help-26

[solemn briar]

I mean in the above svedka gave us our interval i think the [-pi/2 ; pi/2] 5pi/6 is bigger than pi/2 which is outside sinus’ domain

Is this what youre looking for @tight rivet

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[analog venture]

is this the correct way of a v.c by 1/3 ? i get confused on applying vertical c/e to a quadratic / reciprocal function

[half edge]

hmm it looks more like an anti-compression

nvmnd maybe youre right

and its just a language thing

[analog venture]

i think the graph just confuses it with a horizontal transformation

but i meant algebraically

its hard to distinguish whether its a vertical or a horizontal i think..

unless i show in my solution

[half edge]

yea i think youre correct btw

[analog venture]

mhh okay thank you

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[terse trellis]

$\int{\frac{1}{4\cos^2(x)+9\sin^2{x}}} dx$

[thorny flame]

rainy

[terse trellis]

so for similar problems, I used the Weierstrass subst. method

but it doesnt work here

so what should i do?

[terse trellis]

(so t=tan(x/2), ...)

[chilly walrus]

not sure if this will help but the denominator is equal to 4+5sin^2(x)

[terse trellis]

no

but thx

[chilly walrus]

no as in wrong or as in won't help

[terse trellis]

it wont help

becuase

my teacher is doing t=tan x , ... and so on

instead of

t = tan(x/2)

[chilly walrus]

?okay

[terse trellis]

Its like a different type of weierstrass?

[coarse tusk]

you can divide numerator and denominator by cos^2(x)

[chilly walrus]

good point

[ashen nest]

how does that help?

[terse trellis]

She does this

[coarse tusk]

it's the same thing

[terse trellis]

But thats not a typical weierstrass, its different

i want to know why

[coarse tusk]

dividing by cos^2(x) just makes the process easier

[coarse tusk]

why does it have to be weierstrass?

i mean, that would work with some algebraic manipulation

but usually it's a last resort

[terse trellis]

we have to

[coarse tusk]

this isn't a weierstrass though

[terse trellis]

Thats what im asking

This is

right?

[coarse tusk]

yeah

[terse trellis]

Well, for this particular question, instead of letting t = tan(x/2) and calculating sinx, cosx, dx, ... based on that (typical weiserstrass),
she let t = tan(x)

Think i've found an answer (somewhat).
if R(-sin, -cos) = R(sin, cos) -> use t = tan(x)
if R(-sin, -cos) =/= R(sin, cos) -> use t = tan(x/2)

ok that was it

bruh

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[timber sphinx]

Hey, im having hard time taking the derivative of this

[timber sphinx]

i mean

more like just the top side

[pastel salmon]

you must consider all fraction

[timber sphinx]

i mena yes

[pastel salmon]

since numerator goes to e

[timber sphinx]

its just that its taken from a question about delhospital

[pastel salmon]

x->0

i know this exzmple

from th ebook of my friend

in Pl

[timber sphinx]

whats her name

or his

[pastel salmon]

Marian Gewert

[timber sphinx]

Okay then idk him haha, thought it would be my lecturer

[pastel salmon]

i also work in same insitute where MArian works

[timber sphinx]

Amazing u actually help others here

But going back to the questiion

[pastel salmon]

anyway, the hint for yoru limit is: chaneg to exponent

adn then play with de Hospital

twice as i remind

such thing you should get:

$exp\left{ \lim_{x \to 0} \frac{\ln\left( 1+x \right)^{\frac{1}{x}}-1}{x} \right}$

[thorny flame]

Joanna Angel

[timber sphinx]

never done such example

but i assume its from this

[pastel salmon]

insied th elimes, you have 0/0

the formaule i used is:

a^b = exp [ b lna]

to compute derivative, you wil need such a formula too :

$\left{ \left[ f\left( x \right) \right]^{g\left( x \right)} \right}'=\left{ exp\left[ g\left( x \right)\cdot \ln\text{}f\left( x \right) \right] \right}'$

[thorny flame]

Joanna Angel

[timber sphinx]

thats some black magic right there

[pastel salmon]

that is exampel in a book with star symbol haah ) for Polytechnique

[timber sphinx]

yea no wonder, never seen such formulas

i didnt know i study maths

will try to tackle it

[pastel salmon]

eys write all hints i gae you and play )

[timber sphinx]

will do, ty

[pastel salmon]

yw

[timber sphinx]

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[timber sphinx]

for now

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[neon iron]

How do I integratre this using partial fractions?

[hushed ore]

x^2 - 1 = ?

[neon iron]

What I have so far is

A/(x-1) + B(x+1)

[neon iron]

But I have no clue what the value of x is and what the constant value is... 😦

[reef fjord]

https://tutorial.math.lamar.edu/classes/calcii/partialfractions.aspx

[neon iron]

Hmm, so I see this

But mine is x^2-1, not X^2+1

[reef fjord]

there's more

[neon iron]

Ah, I see, sorry about that!

[reef fjord]

above

and below

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[wispy rivet]

Let X_i(n) = {A subset {1,2,...,n} where the cardinality of A, |A| = 3k+i for some integer k, 0≤i≤2.

Prove |X0(n)|+|X1(n)|+|X2(n)|=2^n. This was somewhat easy for me as they are disjoint unions and i get the set of all the subsets of {1,2,...n} which is just the power set of that set.

This though is tricky to me, I don't know where to approach such a thing:
Prove that |X_0(n)|=|X_0(n-1)|+|X_2(n-1)|

[wispy rivet]

my idea is to express X_i(n) as a binomial sum, and then inducting over n. However, I don't know how to deal with boundary cases...

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[wispy rivet]

@knotty finch

.reopen

[topaz sinew]

✅

[reef fjord]

i think it should be |X_1(n-1)|?

nvm, i aint sure

[topaz sinew]

@wispy rivet Has your question been resolved?

[reef fjord]

wydm boundary cases?

think about the case where n is various mod 3s, that may help

i.e. if n=1 mod 3 then the sum to get |X_0(n)| will be nCr(n,0)+nCr(n,3)+...+nCr(n,n-1)

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[proven narwhal]

I'm a bit confused, why did they find the dot product of the two direction vectors?

[sweet shard]

Where do they calculate the dot product

[proven narwhal]

is that the cross product?

and also where did they get the p1p2 from? (-1,0,1)

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@proven narwhal Has your question been resolved?

[gray ridge]

p1p2 is from p1O + Op2
where O is the origin

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[strange narwhal]

Van U help md

[strange narwhal]

Can

[torpid matrix]

dont ping random people, and do you have a question

[strange narwhal]

Ok, can U help md

[torpid matrix]

how could I know?

I dont know what your problem is

ask your question and someone can help

[strange narwhal]

Exercise 2 (3 points): Consider the quadratic function ( f ) below in an orthonormal coordinate system, defined on ( \mathbb{R} ) by ( f(x) = \frac{1}{4}x^{2} + \frac{1}{2}x - \frac{15}{4} ) and represented by the parabola.

1. Solve graphically in ( \mathbb{R} ) the equation ( f(x) = 0 ) and the inequality ( f(x) > 0 ).

2. Solve algebraically the inequality ( f(x) > 0 ).

[thorny flame]

moon4king

Exercise 2 (3 points): Consider the quadratic function \( f \) below in an orthonormal coordinate system, defined on \( \mathbb{R} \) by \( f(x) = \frac{1}{4}x^{2} + \frac{1}{2}x - \frac{15}{4} \) and represented by the parabola.

1. Solve graphically in \( \mathbb{R} \) the equation \( f(x) = 0 \) and the inequality \( f(x) > 0 \).

2. Solve algebraically the inequality \( f(x) > 0 \).

[strange narwhal]

@torpid matrix

[torpid matrix]

yeah I was busy trying to use my romance language knowledge to translate it (i got it right wahoo)

sure, so the first part is "graphically", do you know how to tell when f(x) = 0 and f(x) > 0 from the graph

if not, which one (or both?) do you not know

[strange narwhal]

I wasnt i' school when they doit this part

I sont understand all

Dont*

[torpid matrix]

do you understand what a function is?

[strange narwhal]

Yes

[torpid matrix]

so when f(x) = 0, this means that you want to find the x-values where the y-value equals 0

does this make sense?

[strange narwhal]

Yes

[torpid matrix]

so on the graph, what are these points?

[strange narwhal]

-5 and 3?

[torpid matrix]

yes

[strange narwhal]

Ok

[torpid matrix]

how about the next part?

when f(x) > 0 is the inputs that make the output positive (greater than 0)

in other words, when is it above the y axis

[strange narwhal]

Yes I know how to doit this

Its whit ∆

[torpid matrix]

what

[strange narwhal]

For next part

[torpid matrix]

you are stuck on the next part?

[strange narwhal]

What next?

[torpid matrix]

$\frac14x^2+\frac12x-\frac{15}4 > 0$

[strange narwhal]

Yes

[torpid matrix]

for part 2, you are asked to solve this algebraically

[strange narwhal]

I need ∆=b^2-4ac

[torpid matrix]

no

well like not really

[strange narwhal]

Why

[torpid matrix]

that determines how many solutions you have

it doesnt figure out where they are

you need the entire quadratic formula to find the roots, then you need to check to the left and right of them to see when its > 0 and < 0

[strange narwhal]

So how to do

[torpid matrix]

sorry i have to go now, i hope this is enough:

for solving my equation above, i recommend multiplying the entire thing by 4 first, then you can solve the inequality

[strange narwhal]

Thanks budy

Who can help

[sweet shard]

Just wait and stop flooding your own channel with emojis

[strange narwhal]

Ok

[strange narwhal]

Can U help

[sweet shard]

Just

Wait

[strange narwhal]

Ok

[sweet shard]

Use this hint and then factor your quadratic

[strange narwhal]

I sont understand

[sweet shard]

Why not

Did you understand "Multiply by 4"

[strange narwhal]

But in class we sont do like this

[sweet shard]

Well you're not in class and we're not in your class

[strange narwhal]

I fous this

[strange narwhal]

À friend send me what they do

[sweet shard]

Then listen to your friend?

[strange narwhal]

Ok

Thanks men for helped me☺

[strange narwhal]

I have another question

Its à bonus by our techer

[sweet shard]

What

[strange narwhal]

Question 4: Is there a real number ( b ) such that the equation ( 2x^{2} + bx - 5 = 0 ) has a unique solution?

[thorny flame]

moon4king

[strange narwhal]

Fait this Is the wrong thing

The quadratic function is given by ( f(x) = -3x^{2} + 6x - 4m ). Under what condition on ( m ) does the equation ( f(x) = 0 ) have no solutions? (Any partial attempt at a solution will be considered.)@rie.mann

[thorny flame]

moon4king

[strange narwhal]

@sweet shard

[sweet shard]

Use the discriminant

[strange narwhal]

Ok thanks

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[dull wraith]

i need someone help in limited development 😭

[last cypress]

.help

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Commands:

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[last cypress]

.open

is this working out right?****

[hollow bone]

open your own channel @last cypress

[last cypress]

how

/open

[hollow bone]

wrong guy my b

o you change profile picture

[last cypress]

no i changed my pfp

lmao

yh

[hollow bone]

go there

and just post

no need for a command

[dull wraith]

idk where did i wrong

[hollow bone]

crawber i'm checking wait a bit (aussi on dit assymptotic expansion en anglais pour DL)

[dull wraith]

ah ok

[hollow bone]

,w limit of (x(1+cos(x)) -2tan(x))/(2x-sin(x)-tan(x))

[thorny flame]

Results provided by WolframAlpha

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

Requested by b__n

[hollow bone]

sin(x) = x - x^3/3!

pas +

[dull wraith]

ah

tnx

merci

ta un ide pour la deuxiem ?

[hollow bone]

exp(ln(...))

et arctan(x) equivalent a x

enfin = x+ o(x)

[dull wraith]

ahh

merci bcp !

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[merry citrus]

Hello! Could someone help me with this problem? At the moment I have the letters a, b, c. But now I don't know how to get the common equation tangent line. Someone know? Thanks!

The values are a=-3 b=-13 c=1

[merry citrus]

@hushed ore

Yes, I find the derivate of each

[reef fjord]

yes and find the tangent line equation for each, set them equal :)

[merry citrus]

you mean the tangent line of equation line like this y=mx+n of each?

[reef fjord]

ye

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@merry citrus Has your question been resolved?

[merry citrus]

.close

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[woeful depot]

how to proceed to get that?

[woeful depot]

i got [e^(-x^2)/-x^2] wich is not possible and even not define in 0

[gleaming thunder]

it has no elementary antiderivative

the most common, elegant proof (afaik), is to move to 3D to allow for a very convenient polar substitution. That's why pi appears. And the fact we square the integral to go to 3D is why the sqrt appears

[woeful depot]

i thought that : $e^{f(x)} = f'(x)e^{f(x)}$

[thorny flame]

Max.cc

[gleaming thunder]

let f(x) = 0

[woeful depot]

ye false

[gleaming thunder]

exp is its own derivative
Not exp composed with whatever

[woeful depot]

ok thx

[loud oasis]

it's a known result that $\int_{-\infty}^{\infty} e^{-x^2} , \dd x = \sqrt{\pi}$

[gleaming thunder]

I think not in this case

[odd pagoda]

(lets not forget the minus)

[thorny flame]

cloud

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[neon iron]

how do i find min and max for that function

[pastel salmon]

you need to solve the system of equations:

$f'{x}=0\f'{y}=0$

[thorny flame]

Joanna Angel

[pastel salmon]

when you find solutions of this system , there will be oen or remo points, probably, then you find the hessian,and determine the character of those points

that is all

[neon iron]

yeah but

this is what ive done so far

[noble laurel]

that is a seperate question

[neon iron]

oh yeah my bad

i found out

[pastel salmon]

🙂

[neon iron]

i was doing the wrong question

this is the "right" question

[pastel salmon]

so you calcullated two derivatives of second order and you nedd to find mixed ones too

at least oen of them

since they are equal

[neon iron]

ok, 1sec

[pastel salmon]

k

[pastel salmon]

let me check quick

k

and now

look:

hessian is

$\begin{vmatrix}
f''{xx}\left( 0,0 \right) &f''{xy}\left( 0,0 \right) \
f''{yx}\left( 0,0 \right) & f''{yy}\left( 0,0 \right)
\end{vmatrix}$

[thorny flame]

Joanna Angel

[pastel salmon]

calculated it

[neon iron]

now i need to insert the 0 in?

[pastel salmon]

yes

[neon iron]

so

[pastel salmon]

how much you got ?

[pastel salmon]

very well, but it is determinant

so computedd too

its value

[pastel salmon]

yes

so

$\begin{vmatrix}
f''{xx}\left( 0,0 \right) &f''{xy}\left( 0,0 \right) \
f''{yx}\left( 0,0 \right) & f''{yy}\left( 0,0 \right)
\end{vmatrix}=12\\f''_{xx}\left( 0,0 \right)=2>0$

[thorny flame]

Joanna Angel

[pastel salmon]

if hessian is positive 12 > 0

and left top corner of it

is 2 > 0

[neon iron]

minimum

[pastel salmon]

then the f reaches minimum local

yes

if f''xx wuid be negative <0

then maximum

if all hesisan <0

then no extremeum

if = 0 then yo need to sue soem other methods

formthe edifnion

that is all

[neon iron]

yh 0 = its useless

[pastel salmon]

fianlly find: f(0,0)

[neon iron]

1

[pastel salmon]

to find vlaue of this minimum

[neon iron]

e^0 = 1

[pastel salmon]

ok

🙂

very well

[noble laurel]

violeteyes, do you know why we are finding the determinant of the hessian

[neon iron]

depends what you mean by why

[noble laurel]

I mean, why does it tell us what we want to know about our critical points

like what is your statement of the 2nd derivative test theorem

[neon iron]

no

i dont have the deeper understanding of it, no

[pastel salmon]

yes

I doubt they told them about the definiteness of the quadratic form and Sylvester's criterion, which is what it comes down to here 🙂

[neon iron]

i have read about 5 pages about it

[pastel salmon]

what faculty are you at ?

[neon iron]

what?

like year?

[pastel salmon]

what do oyu study

[neon iron]

physics

but this is a math course

[pastel salmon]

i see, yes, well that depends on instructor

on physics they cud tell more, but

whatever

[noble laurel]

Really the determinant of the hessian only works in a 2d case

and it is because we dont actually care about it

we care about the signs of the eigenvalues

in 2D it is the case that we can find them from our determinant

[neon iron]

yh

that is next week

[pastel salmon]

that is only required on math faculy

[neon iron]

this course has nothing to do with the physics course

it is run by mathmaticians

[noble laurel]

Well I'd just be careful then with the 2nd derivative test incase you hve to do it on non 2variable functions

[neon iron]

what do you for for f(x,y,z)

[pastel salmon]

there is also method on determinants

[noble laurel]

if all of the eigenvalues of the hessian are positive, local min, all negative, local max, neither, saddle

finding the signs of the eigenvalues just becomes trickier than just computing the determinant though

[neon iron]

yh i read some quora post on the eigenvalues if they switched from negative to psotive

or if they remained equal

then you could determine something about the min, max or saddle

[pastel salmon]

yes but, now you do not need it

that is all

[topaz sinew]

@neon iron Has your question been resolved?

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[wintry geyser]

If F(x) = x^3 sinx then

A. Local maxima is 0
B. Local minima is 0
C. X=0 no maxima no minima
D. Maxima is 1

[wintry geyser]

I found f'x and f"x and both are 0

[hollow bone]

straight up 0?

show your working out

[loud oasis]

it may help to use the first derivative test

[wintry geyser]

F(x)= x^3 sinx
F'(x)= 3x^2 sinx+ x^3 cosx

At x=0, f'(x)=0

[wintry geyser]

So i can say option 3?

[hollow bone]

you know that there's an extrema at x= 0

for now

but not much more

you need the second derivative or the sign test

[topaz sinew]

@wintry geyser Has your question been resolved?

[wintry geyser]

What is that actually?

I have wrote second derivative

[vernal matrix]

Sign test is this

[wintry geyser]

But this is not happening for x=0 i got f'(x) and f"(x) both 0

@hollow bone @hollow bone

Sry for ping

[wintry geyser]

@vernal matrix

[vernal matrix]

You’re doing the second derivative test - which of course you’re finding inconclusive

[vernal matrix]

Use this one - take a look at the sign of the first derivative at points close to and either side of zero

[wintry geyser]

0.5 and -0.5 like this?

3 (0.5)^2 sin(0.5)+(0.5)^3 cos(0.3)

3 (-0.5)^2 sin(-0.5)+(-0.5)^3 cos(-0.3)

@vernal matrix

So here both values are different

[vernal matrix]

Yea one should be positive, the other negative (this one should be (-0.5)^3 of course)

[wintry geyser]

Yess

So no maxima no minima right

[vernal matrix]

What’s the sign of this one?

[vernal matrix]

And of this one?

[wintry geyser]

,w 3 (0.5)^2 sin(0.5)+(0.5)^3 cos(0.3)

[thorny flame]

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[wintry geyser]

,w 3 (-0.5)^2 sin(-0.5)+(-0.5)^3 cos(-0.3)

[thorny flame]

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[wintry geyser]

See one is positive and one is negative@vernal matrix

[vernal matrix]

So which of these, if any, do you match with?

[wintry geyser]

It doesn't match with both

[wintry geyser]

No match

[vernal matrix]

...-0.5 is to the left, and gives you something negative...

[vernal matrix]

...and 0.5 is to the right, and gives you something positive...

[vernal matrix]

are you sure?

[wintry geyser]

Yess sure

[vernal matrix]

So it's not negative to the left, and positive to the right then...?

[wintry geyser]

Yes

So no maxima no minima

Am I wrong?

But how

[vernal matrix]

see these...

[wintry geyser]

I have seen

What is your conclusion?

[vernal matrix]

well, as per before, you're positive to the right of 0, and negative to the left of 0, which is what the minimum one says, right?

[wintry geyser]

Nope

Left is positive,right is positive

[vernal matrix]

this is negative, and on the left?

[topaz sinew]

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[jagged wigeon]

I am going crazy.
What's the derivative of f(x)=e^(2x)?
f'(x)=e^(2x) or f'(x)=2e^(2x)?

The first one is the result of the definition of the derivative of e^x which is e^x, and so i define t=2x and so the derivative of e^t=e^t -> e^2x.

The second is the result of applying the definition of composite functions. f(x)=e^2x and g(x)=2x -> e^2x * 2.

[vernal matrix]

second answer

[jagged wigeon]

But the first one seems okay to me

[vernal matrix]

For the first one, you found the derivative of e^t with respect to t, not x

[jagged wigeon]

so i cannot 'define' t=2x?

[vernal matrix]

If you do you have to remember that chain rule would apply still

[jagged wigeon]

what do you mean by 'apply again'?

[wary tulip]

@main mauve

[jagged wigeon]

?

[main mauve]

[gusty bane]

@main mauve

[reef fjord]

@main mauve

[wary tulip]

[main mauve]

get back to work

[gusty bane]

you can define this, but what you found is (\dv{}{t}e^t), not (\dv{}{x}e^t)

[thorny flame]

maximo

[woeful drift]

Most serious helpers

[wary tulip]

very serious helper here

[jagged wigeon]

but t depends by x

[gusty bane]

yes

but again

you found d/dt

not d/dx

quantum do you recall the chain rule

[jagged wigeon]

What is it (the translation of the phrase might not be the same in my language)?

[gusty bane]

if (y = y(x)) and (x = x(t)), then (\dv{y}{t} = \dv{y}{x}\dv{x}{t})

[thorny flame]

maximo

[gusty bane]

it is also written as (\dv{}{t}y = y'(x)x'(t))

[thorny flame]

maximo

[jagged wigeon]

Ok, it seems to get more sense.
Also, another issue : what's the derivative of f(x) = 1/(ln(2x+1)) ?
To do that i use the definition of compositive functions :
there are 3 functions there :
h(x) = [g(x)]^a
g(x) = ln(2x+1)
z(x) = 2x+1
Right?
So, to get h'(x) i have to do a*[g(x)]^(a-1)*g'(x) right?
Then multiply it by g'(x) = 1/(2x+1) * lne and for z'(x)=2.
But this is not correct, where did i do wrong?

[jagged wigeon]

Nono, i've not studied it yet i think

[jagged wigeon]

I mean, in the formula there is that g'(x) multiplicated to a*[g(x)]^(a-1) but if i remove that multiplication, then the result of the Ex is correct.

i think that the real formula is just a*[f(x)]^(a-1) and then it multiplies it by f'(x) because f(x) is the inner function of the exponential function D[f(x)]^a but in my Exercise the inner function is already derivated and multiplied (ln(2x+1)), so i don't have to follow the multiplication of that formula.

(What i wrong might be unreadable, if so tell me, i need to understand)

🫠

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[solemn jolt]

@vernal matrix help-26

ok so

for the part a

I don't get how to write the Π product out for x_i or x_k

[vernal matrix]

as in out here?

[solemn jolt]

wdym

[vernal matrix]

Oh misunderstood I think

[solemn jolt]

i remember the other dude saying that i need to write the entire product and I cant just plug in x_i and x_k for t

and show the results

but idk how I can write out the product of g_k(t) = Π when it's all just a bunch of variables

[vernal matrix]

As in you need to write all the terms in the product, as I was saying before

[solemn jolt]

yeah exactly

how can you write the terms when working with variables

[vernal matrix]

You did it like just for one of the terms in the product rather than everything that the product was made out of

[solemn jolt]

oh so that was enough?

just writing one of the terms out

with the Π behind it

[vernal matrix]

Yea - if you have the product sign to indicate you have all the terms in the product

[solemn jolt]

oh ok got

i got confused i though i had to show it like with numbers going upto n

but that wouldn't be possible anyways

[vernal matrix]

Well I mean the most important cases for $g_k(x_j)$ is when $j = k$ and when $j\neq k$

[thorny flame]

@vernal matrix

[vernal matrix]

Well yea but of course you can write the general product anyway haha, as before, argue case by case!

[solemn jolt]

ok got it

and

I don't understand part b

like tell me if I'm wrong

are we looking for a single and only a single result of g_k such that f(x_i)=y_i

is that right?

[vernal matrix]

More that you want some polynomial, and it should be of degree (n-1), such that it passes through all the points (x1, y1), (x2, y2), ..., (xn, yn)

But said polynomial, you also need to show it's unique, that there are no other ones

[solemn jolt]

hmmm

is this polynomial a result of the product g_k?

[vernal matrix]

You can use the g_k's to make that polynomial, and they set it up so that you do

Of course as per before you know that g_k(x_j) is zero when j isn't k, and 1 when it is, so you can use that to create a polynomial that passes through each of those points

[solemn jolt]

hmm

[solemn jolt]

how does that help

and how are y_i's described using x_i's

[vernal matrix]

The idea is that you want the f(x_i) to be y_i, and you set the polynomial based on that

They're just random points, like you could have (1, 2), (2, 5), (3, 6) etc etc

[solemn jolt]

ok

[vernal matrix]

Like they can be whatever, but you want a polynomial such that e.g. f(1) = 2, f(2) = 5, etc etc

[solemn jolt]

so

I need a g_k that takes x_i and plugs it in for t and the product is y_i

[vernal matrix]

More that the g_k, when you put x_i in, it turns into either 0 or 1

[solemn jolt]

yeah

true

[vernal matrix]

The fact that you know that g_k (x_k) is 1, you can make use of that to make something such that when you put x_k in, you get y_k

[solemn jolt]

a function that produces a unique y for every unique x

?

[wintry geyser]

Ahh my question is closed 😭

[solemn jolt]

i dont quite get how that fact can help me here

i cant make the connection

[vernal matrix]

Well polynomial* (there is a difference between them as was pointed out), and you don't need the y you get to be unique for each x

[solemn jolt]

im so lost

[vernal matrix]

For example, you could take some of the y_i's to be the same, if you wanted

[solemn jolt]

so there exists a single polynomial that is the result of g_k. the polynomial's degree is smaller than n-1 and when we plug x_i in this polynomial we get y_i (which is just a random value) out. and to find this polynomial, we need to use what we did in part a

[vernal matrix]

The polynomial you can find in terms of g_k sure - the degree is (n - 1), and yep

[solemn jolt]

ok i can't think of anything to do here

initial guess would be to manipulate g_k so that i get a function with degree =< n- 1

[vernal matrix]

Alright, you have that $g_k(x_k) = 1$, can you think of an polynomial, say $p$ such that $p(x_k) = 2$?

[thorny flame]

@vernal matrix

[solemn jolt]

p = 2x

wait hold up

thats wrong

2g_k ?

that would still be a polynomial right?

[vernal matrix]

Yea, that works - now, what about if I wanted a polynomial p such that p(x_k) = 50?

[solemn jolt]

50g_k i guess

[vernal matrix]

Yep - now, a polynomial such that p(x_k) = y_k?

[solemn jolt]

oooo

y_k * g_k

[vernal matrix]

That's it - but, remember that's only for one pair (xk, yk)

[solemn jolt]

oh shit

we need to find it for x_i, y_i

hmm

can we do the same thing

but instead of multiplying g_k(x_i) we add to it?

since g_k(x_i) is just zero

[vernal matrix]

Yep, I think you see the general idea of what you're to do!

[solemn jolt]

so like y_i + g_k would be the polynomial

[vernal matrix]

You have each $y_k g_k$, but like you know that $y_k g_k(x_k) = y_k$ and $y_k g_k(x_i) = 0$ when $i\neq k$

[thorny flame]

@vernal matrix

[vernal matrix]

Ohh, not like that!

[solemn jolt]

ok

honestly

I think I'm missing some crucial knowledge that I need to know for these type of questions

I think I'm gonna give up on this for now and study the base material again

[vernal matrix]

it's more of a noticing what you can do

[solemn jolt]

It's just the more you explain the more I realize I don't understand it

like I don'T even know how that degree of the polynomial play a role in a function such as y_k*g_k

or how to show that only a single function holds a certain property

[vernal matrix]

for example, with the first one, you're to notice that with the function we found, say if we're working with $k=1$, then $y_1 g_1(x_1) = y_1$, and $y_1 g_1(x_i) = 0$ if $i$ isn't 1, then similar, with $y_2 g_2(x_2) = y_2$ and $y_2 g_2(x_i) = 0$ when $i$ isn't 2, etc etc

[thorny flame]

@vernal matrix

[vernal matrix]

But fair I guess, definitely worth reading stuff up really(!)

Remember that each g_k is a product of linear factors, and how many there are!

[solemn jolt]

can I add you as a friend and ask you the question later?

[vernal matrix]

Sure if you'd like

[solemn jolt]

OK thank you so much for taking the time to explain this to me

I really appreciate it

I'll try to leanr the stuff and I'll be back at some point

[vernal matrix]

No problem, all good definitely go through stuff!

[topaz sinew]

@solemn jolt Has your question been resolved?

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[twin pollen]

Hi

[twin pollen]

Im not really sure what this question is asking

[hallow depot]

.close

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[quiet apex]

i am having trouble answering this question, tried various different methods but still getting the answer wrong:

[quiet apex]

here is my attempt:

here is another attempt that also is wrong:

what am i doing wrong??

[topaz sinew]

@quiet apex Has your question been resolved?

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@quiet apex Has your question been resolved?

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@quiet apex Has your question been resolved?

[pine sentinel]

Isn’t it just 500 times 1.01 ^ 200

P(1 + r/4)^(4y)

500 times (1 + 0.04/4) ^ 200

= 500(1.01)^200

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[trim jasper]

Can someone explain why

mod x < 1 = -1 < x < 1

[drifting swift]

bad typesetting

you are asking why |x| < 1 is equivalent to -1<x<1, yes?

[trim jasper]

yes yes

[drifting swift]

is this more of a "wtf is going on?" or "i think it should be something else" kind of doubt?

[trim jasper]

kind of the first one

like

why

[drifting swift]

ok. do you know how to view modulus as distance?

[trim jasper]

elaborate a bit

[drifting swift]

|x| is the distance from x to 0 on the number line.

[trim jasper]

yes

-4 and 4 etc

yeah i know this

[drifting swift]

so

|x|<1 reads as "x is closer than 1 unit to 0"

[trim jasper]

this is new

go on

[drifting swift]

the distance from x to 0 (|x|) is smaller than (<) 1

[trim jasper]

that solves it tbh

so

[drifting swift]

as it was supposed to.

[trim jasper]

|x| > 1

yeah i got it

thank you

.close

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[midnight sinew]

Having a very hard time proofing this statement. Just the group part first:

I have already show that multiplication modulo n is a binary operation. I have shown associativity of the binary operation. I have also shown that the identity(1) exists but Im struggling to show that there exists inverses in Z_n

[odd pagoda]

do you know bezout

[midnight sinew]

nope very unfamiliar

[odd pagoda]

do you know the euclidean algorithm

[midnight sinew]

yep

[odd pagoda]

do you know the extended euclidean algorithm

[midnight sinew]

nope.

Would i need the extended euclidean algorithm for this? i can try reading up on it

[odd pagoda]

at least currently I dont know how to do it without that

[midnight sinew]

alright thanks alot, will take a look in that direction

.close

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[forest bloom]

Hi guys, could someone help me understand a solution to this question

[gleaming thunder]

Show the solution then

[forest bloom]

The solution is this, but i don't get this part

sorry by the way i was trying to find the solution

[ivory sorrel]

actually, I was thinking the sin rule would be sufficent $sin(z)/2x=sin(y)/x$

[thorny flame]

Why am. I here

[forest bloom]

not allowed to use sine rule but i think i got it with sine rule

sin(z) = 2sin(y)

sin(z) > 2sin(y) cos(y)

sin(z) > sin(2y)

[ivory sorrel]

How did you conclude the second statemenrt, surely you meant y/2?

[forest bloom]

-1 < cos(y) < 1

cos(y) is never 0 for this triangle

[ivory sorrel]

ah yes, OK

[forest bloom]

unless degenerate but we don't consider that as a triangle in my course

[ivory sorrel]

looks right then

[forest bloom]

yeah but i need 😭 help understanding the geometry thing

just the last line

[ivory sorrel]

wait, are they trying to say angle ADB is equal to z?

[forest bloom]

ADB is z yeah

[forest bloom]

that's in the question as well

wait

[ivory sorrel]

D isn't in the question. Is some information missing?

[forest bloom]

they mirrored the triangle across AC

so i guess they must've messed up the letters?

[ivory sorrel]

maybe.

[forest bloom]

anyway

wait no they didn't

ABC is the original triangle

and then they mirrored it

that's why you have D

[ivory sorrel]

if they have mirrored it, you'd first have to prove angle ABD is also Z as then ABM is congruent to ADM

[forest bloom]

and how do i do that?

[ivory sorrel]

I feel the solution is wrong but again my geometry is a but rusty so I'm not too sure

[forest bloom]

😭 idk either

i don't think it's wrong though (but idk)

"by construction"

I just don't understand how they constructed it then

[ivory sorrel]

If ADC is z as they claim, and they've mirrored ABC, the triangle has to be isosceles

[forest bloom]

wait i think i see it

wait ACB and ADC are the same triangles right?

ACB = z then ADC also = z

since they mirrored it then they are congruent triangles by construction

yeah i think that works(?)

yeah

wait

oh it's reflected?

so AC is analogous to the left triangles 2x

idk geometry so i need a sanity check

[ivory sorrel]

They may be congruent by construction, yes, but if ADC is equal to BCA , then it follows that ABC is too, making the og triangle isosceles

which doesn;t seem to be given anywhere

I think I'm digressing , sorry

yeah, the solution makes no sense to me, sorry

[forest bloom]

[ivory sorrel]

ping other helpers imo

[forest bloom]

okay i'll try

[forest bloom]

<@&286206848099549185> , could someone explain this last line of the solution

[neon iron]

last line means angle ADC is equal in measure to the angle ACB and the equality is a result of the construction of the figure

[forest bloom]

yeah lol i mean it doesn't work or why does it work

is the problem

[neon iron]

im trying to figure that out wait

[topaz sinew]

@forest bloom Has your question been resolved?

[topaz sinew]

@forest bloom Has your question been resolved?

[neon iron]

what is"<" used to denote?

[neon iron]

i see 2BC denoted as 2x , then it says BD < 2x

what is this

and then 2x is on AB

[forest bloom]

i forgot this question existed lol

[pastel oracle]

looks like less than

[forest bloom]

I mean BM < BC

[neon iron]

ok so it is less

but the solution itself looks wrong

[forest bloom]

so BD < 2x

[neon iron]

but 2x is measure as AB

look on the drawing

haha

[forest bloom]

BC is x and CD is also x by construction

BM < x and MD < x

BM + MD = BD < 2x

BD < 2x

[neon iron]

so BD < AB

[forest bloom]

yes i guess lol

the 2x is made by angle D which is apparently z (?)

and BD is made by 2y

so 2y < z

[neon iron]

are you sure the solution is correct?

i really doubt it

[forest bloom]

No clue

it works out though

maybe

[neon iron]

it does?

[forest bloom]

but the issue is in the construction

which was the whole qualm 😔

not sure if D is even z

if it is then it looks like it works

[neon iron]

still doesnt make sense to me

[forest bloom]

ABC is the original triangle

it's reflected across AC

so you get that same triangle reflected but now it's ADC

(that's what i understand)

looks like it's translated not reflected lol

[neon iron]

if the triangle is translated then it will be same length

yeah this makes sense now

since ABC is translated to ADC , the lengths will be the same

[forest bloom]

the ABC is translated so that AB of the new triangle is AC

[neon iron]

yes

[forest bloom]

so <D is z just like <C is z

[neon iron]

and AC is AD

yes

[forest bloom]

yeah

but @ivory sorrel (sorry for ping) said all this doesn't work

[neon iron]

why doesnt it work?

[forest bloom]

idk i just mentioned what i understood

makes sense to me a bit now

idk geometry so i was just trying to make sense of stuffs abstracting the geometry lol

[wary tulip]

snow

[neon iron]

?

[forest bloom]

[ivory sorrel]

what did I say again? I've lost track.

[forest bloom]

Lol

Idk what you said 😭

[wary tulip]

it is just that @pseudo jetty has been a significant part of my mathcord experience so it is funny to see someone else named snow

i have nothing helpful to add as usual

[forest bloom]

is that snow a better version?

[wary tulip]

which snow do you mean

[forest bloom]

your snow

lol

[wary tulip]

i do not know snow yet i cannot say

[forest bloom]

no i mean the snow you know

[neon iron]

[wary tulip]

i do not know snow so i cannot say whether snow or snow is better

[forest bloom]

Oh yeah my brain doesn't function after this geometry mess

okay so i think i figured that this thing works

[ivory sorrel]

I was saying if what the solution claims is true, woudln't ABC be isocles ? (ABC is congruent to ADC by construction and ADC is apparently z?, so it must be isoceles which isn't given, I don't think I said anything more than that

[neon iron]

yes you are right

but i dont see any problem in ABC being isocles

[forest bloom]

but it can be anything though right

why should it be 2x

anything from triangle inequality

x < AC < 3x

[vale furnace]

Looks wrong I dont like this solution.

[ruby tree]

Angle ADC is definitely not congruent to angle ACB

[vale furnace]

I don't like geometry.

[forest bloom]

same

[ivory sorrel]

yeah OPs solution using sine rule was much better

[forest bloom]

Oh

[neon iron]

yeah solution is wrong

[forest bloom]

my teacher is an elitist

in the regressive way

if i use trig then i get 0 lol

[neon iron]

lol

[vale furnace]

[forest bloom]

if i use calculus i get 0

if i use number theory i get 0

so i have no option :V

[wary tulip]

[neon iron]

use words then

[forest bloom]

Anyway, thank you guys!!

I'm guessing that stuff is just all wrong lol

buh bye

[ruby tree]

I just realized this is blatantly wrong too

Maybe just a typo, so like BD < 2x = AB

@forest bloom I have a geometric proof if you want

[forest bloom]

ohh

[forest bloom]

can i ask you later? I don't really want to do geometry now... whenever you're free :V

[ruby tree]

No

[forest bloom]

i sacrifice the geometric proof lol

Thanks anyways~

[ruby tree]

You sure? It's pretty easy

[forest bloom]

right i guess i'll just consider all of this wrong in general

[forest bloom]

easier than this?

(if it was correct)

[ruby tree]

Yeah if you know a bit of basic geometry

[forest bloom]

Okie let's goo

Idk geometry but i'll try~

lol

[ruby tree]

So the small circle is here just for construction, so AB = 2BC

The big circle is the circumscribed one

So the center O lies on the perpendicular bisector of AB (with midpoint P)

So far so good?

[forest bloom]

wait is this supposed to follow from some circle theorem?

I see they are from the same arc

[ruby tree]

Yeah

[neon iron]

yes

[forest bloom]

so inscribed angle thing?

oh

[ruby tree]

2BAC = BOC

[forest bloom]

yeah i get the angle relationship

but wait where did you get the side relationship

[ruby tree]

Wow hold on I don't see some messages anymore

Ok it's good now, just restarted Discord

What side relationship?

[forest bloom]

AB = 2BC

[ruby tree]

That's given

[forest bloom]

oh okay by construction

sure

[ruby tree]

Ok let's call the angle BOC 'w'

So 2y = w

[forest bloom]

right yeah

[ruby tree]

Now you can see the triangle OBC is isosceles

OB = OC

[forest bloom]

mhm two radii's

[ruby tree]

And OB > BC because O is outside the small circle

(because it lies on the line, which is tangent to the small circle)

[forest bloom]

oh yeah i can see that

was that the intent of the small circle?

[ruby tree]

Kinda

[forest bloom]

Okay okay i get it so far

[ruby tree]

Hmm did I make a mistake

[forest bloom]

[ruby tree]

No I'm just missing a segment in my screenshot

Ok now let's call the angle AOB 'u'

[forest bloom]

what's the point of this segment

[forest bloom]

mhm yeah

[ruby tree]

So we have 2z = u just like 2y = w

Hold on I'm still missing a segment, I think it was slightly more complicated than I wanted but it's still pretty simple

Let me label the angles too

[vale furnace]

Ofc, you draw a circle

Like every other geometry problems

I hate geometry

[ruby tree]

There you go

Let's not use w and u actually

So we have BOA = 2BCA because same chord, and 2BOP = BOA because bisector

So BOP = BCA = z

Makes sense?

[forest bloom]

yes kinda

[ruby tree]

This is the final part

[forest bloom]

oh yeah i didn't see the bisector part

sure makes sense

[ruby tree]

If you look at z and 2y around O, you can see that z > 2y because BH > BC

[forest bloom]

lol is it dumb to ask why do you know BH > BC?

oh i see

[ruby tree]

H is outside the small circle

[forest bloom]

one is outside the circle and one is inside the small circle

okay yeah

[ruby tree]

C is on it

[forest bloom]

thank you!!!

[ruby tree]

H approaches P when A, B, C become colinear

[forest bloom]

I just want to say that i would've definitely thought of all this if i had the question to do on my own

[ruby tree]

but then it's no longer a triangle

Maybe there is a simpler proof but I kinda like this one tbh

[forest bloom]

well for people like me:

sin(z) = 2sin(y)
sin(z) > 2sin(y) cos(y)

sin(z) > sin(2y) is the only thing i understand lol

[forest bloom]

btw what is the point of this?

[ruby tree]

Yeah but that requires knowing the rule of sines

[ruby tree]

Just to say that BH > BC, not just BH >= BC

H is always on the tangent line, but it could coincide with P

In which case it would give you 2y = z

But that limit case makes it 2y = z = 0

[forest bloom]

yeah i think i get it

[forest bloom]

how are you making it collinear btw

like which points are you pulling or something

to make it colinear

[ruby tree]

C

O is at infinity

[forest bloom]

i kinda see 😭 wow okay

okay okay thanks nel!!

I understand it

[ruby tree]

O can also be on the other side btw

[forest bloom]

[ruby tree]

But the proof still works because even though the z's aren't the same anymore, BCA is clearly larger than BOP

[forest bloom]

yeah fair enough

btw would you still make this construction on paper

if u had to do this question?

[ruby tree]

If I only had paper, probably yeah, but it would be messy

[forest bloom]

okay okay fair enough tyy

i think that's all from me for now lol

[ruby tree]

[vale furnace]

.close

[topaz sinew]

Channel closed

Closed by @vale furnace

Use .reopen if this was a mistake.

[vale furnace]

[forest bloom]

oh yeah i forgot to close whoops

sorry

[vale furnace]

Saul Goodman

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
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• Be polite and have a nice day!

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

[neon iron]

the second part confuses me

[neon iron]

Here is the answer to a), but I've no clue on what to do with b

[drifting swift]

well what's the minimum value of |a+b| according to part (a)?

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

2?

[drifting swift]

yes

so what should we take for b so that |a+b| = 2?

think geometrically

[neon iron]

didn't see you respond mb

[neon iron]

but arent there many values, like isn't it sqrt((12+x)^2+(y-5)^2)? =2?

I'm probably making this more confusing

[drifting swift]

yes you are

what's the length of a?

[neon iron]

13

grr bad wifi

[neon iron]

13 right?

[drifting swift]

yes

and the length of b is fixed at 15

how should we orient b so that |a+b| ends up as 2

[neon iron]

part of me wants to say -15 but howwwwww

isnt there a seperate x and y value

[drifting swift]

i don't want to hear about x and y values

just tell me how to orient b relative to a

[neon iron]

|13-15| =2 is this what you mean

[neon iron]

I'm missing something grr