help-28

[keen spruce]

@inland latch f(x) has a critical value when ___ = ___?

[inland latch]

i just need the points of critical value on the graph f(x) based off the graph

[keen spruce]

yes

when does f(x) have a critical value

[inland latch]

when the graph changes direction?

[keen spruce]

yes... getting closer

when f'(x) = what

[inland latch]

CROSSES THE X AXIS

[keen spruce]

when f'(x)=0

and change of sign

[inland latch]

but how can i know if the point is a global or local point

[keen spruce]

we need information about f(x)

[inland latch]

b is use the f'(x) to know when f(x) changes directions

no need to find the exact spot just be able to explain why

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[torn jolt]

How to integral right-bottom part?

[oak seal]

change the variable x

[torn jolt]

to u?

[oak seal]

take u = ln(x)

[torn jolt]

yeah i did but

how to get dv

oh

[oak seal]

u= ln(x). du=1/x dx

[torn jolt]

mis typed wait

[keen spruce]

yea this is u sub not ibp

right

[torn jolt]

oh

[keen spruce]

@oak seal right

[torn jolt]

ignore that x after sin on right hand side

ok I am now trying

I got sth like "intsymbol sin(u)dx" and I have (1/x)dx = du, so how to simplify further

[oak seal]

?

you should have integral sin(u) du

[torn jolt]

how did you get du

[oak seal]

.

[torn jolt]

is there any 1/x dx

sorry btw i really suck

[oak seal]

it’s clearly sin(ln(x)) / x dx

the 1/x dx = du

and you replace the ln(x) with u

i’m gonna leave you

[torn jolt]

okay thx

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[sturdy trail]

Need some help

[keen spruce]

,rotate

[keen spruce]

@sturdy trail with what

[sturdy trail]

i don’t understand what it means by sequence

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[robust cave]

what goes wrong here:

[robust cave]

$e^{\pi i} = -1$

[glossy valve]

Suro

[robust cave]

$(e^{\pi i})^{\frac{1}{2}} = (-1)^{\frac{1}{2}}$

[glossy valve]

Suro

[robust cave]

$e^{\frac{\pi i}{2}} = i$

[glossy valve]

Suro

[warm abyss]

Where's the problem?

[robust cave]

thats my q

[warm abyss]

That equation is true?

[robust cave]

oh

ok thx

oh forgot

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[dark escarp]

How can I differentiate (log x)^2?

[limber flicker]

⛓️

[torn jolt]

⛓️

[dark escarp]

@limber flicker @torn jolt 2log x?

[limber flicker]

close

forgetting to multiply by the derivative of the inside

F(x) = f(g(x))

F'(x) = f'(g(x)) * g'(x)

[dark escarp]

@limber flicker fine now?

Many thanks @limber flicker @torn jolt

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[worthy gull]

Hello

[worthy gull]

How do I find the charge in a capacitor without a given Volt?

[errant juniper]

i need help finding my test statistics

@charred jolt

[worthy gull]

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[elder cape]

What will the lowest common denominator here be and why?

[keen spruce]

well

we have a factor on that right fraction that the left doesnt have

so let's multiply by a giant one to make the denominators equivalent

[elder cape]

Yes which is (1+cos(x))

[keen spruce]

multiply the numerator and the denominator by it

(thats really a giant 1)

[elder cape]

So the common denominator is (1-cos(x)) × (1 + cos(x)) and the numerator is (1+cos(x)) - 2 ?

[keen spruce]

yes

[elder cape]

Also, is 1-cos(x)^2 a difference of squares?

Which is just become 1+cosx × 1-cosx

[keen spruce]

maybe 😉

[elder cape]

?

[rapid rain]

I mean it was already factored

[keen spruce]

,w (1+cos(x))(1-cos(x))

ignore

[elder cape]

This:

Becomes this:

Via common denominator as we discussed, right?

Yes? No?

<@&286206848099549185>

[rapid rain]

you can simplify the numerator a little more

and then magic

[elder cape]

Yes I can further simply but this step is correct right?

[rapid rain]

yep

[elder cape]

Also, is 1-cos(x)^2 a difference of squares?

[rapid rain]

it is 1^2 - cos(x)^2, so I think you'll agree that yes

[elder cape]

Yes ty lol

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[tepid basin]

how do i get third term?

i know a1 is 2

but i dont know what a3 is

i know third term : a(n-1) = a(3-1) = a2

[quaint prawn]

the third term would be a3 right

[tepid basin]

yeah

[quaint prawn]

where n=3

[tepid basin]

yep

[quaint prawn]

so then what would be a(n-1)

[tepid basin]

a2

[quaint prawn]

and don't you know what a2 is

[tepid basin]

but i dont know what a2 is equal to

well

maybe i'm confusing myself

[quaint prawn]

don't you have it entered in right up there

the second term

is a2

[tepid basin]

i think it wants the solved value

[quaint prawn]

yes

and you find that by plugging in what a(n-1) is

which is a2

which you said above is 0

[tepid basin]

wait a min

[quaint prawn]

so plug in 0 for a_(n-1)

[tepid basin]

oh

so its like a chain thing...

i see

[quaint prawn]

⛓️

[tepid basin]

me

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[grave moon]

how is rcostheta equal to x?

[grave moon]

what if the radius is equal to 2?

then cos(theta)*2 is not equal to x

right?

does anyone follow it?

[rocky vale]

cos(theta) = x/r, yes?

if r=2

cos(theta) = x/2

I think I follow lol

Isn't this true? I don't see the problem

Make a triangle with x, y, and r in the first quadrant

and just do SOH CAH TOA

and you get cos(theta) = x/r

[grave moon]

cos(theta) would equal sqrt(3)

and that is not equal to x/2

[rocky vale]

yeah hang on

It seems like you're talking about a specific value of theta here?

[grave moon]

yeah

in a 30 60 90 triangle

if the radius(hypotenuse)

is equal to 2

[rocky vale]

ok sure

yes

[grave moon]

then what would the second longest side be

sqrt(3)

right?

because the shortest would be 1

[rocky vale]

sqrt3, yeah

[grave moon]

but if we follow that formula rcos(theta)

r is equal to 2

and cos(theta) is equal to

sqrt(3)

so we get 2*sqrt(3)

2*sqrt(3) != sqrt(3)

is my conclusion

[static bramble]

sqrt(3) > 1 so cos(theta) cannot possibly equal that

[rocky vale]

It sounds like you're describing this triangle, right?

[grave moon]

yeah

so what is cos(30 degrees)?

[rocky vale]

following the triangle, this shows cos(30) = sqrt(3)/2

which is true

[grave moon]

wait why is it over 2

[rocky vale]

in the triangle?

Just SOH CAH TOA

[grave moon]

but you wrote sqrt(3)

in the image

[rocky vale]

cosine = adj/hyp

sqrt(3)/2

The cosine is equal to the x value ONLY in a unit circle

where the radius is 1

[grave moon]

oh..

and the formula

rcostheta is not bound to the unit circle?

it can be outside of the unit cicle?

[rocky vale]

nope, if it's a unit circle then r is just equal to 1

so it's just costheta = x in that case

[grave moon]

i see

thats exactly what i got confused on thanks so much

[rocky vale]

no problem 👍

[grave moon]

sorry just wanted to ask one more

so rcos(theta) is from the unit circle?

[grave moon]

oh yeah

you said that lol nvm

[rocky vale]

haha no problem

yeah, just thinking of the cosine as the x value of a point on the unit circle is a good way to interpret it

so if it's a bigger circle, it's just scaled up by a factor of r

same for sine

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@grave moon Has your question been resolved?

[grave moon]

oh i see why i was getting the wrong value

is it a common mistake for people to think of cos as just adjacent side?

i had the habbit of thinking as the adjacent side because of the unit circle

but in reality it's adjacent over hypotenuse

when i think of a 30 60 90 triangle

i think of it as 30 degrees on the unit circle

and just think of adjacent side lol

which skips the whole process of "over hypotenuse"

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[last scaffold]

can someone explain example 8 and use it to solve question 2

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[vast fossil]

!status

[full forum]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[copper harbor]

to be more clear i want to know how to get to the answer

[hallow walrus]

try using different x values

[vast fossil]

Alright, you basically need to consider 3 cases: 0 < x < 1, x = 1, x > 1

Assuming this is not complex analysis

[spice orchid]

Negative x too

[copper harbor]

nah its fairly simple

i just forgot how to do it 💀

do i just test the 3 cases

[vast fossil]

Won't (-3)^n for example go through complex numbers as n gets bigger/generally changes?

[vast fossil]

Yeah

[spice orchid]

Which would make the limit undefined imo

[copper harbor]

the limit is undefined

but there is a range of x that allows it to be defined

is there some formula i can use

[vast fossil]

Fine, consider |x| < 1, x = 1, x = -1, and |x| > 1

[copper harbor]

ok thats helpful thanks

in theory it can negative to -1 right

well not -1 but up to -1

[vast fossil]

Limit of (-1)^n doesn't exist

If that's what you are talking about

[hallow walrus]

the limit wont exist for any negative x

[copper harbor]

i got -1 < x <= 1

[vast fossil]

x = -0.1?

[hallow walrus]

how

[vast fossil]

It still goes to 0

[copper harbor]

yeah

lim n-> infinity

[vast fossil]

Some complex numbers getting in the way but still

[copper harbor]

i mean its calculus so we don't have ot think that far

[vast fossil]

Yeah that's it

[hallow walrus]

does that work?

[copper harbor]

ok thanks

[hallow walrus]

ah yes

it does

[copper harbor]

how do i close this lol

[hallow walrus]

.close

[vast fossil]

It's .close

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[copper harbor]

ok thanks

[vast fossil]

[hallow walrus]

where'd your colour go

[vast fossil]

It's hidden cuz I took the "Not very ppl" role

[hallow walrus]

ah

[vast fossil]

I still have Very Active and Helpful

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[fossil mason]

I was wondering how to get this problem started

[fallen loom]

so f(6)=5

let the function be y=f(x)

that means when you put x=6, you get y=5

hence, (6,5) is one point on the graph of the function in x-y plane

similarly you can find the other point

[fossil mason]

Ok then put that into the slope formula and find the slope of the line

[fallen loom]

and using the two points find the slope of the line

[fallen loom]

yes!

[fossil mason]

Then put it into slope intercept formula then that’s the answer

Thx

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[stray rover]

what am i doing wrong here

I integrated

and i got -1/3+x

then a/1-r

-1/3-(-x)

AR^n

-1*(-x)^n

is that not correct?

[rapid rain]

You integrated 1/(3+x)² to get to -1/(3+x), which give you its power series, and then you... do what exactly?

Basically you want to use the fact that :
$\frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^nx^n$

[glossy valve]

rafilou2003

[stray rover]

a/1-r

=

series of X^n

[stray rover]

im using a/1-r

[rapid rain]

Ok, so then

[stray rover]

where a is constant

[rapid rain]

$\frac{a}{1-x} = \sum_{n=0}^{\infty}ax^n$

[stray rover]

?

[glossy valve]

rafilou2003

[stray rover]

ye

-1/3-(-x)

is = to

-1/3+x

[rapid rain]

Write it down

[stray rover]

i have

our X

is (-x)

a is -1

so -1*(-x)^n

which is (-1)(-1)^n(x)^n

[rapid rain]

But that doesn't give you the good stuff

[stray rover]

wym

[rapid rain]

You want $\frac{1}{3+x} = \frac{a}{1-r}$

[glossy valve]

rafilou2003

[rapid rain]

What did you pick for a, and what did you pick for r? And does it make sense?

[stray rover]

a is -1

r is (-x)

[rapid rain]

But then this gives you $\frac{-1}{1+x}$, which is not it

[glossy valve]

rafilou2003

[stray rover]

how

this is the same thing

is that

its in the form of a/1-r

[rapid rain]

But the formula is for a/1-r, not a/3-r

[stray rover]

ohh

i forgot

lol

mb

multiply

by 1/3

[rapid rain]

There we go

Now we can apply :
$\frac{-1}{3+x} = \sum_{n=0}^{\infty}\frac{-1}{3}(\frac{-x}{3})^n$

[glossy valve]

rafilou2003

[rapid rain]

So $\frac{-1}{3+x} = \sum_{n=0}^{\infty}(-1)^{n+1}\frac{x^n}{3^{n+1}}$

[glossy valve]

rafilou2003

[rapid rain]

Got it @stray rover ?

[stray rover]

yes

[rapid rain]

So now, to get back to 1/(3+x)², what do we do...?

[stray rover]

differentiate

[rapid rain]

Yep

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@stray rover Has your question been resolved?

[stray rover]

ok im confused

[stray rover]

@rapid rain how did u get this

[rapid rain]

from this @stray rover

[stray rover]

yea but how

is it not (-1/3)(-1/3)^n(x)^n

[stray rover]

how did u get the 1

[rapid rain]

it is exactly what i wrote

$(\frac{-x}{3})^n = (\frac{-1}{3})^nx^n$

[glossy valve]

rafilou2003

[stray rover]

the derivative

of that would be

(-1)^(n+1)n*(x)^(n-1)

/

3^(n+1)?

[rapid rain]

yes

though you will need to reindex the sum, not sure if x^(-1) is pretty

[stray rover]

how did they get this then

[stray rover]

how would i do this

[rapid rain]

reindex the sum : take k = n-1

[rapid rain]

so this expression becomes, in terms of k

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

[glossy valve]

rafilou2003

[stray rover]

how did u get k+2

im confused for that part

[rapid rain]

see what i did there ? everytime there was an "n", I replaced it by "k+1"

[stray rover]

ooh

[rapid rain]

for example, 3^(n+1) becomes 3^((k+1) + 1), which is 3^(k+2)

[stray rover]

i see

why do we have to do this tho

[rapid rain]

we always need to have x^n in our series, not x^(n-1) or anything else

[stray rover]

H

ah i c

or (x-c)^n only?

[rapid rain]

Yes

[stray rover]

ok then for this

we just take antidervative

of this and we get -2/u^2

which is what we did basically in the part 1

but multiply by -2

i get

now we do like m=k-1?

[rapid rain]

Yes

Also, it's divided by -2

Not multiplied

[stray rover]

why divided

oh wiat

yea

right

[stray rover]

boom chakala

[stray rover]

and for this its the same thing multiplied by x^2?

or different

[rapid rain]

Well now yes we multiply by x^2

And it's done

So x² * x^n = x^(?)

[stray rover]

x^2+n

[rapid rain]

yes

[stray rover]

and for this one we take the dervative

and get

we find series for that

then multiply by -2x?

then take the integral of that

right?

[rapid rain]

you don't know the power series of ln(1-x)?

[stray rover]

i took dervative of ln(1-x^2) here

[stray rover]

its 1-x^2

and no

[rapid rain]

$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$

[glossy valve]

rafilou2003

[rapid rain]

integrating this:

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

[glossy valve]

rafilou2003

[rapid rain]

replacing x by x^2 :

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

[glossy valve]

rafilou2003

[rapid rain]

easy, isn't it?

[stray rover]

how would we do it using dervatives

and integrals

i assume thats how my teacher wants me to d o it

[rapid rain]

well, that's pretty much how i did it:
step 1: derivate ln(1-x) (easier expression) to get 1/(1-x)
step 2: develop the power series of 1/(1-x)
step 3: integrating the power series, ln(1-x) - ln(1) = ...
step 4: replace x by x^2 to get the result for ln(1-x^2)

[stray rover]

yeah

but

how would go if u started from ln(1-x^2)

[stray rover]

antiderivative of this is -ln(x-1)

no?

[rapid rain]

ok in that case, derivate $ln(1-x^2)$ to get $\frac{-2x}{1-x^2} = \sum_{n=0}^{\infty}-2x*x^{2n}$

[rapid rain]

yeah, let me redo it

[rapid rain]

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

[glossy valve]

rafilou2003

[rapid rain]

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

[glossy valve]

rafilou2003

rafilou2003

[stray rover]

that would be wrong tho?

a has to be a constant

and we have -2x

thats what i was asking here

if we do it this way

[rapid rain]

not necessarily

[stray rover]

how come

[rapid rain]

well, since 1/(1-x) = sum (x^n), if we just multiply by -2x on both sides we get
-2x/(1-x) = sum (-2x*x^n)

and there's nothing wrong with that

[stray rover]

thats what i meant here

[stray rover]

.

get the series of 1/(1-x)^2

after factoring out the -2x

then multiply by -2x

[rapid rain]

so you agree that "a" can be anything, not necessarily a constant

[stray rover]

yea

[rapid rain]

alright

[stray rover]

so i get this

-2x^2n+1

which is the derviative

of ln(1-x^2)

[rapid rain]

Yep

[stray rover]

simplify to this?

[rapid rain]

i don't know why the 2n+2 became 2n+1 in the exponent

[stray rover]

antidervative rule

x^2n+1 + 1/2n+1 + 1

[rapid rain]

you wrote -2x^(2n+2)/(2n+2) which you simplified into -x^(2n+1)/(n+1)

I assume you meant -x^(2n+2)/(n+1)

[rapid rain]

which is what I got here

[stray rover]

yea

-x^(2n+2)/(n+1)

which can be further simplified to

[stray rover]

.

[rapid rain]

no

[stray rover]

why not

[rapid rain]

for starters, you get a non zero coefficient in 3

then, the coefficient in x^2 is positive

perhaps you meant $\frac{((-1)^n-1)x^{n+2}}{2n+2}$

[stray rover]

nvm

no

Idk why u cant do that

[glossy valve]

rafilou2003

[stray rover]

so

our An

is -1/n+1

to find coefficients

[rapid rain]

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

[glossy valve]

rafilou2003

[stray rover]

we still have to further simplify??

[rapid rain]

nope, you should be able to find the coefficients from here

[stray rover]

no i mean from -x^2n+2/n+1

we have to simplify that?

[rapid rain]

not needed

[stray rover]

how the heck

R is x^2n+2

the rest left is -1/n+1

[rapid rain]

you can use the ratio test to show that $\frac{\frac{r^{2n+4}}{n+2}}{\frac{r^{2n+2}}{n+1}} \longrightarrow 1$

[glossy valve]

rafilou2003

[rapid rain]

so r = ...

[stray rover]

wait r = x

because they dont change right

if we integrate or differentate

and from -2x/1-x

r is x

[rapid rain]

if you want to determine the convergence radius, you replace x in the formula by r, and you put everything in absolute value

[rapid rain]

giving us this

[stray rover]

well we dont have t odo that tho

since we know r = x

right?

-1<x<1

R = 1

[rapid rain]

Ok good

[stray rover]

i got r

from -2x/1-x

equation

because r never changes

so that is correct?

[rapid rain]

by r i meant R

[stray rover]

oh

i was taking about

[rapid rain]

so replace "r" here by "R" and you get what i meant

[stray rover]

yea

I was talking about something else here

lol i think u got confused

I was talking about the geometric Ar^n

to get the coefficients

[stray rover]

isnt a = 1/n+1

is what i meant

to get coefficients

[rapid rain]

what coefficients c0,c1,c2,c3,c4 did you get?

[stray rover]

1,1/2,1/3,1/4,1/5

[rapid rain]

you almost have the idea

ln$(1-x^2) = \sum_{n=0}^{\infty}-\frac{x^{2n+2}}{n+1}$, but from here we have to deduce ln$(1-x^2) = \sum_{n=0}^{\infty}c_nx^n$

[glossy valve]

rafilou2003

[rapid rain]

do you see what I mean? @stray rover

[stray rover]

yea

an is what i meant is cn

so we have to get

x^n

by itself in the series we have?

[rapid rain]

we know ln$(1-x^2) = \sum_{n=0}^{\infty}-\frac{x^{2n+2}}{n+1}$, so know we have to write it out : ln$(1-x^2) = -\frac{x^{2}}{1} - \frac{x^{4}}{2} - \frac{x^{6}}{3} + ...$

[glossy valve]

rafilou2003

[rapid rain]

do you get it @stray rover ?

[stray rover]

no not really

[rapid rain]

we know ln$(1-x^2) = \sum_{n=0}^{\infty}c_nx^n$ = c_0x^0 + c_1x^1 + c_2x^2 + c_3x^3 + c_4x^4 + ...

[glossy valve]

rafilou2003
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rapid rain]

we also know ln$(1-x^2) = -\frac{x^{2}}{1} - \frac{x^{4}}{2} - \frac{x^{6}}{3} + ...$

[glossy valve]

rafilou2003

[rapid rain]

so from this, can you identify the values of c0,c1... ?

It should be easy once I write that $c_0x^0 + c_1x^1 + c_2x^2 + c_3x^3 + c_4x^4 + ... = -x^{2} - \frac{x^{4}}{2} - \frac{x^{6}}{3} + ...$

[glossy valve]

rafilou2003

[rapid rain]

@stray rover ping me if you need more help

[stray rover]

-1,-1/2,-1/3

[stray rover]

-1,-1/2,-1/3?

[rapid rain]

what is the coefficient in x^0?

[stray rover]

-1

[rapid rain]

no, that's the coefficient in x^2

[stray rover]

oh 0

0

[rapid rain]

yes !

the coefficient in x^1 is...

[stray rover]

-1/2

[rapid rain]

look again

[stray rover]

how are u getting that?

the equation is -x^2n+2/n+1

c_0

-x^2/1

[rapid rain]

@stray rover

[stray rover]

yea when n=0

-x^2/1

-x^4/2 when n = 1

[rapid rain]

c_0 is not when n=0, it's the coefficient in front of x^0

here, there is no term in x^0, so c_0 = 0

[rapid rain]

if you really want to get the general expression of c_n, use this

[stray rover]

Oooh

thats what

you mean

when c_0 they mean what is when x^0

C_1

when x is x^1

?

[rapid rain]

yes

so, by that logic, what are the values of c0,...,c4?

[stray rover]

0 0 -1 0 -1/2

but i had 1 question for that

yesterday

i did

this question and i get n10^n*(-1)^n+1(x^n-1)

and i used for C_0 = n10^n*(-1)^n+1

[rapid rain]

yeah, it won't work this way

[stray rover]

it did tho

[rapid rain]

it works if it is ... * x^n

[stray rover]

i just plugged in 1 for C_0

[stray rover]

ah i see now

[rapid rain]

oh yeah if you plug n+1 for c_n

but that's the same as taking (...)x^n

[stray rover]

,d/dx

[stray rover]

in here its asking what is the lowest degree coefficient we can find?

that isnt a 0

would be coefficient of 12

and c_12 is 70/12

[rapid rain]

It's asking about the lowest n such that c_nx^n is not zero

So It's 12

[stray rover]

so i did it correct

but it wasnt asking for coefficient

x^12 which is C_12

when n = 0

also

the first part is correct

but idk why when i simplify it gets wrong

[rapid rain]

Because you have to distribute the exponents

(-2x⁷)^n = (-1)^n 2^n x^(7n)

[stray rover]

oh

we have to do the 2 as well

i didnt know

[rapid rain]

So x⁵(-2x⁷)^n = (-1)^n 2^n x^(7n+5)

What did you find for convergence radius?

[stray rover]

bruh idk how to check it

its wired

r = -2x^7

|-2x^7|<1

-1/2<x^7<1/2

sqrt of 7(-1/2) < x < sqrt of 7(1/2)

how am i supposed to check convergence

[rapid rain]

It works, don't worry

[stray rover]

yes but we are supposed to check

the endpoints

💀

[rapid rain]

Looking at $f(-⁷\sqrt{\frac{1}{2}})$, you can immediately see the problem

[glossy valve]

rafilou2003

[rapid rain]

@stray rover does this answer your question?

[stray rover]

no not really

what do u see that i dont

[rapid rain]

Go back to f(x) = x⁵/(2x⁷ + 1)

What happens when x = -7throot(1/2) ?

[stray rover]

0

but 7throot(1/2)

[rapid rain]

0???

[stray rover]

x^5/0

=0

[rapid rain]

1/0 = 0 now?

[stray rover]

2*root(-1/2)

is -1

+1

is 0

x^5/0

[rapid rain]

I think you're confusing yourself

Because 1/0 is most certainly not 0

Other wise, 1=0 and that's definitely not true either

[stray rover]

1/0 DNE

[rapid rain]

Yes

So f(-7throot(1/2)) is DNE, so the convergence radius IS 7throot(1/2)

[stray rover]

why are we plugging in

F(X)

when we need to check the convergence

in the series

thats how we do it

or my teacher does

so would it be (-1)^n(2)^n(7root(-1/2))^7n+5

and we would have to do AST

[rapid rain]

Well, if f(-7throot(1/2)) is DNE, then obviously sum((-2)^n (-7throot(1/2))^(7n+5)) does not converge

[stray rover]

ok what about

[rapid rain]

If you want to expand, it just becomes -7throot(1/2)^5*sum(1)

[stray rover]

7throot(1/2))

we have to check that as well

[rapid rain]

You're going to get 7throot(1/2)^5*sum((-1)^n)

Does this converge or diverge?

[stray rover]

why are putting in f(x)

[rapid rain]

Oh we're just looking at the power series and inputing 7throot(1/2)

Nothing to do with f(x) here

[stray rover]

teenw why u write that

it would be

(-1)^n(2)^n(7root(1/2)^7n+5

[rapid rain]

@stray rover

[stray rover]

where did 7n go

can u write that in latex

hard to understand

[rapid rain]

$2^n * ⁷\sqrt{\frac{1}{2}}^{7n} = (2*⁷\sqrt{\frac{1}{2}}⁷)^n = 1$

[glossy valve]

rafilou2003

[rapid rain]

Clearer @stray rover

[stray rover]

no because where did ur 5 go

why not just say this

im confused

ur just plugging in 7root(1/2)

into x

[rapid rain]

$(-1)^n2^n × ⁷\sqrt{\frac{1}{2}}^{7n+5} = (-1)^n × ⁷\sqrt{\frac{1}{2}}⁵(2× ⁷\sqrt{\frac{1}{2}}⁷)^n = (-1)^n× ⁷\sqrt{\frac{1}{2}}⁵$

[glossy valve]

rafilou2003

[rapid rain]

Clearer?

[rapid rain]

Thus this @stray rover

[full forum]

@stray rover Has your question been resolved?

[stray rover]

oooh yes

[stray rover]

i think the placement of 7 here confused me

[stray rover]

that would mean it converges?

right

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✅

[stray rover]

i have qquestion for interval of convergence
How are we supposed to check it
the endpoints
i get like some absurd number
7root(-1/2) < x < of 7root(1/2)
are endpoints

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[torn jolt]

I have a test tomorrow, and want to be able to check all my answers to make sure they are correct.

I know that if I have a number like 2, I could just plug it back into the original equation to see if it’s true.

If its true the answr is correct, right?

But how could I check my answer when I get an answer like this?

[hot herald]

not practical to go back and check your answer for everything due to time constraints,
you can plug your values like any other, its just more tedious to do the expansion

but ideally make sure each individual step is done properly

[torn jolt]

But I want to make sure I am correct

[hot herald]

you can plug your values like any other, its just more tedious to do the expansion

[torn jolt]

Oh, let me try. But idk how I’m going to plug that “number” in

[torn jolt]

Bro, what is even that

🤯

[hot herald]

using conjugates may help,

though you could multiply both sides by the lcd just like before

[torn jolt]

What is a “conjugate”? The opposite of something you mean?

[hot herald]

a+b and a-b are conjugates of each other

opposite is a vague word, try not to use it

[torn jolt]

Ok

But is this how I have to do it then?

How would I even do this

@hot herald

[torn jolt]

<@&286206848099549185>

Anyone ??

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[grizzled ridge]

hello can i get a help from this one?

[grizzled ridge]

im stuck with the integration part

this is where im at

knowing this is a trigo sub im still confused with the later process on solving, please help me, also im having trouble finding the new limits for this one consider that t=3 and t=0

[pulsar gale]

,rotate

[pulsar gale]

boi is that an integral symbol

[grizzled ridge]

sorry for the symbol

but yes

[grizzled ridge]

sorry sir

[devout inlet]

Your set up looks correct. Evaluating it seems very difficult though.

[pulsar gale]

can you show the rest of your work

[grizzled ridge]

here, sorry if its quite messy

@pulsar gale

[pulsar gale]

jeesus

thats one question?

[grizzled ridge]

yessir

[pulsar gale]

holy

[grizzled ridge]

thats why im totally jdjsjdjsjsjsjsjs

idk what to do now

[pulsar gale]

im guessing no calculator?

[grizzled ridge]

yeah, no calculator

plus, our prof wont allow us to answer in decimals so it has to be in simpliest form without decimals

please help me, to anyone here 🥹

[pulsar gale]

looks like pain

prob will take a few hours to find someone ngl

[grizzled ridge]

but,can u do it sir?

[pulsar gale]

nah i never learned trig subs 💀

[devout inlet]

Use: https://www.integral-calculator.com. It provides the full steps.

[pulsar gale]

yeah I was just gonna say

use that

@devout inlet what is that pfp

[devout inlet]

It seems to complicated to do reasonably.

[grizzled ridge]

thank yooooooou so much, lemme try this

[devout inlet]

The deputy head USEC (one of the two factions you can pick from) as I call him, the black beret is the head USEC. It's a bald USEC in the game Escape from Tarkov with a blue beret on. It's just a meme between my friends really. 🙂

[pulsar gale]

gotcha

[grizzled ridge]

uhm sir, i a, having a difficulty using this one

can u please help me h

[grizzled ridge]

can u please show me how they got the limits with the new function

[grizzled ridge]

but it did not show how did the limits changed hehe sorry for asking to much

im just confused how they got new limits

[devout inlet]

Maybe leave out the bounds till the end and put everything back to t.

[grizzled ridge]

ow so ymou suggest that i use the original t=3 and t=0?

until i got back the values?

[devout inlet]

Just don't include bounds. Go through the process to solve

[grizzled ridge]

okay lemme try, thank you sir!

[devout inlet]

$\int \sqrt{\frac{1}{2} t^2 + 1} \dd{t}$.

[glossy valve]

stabulo

[grizzled ridge]

it is not giving me the correct answer :((

[devout inlet]

The calculator will almost certainly give the correct answer.

[quiet stirrup]

This isn't really taught in most calculus classes, but I would use an Euler Substitution for this kind of integral

https://en.wikipedia.org/wiki/Euler_substitution See the first Substitution of Euler.

[grizzled ridge]

well hehe i tried manullay computing

[quiet stirrup]

So you would in this case factor out the constant sqrt(2). Then use sqrt(x^2 + 2) = x + t as your substitution

[grizzled ridge]

lemme try this sir

[quiet stirrup]

The first example in the wikipedia entry, despite looking quite different from yours is basically the same algebra

It's not the most useful exercise tbh. I use an euler substitution maybe once every 3 years.

[devout inlet]

I feel like you could learn many tricks like this form the old calculus books. I should probably look to study them at some point.

[quiet stirrup]

I was taught this in honors calculus. I think Stewart probably has it somewhere. He has good exercises that are only appreciated by people that already know Calculus

All the advanced tricks boil down to Euler sub, Half angle sub, Differentiation under the integral sign, or the residue theorem. The old books are not like a secret repository of black magic

Oh and just using the definition and knowledge of series

[devout inlet]

I don't like Stewart personally, I'm not even sure why anymore.

[quiet stirrup]

Eh, he earned his fortune, no reason to promote it for his heirs.

[devout inlet]

True. He's cashed in for sure.

[grizzled ridge]

i still couldnt get it :((

[quiet stirrup]

A fun fact is that he has the half angle substitution in his book. He says it is invented by Karl Weierstrauss with no citation. To this date all citations end up at Stewart and there is no evidence that Weierstrauss actually knew this technique.

[grizzled ridge]

my answer would be correct if i remove yhe one that i encircled

but it is part of the limit

can u guys help me how to get this correct, please 🥹

guys, @quiet stirrup @devout inlet 🥹

[devout inlet]

I don't know about the Euler trick.

[grizzled ridge]

me neither 😭

[devout inlet]

The book I study from has something called Duhamel's theorem and there's no mention of it anywhere on the internet except from like PDE thing.

[full forum]

@grizzled ridge Has your question been resolved?

[quiet stirrup]

I don't really want to grind out the calculation that I described using the Euler substitution, but you can see that if you use the substitution I gave you, that the integrals are the same

https://www.wolframalpha.com/input?i=(1%2Fsqrt(2))*(integral+(1%2Fw+%2B+w%2F2)(-1%2F(w^2)+-+(1%2F2))dw)+from+w+%3D+sqrt(2)+to+w+%3D+(sqrt(11)-3)+-+int_0^3+sqrt(t^2%2F2%2B1)dt

If you use the substitution sqrt(t^2 + 2) = t + w and do the "u" substitution correctly, you will get the integral on the left hand side which is easy but annoying to integrate, and you can see it agrees with the integral you are trying to calculate since their difference is zero

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@grizzled ridge Has your question been resolved?

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[modest sail]

Just want to make sure the math is correct here

So the equation is

This

And we need to find the x and y intercepts

I got -12 for x intercept and -4sqrt(3) for y intercept

Which is the opposite sign of what they have in the image

Is the image wrong or am i wrong?

Yeah they should be negative

I used symbolab to confirm

Why is the teacher uploading the answer key that is wrong

😦making me confused all the time lol

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[amber moss]

i get why min(X,Y) < a implies x < a or y < a

[amber moss]

but why does max(X,Y) < a imply x<a and y < a

[shy tinsel]

It doesn't

[amber moss]

sorry i corrected

[shy tinsel]

But min(X,Y) > a does

[amber moss]

hmm

im confused

okay so min(X,Y) > a implies X >a Or Y>a

[shy tinsel]

Wait what

Nvm

I mixed it up

[amber moss]

idk im correcting what u said

not correcting

repeating

[amber moss]

yeah xD i did too 10 mins ago and im trying to clear it

[shy tinsel]

This implies both X>a and Y>a

[amber moss]

lets say X = 3

[shy tinsel]

Oh yeah and what u said is correct

[amber moss]

Y = 1

a = 2

hmm

yes ic ic

[shy tinsel]

Wait nvm lol

[amber moss]

min(1,3) is 1

[shy tinsel]

Wait yes

[amber moss]

2

[shy tinsel]

Yes

[amber moss]

but Y isnt bigger than a here

it should be Or not and

i think

?

[shy tinsel]

What

[amber moss]

okay lets

im gonna try to translate into enlgih that helps

[shy tinsel]

Wait i am messing this up so bad

You are correct

[amber moss]

min(X,Y) > a that means the smaller of X and Y is bigger than a

[shy tinsel]

It only implies one of them

[amber moss]

which means the second letter which is bigger

also has to be bigger

OHHHHHHHHHHHHHHHHH

so min(X,Y) > a implies X > a and Y > a

max(X,Y) > a

i dont think this implies anything

because the bigger of the two is bigger than a but the second can be smaller

but max(X,Y) < a should

the bigger of the two is smaller than a that means the second that is smaller has to also be less than a

right right right

english alwyas helps me

i can close this now

.close

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[torn jolt]

I’m trying to do partial fractions set up but u can’t see where I went wrong

[willow sedge]

in line 4

it is (3c-3b)x

but you wrote (3b+c)x

and then -11=-9A

here A will be -11/-9 which is simply 11/9

but you wrote -2

[torn jolt]

Ty

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[slate jetty]

hi, how can i solve it?

[slate jetty]

$2=x +e^x$

[glossy valve]

gelmemeti

[grave elm]

do you know lambert w function?

[twilit leaf]

Not gonna work here

[slate jetty]

nop

[twilit leaf]

Thats xe^x

[slate jetty]

i need this?

[grave elm]

yeah

[grave elm]

1=(2-x)e^(-x)?, then substitution and u=2-x ue^u=e^2

it would be possible to solve it with lambert w

[slate jetty]

no idea, i'm going to study that

thanks

[grave elm]

solution of a=x*e^x is defined to be W(a)

[slate jetty]

mmm okay, if i don't understand "lamber w function" i'll ask again, thanks 🙂

literally i have no idea about this concept

.close

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[upper badge]

why do we want to solve for this? ^^

[shadow oyster]

?

what do u mean by that

[upper badge]

like why are we solving that?

tbh I have no idea whats going on in that explaination lmao

it seems the r^n equation came outta nowhere

[shadow oyster]

because r is the unknown here

[shadow oyster]

so ur teacher made it

[upper badge]

whats the point of putting n in the power though?

don't I already have this equation

[shadow oyster]

its not power

it means the n-nth term

[upper badge]

my bad I meant in here

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[calm field]

can someone pls explain how to solve this?

[vast fossil]

I would start by identifying the intersection points of x = y^2 and x = y^3

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[stuck island]

how would u solve $4x^3-2x^2+4x-2=0$

[stuck island]

i forgot how to do this

[glossy valve]

Hehehehaw

[astral ermine]

You can try hit and trial method