help-38

[violet gust]

use this: 2z+KLC=180

[proper saffron]

do i need to solve it or use it somewhere else to figure out another angle

[violet gust]

can you find 180-KLC from that?

[proper saffron]

2z

so exterior angle is 2z

[violet gust]

YES

Now, the next is the most critical step for this problem

Join the points K and B

[proper saffron]

sure

k done

[violet gust]

now what can you say about KLB and KAB triangles

[proper saffron]

they are the same right?

[violet gust]

proof?

[proper saffron]

uh

each of them has one y side

and 1 x side

im not sure if thats how you do it

oh wait nvm

because angle BLK is 2z

and thats not the same

[violet gust]

Do you know congruence of triangles?

[proper saffron]

no

[violet gust]

Like can you name the tests for congruency?

[proper saffron]

not entirely

isnt it like when there is 2 triangle

and they are proportional

in angle

[violet gust]

proportional is similarity

[proper saffron]

oh

[violet gust]

exact match is congruence

congruence is stricter than similarity

[proper saffron]

yeah idk congruence tests

[violet gust]

Congruence is like 8th grade maths. you sure about that you dont know it?

[proper saffron]

no

im just now in grade 8

[violet gust]

Ohh

without congruency, its gonna be tough

[proper saffron]

how do you find out if something is congruent

[violet gust]

theres test for congruence

like if all three sides are same, then all the properties of the triangles are the same

[proper saffron]

oh

[violet gust]

or if two sides and the angle between the two side is same, then also all the properties are same

[proper saffron]

thats what i said before right?

like they are congruent

[violet gust]

Yea, somewhat. Bu you only talked ablut 2 sides. So thats not sufficient

[proper saffron]

i talked about 3

[violet gust]

But yea

[proper saffron]

oh 2

right

nvm

[violet gust]

thats the point. The twi triangles are congruent

[proper saffron]

wait

[proper saffron]

only from 2 sides?

[violet gust]

coz the 3rd side is the common side so its also the same

[proper saffron]

ohh

yeah

so that means

CAB=BLK

[violet gust]

That means the angles KAB and angle KLB are equal

[proper saffron]

ok so 2z=a

and then

z=1/2

a/2

[violet gust]

mhm

[proper saffron]

a+a+a/2

=180

[violet gust]

yep

[proper saffron]

5/2a

=180

5a=360

a=72

or so

and since top angle is 1/2a

ITS 32

36

YES

I GOTIT

tysm

[violet gust]

[proper saffron]

i js realized its been 1:30 hours

crazy

[violet gust]

yea lol

[proper saffron]

anyways tysm i learned sm new

gotta practise more

[violet gust]

npnp

good luck

[proper saffron]

ty

i close this now right?

[violet gust]

yea

[proper saffron]

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[high wraith]

Can someone tell me why the bottom one is right and the top one is wrong. And why on other questions both methods work?

[high wraith]

,rotate

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[crimson pebble]

$x(x+h)\neq x^2+h$

[solid kiln]

CST (reply ping for help)

[vocal epoch]

oh yea

[vocal epoch]

am i allowed to factor out h from that top fraction in the 2nd last step the way i did

[vernal palm]

ussualy yes

[crimson pebble]

Oh please put lim_(h->0) everywhere

[vocal epoch]

yea i did that on my paper

i just didnt write it on that screenshot thats my bad

[vernal palm]

but this time you removed the h from x^2+h

[vocal epoch]

ive got this rn

can i factor out h like this to cancel out the h in the denom

[vernal palm]

yes but you need to still multiply x by h at x^2+x it should be x^2+x*h

[vocal epoch]

so it should be like this?

[vernal palm]

yes

[vocal epoch]

i got this

is there anything else i gotta do or is that it

[vernal palm]

now you just need to "take"/"solve"/"complete" the limit

[vocal epoch]

what

[grave monolith]

When h = 0 ,xh = 0 leaving -2/x^2

[vernal palm]

remove the h

[vocal epoch]

isnt that what i had before anyway

[vernal palm]

!noans

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[vocal epoch]

-2/x^2

[vocal epoch]

in this

[vernal palm]

yeah you can get the right answer for the wrong reasons

[vocal epoch]

fair enough

does it matter if the derivative function i get is a rational function like this

questions that ive done for this stuff before usually give a linear function

[topaz heart]

No

You can do -2/x^2 or -2x^(-2)

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[tight mantle]

how would i go about finding y, i believe i found x correctly, i used tan inverse and got 29.74 degrees

[burnt mulch]

Angle sum of a quadrilateral = 360 deg

[tight mantle]

so would i minus x from 360 to find y?

[burnt mulch]

What about the two right angles?

[tight mantle]

and those

[burnt mulch]

then yeah

[tight mantle]

thank you

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[thin vine]

could someone help explain why the answer is d

im going to upload my work in a sec

ok here it is

so i managed to factor it into this form

i just set x=log2x

sorry if thats confusing

[austere cedar]

Make sure to keep a table of log identities nearby as you work

[zinc ginkgo]

,tex .log rules

[solid kiln]

riemann

[thin vine]

uh

so should i do like

x-log2(8)

and like x-log2(2)

and then divide

nvm

[austere cedar]

I'm not sure why they use f(x) when they just mean "log2", maybe to trip you up? It doesn't really do much

You just want to simplify. You'll get one of the answers

[thin vine]

ohh ok im gonna do that real quick

uh i think i did something wrong

[thin vine]

I did x^2/4x to get log2(x/4) and then multiplied by 8 but that gave me log2(2x)

[austere cedar]

log2(2x)
= log2(2) + log2(x)
= 1 + log2(x)

[thin vine]

oh bruh im so dumb

thank you for helping me

[austere cedar]

You're good haha, almost there

[thin vine]

i appreciate it

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[urban copper]

gauss plane and argand plane refer to the same thing?

[urban copper]

I know it as Argand plane, the plane that has like x = Re(z) and y = Im(z)

x-axis and y-axis respectively I mean

[dapper swift]

yes

Wikipedia says that Gauss plane is another name for the Argand plane

but I've never heard of it being called the Gauss plane

the most common term in English is just the complex plane

[urban copper]

exactly my thoughts until I saw a maessage in #math-pedagogy

[dapper swift]

translated from their own language, lol

[urban copper]

I appreciate the help, no need to create more work for u guys, I can actually google this, sorry for this

[urban copper]

thanks

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[clear ember]

I’m stuck at this question:
“Imagine you throw a bouncing ball at a train. You throw the ball at a speed of 20 m/s directly towards the train, which is moving at a speed of 180 km/h towards you. What speed does the ball have if we assume the bounce is perfectly elastic?”

I checked and the solution in the answer key is:
“v_train = 180 km/h = 50 m/s
The direction of the train's movement is the positive direction.
Towards the train, we imagine a thrown ball with
a velocity of v_ball before = -20 m/s (negative direction).
We are looking for the ball's velocity after the bounce towards the train and call it v_ball after. The train's velocity after the bounce is still
50 m/s. The train's velocity is hardly affected by a small ball hitting it.
The bounce is perfectly elastic. The following equation holds:
v_ball after - v_train = v_train - v_ball before
v_ball after - 50 = 50 - (-20)
v_ball after = 120 m/s (432 km/h)
Answer: 120 m/s"

I don’t really understand how the solution arrived at that answer. I tried using the formula: mAvA1 + mBvB1 = mAvA2 + mBvB2. I thought that the mass in this case should be irrelevant, so I wrote it as:
-20 + 50 = x + 50
-20 = x
Answer: The ball's speed is -20 m/s. (Incorrect answer)

How should I have solved the problem instead? Am I using the wrong formula, and if so, which formula should I have used?

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[clear ember]

<@&286206848099549185>

[supple copper]

,w 180kmph to m/s

[solid kiln]

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[supple copper]

So looks like you’re using this

m_A = train >> m_B = ball

Hmm it’s strange

[clear ember]

Yea I think so?

[wraith hinge]

wouldnt the velocity of the ball be -70m/s?

[supple copper]

Ok if you’re on the train

The ball comes at you at 20 + 50 m/s

This means from the perspective of the train it should bounce and go forwards (away from the train) at 70 m/s

[wraith hinge]

my english is not englishing, where is the ball thrown from?

[supple copper]

But you’re on the train that’s also moving at 70m/s so from a stationary observer the ball would be moving 70m/s + 50m/s of the train’s frame of reference

So 120m/s in total

[wraith hinge]

are you throwing the ball at the train while you being inside the train or outside the train?

my english level is very poor

[clear ember]

As I understood the text I am trowing the ball on the train from the outside but yea it's a bit unclear so I might be on the train?

[supple copper]

No the ball is thrown from stationary towards the train

[clear ember]

So I'm on the train then? Sorry I am quite lost lol

[supple copper]

No you aren’t

You’re on the tracks

[clear ember]

Alright!

[clear ember]

If I'm supposed to use this formula then why do they switch to - instead of using + in the answer?

[supple copper]

This I don’t think is right

This assumes that the bounce will take kinetic energy away from the train

[clear ember]

Alright so what formula am I supposed to use?

[supple copper]

What they gave is fine

I think this question is more about frames of references

[clear ember]

Alright thanks!

[clear ember]

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[inland bear]

Can someone see for me if my working for part B is correct?

[west sleet]

No the tangent is a straight line

Sub x=4 into the derivative function to find the slope of the tangent

Before using point line formula

[dapper swift]

you mean x = 4

[west sleet]

Oh yea

[dapper swift]

yeah it's $y - 2 = f'(4) (x - 4)$

[solid kiln]

south

[inland bear]

Oh okay thank you

[dapper swift]

no worries!

[inland bear]

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[true canyon]

i'm not sure how to deal with my a.b term here

[true canyon]

i cropped it out but p = λa + (1-λ)b (or so i think)

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[true canyon]

nevermind sorted it

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[glad patio]

ive done a b c

for d do they want me to use 3-13x to find 2.9998^6?

im kind of confused

also for c does my value of x seem to be a suitable value

[split chasm]

where's -13x coming from

ah

[glad patio]

i dont know which expansion they want me to use from

i assumed the original one from a

[split chasm]

you shouldn't have done those last two lines in a)
if you want to factor out a value, indicate it as such instead of just dividing

you x value is fine

[glad patio]

oh i only did it because in the question it asked for the simplest form

[split chasm]

the point still stands

729 isn't the same as 3 etc

whether the factored form or standard form is considered simpler is a different matter
in this case, they probably want expanded

[glad patio]

so for a) should i have just left it like that without dividing it?

[split chasm]

yes
but

if you want to factor out a value, indicate it as such instead of just dividing

[glad patio]

wdym indicate it as such

[split chasm]

instead of going from
pq + pr
to
q + r

you should instead have p(q+r)

[glad patio]

ah

[split chasm]

also 36/3 is 12, not 13

[glad patio]

oh yes

so like this?

[split chasm]

yes

[glad patio]

okay thank you

but for d

do i use 729-2916x?

[split chasm]

yes

[glad patio]

so 729-2916x=2.9998^6?

[split chasm]

not = and not quite

729 - 2916x when x=0.0001 (the value you found in c)
will give an approximation of 2.9998^6

[glad patio]

i got 728.7084

[split chasm]

,w 729 - 2916x when x=0.0001

[solid kiln]

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[glad patio]

that should be right then cause they asked for 4 dp

[split chasm]

grr, wolfram cut off that last digit

but yeh

[glad patio]

alright thank you for the help

[split chasm]

,calc 2.9998^6

[solid kiln]

Result:

728.70844859568

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[limber heath]

Can someone help me with this?

[limber heath]

I understood the part where sad face means subset relation and smiley face means set equality

I just cant understand the idea of x and y

like if x and y are in sigma then x and y are elements ? or are they some powerset or subset of sigma?

[silver glade]

if x and y are in sigma, they are sets of the form
{1,2,...n}

[limber heath]

Also in discrete math I just dont understand the idea of writing a proof in general...
like I can explain it in words... but how do i make it into math expressions

[limber heath]

yea but if the form is the same, doesnt that means that x and y are the same ? or is n different for both of them ?

[silver glade]

n is different

potentially

[limber heath]

so Sigma = { {1}, {1,2}, {1,2,3}, ... , {1,2,3,...,n}} ?

[silver glade]

sigma = { {1}, {1,2}, {1,2,3}, ... }

it has infinite cardinality

i mean it doesn't have a final element

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[limber heath]

so now I known this
and I understand that x and y are gonna be unique in the sense of the number of elements and what they are
and I also understand that if x and y are unique one will have more elements than the other
and as they have repeating elements, one has to be a subset of the other

BUT HOW DO I Write it as a proof

@twilit nacelle

[silver glade]

x and y don't have to be unique

i'm not sure exactly what was covered in your class, but i would try to say something like this, for part (a):

$x = {1,2,\ldots m}$ and $y = {1,2,\ldots n}$ for some $m,n \in\mathbb{Z}^{+}$. Either $m>n$, $m=n$, or $m<n$. In the first case, for all $y'\in y$ we have $1\leq y'\leq n < m$ so $y'\in x$. This means that $y$🙁$x$.

[solid kiln]

Axe

[silver glade]

and so on, for the other two cases (m=n and m<n)

[limber heath]

oh okay

wait so proofs can be written like this??
we have been taught that it has to have some amount of propositions or some kind of infereicng or something to show. you cant just say stuff

[silver glade]

could you give an example?

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[solemn phoenix]

Die Gleichung (\mid x-2\mid =5) besitzt mit (x \in \mathbb{R}) folgende Lösungsmenge (L:)

[solid kiln]

Subniif

[solemn phoenix]

this question is hard but what do these lines mean

[clear cloud]

You have two possibility for abs value

if x > 0 , |x| = x

If x < 0, |x| = -x

[real rose]

Absolute value. Takes the positive output of any real number.

i.e., turns something like -3 into 3.

[clear cloud]

,w graph abs(x)

[solid kiln]

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[solemn phoenix]

im insanely confused

[real rose]

|-3|=3.

[solemn phoenix]

but why

[real rose]

It is basically the distance a number is from 0.

Can distance be negative?

[solemn phoenix]

ahhhh

[solemn phoenix]

ye right

in physics they can

[real rose]

Ok, in numerical analysis.

Not in physics.

[solemn phoenix]

no

[real rose]

Yeah.

So, if |x| is defined as the distance of x from 0, wouldn't the output always be positive?

[solemn phoenix]

ye

[real rose]

Do you see why I said what I said earlier?

[solemn phoenix]

yes

[solemn phoenix]

so the answer of this should be 7

[real rose]

For the positive branch.

[solemn phoenix]

just 7 right

wait im a touch confused now

can you walk me through the question just so i can understand the thinkng behind it

[real rose]

Ok.

[real rose]

Try to understand this.

[solemn phoenix]

ok

so if x between those lines (im sorry idk how to put that w my keyboard lol) only has a positive answer then x stays unchanged

[real rose]

Yes.

[solemn phoenix]

and if x between those lines is negative then x inverts and becomes positive

oh i found the lines on my keyboard ||

[real rose]

Yep.

[solemn phoenix]

is that all?

anything im supposed to know about it

[real rose]

Yes

[solemn phoenix]

ok thank you

idk how to close the chat

[zinc ginkgo]

.close

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[spice fulcrum]

I think it wants me to do it with vectors/linear algebra if possible (calculators allowed)

I solved it through equating magnitudes AB=AC=BC, then adding in another equation relating the angle ABC using dot product

But it took me a whole page. In the exam its only 2 marks and theres only 7 lines, so i'm asking for ideas on probably the most efficient method

[quick tusk]

Distance formula?

All sides of the triangle are the same length, you can get this length by calculating AB, then equate the distance BC and AC into this value.

There isn't any reason to add the third equation with the dot product, AC = AB and BC = AB are two independent equations.

[spice fulcrum]

When it wants me to verify, can I just substitute it in or do I have to show it by solving it myself?

[quick tusk]

idk how your teacher marks it but it should be sufficient to sub in

[spice fulcrum]

Ahh okay, tysm

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[wraith hinge]

yk slopes and stuff..?
(if yes use tan theta = |m2-m1/(1+m1m2)|)

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[neat scarab]

this is super open-ended, but is anyone here familiar with actuarial math/financial math/working with data?

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[outer mirage]

Hello, how do you solve for question 2c? Is my method correct? Because the Ans sheet said is ‘5’

[ivory dawn]

Expand and solce

,wolf expand 3(7-4x)=1-4(x+5)

[solid kiln]

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[ivory dawn]

Move all x terms to one side

21=8x-19

8x=40

x=40/8

x=5

[split chasm]

you messed up in the first step

by not following the order of operations

[outer mirage]

So, when you do the brackets 1-4(x +5) you don’t get -3?

[split chasm]

the multiplication takes priority over the subtraction

[ivory dawn]

do u mean the wolf thing

bc i aint do no multiplying

just rearanging

[split chasm]

i mean what op did in their work

[ivory dawn]

ohh

[outer mirage]

Oh, so there isn’t -3(x+5)?

[split chasm]

no

[outer mirage]

Ok

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[stone briar]

every point on a plane can be written as linear combination of the other points on the plane? regardless of the position vector?

[stone briar]

say i used b as my position vector

i can choose any linear combination of a,b,c in my direction vector?

[molten comet]

wdym lc of a, b, c?

You can add lc of vector BC and BA

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[young reef]

how this step change

[split chasm]

multiply by $\frac{\sqrt{2}}{\sqrt{2}}$

[solid kiln]

ℝαμOmeganato5

[young reef]

Ohhh

Must be multiply?if need to simply

[split chasm]

at your level, its what's preferred

but as you get higher, noone really cares about roots of integers in denominators

[young reef]

Ok!

Thank youuu

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[fallen oyster]

Find the area of inscribed circle in pi unit sq

[exotic pine]

hint:

[fallen oyster]

(r×15/2) + (r ×14/2) + (r×13/2) = √21(8)(7)(6)

[exotic pine]

continue it

[fallen oyster]

15r/2 + 14r/2 + 13r/2 = 84 sq units

21r = 84

r = 4

πr^2 =

16π sq units

[exotic pine]

good job (it should be right)

[fallen oyster]

how to close it?

[exotic pine]

write ".close"

[fallen oyster]

.close

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[exotic pine]

fyi with the method you can derive the formula for the inradius of the triangle, which would be r=A/s (A=area, s=semiperimiter)

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[slender tartan]

Hi I'm trying to understand what being a 'normal number' actually means --> from what I can gather it means that any digit in the decimal expansion in that particular base is equally likely to occur, but does that guarantee that some particular combination of digits is going to occur?

[slender tartan]

I'm a little confused by the wikipedia page

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@slender tartan Has your question been resolved?

[shrewd ridge]

yes

any digit is called "rich"

any combination

or idk

wikipedia is very unsure what it means

@slender tartanit's probably the same thing

if a digit is random, independently of other digits, then inevitably every sequence is equally likely to occur

yeah, i'm confused too nevermind

[slender tartan]

me too 💀

[shrewd ridge]

@slender tartanit's certainly true though

ok, i misunderstood the question

yes, anything you could think of is guaranteed to occur

there's a high res jpeg of eiffel tower in a normal number

and that's weaker than normal

if you can find infinite amount of anything you want

that's only rich

[slender tartan]

wdym by rich

[shrewd ridge]

it's from wikipedia

normal number page, tells you about rich numbers, where anything occurs at least once

[slender tartan]

ok

i mean im kind of going down a rabbit hole lol

[shrewd ridge]

but that means it occurs at least any amount of times

[slender tartan]

i was initially trying to prove there are 2 consecutive zeroes in the base 10 decimal expansion of sqrt(3)

[shrewd ridge]

i don;t know how to prove it

[slender tartan]

but i just realised sqrt(3) hasn't been proved normal yet afaik

but thanks

this is kinda interesting

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[covert spindle]

Use maclaurin series for $f(x)=\frac{x}{\sqrt{4+x^2}}$

[solid kiln]

BrettASUS

[covert spindle]

so i thought

$x*(4+x^2)^{-1/2}$ ?

[solid kiln]

BrettASUS

[covert spindle]

$g(t)=(4+x^2)^{-1/2} <=> \ln{g(t)}=-1/2\ln{(4+x^2)}$ ?

[elder linden]

what's exactly the question
you need to expand the function using the Maclaren series?

[covert spindle]

yes

[elder linden]

ohk

[solid kiln]

BrettASUS

[elder linden]

i think you can do it separately
1st expand the $\frac{1}{\sqrt(x²+4)}$

😕😕.

[covert spindle]

\sqrt

and not x² but x^2

[solid kiln]

T&C

[elder linden]

okayy sorry for the inconvenience

[covert spindle]

no worries

$\frac{1}{\sqrt{x^2+4}}$

[solid kiln]

BrettASUS

[elder linden]

yeah
id find the 1st few derivatives

[covert spindle]

of this?

$-\frac{x}{(4+x^2)^{3/2}}$ ?

[elder linden]

you know the expansion formula no?

[red mountain]

keep it as ^(3/2) PLS

[elder linden]

yeahh

[covert spindle]

OOPS

[solid kiln]

BrettASUS

[covert spindle]

okay

[red mountain]

ty ❤️

[covert spindle]

you welcome ❤️

this does not look fun ;-;

[limpid dawn]

sheep in wolfs clothes

[covert spindle]

really?

hmmm

[limpid dawn]

maclaurin series centers around c = 0

[covert spindle]

yup

[limpid dawn]

so some derivatives will be 0

[elder linden]

yess
now put x=0

in each of the derivatives

[covert spindle]

that will give 0?

$-\frac{0}{(4+0^2)^{3/2}}$

[elder linden]

ohh yeah shit

[solid kiln]

BrettASUS

[limpid dawn]

you are literally multiplying 0

with something nonzero

[covert spindle]

so $-\frac{1}{4^{3/2}}$ ?

[solid kiln]

BrettASUS

[covert spindle]

i dont understand ;-;

stupid taylor series

[red mountain]

ong fr no cap

[red mountain]

no.

$\left.\begin{aligned}\frac{-x}{(4+x^2)^\frac32}\x=0\end{aligned}\right}\implies\frac{0}{(4+0)^\frac32}$

[covert spindle]

so it is zero?

[solid kiln]

Bonk

[red mountain]

yes

[covert spindle]

but I said that

[red mountain]

g'(0)=0

[elder linden]

omg okay i got it

you need to transform the denominator in the (1+u²) form

[covert spindle]

not arctan ;-;

hmmm

i mightk now how to do that

wait please

[elder linden]

so take out the 4
you'll get
$2\sqrt(1+\frac{x^2}{4})$

[limpid dawn]

what is it with the sqrt(()

[covert spindle]

$\frac{1}{\sqrt{(4+x^2)} = \frac{1}{2} * \frac{1}{\sqrt{(1+\frac{x^2}{4}}}$

[red mountain]

use \sqrt{}

[elder linden]

not quite but you get it

[solid kiln]

T&C

BrettASUS
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[covert spindle]

okay i give up

but i understand

I am lowkey hurting latex at this point latex on IC right now

okay what now?

[solid kiln]

$\frac{1}{\sqrt{(4+x^2)}} = \frac{1}{2} \cdot \frac{1}{\sqrt{(1+\frac{x^2}{4})}}$

[red mountain]

stop trying to do latex 😭

just write it out

[limpid dawn]

in latex

texit

[elder linden]

expand ( 1+x²/4) ^(-1/2) with binomial expansion

[covert spindle]

aaah okay

was it also possible with ln?

[limpid dawn]

what?

[red mountain]

you can prob so some chain rule thingy, yeah

[stable crescent]

whats the question

[red mountain]

check the pins

[covert spindle]

Use maclaurin series for $f(x)=\frac{x}{\sqrt{4+x^2}}$

[solid kiln]

BrettASUS

[covert spindle]

gothendrieck casually spawns

[stable crescent]

this ur alt account?

[limpid dawn]

sus account

[covert spindle]

yes

[stable crescent]

what u tried?

[covert spindle]

something that looked really ugly

wait ill show

[elder linden]

the maclaurin is related to the binomial expansion of (1+x²)

[covert spindle]

this is what I thought honestly

but then I came here and we did something else

[elder linden]

if you can expand the denominator with binomial formula you'll get through 90 of the question

[stable crescent]

f(x) = x/2 *1/sqrt(1+u) ; u = x²/4 apply mcl to the u-expression

[covert spindle]

so this is what T&C said

[elder linden]

yess

[limpid dawn]

bro

[covert spindle]

ye that is what T&C said

[limpid dawn]

-1/2 is not an integer

[stable crescent]

ye go into that

[limpid dawn]

use your mind

and it's negative

[stable crescent]

no problem continue wit it

[covert spindle]

why does it have to be an integer?

[limpid dawn]

cause you are not supposed to work with some gamma function

[covert spindle]

what do I do then ;-;

[limpid dawn]

look up the binomial coefficient

[covert spindle]

I did you have two

one with k element R

and one with k >0

but one is for a finite series and one is infinite

[stable crescent]

u look up the generalized binomial coeff

[elder linden]

yess

[covert spindle]

we got told to use the k element R one

so my answer is correct?

more like T&C's

[covert spindle]

I did right?

[stable crescent]

idk didnt check, but follow what T&C said

[stable crescent]

u gotta expand that even more

[elder linden]

$\binom{-\frac{1}{2}}{k} = (-1)^k \frac{(2k-1)!!}{2^k k!}$

for n=-1/2

[covert spindle]

i need to expand the (-1/2 n) i suppose?

alright and add the 1/2 to the sum

okay wait please imma try

[solid kiln]

T&C

[covert spindle]

how did you figure that out?

[elder linden]

um the terms fall into a pattern

[covert spindle]

so uhm

it will be like

$\frac{\frac{(-1)(-3)(-5)....(2n+1)}{2^n}}{n!}}$

[solid kiln]

BrettASUS
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[covert spindle]

idk where the error is anyways

how can I like simplify the top part

(-1)(-3)(-5)?

i suppose that is the (-1)^n (2n-1)!!?

[red mountain]

dont latex it

just write it down 🙂 (on paper)

[elder linden]

yup
now this n was inside the summation
put n=0,1 and so on

[elder linden]

the last term should be (1-2n) btw

i think

[covert spindle]

-1, 3/2, -15/48

so it is alternating

[elder linden]

yupp doesn't matter
you just figured out the coefficients for the 1st 3 terms

now use these in the expansion of the summation

[covert spindle]

what do you mean?

[elder linden]

here

[covert spindle]

this?

[elder linden]

simplify a lil bit more
4=2²

[covert spindle]

you mean with the 1/4 and 1/2?

[elder linden]

yeah

[covert spindle]

alright

[elder linden]

it'll be easier then to calculate the net coefficient

[covert spindle]

so like this?

so (1)(3)(5)(7)(9)....(2n-1) = (2n-1)!! ?

[elder linden]

yess!

calculate for a few terms. how many terms do you need to expand up to??

[covert spindle]

i just had to give the maclaurin series of the function

so i think this is enough

[elder linden]

ahh yeah then this form will be enough.
btw did you multiply with x? becus we were figuring things out of the denominator. so dont forget there was a x already in the numerator

[covert spindle]

the x from x/sqrt(4+x²)?

[elder linden]

yeah

[covert spindle]

ye i did add that already

oh i made a mistake at the end

it should be 2n+1

idk why i did that there

[elder linden]

yeahh whats wht I was thinking also check the power of the negative 1

[covert spindle]

but for n=0 it will be negative right?

cuz the binomial gave n=0 -1?

[elder linden]

oh yeah

[covert spindle]

alright thank you very much 🙂

[elder linden]

okay!! it was a pleasure
make sure you verify the solution though 💜

[covert spindle]

how?

[elder linden]

with your professor

[covert spindle]

aaah ye thank you!

have a nice day or night

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[sonic moss]

how should I try to solve this? its very vague

[clear cloud]

Row reduce it

[willow urchin]

try finding a theorem in your notes about when a unique solution exists

[clear cloud]

And get a compatibility eqn

[sonic moss]

sorry what do you mean?

[trim lichen]

gaussian elimination

[sonic moss]

Like this?

[trim lichen]

,rccw

[clear cloud]

,rotate

[trim lichen]

that's a start

sorta

[wraith hinge]

https://www.youtube.com/watch?v=eDb6iugi6Uk

watch this video

it explains everything

You would be able to solve these kinds of questions easily

[sonic moss]

thanks man!

[wraith hinge]

Good luck

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@sonic moss Has your question been resolved?

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[sullen swift]

Could someone explain what Im missing here? Im trying to understand the area of a circle formulas, but some things are weird

like, why does the circumference times the radius = twice the area? shouldnt it just = area? how do you get area(2) from that?

like, im getting a mental image of (radius)(circumference) just being the the radius taking a revolution around the center of the circle, giving you the area. if anything, id assume the diameter times the circumference would be area(2)

[lime magnet]

Think of the area of the circle as made up of concentric circles

[sullen swift]

okay, i can do that, yeah

[lime magnet]

Notice how the concentric circles slowly decrease in radius

[sullen swift]

okay

[lime magnet]

So consider each concentric circle as a strip of very small width h

What will be the area of that strip?

[sullen swift]

itd be the very small width times the length of the strip, right

[lime magnet]

Correct

What will the length of strip be?

[sullen swift]

its a circle so pi r 2, right

[lime magnet]

Not quite

It will depend on which concentric circle we are taking area of

The concentric circle closer to the centre will have the length equal to a smaller circle's circumference

[slender shard]

If I'm reading this correctly, that only applies to a unit circle.

[lime magnet]

No? It applies to all circles.

[sullen swift]

yeah, because circumference is just perimeter of a circle

[slender shard]

"why does the circumference times the radius = twice the area?"

[lime magnet]

Right, so when we are making up the area, we are adding up smaller and smaller strips yes?

[slender shard]

There is a difference in how the two things are measured. One is a unit of length, R^1, while the latter is a unit of area, R^2.

[sullen swift]

yes, thats true

[lime magnet]

2pi r * (small width) + 2pi (r - small width) * (small width) and so on till 0

[lime magnet]

No? We are multiplying two length measures to get the area measure

[slender shard]

"why does the circumference times the radius = twice the area?"

Read what OP asked.

[lime magnet]

So if we sum over all of them, what can we take as common factor?

[lime magnet]

That statement is true for all circles

[sullen swift]

2 pi?

[lime magnet]

And also the small width

So what do we have as a sum?

r + r-small width +.. so on till 0

[slender shard]

2pi r * r = 2 pi r^2. Ok, now I'm on the same page.

[sullen swift]

i mean, its got to sum to the area, right

[lime magnet]

Since we are summing smaller and smaller radii

Going from r to 0

The avg is gonna be r/2

[sullen swift]

yeah, thats true

[lime magnet]

https://youtu.be/WUvTyaaNkzM?feature=shared

Watch this video

He starts by explaining exactly ur problem

In the first half

And in the second half he introducea basics of calculus

So u get two birds in one stone

[flint sluice]

love 3b1b!

[sullen swift]

okay
i think im starting to kind of see where youre leading although im still a little lost, ill watch this right now, yeah

thanks for the help both of you

[trim jolt]

@sullen swift Has your question been resolved?

[sullen swift]

wait a second, i think i get something

like, i havent finished the video (showering to help me think) but i think i understand like, why radius * circumference is 2x area

if you took the circumference of a circle and then laid it out as a flat length, you could multiply it by the radius to get an area, yes

but youd end up with more area than expected, because a circle has less area than a square of the same height and width

a circle is made of cuts of triangles compared to a square which is cuts of squares, and since the area of a triangle is half of the area a square in the same dimentsions would have, the area youd get from radius * circumference is twice as much as is accurate

so the area of a circle would be the radius * circumference divided by 2

at least thats what im currently thinking, ill continue the video yeah

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[real herald]

instead of doing a case 3, can i just write WLOG on case 2?

[marble wharf]

yes

[real herald]

@marble wharf thank you

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[low cloud]

I think that this is where i need to use l'hospital's but im also not sure

[limpid dawn]

you can

but with some algebra you can already tell the limit as it is

[low cloud]

sick then lets try algebra first

[limpid dawn]

just factor 2^n

[low cloud]

6^n?

[limpid dawn]

how

[low cloud]

i was thinking 2*3=6

[limpid dawn]

doesnt work because it's just 3

not 3^n

[limpid dawn]

factored 2^n

[low cloud]

ohh okay

[limpid dawn]

we can rewrite it a bit better taking 4^n

[red mountain]

u-sub

[limpid dawn]

at first glance it might look like inf/inf the left

making u think lhopital

but with one trick you can tell what it really results to

[low cloud]

how do i know this trick?

[limpid dawn]

exponent rules

2^n/4^n can you simplify it more

[low cloud]

1^n/2^n => 1/2^n?

[limpid dawn]

yes

[limpid dawn]

any idea now what it could be

[low cloud]

is it 0?

[limpid dawn]

can you justify why

[low cloud]

im pretty sure its because 1/2^n is a number getting really large quickly that is dividing 1 which is basically approaching 0 and then you have the same case in the numerator of 0/15

[limpid dawn]

the right goes to 3/15 but yea

it's basically 0 times a constant which is 0

2^n becomes very large quickly making 1/2^n become very small quickly

[low cloud]

ya this is what i was trying to say but mine explanation was just super confusing

[low cloud]

i dont know why all of these questions are so hard for me to understand when they also seem like simple concepts, like doing this question was so simple and im like ya i can totally do this now but then i look at the next problem and i have no idea what to do again! its so frustrating

[limpid dawn]

i guess that's life

[low cloud]

so philosophical

[limpid dawn]

only way is to try and fail and try again

if you wanna learn

i just guided you maybe that's why it felt kind of easy

[low cloud]

ya maybe

okay but for this one isnt it always going between 1 and -1 so wouldnt it be divergent?

but it said that was wrong

[limpid dawn]

hmm okay isee your thinking

it oscillates that's what you are saying

but what happens with sin(14/n)

like yea it goes back and forth

but?

[low cloud]

isnt sine bounded by -1 and 1 so it also doesnt approach anything

[limpid dawn]

no

you have sin(14/n)

as n grows really big

what happens with 14/n

[low cloud]

it approaches 0

[limpid dawn]

yes

it becomes infact smaller and smaller

which means for sine

what does sine become for small values near 0

[low cloud]

sine of 0 is 0

[limpid dawn]

yes

sine becomes therefore also really small near 0 as n grows large

so yess it keep going back and forth

but the oscillation becomes smaller

it's like dancing around the x-axis

[low cloud]

damn ya i see what you mean

[low cloud]

the spinning cat is so random it makes me laugh every time

i can never tell what its supposed to mean

im dean rn

[limpid dawn]

my trade mark

[low cloud]

i literally have no words

its so funny

[low cloud]

okay last math problem for today this one i also think that i need to do l'hospital's

[limpid dawn]

just observe what's happening inside the ln

for large n

actually l'hopital is not quiet applicable

[low cloud]

in my head all i see is inf/inf

[limpid dawn]

yea inside

you can factor n from the numerator and denominator as a hint

[low cloud]

like this?

[limpid dawn]

yes well done

can you now tell what happens inside

[wraith hinge]

u can take the lim inside of the ln

if it helps you

[clear cloud]

Better thing to do before

[low cloud]

the 7/n and 4/n go to 0

[wraith hinge]

correct

[limpid dawn]

yes

[wraith hinge]

therefore..?

[low cloud]

well then i have 3n/6n which is basically what i had before

[limpid dawn]

n cancels

[wraith hinge]

u already factored it why not cancel it lol

[low cloud]

ohh oops okay then its just ln(1/2)

[wraith hinge]

yes

[limpid dawn]

if you even just need to tell the limit of some rational function then you can do it instantly with this trick

the 3rd applies to you since both numerator and denominator are linear

so you instantly can tell 3/6

nice to know

[low cloud]

okay sick i added this into my notes

thank you

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[cyan furnace]

Problem is to find the particular solution to the following differential equation using the inverse differential operator and noting the following.

[cyan furnace]

I understand that i need to rewrite the particular solution in terms of this operator and the right hand side. y_p= L^{-1}(D)[tcost] then go from there but i dont really understand how im supposed to get some quadratic in terms of differntial operators out of this.

the bottom equation is when you have some L(D) =a_0+a_1D+a_2D^2

currently i have the following

Like i have a similar example from my prof but like it has a crazy jump that i don't really get and would like help with.

Like i know the next step is to do L(D+-i) = D(+-i)^2 +4 but i would like to understand why this works.

Never mind i got it just have an issue with the notation

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[meager hornet]

Hello can someone help with regression analysis

[empty orchid]

!da2a

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[meager hornet]

Time series regression analysis

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"Don’t ask permission to ask a question, or ask if anyone knows how to solve a type of problem. Just ask your question."

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[zinc ginkgo]

.close

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[lost urchin]

how do i evaluate this

[cunning cliff]

use series for cos

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[left oriole]

could also try multiplying and dividing by 1+cos(4x)

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[high raptor]

Hello, can someone use rk4 to approximate y(0.2) , i just want to check my work im getting couple different answers

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[simple haven]

@high raptor using an online calculator I arrived at about 1.24

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[simple haven]

(note this agrees well with the actual value of the IVP, 4/(exp(4/5)+1))

[high raptor]

thats what i got my prof had wrong ans

thanks

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[carmine skiff]

Problem 4: A woodworking factory uses mahogany to produce 5 tables per day. The cost for each raw material shipment is 5,800 USD, and the storage cost for one unit of raw material is 10 USD per day. Each table requires one unit of raw material.

The factory must order raw materials in batches, and the goal is to determine the optimal order quantity that minimizes the average daily cost, which includes both ordering costs and storage costs.

Question: How many units of raw material should the factory order each time?

[carmine skiff]

5800/(5m) + 10m

m is supposed to be the amount of days in a cycle

m's value is 215,4

Therefore, raw materials's value is 215,4*5

Answer is 1077

Is there anything wrong with my solution

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[tepid condor]

Is the greedy choice property basically saying that there is only one local minima and that is therefore the global minima

[tepid condor]

How do I relate this idea of there being just one minima

To something like Dijkstras

If someone knows please confirm this

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[radiant portal]

I literally have no idea what this question is asking 💀

I understand everything before this part

This is the bit that I dont understand

Wtf is a

[covert spindle]

alpha is just a variable

to a point Pa is just a point where t=alpha

[trim lichen]

a, not alpha

but yes

[covert spindle]

Sorry i see now too

[radiant portal]

And what line do they mean

[covert spindle]

dont want to give it away immediately

choose a random value a

[radiant portal]

1

[trim lichen]

the line passing thru the points at t=a and at t=pi-a

[covert spindle]

okay so use t=1 and t=pi-1

[trim lichen]

!noadvert

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[covert spindle]

sorry

Any idea what you have to do @radiant portal ?

[radiant portal]

Not at all 😭

[covert spindle]

okay so P_a and P_pi-a are just two points on the curve

the subscript gives a certain time

so for example P_1 would mean the point (x,y) that comes from t=1

you got me so far?

[radiant portal]

Yes I think so, but im still not really sure what to do

This question seems really hard

[covert spindle]

well instead of t=1 now we have two t's

t=a and t=pi-a

What can we do with t?

[radiant portal]

What do you mean instead of sry

They are still equal 1, no?

Oh wait nvm

I get you

[covert spindle]

alright

okay so we have t=a and t=pi-a

what can we do with that?

[radiant portal]

I mean we can sub them in to x(t) and y(t) but im not sure how that would help us prove theyre in a vertical line

[covert spindle]

in math sometimes you just must try stuff

it sometimes makes sense if you are halfway through it

so i recommend doing exactly what u just said

and it will probably make sense when ur done

[radiant portal]

x(a) = cos(a)sin(2a)

y(a) = cos(a)

[covert spindle]

okay so P_a (cos(a)sin(2a),cos(a))

[radiant portal]

x(pi - a) = cos(pi - a)sin(2pi - 2a) = -cos(a)sin(-2a)

Is that correct? I tried to use identities

[covert spindle]

almost

u have 1 more to go

it is a pretty simple one

[radiant portal]

y(pi - a) = cos(pi - a) = -cos(a)

[covert spindle]

now for the x(pi-a)

we are not there yet

-cos(a)sin(-2a) = ?

[radiant portal]

-cos(a)sin(2a)

Chartbit ❤️ ❤️

[covert spindle]

try again