help-42

[upper pilot]

I need help in this

If $f(5x) = x^2+x$ then then $f'(2)-f''(2)=$

[potent lotus]

Joeller

[upper pilot]

how do I get rid of the x in f(5x) ?

[atomic idol]

use chain rule

and then just plug in x = 2/5

[upper pilot]

f(2)'' is going to be zero right ?

[atomic idol]

I don't agree

did you apply the chain rule?

[upper pilot]

$f(5x) = x^2 + x$ \newline $5f'(2)= (0.4)^2 + 0.4$ replaced the x with 2/5

[potent lotus]

Joeller

[upper pilot]

then divide by 5

[atomic idol]

yes I agree with that

[upper pilot]

f'(2)=14/25

how do I get f''(2) ?

[atomic idol]

wait

[upper pilot]

ok

[atomic idol]

not exactly

5f'(5x) = 2x + 1 so f'(2) = (2/5)(2/5)+1/5

[upper pilot]

I see

[atomic idol]

you should find f''(5x) with chain rule and plug in 2/5

[upper pilot]

I want f''(2) tho

[atomic idol]

you already have f'(5x) so you use that to get f''(5x)

[atomic idol]

yes you need to find both

[upper pilot]

then replace the x with 2/5 ?

[atomic idol]

you already have f'(5x)

now you get f''(5x)

[upper pilot]

aha

ok wait a sec

so its going to be 25 f''(5x)=2 ?

[atomic idol]

yes that's correct

[upper pilot]

so now divide by 25 and done ?

okay yeah

9/25 - 2/25 = 7/25

thanks

have a good day and thanks so much for your help

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[calm coral]

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[calm coral]

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[remote glen]

quick question- what does the notation $\mathbb{Z}^{\times}_{12}$ mean? is it the set of integers that are coprime to 12?

[exotic falcon]

it is the multiplicative group Z_12 without 0

[remote glen]

so 1-11?

or 1 5 7 11

[exotic falcon]

mod 12

Should be 1-11

over multiplication

mod 12

[remote glen]

ty

that makes sense

im jsut making amulitplication table and this notation hadnt been used yet

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[exotic falcon]

the * just indicates both that it's multiplicative and that we aren't including 0 (because it wont have an inverse in the multiplicative group)

[calm coral]

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[orchid palm]

https://cdn.discordapp.com/attachments/1194738644976672790/1231659742636539944/image.png?ex=6637c385&is=66254e85&hm=0becd6da8096bcb110979f08170eb4ee1d11ee125da147f345152234b4831fcd&

[orchid palm]

I need to find the radius of convergence and interval of this series, along with the convergence or divergence of the series at its endpoints

I will post work one sec

So I got to here

[eternal shard]

Your sequence is: $a_n = \frac{(-1)^{n+1}}{n\cdot 9^n}$

[potent lotus]

𝔸dωn𝓲²s

[orchid palm]

I just exclude the (x-4)^n

?

[eternal shard]

Now you do $\lim_{n \to \infty} \abs{\frac{a_n}{a_{n+1}}}$

[eternal shard]

yes

[potent lotus]

𝔸dωn𝓲²s

[orchid palm]

its flipped

this one

other way around

[eternal shard]

you will flip it anyway the limit for convergence of radius

you probably got something like R = 1/a

[orchid palm]

ok I will work it out with this in mind, one sec

mmm

for a_n

I pull out the -1^n+1

And just leave 1/n*9^n

I mean I get 1/9 for the limit result

Which is < 1

which is convergent

But where does the (x-4)^n come in

[eternal shard]

hold on

your (-1)^(n+1) is gone due to absolute value

[orchid palm]

mmm true

[eternal shard]

$\lim_{n \to \infty} \frac{1}{n\cdot 9^n} \cdot (n+1)\cdot 9^{n+1}$

[orchid palm]

you're right

[potent lotus]

𝔸dωn𝓲²s

[orchid palm]

that evaluates out to 1/9

I believe

[eternal shard]

9*

[orchid palm]

how

[eternal shard]

If you did a_(n+1) / a_n then yes

[orchid palm]

nvm

I see

[eternal shard]

I flipped it in the beginning already

That's what I meant

[orchid palm]

Why did you flip it in the beginning though?

[eternal shard]

If you flipped the limit a_n / a_(n+1) you would get 0

[orchid palm]

I got 1/9

not 0

[eternal shard]

I meant your example

anyway doing your method

you end up with R = 9

[orchid palm]

I get what you mean

[eternal shard]

now (x-4) comes into play

[orchid palm]

now

[eternal shard]

we identify a = 4

[eternal shard]

Let's call your limit L

L = 1/9

then 1/L = 9

ok?

[orchid palm]

yes

[eternal shard]

Convergence radius is: $\abs{x-a} < \frac{1}{L}$

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

So here |x-4| < 9

[orchid palm]

so x - 4 < 9

x < 13

?

oh wait

abs

value

[eternal shard]

which is the same as -9 < x-4 < 9

now +4

[orchid palm]

-5 < x < 13

[eternal shard]

-5 < x < 13

[orchid palm]

?

[eternal shard]

yes

[orchid palm]

hmmm

[eternal shard]

now

[orchid palm]

and now I plug in -5 for x and 13 for x?

[eternal shard]

you need to check the borders

yes

[orchid palm]

into the original sequence right?

and resolve for convergence / divergence

[eternal shard]

into the series

and then with x = 13 and x = -5 you see if the series converges or diverges

[orchid palm]

?

These two

With -5 and 13 respectively

[potent lotus]

𝔸dωn𝓲²s

[orchid palm]

both divergent?

I think both would be divergent, at the -5 endpoint and 13 endpoint for x

right, because 1/n and p = 1 so for p <=1 we get divergence?

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

hold on

[orchid palm]

this simplifies to this right? https://cdn.discordapp.com/attachments/948770026914213909/1231702138569359381/image.png?ex=6637eb01&is=66257601&hm=5b3beae0471f040baac74ac3bdbd9fb5b84ea4eb4050242c38a5b050ab465322&

[eternal shard]

the left one is a leibnitz

[orchid palm]

shiet we haven't been taught leibnitz

is it something specific or just a general idea?

[eternal shard]

like alternating series test?

[orchid palm]

ah AST we were taught

that's fine

didn't know Leibnitz came up with it

[eternal shard]

yes

[orchid palm]

yeah I just take the limit

and see if a_n+1 <= a_n

[eternal shard]

$\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n}$ and $\sum_{n=0}^{\infty} \frac{(-1)^{2n+1}}{n}$

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

You can make the argument that both converge but (not absolutely) due to AST

[orchid palm]

both converge hmm

oh you're right

I just took the limit straight up

[eternal shard]

but oh I just realized n = 0

lol

[eternal shard]

n cannot be 0

otherwise division by 0

[orchid palm]

well that's the problem my teacher gave me idk

maybe there is some other way?

or my teacher wrote it incorrectly

that's possible

but if we assume n = 1 then it's fine right?

[eternal shard]

yess

it's probably a typo

[orchid palm]

ok just one thing I don't fully understand

1 sec

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

notice how

(-1)^(2n+1) is the same as -1

because the exponent is always odd

[orchid palm]

I see that

[eternal shard]

so the right one diverges

and the left one converges due to AST

[orchid palm]

mmmm wait

what

1 sec

Why would it not just be this

I don't worry about the -1^n+1 with AST

[eternal shard]

(-1)^n+1 alternates

[orchid palm]

just a_n

[eternal shard]

you are doing the ratio test or what?

[orchid palm]

Nah AST

[eternal shard]

oj

[orchid palm]

But it shows that a_n is separate from -1^n

and I don't use that for the test itself

it's irrelevant at least from what I can see in the theorem

[eternal shard]

yes it is

(-1) just makes the sequences alternating but still converging to 0

[orchid palm]

so my conclusion would then not be divergent for the right one

it would still be convergent

[eternal shard]

no

[orchid palm]

does it fail the second test?

with a_n+1

<= a_n

[eternal shard]

(-1)^n or (-1)^(n+1) is not the same as (-1)^(2n+1)

the right one is appears to be alternating but it isn't

it's the same as -1/n

then there you can use that p series test

or what it is called

[orchid palm]

huh

[eternal shard]

This

can be used for the right series

[eternal shard]

look from here how both series were simplified

[orchid palm]

You pulled the neg. out?

[eternal shard]

yes

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

and

[orchid palm]

hmmm I see

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

because the exponent is always odd

[orchid palm]

ok so because you pulled the neg. out the right side function was no longer alternating

[eternal shard]

yes

[orchid palm]

and so you applied a regular test

like p-series

[eternal shard]

yes

yes

test it

[orchid palm]

left side was alternating so you just did AST

ok I see

[eternal shard]

yess

because the left actually alternates

[orchid palm]

ok let me write out the whole thing now

[eternal shard]

(-1)^(2n+1)
n=1: (-1)^(3) = -1
n=2: (-1)^(5) = -1

[orchid palm]

I will know if I get it by the end of it

give me like 2-5 mins tops

ok I did it but

here lemme just post the picture

https://cdn.discordapp.com/attachments/1161035250676285512/1231707908962652240/470301F1-F105-4AB2-95AA-493D73F23BF3.jpg?ex=6626cce1&is=66257b61&hm=0c1a42c777476fcb3050051b8991adf5318857bab904820ef3c93803d1855e3c&

I think it needs to be opened in browser

to see the full thing

[eternal shard]

huh

[orchid palm]

I got that the endpoint for 13 is convergent

-5 is divergent

[eternal shard]

yes

so what is the interval

[orchid palm]

well -5 to 13

-5 div. to 13 conv.

ah yeah

in an inequality

I suppose (-5, 13]

works as well

[eternal shard]

yes

[potent lotus]

𝔸dωn𝓲²s

[orchid palm]

ok I solved it fully

thanks a bunch for the help

I am sure this will help me for future problem solving

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[calm coral]

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[calm coral]

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[mild jetty]

I dont know how to do this question. Any help would be appreciated.

[mild jetty]

So I guess I can start by some things I can see. Since ABE and DEC are equilateral and also E is the midpoint, we can conclude the six angles are all equal. Furthermore, that |AB| = 2 = |DC|.

[glacial nacelle]

have you tried using the hint?

[mild jetty]

i have, but im not very sure my labeling is helpful

So i centered everything around B

B(0,0)

[glacial nacelle]

yes, that should work

[mild jetty]

Since |AB| = 1, then A(2,0)

[glacial nacelle]

no

|AB| is 2

so A is at (0,2)

[mild jetty]

oh yes, typo ty

[glacial nacelle]

keep going

[mild jetty]

Ok since the triangles are equalateral, and E is the midpoint of AC, E should have a height of 1. So something like E(x,1)

[glacial nacelle]

yes

how can we find x?

any ideas?

hint

what is the height of an equilateral triangle

[mild jetty]

Hmm ok. So the triangle ABE is equilateral, so all three sides are equal. Hence |AB| = |BE| = 2. Also |EG| = 1. Then the triangle BEG is a right angled triangle, so by pythagorean should be root 3

[glacial nacelle]

nice

so what are E's coords?

[mild jetty]

E(sqrt 3, 1)

[glacial nacelle]

In order to find the area of the quadrilateral, a simple intermediate step would be to find the coords of all four points

We have E

So we need H,G,F

What is G?

[mild jetty]

G(sqrt(3), 0)

[glacial nacelle]

any ideas on how we can find H or F?

[mild jetty]

no

I dont think we were given anything to work with here

[calm coral]

@mild jetty Has your question been resolved?

[glacial nacelle]

try to find the equation of the lines

[mild jetty]

$y = \frac{1}{\sqrt{3}}x$

[potent lotus]

WhoTao

[mild jetty]

but still i dont really see a way to find either x or y coord

[glacial nacelle]

Which two lines intersect at point H?

You found the equation of one line

find the equation of the other line

[mild jetty]

Ah ok. So I found their intersection

x = 2/3sqrt(3)

Then H(2/3sqrt(3), 2/9)

As for F, I should find D and C first

[mild jetty]

C(2sqrt(3), 0)

D(2, 2sqrt(3))

so then I can use a similar method to find F

Finally, I have E,H,F,G

[mild jetty]

Then I can just find |EH| and |HG|, or |FG| and |EF|

Then multiply the two

thanks @glacial nacelle

[glacial nacelle]

np

[mild jetty]

.close

[calm coral]

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[mossy wigeon]

for this

is the integration this

[quaint sphinx]

no lone x term

[mossy wigeon]

oh yeah forgot to remove that

ty

.close

[calm coral]

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[quaint sphinx]

yw

[calm coral]

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[leaden marsh]

so right angle time are of the form A:(5A+15)

looping back around to 5A-45 when 9<A<1pm

bc the minutes hand is 5 times the hour markings on the clock

we add 15 for the 90 deg angle

12:15 and 12:30, the minute hand moves 90 degs

[leaden marsh]

this is bc like if the minute hand is pointing towards 3 it’s :15

no

I’m saying the times

1:20 pm is the time you get when you set a=1 for example

then 2:25 then 3:30

(actually multiply the number u get by 2 for right angles the other way)

[marsh valley]

I'm not sure you're accounting for the fact that the hour hand moves continuously.

At 3.30, the angle isn't 90 degrees, the hour hand is already halfway through the 4th gap

[leaden marsh]

ok I actually thought about this on paper and I got 30h+11m where h and m the hour and minute

we’re looking for when this is equal to 90 or 270

were able to get m in terms of h

by solving for m

ummm

can u solve for m?

30h+11m=90 first

then =270

bc the hour hand moves 0.5 degrees every minute

the number of minutes is 60h+m

so theta_h = 30h+m/2

(in degrees)

the minute hand moves 360 degrees per 60 minutes

so 6 degrees per minute

so theta_m = 6m

so the angle between the hands is theta_h - theta_m

oh wait I wrote this wrong, 30h-5.5m is the difference

the number of minutes since 0°, i.e. clockwise from the 12:00 position

yes

the minute hand is 6 degrees, the hour hand is 0.5 degrees

and the angle between them is 5.5 degrees

oh yeah

I was answering at 12:01 like u asked

ok so

our equation is $30h+\frac{11}{2}m = [90, 270]$

[potent lotus]

blahaquil

[leaden marsh]

ok

to find the angle between hands we can find the angle between each hand and an anchor point (we will choose 12:00) and subtract them

the hour hand moves 0.5° every minute

(tell me if u get lost at any point btw)

so we can count the minutes by taking 60h+m

where h and m represent the hours and minutes respectively, expressed as natural numbers

because

use a free channel pls

[remote mural]

alr

[leaden marsh]

if its 1:05 its been 65 minutes

if its 4:10 its been 370 minutes

we can generalize this to 60h+m, meaning every 60 minutes we add it the new hour

when h=12 its gets weird

because the angle is more than 360° degrees but they’re equivalent

do u understand?

ok

so the hour hand moves 0.5° * 60h + m

so theta_h is (30h + m/2)°

the minutes hand moves 6° per minute

so theta_m is (6m)°

we’re just keeping units

if u input h and m you get the angle of the hand from 12:00

yaaaaaa

min*(deg/min)=deg

dimensional analysissss

[leaden marsh]

subtract the equation for theta_m

[leaden marsh]

from the one for theta_h

and u get theta_h - theta_m = 30h - 11m/2

this is the angle between the hands

we want right angles

so we equate this to 90 and 270

case 1: 30h - 11m/2 = 90

case 1: m = (60h - 180)/11

yes

soooo solve for m

yes

so in general, m = (60h-2*theta)/11

where theta is the angle you want

the actual minute number

so like at 7:30, h=7, m=30

now u just count values of m given h that satisfy the two equations

within the time delimitation ofc

does that make sense?

kinda

there are 7 possible values of h

so 14 possible right angles

(although check 8 and 3 to check if you need to subtract any for being outside the limit)

you can plugin 1-12 for h and the corresponding m value will show up

I think that’s easier than the other way, unless ofc you’re given an m value and are expected an h value

24

right angles can also happen the other way

yes

so (as I said) 7 poss hours, 14 poss right angles

I’m sure there are some on the edges

[leaden marsh]

check h = 8 and 3 in both eqs to see how many you need to subtract from 14

8:27 or 8:55, the first one is outside the limit

and then 3:00 and 3:30 are the ones are the right angles at h=3

so 14-1=13 right angles

sry ya you’re right

when that happens you can just take mod 60

clocks are circles 😄

[leaden marsh]

when that happens you can just take mod 60

in our example, -5 mod 60 = 55

if we got -25, -25 mod 60 = 35

js add 60 ig, we don’t get anything <-60

because the minute hand loops around every 60 minutes

much like the mod operator

do u understand

if u want you can also just use the 30h+11m/2 = (360 -) theta in an actual competition, it’s more general use and might be simpler

do they explain why

oh gd it

ya that’s my fault

I counted 3:30

Yes

it moves 30h+m/2

it was because I did it in my head and not with the equation

so just remember the formula and you should be good

yes, this is the form we later used to calculate the angle between the hands

yes

does that make sense?

[scarlet badger]

what question is in play right now

[calm coral]

@remote mural Has your question been resolved?

[cedar sinew]

bro

The answer is 30 degrees

[marsh valley]

Right so. The angle between the minute and hour hands increases at rate of 11/2 deg/min. We want to find out when that angle is 90 deg, or 270 deg (depending on whether the minute hand is after the hour hand, or before it.
I will use t to determine the time passed after midnight.
That means that we want 11t / 2 = 90 for the first time it happens, that is t = 16.36.
and then the one after that is 11t / 2 = 270, which yields t = 49.09. Between those, there is 32.73 minutes, and it's periodic, so the hands are 90 degrees from each other every 32.73 minutes.

Starting from 9, we can see that we have a total of 6.5 hours, meaning this can happen 6.5 * 60 / 32.73 = 11.9 (round down to 11 times) PLUS the initial position at 9. That gives a total of 12, as you were required.

If the angle grows at a steady rate of 11/2 deg/min, then after t minutes, the (total) angle between the hands is 11/2 * t

We want that angle to be 90 degrees, at least for the first one.

Do you mean why I use 6.5 in lieu of 7 hours?

It's actually the opposite : I'm disregarding the first 30 minutes because we know the first 90 degree moment after 8.30 is at 9. Then we can just start counting from there until 3.30

Since the 90 degrees happen every 32.7 minutes, we start adding 32.7 minutes from 9h00, instead of 8h30

Yeah but we're just counting the number of times we can add 32 min after 9

[marsh valley]

We don't know that unless we count

The point is, we know the "cycle" starts at 9 and repeats every 32.7 minutes. Since we have 6.5 hours until 15:30, this adding can only be done 11 times before going over.

Yes but you have to add up to 3 to know it ends there.

The only difference is it'll give an actual integer.

Yeah I suppose so. Again the only difference is that you'd get 11 straight away instead of 11.91 or something

It's just we're 9/10 of the way to the next 90 deg

No, if you use the actual exact value you get exactly 11 :

360 / 32.727272... = 11

[calm coral]

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[brisk pelican]

can i use this if i change it to dg/du and dg/dv ?

dont know what to do

[spark lotus]

can somebody please help?

[swift laurel]

that's the formula for a surface given by $z = g(x,y)$. you instead want to use the formula for parametric surfaces: [ \int \va{F} \cdot \odif{\va{S}} = \iint_D \va{F} \cdot (\va{r}_u \times \va{r}_v) \odif{u,v} ]

[potent lotus]

pnoןɔ

[brisk pelican]

oh that makes sense actually

cause ru x rv is the normal vector i think

[swift laurel]

we're supposed to dot with the unit normal vector but the scale factor of |ru x rv| cancels out (from https://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx):

[brisk pelican]

whats the difference between dS and dA again?

dA is in terms of the parametric variables right

but why is it really any different

[swift laurel]

dA here is just du dv

it's the area differential in the uv-plane

[brisk pelican]

and dS is in the original surface, which is different by a factor of the magnitude of the cross product?

[swift laurel]

yes

[brisk pelican]

i got the opposite of the answer

which i half expected cause i chose the cross product kind of arbitrarily

[swift laurel]

yeah you have to pick + or - the answer based on orientation

[brisk pelican]

how do you choose ru x rv or rv x ru though

[swift laurel]

calculate one of them and see if it matches the description of the orientation of the surface

[brisk pelican]

oh i see

thank you for the help

much appreciated 🙏

.close

[calm coral]

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[orchid narwhal]

can someone help on the ones except a

[marsh valley]

So a is ok?

[dawn sparrow]

.reopen

how the hell do i make it

[wanton spear]

?

[dawn sparrow]

how do i make a help channel

[wanton spear]

Did you read #❓how-to-get-help ?

[dawn sparrow]

ye doesnt say

[wanton spear]

In the list of channels, there's a few help channels under help(available)
Send your question in one of those

Those are the channels with no name in the channel name

[dawn sparrow]

k il try thanks

[wanton spear]

How far did you get on your own?

[dawn sparrow]

thanks bro

[orchid narwhal]

I did the first one but im confused on how to do the rest

[wanton spear]

How did you do the first one?

[wanton spear]

So you have two steps,
You'd determine the equation of the line

Then you determine the direction of the inequality by testing a point

The same applies to b,c, and d

For b, do you have an equation?

[orchid narwhal]

nah, that's what im confused on, I got the y-int of 8 but idk about the rest

[wanton spear]

Well, flat lines always have a slope of....

[orchid narwhal]

0?

[orchid narwhal]

yo? it would be 0?

[wanton spear]

Yep
So the equation for b would be...

[orchid narwhal]

y = -3

but I got a question, how would I know which which way the inequality sign will be pointing?

cuz I know now we pick any point in the shaded area and substitute but does it matter which way the sign is or is there a way to tell which way the sign should be?

[orchid narwhal]

y = -3

[wanton spear]

yeah, the direction of the sign matters

There's two ways to figure out the direction of the sign,
either you test a point
or you just figure it out from if the shaded area is above or below the line

if it's below the line, y < ...
if it's above, y > ...

[calm coral]

@orchid narwhal Has your question been resolved?

[calm coral]

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[nimble field]

Why does (3-x) change to -1

[keen wraith]

you see we are trying to divide (3-x) by (x-3)

so we need to multiply -1 to (x-3) to get (3-x)

[nimble field]

Yeah but why doesn’t it just disapear

[keen wraith]

umm

imagine this scenario

$$
\frac{3-1}{1-3}
$$

[potent lotus]

notwish

[keen wraith]

what do you get?

-1 right?

[nimble field]

I don’t know the term in english

But i would just cross them out

[leaden marsh]

but they don't cancel each other out

3-1≠1-3

[nimble field]

Oh

So it would be 2/-2

So -1

[leaden marsh]

yes

replace 1 with x

you also get -1

[calm coral]

@nimble field Has your question been resolved?

[calm coral]

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[remote mural]

No matter how fast I throw the ball vertically upward
Acceleration is still -9. 8?

[bronze adder]

basically yeah

it doesn't depend on how fast you throw it up

Just know that 9.8 is an approximation for when you're very close to the surface of the earth

[remote mural]

But if I throw a ball
Initially it's goes very fast
So that acceleration is provided by me Or is it just initial velocity I am giving!

[remote mural]

Ye

[bronze adder]

Just the initial velocity

You may impart an impulse to the ball, but that's a whole different story

For simple kinematics it's only velocity

[remote mural]

Oooooooo
Big brain big brain
And 1 more question
Is resnick halliday krane good for understanding concepts of phy?

[bronze adder]

never heard of it

couldn't say

[small shuttle]

reading a textbook is probably one of the best ways to learn physics outside of a classroom

or only ways tbh

[remote mural]

And yeah last question
Integration of 2x is x + c right
But if u derive it through geometry that is by area of triangle it's just x, so what is the extra 'C'

[bronze adder]

x^2 + C

[bronze adder]

well first of all when you're computing an indefinite integral its giving you the antiderivative

[remote mural]

?
2 × x^2/2
So it's x no?

[bronze adder]

There's no relation between an anti-derivative and the area under the curve except for the fundamental theorem of calculus

[bronze adder]

$\int x^n \dd{x} = \frac{x^{n + 1}}{n + 1} + C$

[potent lotus]

uoǝu

[bronze adder]

you divide the term by the power

not the power by the power

[small shuttle]

$x^2 + C$ is the antiderivative of $2x$ because $\frac d {dx} x^2 + C = 2x$ for any value of $C$

[potent lotus]

b

[small shuttle]

which i guess answers both questions lol

[bronze adder]

the C just gives you all curves that possess the gradient

[remote mural]

Take some values and kindly tell how C is useful

[bronze adder]

$\int \sin^2 x \dd{x}$

[potent lotus]

uoǝu

[remote mural]

And what should the constant '2'?

[bronze adder]

Do you know how to work this out

[bronze adder]

nothing jut keep it

integrals are linear operators

[harsh siren]

theres two types of integrals, indefinite and definite
indefinite is used to reverse differentiation, and definite is used to find areas under functions

[remote mural]

Oh yeah shit
2 × x^2/2 is x^2 not x
Shit

[harsh siren]

theres no +c in definite integrals

[bronze adder]

(technically there is it just cancels out when you FTC :P)

[remote mural]

Example pls

[harsh siren]

(no one cares about that shh)

[small shuttle]

from the last time someone asked🤓

[bronze adder]

wonderful notation

😭

[small shuttle]

abuse of notation is one of my favourite things

i think it's close enough to be helpful though

[loud fiber]

Chat

Who here is a perms and coms go

God

[remote mural]

WHAT? 😭😭😭😭

[bronze adder]

open a channel

youll find someone

[loud fiber]

No one wanna answer fr

😔

[small shuttle]

lmao disregard that actually it's not really relevant
and also wrong (integral of x is not x)

[remote mural]

Kindly solve a short little problem for both definate and indefinite so I can understand use of C

[small shuttle]

if i know where this is going try to solve this

[bronze adder]

@remote mural

[harsh siren]

like i said, indefinite integrals is used for reversing differentiation

[remote mural]

Hmmm.. I didn't study integration of trig functions yet

[harsh siren]

say i give you a function f(x) that is resulted from differentiating g(x)

now i want you to find g(x)

[loud fiber]

-1/2cos2x + c right

[remote mural]

So I will integrate it

[harsh siren]

yes

and you might ask, why is there a +C?

well the original function might have a constant somewhere

[remote mural]

Yea was typing that only

[harsh siren]

but differentiating a constant gives 0

[loud fiber]

Last calc test I lost so many marks for forgetting + C

It’s depressing

[remote mural]

Oooooooooooooooo
Big brain big brain

[harsh siren]

so by adding that +C, you essentially cover all possible constants that might be there before it was differentiated

could be 7

[bronze adder]

uh

[harsh siren]

could be 5

could be 1 billion

[bronze adder]

not quite

[loud fiber]

Cos2x

Mb

Oops

!!

[remote mural]

Real real

[harsh siren]

now the way that you said by using geometry

essentially you're finding the area under the function

which is definite integrals

[remote mural]

Yea

[remote mural]

Oooo

[harsh siren]

so lets use the example of 2x that you gave

[remote mural]

Yep

[harsh siren]

so you're essentially integrating from 0 to x, of the function f(t) = 2t

why change to t you ask, because you're integrating to x

[remote mural]

Yea

[harsh siren]

the integral of 2t is t²

then plug upper bounds and lower bounds into it and subtract

so x² - 0 = x²

thats the area under 2x

you see that theres no c there

[remote mural]

Yea

[harsh siren]

so thats the difference

indefinite: use to find function before it was differentiated
definite: use to find area under function

[remote mural]

Ohhhh real real

[remote mural]

Thanks a lot man thanks a lot

[harsh siren]

and always remember to add c when doing indefinite integrals or you might cry to sleep your whole life :D

[remote mural]

Thanks @harsh siren @bronze adder @small shuttle

[remote mural]

Lol😂

[remote mural]

I haven't studied integration in maths now
I am studying because I need it in physics

.close

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[remote mural]

This thing is only applicable for projectile or for vertical throw as well?

[cursive linden]

wdym? vertical throw is also considered a projectile motion unless you mean something else by 'vertical throw'

[remote mural]

I mean it doesn't take a parabolic path

Just up and down

[cursive linden]

still a projectile

[remote mural]

So will it be projectile as well?

[remote mural]

Ooooo
Ookok
My bad my bad

Thanks

[cursive linden]

projectile motion is when only gravity acts after the initial force that launches it

[remote mural]

Oh.. I thought it's defined by it's shape of trajectory

[cursive linden]

actually, 'projectile motion' is the motion that happens in two dimensions (curved path yeah) but the word 'projected' well you already know what it means, it'll be applicable for vertically thrown object as well

but to answer your question, the thing is applicable to the motion that occurs along y-axis

as gravity only acts along y-axis

so for a projectile motion, that thing applies to its y motion and for a vertical throw, well its whole motion is along y-axis

[remote mural]

Ohhhhhh thanks thanks

BTW the thing I am referring to is same magnitude of velocity at same heights with opposite direction

[cursive linden]

yeah as i said, that applies to the velocity along y-axis

the magnitude of velocity along y-axis remains the same at same heights

[remote mural]

Real
Thanks man!

[remote mural]

Yep

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[stiff pagoda]

I do not know how to find f(x) from the table

[calm coral]

@stiff pagoda Has your question been resolved?

[calm coral]

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[snow hare]

how do I find general term for sequence i

this is like fibonacci or something

[cloud nexus]

Which one

[vivid vine]

you have 2 subsequences 1, 1/2, 1/3... for odd and 1,2,3... for even

[cloud nexus]

If n is even: n/2

Else: you can find thwt

[snow hare]

true

mmm

so its divergent

?

nooo

but

[cloud nexus]

Yeah

Its divergent

[snow hare]

no but

I cannot find for odd

I need more help

[cloud nexus]

It approaches 0

Its just like

1/n

But only for odd terms

[snow hare]

check for n=3

1/3

[cloud nexus]

Yeah but

I mean

If you consider the odd subsequence

Its 1/n

It doesnt even matter

The even part diverges

[vivid vine]

i think it is 1/((n+1)/2)

or 2/(n+1) simplified

[snow hare]

lim of 2/(n+1) is 0

but lim of n/2 idk

[cloud nexus]

Mate

Limit of n/2

Whats the limit of n

Just half that

[snow hare]

still infinity, unbounded?

i don’t get it

[cloud nexus]

Yea

[vivid vine]

its infinity

(diverges)

[snow hare]

gotchu

sorry for the hassle

[vivid vine]

dont feel sorry, its no problem

[snow hare]

thanks

.close

[calm coral]

Channel closed

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[vivid vine]

h converges btw

to 1

[snow hare]

.reopen

[calm coral]

✅

[cloud nexus]

Whay

[vivid vine]

the sequence is (n+1)/n

and its limit is 1

you can take large n to see it

[cloud nexus]

No im asking why did it get reopened

[vivid vine]

beacuse i saw here, the answer for 3rd question is wrong

[snow hare]

,rotate

[vivid vine]

ok, so why did you say it diverges?

[snow hare]

I just calculated by what you said

[vivid vine]

o hok

[snow hare]

first time I just did it by eye

im stupid sorry

[vivid vine]

yeah its correct

[cloud nexus]

Also 1,-1,1 … doesnt converge

[vivid vine]

oh yeah

true

[snow hare]

what about that

[vivid vine]

💀

[snow hare]

how do I find lim of (-1)^n

[cloud nexus]

You wrote convergent

[vivid vine]

it doesnt have it

[cloud nexus]

Doesnt exist

[vivid vine]

you can proove that doing a subsequence

[snow hare]

how

[cloud nexus]

Its an oscillator

[vivid vine]

1 for odd and -1 for even

[cloud nexus]

It never converges to a single value

Its also trivially easy to prove its not cauchy

[snow hare]

?

cauchy?

[cloud nexus]

Wait nvm

Ignore

Its a property of convergent sequencee

[calm coral]

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[valid zodiac]

hello

[valid zodiac]

for this equation

i got answer to be

y(x) = k * sqrt(|x|)

however, the answer is suppose to be:

y^2 = C*x

so im confused

if you square my answer, you would get k^2 * |x| right?

k^2 = C because both is constants sure

but |x| is not the same as x

[cerulean cradle]

Hii

dy/y = dx/2x

Integrate

lny = 1/2lnx + lnc , where lnc is integration constant

Using properties of log

ln y = ln c x½

So y =|cx½|

[valid zodiac]

when i integrate this, i get ln(|y|) + c1 = 1/2ln(|x|) + c2

where do you get lnc from

[valid zodiac]

but yeah with the method of integrating this i get y^2 = Cx

however

[valid zodiac]

i got this answer when transforming the equation to the standard form:
y' - (1/2x)y = 0
so the answer is ke^(-u(x)), where u(x) is the integral of -1/2x

but my answer is different from the correct answer, why?

[cerulean cradle]

We can choose the constant of our choice

To get desired result

[valid zodiac]

i know

[cerulean cradle]

Its manipulation just

[valid zodiac]

but i got k * sqrt(|x|)

[cerulean cradle]

I too

Thats the answer

[valid zodiac]

the answer is y^2 = C*x

if you square that u get

y = c * sqrt(x)

but sqrt(x) is not the same as sqrt(|x|)

[calm coral]

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[muted widget]

since centi = 1/100 => percent = %
and per mille = 1/1000 = ‰
does that mean that there are either perdeci for 1/10 and or per ten thousand for 1/10000?

[untold compass]

sure but no one uses those words

per ten thousand actually has the term "basis point"

[muted widget]

is there one for per ten?

1/10?

[untold compass]

don't know of any

[muted widget]

okay

thank you^^

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[dire depot]

Linear algebra question

[dire depot]

Number 12

So I’m unsure on how to do this

I was looking through notes and the eigenvector is typically given

[shadow tendon]

It is $Ax=\lambda x$ for $x \neq 0$

[dire depot]

And if I were to find it the matrix needs to be square either?

[potent lotus]

Nekoyashiki Yuki

[shadow tendon]

The value of lambda is the eigen value and x is the eigevector

[dire depot]

Oh so it’s just asking for the regular definition

[shadow tendon]

Yeah

[dire depot]

Alright that makes sense

Thank you

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[random stratus]

$\prod_{k=1}^n (k + 1/2)$

[potent lotus]

Max.cc

[random stratus]

how can we express this with factorial

[remote mural]

you cant

you have to use the real-extended version of the factorial

called the gamma function

[random stratus]

hum but if i write it like this $\frac{\prod_{k=1}^n (2k + 1)}{2^n}$

[potent lotus]

Max.cc

[random stratus]

i think we could have something

[calm coral]

@random stratus Has your question been resolved?

[random stratus]

<@&286206848099549185>

[hollow granite]

doesnt help

still needs gamma function

[hollow granite]

well actually there is a way to do it

without gamma

[hollow granite]

this one is actually easier

firstly write it as a pochammer symobl

which you can directly convert your product to

using pochammer symols

it is (1.5)_{x}

which we can turn into a gamma function

1.284... * gamma(x+1.5)

thne using the equality

you can have an expression only using factorials

which after some fiddling you get to

which then become this monstosety

@random stratus

the process is painful i am unable to explain in it 1 comment

if you want further explanation you can either dm or ping

[random stratus]

alright thx for explaning

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[coarse minnow]

Given an arbitrary complex-valued polynomial f(z)
is it always possible to construct a matrix A that has f as its characteristic polynomial?

[coarse minnow]

if not, why?

[hollow totem]

How about a matrix with the roots of the polynomial on the diagonal and zeros everywhere else

[coarse minnow]

that seems to solve it very neatly

thanks

[hollow totem]

Np

[coarse minnow]

for all roots z_1, ..., z_n we construct a matrix as diag(1-z_1, 1-z_2, ..., 1-z_n)

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[hollow totem]

ah right

I just did determinant is the polynomial

Teamwork

[glass heart]

.reopen

[calm coral]

✅

[glass heart]

there is also a solution by putting the coefficients of the polynomial into the matrix itself

try finding that one

that also works over fields where not every polynomial factors

|| companion matrix||

@coarse minnow

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[devout yoke]

How do u do this?

[remote mural]

for this you'll have to use the laws of cosine

[glacial nacelle]

,rccw

[calm coral]

@devout yoke Has your question been resolved?

[glacial nacelle]

,rccw

[glacial nacelle]

@devout yoke have you tried using the laws that were posted above?

[devout yoke]

I havent tried yet

[glacial nacelle]

would you like more guidance or are you just taking a break

either is fine

[devout yoke]

couyld you show me how to do it pls😭

[glacial nacelle]

np

so for the law of cosines

we have that a^2 + b^2 - 2ab*cos(c) = c^2

Do you know what that means?

[devout yoke]

yeah kinda

[glacial nacelle]

can you guess where we want to apply that in the figure?

[devout yoke]

well im probably wrong but should we use it for the triangle with G in it?

idk😭 😭

[glacial nacelle]

dont be so pessimistic

thats correct

you want to find the angle

and the only way to find an angle using the formula is by finding what cosx is

so if we can isolate cosc in the formula

and we can evaulate the other side

and take cos^-1

then we will have x

wanna have a shot at it?

if you want, you can post your work every so often so that someone can follow you

[devout yoke]

really?

[devout yoke]

hmm alright

[glacial nacelle]

you want all three sides

and the only triangle including angle x that you know all three sides of is triangle HGF

So try applying law of cosines on that triangle

[devout yoke]

I got 29.5

tysm!!

.close

[calm coral]

Channel closed

Closed by @devout yoke

Use .reopen if this was a mistake.

[calm coral]

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[verbal abyss]

Im not sure how to go about these problems, im not too unfamiliar with the emperical rule but i just dont know where i should be looking first

[leaden marsh]

can u state the empirical rule?

[verbal abyss]

I know that you need to know the standard dev and the mean before you can use it. I also know that the rule is smth along the lines of 68-95-99.7

[leaden marsh]

yes

approx. 68% of data lies within 1 sd, 95% lies within 2 sd, 99.7% lies within 3 sd

we know the sd is 10

soooooo

how much of the data is within each shaded region of the graphs they give you?

[verbal abyss]

erm

would it b the 3rd graph?

[leaden marsh]

wdym

[verbal abyss]

wait no

i dont think i get what ur asking lol

my b

[leaden marsh]

let's define the shaded region of the 1st graph

[verbal abyss]

60-80

[leaden marsh]

the mean is 70, the sd is 10

can u express it in terms of these numbers

[verbal abyss]

is the outline smth like 68% of the data is within 1 standard dev of the mean...?

[leaden marsh]

yes

the shaded region is 68% of the data

similar thing w the 2nd?

[verbal abyss]

95% of the data is within 2 standard dev

and then 99.7 of the data is within 3 stand. dev..?

[leaden marsh]

yes

[verbal abyss]

so i have the mean and the sd, we talked about upper bounds n lower bounds. I think i know how to calculate them but i dont know how we got those bounds...

these are my notes, they make sense on paper but i feel like im missing smth to apply them to the questions given D:

[calm coral]

@verbal abyss Has your question been resolved?

[nimble badge]

Does Z* or Z-critical sound familiar?

[verbal abyss]

z scores?

[nimble badge]

Yeah, when discussing z scores

[verbal abyss]

z score = xi -mean/standard deviation

yeah..?

[nimble badge]

(No but actually don’t worry about it, you don’t need it)

[nimble badge]

Yes