell me if i get it wrong or right

[delicate charm]

chain rule: g(f(x)) = g'(f(x)) • f'(x)

product rule: g(x)•f(x) = g(x)•f'(x) + f(x)•g'(x)

[sage girder]

pretty sure im wrong

[delicate charm]

Your g(x) should be sin x, review f'(g(x))

in f'g(x), you only differentiate the outer function, which is the cosine. So, it should be -sin(sinx)

[sage girder]

oh okay

[delicate charm]

So you won't get lost in the process, its better to assign the inner function to a variable.

f(x) = cos x; g(x) = sin x

find f(g(x))

y = cos(sinx)

let u = sin x; u' = cos x

y' = cos'(u) • u'
y' = -sin(sinx) • cos x

In this way, you won't be confused

[calm coral]

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[hushed crater]

Hey what's the solution to this problemo

[hushed crater]

I don't think n!/(n-r)! Can solve it

[calm coral]

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[stark smelt]

Are these answers correct P. 68 M. 72 R.108 T.70 S.108 N. 108

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[lime solar]

can someone help with 9b

[calm coral]

@lime solar Has your question been resolved?

[lime solar]

<@&286206848099549185>

[leaden thunder]

Try writing the cosine terms as a sum of the ninth roots using demoivre

[lime solar]

So like (z-alpha)(z-alpha^8)?

[leaden thunder]

Should be no z at all if you're doing this

[lime solar]

Could u show me what u mean?

,rotate

[lime solar]

Do u mean these roots?

[leaden thunder]

I don't know why you're setting z^6 = 1

But yes those are the ninth roots of unity

[lime solar]

I thought i had to get them

cuz it had z^6 + z^3 + 1

[calm coral]

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[floral dew]

The following image appeared in a newspaper in 1979. What is missing from the image from a statistical perspective?

Title of the chart: “Purchasing Power of Wage Rates Compared to 100 in September 1978.”

X-axis labels; months.

[calm coral]

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@floral dew Has your question been resolved?

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@floral dew Has your question been resolved?

[tough prairie]

Can someone help me with simultaneous word problems

[calm coral]

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[kindred rapids]

i need help with this question i got 1/n+7, 993, 1/r-7. but its all wrong

[calm coral]

@kindred rapids Has your question been resolved?

[kindred rapids]

<@&286206848099549185>

[flat gorge]

!show

[calm coral]

Show your work, and if possible, explain where you are stuck.

[kindred rapids]

1

2

3

[soft wind]

Looks right to me, maybe try nearby values like 994 or smth in case they messed up some strict inequalities

Also don't forget parentheses where needed

[kindred rapids]

ok let me try

yeah idk i give up

[soft wind]

Bruh the first one should be right

Maybe they want -1/...?

[leaden thunder]

,w simplify abs((n+5)/(2n+14)-1/2)

[potent lotus]

Results provided by WolframAlpha

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

Requested by rie.mann

[soft wind]

Or maybe they want the absolute value in the answer?

[kindred rapids]

im not sure

i only have 1 try left

[kindred rapids]

should i do sqrt(n+5/2n+14 -(1/2)^2)

?

and the 2nd one is not 993 or 994

so i got no clue for that

[soft wind]

Would you know if anyone else got them right?

[kindred rapids]

this is like one of the example work

no

we dont know until the 19th

its just graded hw

[soft wind]

Rip

This is some bad quality hw

Your answers seem right to me at least, so it's just a pb with the website solutions imo. If you can ask any other classmates/prof I'd suggest that

[kindred rapids]

yeah idek what im missing

what about this one

its the same concept

[soft wind]

Just solve for x_N =10 or 20 or K like before

[kindred rapids]

so the last on eis k^4? right

but it says k is not definied\

[soft wind]

Capital K?

[kindred rapids]

no

oh just 4? maybe

[soft wind]

Well this is just dumb

[kindred rapids]

nope wrong

[soft wind]

No not 4

Maybe k^4

[kindred rapids]

alr i used all 3 tries im just too frustrated

i alr did that too

[soft wind]

(no capital)

[kindred rapids]

https://tenor.com/view/frog-lily-lake-water-relax-gif-21614201

[soft wind]

Lol it's not your fault, whoever made this did it really badly

[kindred rapids]

i honestly think its the rounding

but ill see after the results

from previous hw it just looks normal so im not sure

[soft wind]

No the last one is definitely K^4

[kindred rapids]

thanks for ur help

have a great night

[soft wind]

Np, you too

[calm coral]

@kindred rapids Has your question been resolved?

[heady dagger]

@kindred rapids close it

[calm coral]

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[calm coral]

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[stone prism]

I fail to prove the following equation
Also I'd appreciate if somebody could tell me how to recognise the 'patterns' of trigonometric equations

[stone prism]

essentially the goal is to turn the equation into two same parts, like 1 = 1, or more likely in this case, sin 2a = sin 2a(only an example of the result)

[calm coral]

@stone prism Has your question been resolved?

[heady dagger]

Sinc +sind =2sin(c+d/2)cos(c-d/2)

[calm coral]

@stone prism Has your question been resolved?

[stone prism]

?

Can you elaborate, please?

I tried using the sum-to-product identity, but I get stuck later

[calm coral]

@stone prism Has your question been resolved?

[abstract wave]

$4sin(\frac{\alpha}{2})$ $cos(\alpha)$ $cos(\frac{3\alpha}{2})$ $ is $ $2cos(\alpha)[sin(2\alpha)-sin(\alpha)]$

[potent lotus]

Horsi135

[abstract wave]

and that can be opend to $2cos(\alpha)sin(2\alpha)-sin(2\alpha)$

[potent lotus]

Horsi135

[abstract wave]

and that is $sin(3\alpha)+sin(\alpha)-sin(2\alpha)$

[potent lotus]

Horsi135

[abstract wave]

and that is the left side

[calm coral]

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[fading shuttle]

does anyone know how to solve b(ii)?

I try to use y = ax+b (hint from (i)) and I get y = -x -a

Also I figured out as a = 0, infinity ,both (x,y) = (0,0)

Then I tried to use the Green theorem, and the integral became nasty. Should I change the method, or were my assumptions wrong?

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@fading shuttle Has your question been resolved?

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[upbeat shoal]

how would i go about solving part a? so far i have done the base case which is (10^0 - 1) = 0 which is a multiple but idk what to do from here

[floral vessel]

you do the step case and then you modify 10^(k+1)

then subtract 1 in the end and somehow factor a 9 out

[stoic oyster]

well you wanna relate 10^(n+1) - 1 vs 10^n - 1

these are both almost ||geometric sums||, it should get you a useful equality for the problem at hand

[mortal orbit]

otherwise, since they're keen on you solving it by induction

||what's their difference||

[stoic oyster]

||yeah we're essentially doing the same thing, yours is a bit more to the point||

[upbeat shoal]

so can you do (10(10^k)-1)=9m? m being a constant

[mortal orbit]

well that's what you want to prove

I'm suggesting you look at 10^(n+1) - 1 - (10^n - 1)

[upbeat shoal]

where did you get the - (10^n - 1) from?

[mortal orbit]

well

it's your induction hypothesis

10^n - 1 is a multiple of 9

so if 10^(n+1) - 1 - (10^n - 1) is a multiple of 9

you'll get the result you want

[upbeat shoal]

so 10^(n+1) - 1 - (10^n - 1) = 0?

[stoic oyster]

you sure about that ?

[upbeat shoal]

bc i thought you did 10^(n+1) - 1 = (10^n - 1) then just moved it to the LHS

[stoic oyster]

no

the idea is

[mortal orbit]

10^(n+1) - 1 - (10^n - 1) = 9k

so 10^(n+1) - 1 = 9k + (10^n-1) = 9k + 9m

sorry went ahead

that's what we're going for

[stoic oyster]

if A = 10^n - 1 is a multiple of 9 [induction hypothesis]
and B = 10^(n+1) - 1 - (10^n - 1) is a multiple of 9 [which remains to be shown]
then A+B = 10^(n+1) - 1 is also a multiple of 9

and you win

it's fine we wrote essentially the same thing lol

[flat gorge]

Assume P(k) = 10^k - 1 = 9m
Then P(k + 1) = 10^{k+1} - 1 = 9*10^k + 10^k - 1 = 9*10^k + 9m = 9(10^k + m)

With m' = 10^k + m => P(k + 1) = 9m' and you're done

[upbeat shoal]

i dont get where you guys are getting the -(10^n - 1) from,
the base case is (10^k - 1) = 9m
the step case is (10^(k+1) - 1) = 9s
then what did you guys do?

[flat gorge]

Did you not get this either?

P(k) is the base case, P(k + 1) is the step case

[upbeat shoal]

im not sure why you're doing 9*10^k + 10^k - 1

[flat gorge]

to reduce the step case into base case, and since we need to show the step case is divsible by 9 also, pulling out 9*10^k out of 10^{k+1} = 10*10^k kills two birds with one rock

gets me a 9multiple in addition to the base case which we assumed was divisible by 9

[upbeat shoal]

so your turning a 10 into a 9?

[flat gorge]

10 into a 9 + 1

10*10^k = 9*10^k + 10^k

[upbeat shoal]

my bad i missed this

i got it thanks

how about part b? would the base case be P(k) = (x-0)?

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[upbeat shoal]

.close

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[pseudo wedge]

I'm not sure how to go about this

clearly {0} is a two-sided ideal. Then I assume epsilon != {0} and also epsilon != L(V) and try to get a contradiction

I feel like I have to do something with T in L(v) which is not in epsilon

but I can't put my finger on it

[glass heart]

slightly reworded, show that if there is a nonzero map in E, then all maps are in E

like earlier, I would recommend to think about it in terms of matrices

[pseudo wedge]

yes

ahh

of course

matrices

so this amounts to the fact that I can transform any matrix to any other matrix using elementary row and column transformations

but I don't know how to prove that

[flat gorge]

||I'm feeling an urge to revisit linear alg||

[pseudo wedge]

it's great so far I like this

[glass heart]

instead of going from A to B for all A,B, it can be helpful to instead show A to C for some fixed C and then you also have B->C, so by reversing time you have A->C->B

[pseudo wedge]

Right

Wait but if a matrix is not diagonalisable then it can't be converted to a diagonal matrix right?

[glass heart]

you are messing up what you are allowed to do to the matrix

[flat gorge]

you can construct 3 elementary matrices for:

1. row switching transformation
2. row multiplying transformation
3. row addition transformation
   and you're done ?

[glass heart]

to diagonalise you are only allowed to do XAX^-1

here you are allowed to do XA and AX

[pseudo wedge]

Ah so I can do XAY

for arbitrary X and Y

[pseudo wedge]

well

hmm

well yeah I do know what elementary transformation matrices look like

[flat gorge]

does that not suffice to show this?

[glass heart]

also dont forget that the ideal is also a subspace

so you can also add

[pseudo wedge]

It probably does, I just don't see it

[pseudo wedge]

I'm assuming by C you mean the identity?

or any diagonal matrix

[glass heart]

for example

on second thought that isnt actually the best approach because to be able to reverse time you are only allowed to use invertible X

but its still enough to show that the identity is in E

why

[pseudo wedge]

because then T is in epsilon for all T in L(V)

[glass heart]

why

[pseudo wedge]

because TI=T

[glass heart]

yes

(in fact any invertible matrix would be enough)

[pseudo wedge]

I was just about to say that

we need an invertible matrix

but what happens if epsilon is entirely composed of singular matrices

[glass heart]

basically you have to show that that doesnt happen

[pseudo wedge]

right

so I need to prove that there exists E in epsilon so that dim Im(E) = dim V

[glass heart]

try to get the identity

its easiest

as a first step, try to create "simple" matrices

[pseudo wedge]

I can isolate any non-zero element A_{i,j} of A by pre-multiplying it by a matrix X with all 0s except 1 in the (i, i) slot and post-multiplying it by a matrix Y with all 0s except 1 in the (j, j) slot

X isolates the row and Y isolates the column

[glass heart]

good start

[pseudo wedge]

and I can scale this so that the A_{i,j} element becomes 1, so I have a matrix with a single element 1

which means I have an E in epsilon such that E(v_i)=v_j for some i, j and E(v_k)=0 for k != i

where v_1, ..., v_n is a basis of V

ah okay and then I can just duplicate the 1 to make the identity

Now to translate this to the language of linear maps

[glass heart]

how

[pseudo wedge]

I do not

I realised

the determinant can't become non-zero

[glass heart]

.

[pseudo wedge]

yeah

But something like $$\varepsilon=\left{\begin{pmatrix}
a&0&0\
0&0&0\
0&0&0\end{pmatrix}: a\in\bR\right}$$ is also a subspace

[potent lotus]

kheerii

[glass heart]

yes

but you are allowed to do XA, AY and A+B

[pseudo wedge]

That is a lot of stuff

I can't wrap my brain around everything

[glass heart]

yes

take your time

this is not an easy question

[pseudo wedge]

ah wait with any matrix with a single 1 I can get all the matrices with a single 1

[flat gorge]

Is this not sufficient Denascite? what am I missing

[glass heart]

(and I dont understand how axler expects anyone to solve it without the matrix perspective)

[pseudo wedge]

just by changing the rows and/or columns

[glass heart]

yes

[pseudo wedge]

which forms a basis of the whole space

[glass heart]

and then?

yes

[pseudo wedge]

We technically have never established what a basis of L(V,W) looks like though

[glass heart]

you can also construct the identity and follow the previous plan

[pseudo wedge]

That works yeah

[pseudo wedge]

Wait, I think we have

or at least for L(F^n,F^m)

This should be easy enough to appropriate into a basis of L(F^n, F^m)

which in turn should be easy enough to appropriate into a basis of L(V,W)

but I think I'll just do the identity thing

Let me see if I can translate this to linear maps

[calm coral]

@pseudo wedge Has your question been resolved?

[pseudo wedge]

v_q/c_j, not c_k

there's a lot of indices going on here and idk how to make it simpler

[glass heart]

youre rushing a bit

first get the map with only a single entry

and then afterwards get all the other ones you want

[pseudo wedge]

does this look good?

How is someone supposed to do this without knowing what a matrix and matrix transformations are

[glass heart]

technically you dont need it ^^

looks good

although I didnt fully check every index

[pseudo wedge]

Lol

[pseudo wedge]

Yeah the indices should be right, I was talking about the writing of the proof

Thank you so much for your help

.close

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[keen sigil]

help

[keen sigil]

help with b

ive narrowed it down to 13 17 and 19

we know that this number is bigger than 10 and smaller than 20

numbers like 15 dont work as if u multiply it by itself it would always yield 5 as its last digit

however the last digit number is 1

so its either 13 17 or 19

any way to solve this without a calculator without estimating?

[manic grove]

maybe a dumb thought but you can write x^16 as x^4^4, so just calculate 13^4, 17^4, 19^4 (should be doable on paper maybe?) and then just look at the last two digits as you take ^4 again

[dapper sable]

You can do a last digit check.

Find the first four powers of 0 through 9.

[manic grove]

oh man im dumb

but they all have 1 as last digit in 4 steps, 9 in 2, so you get 1 at the end for all 3 after 16

[stray tundra]

11 would also have 1 as last digit, no matter the power

<@&268886789983436800>

sure, but i'm saying that his justification is not enough to discard 11

[sand sky]

Oh sry !

[bold bear]

Honestly I can see a few ways of doing it but none of them are that pretty. You could calculate each of them raised to the 16th power modulo 100 which rules out two of them (but in principle it could only rule out one of them)

16^16 is quite easy to estimate so you can do that to rule out one or two of them

But I would probably just divide it by each of those numbers with long division

It's not very painful

[stray tundra]

it cannot be an even number

[bold bear]

Yes but if 16^16 is smaller then it can't be 15^15

I'm not saying it's 16

sorry 15^16

sorry 13

I'm tired oka

It cuts out options

[stray tundra]

As followup on what Gandhi stated
You can check 13^4, 17^4 and 19^4
Then you can get the last two digits of each of those, and take it to the 4th ower as well, which will give you the last digits of the number

[dapper sable]

x^1: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
x^2: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1
x^3: 0, 1, 8, 7, 4, 5, 6, 3, 2, 9
x^4: 0, 1, 6, 1, 6, 5, 6, 1, 6, 1

Then this repeats every four powers, so x^16 uses the x^4 line.

So, it's some odd number other than 15.

[sand sky]

Check if 13,17,19 divide it by division

😭lol but seems easiest !

[stray tundra]

This is a brute force method, which while will clearly give the solution, but i think the idea is doing it with theory?

[sand sky]

Else we may try to factor

a¹⁶-1

So just factor by division

Until we get the least factor

[bold bear]

It's not "brute force" because you're not actually raising the numbers to the power of 16

Depends how you define it ig but it's much much easier

[sand sky]

(a+1)(a-1)

[dapper sable]

Yeah, long division will work nicely.

[bold bear]

It also relies on the fact that those numbers are all prime so it does still use theory

[stray tundra]

well i guess cutting it down to only three possible candidates is the "enough theory" to cut it down to a reasonable effort by brute force

long division of a 18-digit number three times is not particularly difficult or tedious

[sand sky]

,calc 6+6+5+4+1+6+6+0+9+1+8+3+1+7+9+8+4+0

[potent lotus]

Result:

84

[sand sky]

,calc 84/3

[potent lotus]

Result:

28

[sand sky]

So ig 13 works

13^n mod 3 = 1

We may use similar logic to check the remaining possible combinations

[dapper sable]

16^16 = 2^64, which, if you know the highest value of a 64-bit unsigned number plus one, you can perhaps discard two answers.

Or discard one.

[sand sky]

This doesn't work well !

If 13 is an answer then a¹⁶-1 will have a factor (13+1) = 14 that is we can check if 7 divides a¹⁶-1

[calm coral]

@keen sigil Has your question been resolved?

[calm coral]

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[winged beacon]

What is the difference between f(x) and f(y)?

Here's the part of the problem that's confusing me (from the review chapter in Calculus 1:)

1. Let f(x) = x^3
2. f(y) = ?

What I'm struggling with:
It appears the answer would be simply y^3 just like f(q) would be q^3, but I've not encountered f(y) before and want to confirm this is correct. It doesn't feel right to treat "y" like a regular variable but perhaps I am overthinking it. Thanks!

[wanton wave]

you're right, if f(x) = x^3, then f(y) = y^3

You're probably used to treating y as a dependent variable, but it's just a symbol, no different than x

[winged beacon]

Cool, thanks for confirming!

.close

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[wanton wave]

np 👍

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[slim cove]

should i seperate it first?/

but then derivative of ln is u'/u right

how would i do that if 2 is on the bottom

is it possible to take out 1/2 out of an ln then

[glad parrot]

Write the formula purely and remember some quotient proprety

Such as dividing by a fraction is the same as multiplying by his reciprocical

[slim cove]

wait so

[slim cove]

where do we apply it. .

[glad parrot]

Write u'/u

[swift dragon]

you can also simplify what is in the ln

its a trig identity

[glad parrot]

Indeed, if you know it its also good

[slim cove]

uhh

i ddid it

but

wouldnt the two cancel out

or did i do something wrong

[glad parrot]

What is u'

[slim cove]

2(e^x-e^-x) - 0(e^x+e^-x)

oops i forgot the minus

[glad parrot]

No

Its not a quotient

Its a sum of function

/2 is just a factor

You don't have to use quotient rule here

[foggy crypt]

If you find it difficult to deal with the 2, you can use the laws of logs to get rid of it before differentiating.

[slim cove]

.close

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