help-39

[sharp smelt]

yeah, ok, that works

[pearl pond]

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[sacred knoll]

can someone please help me out with a sequence problem? thanks

[lusty tundra]

help

[sacred knoll]

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[gleaming heath]

turn it into (1-cos(1/n))/(1/n) and then lhopital

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[lost hearth]

How do i do the second part?

[lost hearth]

<@&286206848099549185>

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[hollow pike]

is this correct?

[sharp tree]

Seem correct

Im too lazy to check for sec and csc but for sin cos tan its seem correct

[hollow pike]

how tho?

we dont know what theta is located

[sharp tree]

I get the exact value of theta then use my calculator to check

[hollow pike]

what how?

[sharp tree]

See the angle that with theta make a 180 degree

Thats angle is inside the right triangle so I get it then get theta

[hollow pike]

we dont know angle of theta tho

[sharp tree]

We do

180- that angle = theta

Theta= 121.89 deg

[hollow pike]

what is "that angle"

[sharp tree]

Its the angle marks by u using the red pen

The angle inside the right triangle

[hollow pike]

that small little angle mark is theta?

[sharp tree]

No

Theta =180- that angle

Dont u see theta??

[hollow pike]

idk what that angle is

[sharp tree]

Bruh 😭

[hollow pike]

please explain

[sharp tree]

Cant

[hollow pike]

ohh reference angle

?

[sharp tree]

Going out rn

Sorry

[hollow pike]

alr

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[knotty schooner]

hi

[knotty schooner]

this is a theorem, i required some help on the a part of the proof which i understand now, but the 2nd half of the proof i dont get

so in terms of the red part

how does F'(t) have at least two roots??

and hence at least one root for F''(t)?

[quartz citrus]

apply rolles theorem to $F(x_0) = F(\hat x)$ and $F(\hat x) = F(x_1)$

[jolly parrot]

chebyshev's infinite pee norm

[knotty schooner]

cant it also be applied to F(X_0) = F(x_1)?

[quartz citrus]

then apply it again on the first derivative

x0 < hat x < x1

if you apply over [x0, x1] you're just going to get 1 zero for F

[knotty schooner]

yh that makes sense thank u

[knotty schooner]

if two functions are the same at an x value, does it auto mean that its derivative is the same at the x value?

im a bit confused on how u apply the theorem again on the first derivative

[quartz citrus]

like f(a) = f(b) implies f'(a) = f'(b)?

[knotty schooner]

ye

[quartz citrus]

take x^2, x = 1 and x = -1

[knotty schooner]

yh nah

its different then

[quartz citrus]

rolles comes from ivp

just think about that

[quartz citrus]

from here

for F(x0) = F(hat x)

apply rolles

[knotty schooner]

ok

[quartz citrus]

you get some q in (x0, hat x)

with F'(q) = 0

then on F(hat x) = F(x1) apply rolles

you get some r in (hat x, x1) with F'(r) = 0

so on [q,r], F'(q) = F'(r)

apply rolles again on the first derivative

[knotty schooner]

thank u

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[flint musk]

How does $e^3-((3)(e^3))$ become $-2e^3$

[jolly parrot]

guy

[flint musk]

the most simplified i got it to was $e^3-3e^3$

[jolly parrot]

guy

[wary bane]

what is x - 3x

[flint musk]

-2x

[wary bane]

what if you let x = e cubed

[flint musk]

ok i see what u mean now lmao

-2e^3

ok thanks

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[delicate ibex]

Hi, im having a test tmr and im working on parabolas

[delicate ibex]

one of the questions are that we get a transformation

and we need to come up with a equastion for it

ill use the example, something goes up by 2 and to the left 7

so for that, would the equastion be y= (x-7) +2?

[glad falcon]

The thing inside the bracket is basically the opposite of what you expect

So a left shift of 7 is actually (x+7)

[delicate ibex]

oh yea i forgot abt that

[glad falcon]

The other problem is if you're shifting a parabola, it's not y=

You'd have a parabola which you'd call a function, say f(x)

So to transform that the way you want, it would be f(x+7) + 2

As y = x-7 + 2 is actually just a straight line you can simplify to y = x - 5

[delicate ibex]

ah ok

so only whats in the bracket

you switch the signs?

[glad falcon]

Ye, the up/down shifts outside the bracket work as expected

Same thing with stretching

If you stretch up and down the numbers sorta work the way you'd expect, like 2f(x) is just a stretch scale factor 2, but f(2x) is a stretch scale factor 1/2 (so you're compressing it)

[delicate ibex]

speaking of stretching, how would you know if somethings a aos, stretch, compress etc

[glad falcon]

That is all there is to it

Number A outside the function like Af(x) = up/down stretch scale factor A, number inside the function like f(Ax) = left/right stretch scale factor 1/A

up/down and left/right are more formally known as a 'stretch in y' and a 'stretch in x' if your teacher is going to be on you for wording

[delicate ibex]

oh ok

uh one last thing

more of a clarification

for this i got

oh wait im supposed to discribe the sequence of transformation

for the record

of like y=x2 which idk what that means but anyway

i got sh 1 unit up

reflection in the x axis by a factor of 3

is that correct?

[glad falcon]

Not quite, so f(x) = x^2 so you now have -3f(x-1)

That -1 is inside the bracket

So that's going to be left/right

Which one?

[delicate ibex]

so sh 1 unit right

[glad falcon]

Ye exactly, it's addition/subtraction so it's a shift rather than a stretch, and it's inside the bracket so minus = right

Now for the outside bit

Do the sign and the number separately

Minus as you say means reflect in the x axis

The 3 is a stretch in the y scale factor 3

Reflecting is just reflecting and doesn't have a scale factor

So write that bit separately

So your final answer would be

• Shift in x by 1 unit right
• Reflection in the x axis (or along y=0)
• Stretch in y scale factor 3

[delicate ibex]

ok wow

ty

[glad falcon]

You could in theory say stretch in y scale factor -3 to be fair

But they might get a bit antsy about that lol

[delicate ibex]

hes chill

we dont even need a concluding sentance

[glad falcon]

In which case a stretch SF -3 is perfectly mathematically valid, it's just something that sometimes examiners in early years get annoyed about XD

[delicate ibex]

ok well thank you

thats all

[glad falcon]

Np

Good luck have fun haha

[delicate ibex]

i wont

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[thin sigil]

Is there a general algorithm for solving these kinds of quadratic systems?

[thin sigil]

<@&286206848099549185>

Oh sorry, 3 more minutes...

[limber oasis]

Didn't find anything satisfactory on google?

[thin sigil]

No, english isn't my first language so I'm probably looking up the wrong terminology

[limber oasis]

It seems you can abstract-algebra yourself to a solution
https://math.stackexchange.com/questions/2119007/how-do-you-solve-a-system-of-quadratic-equations

[limber oasis]

I've never studied this but a trivial idea (though not a trivial method) I just thought of with just 2 equations is to combine them into an equation with no xy term

Then y is given as a function of x by the quadratic formula
Then plugging it into the other equation (conveniently removing the y^2 term for convenience I presume) should yield an equation that results in a 4th degree polynomial...

Ok nvm the first answer is also reducing the number of variables to end up with a high degree single variable polynomial

[thin sigil]

That sounds like something which can't be done in the general case... I don't think I could apply that to the system I have at hand at the moment either...

[limber oasis]

You can
x = 1/2 sqrt(y^2 + 6y + 212)
Then insert that into the first equation and you get an equation in y only
Slight issue is actually the xy term being annoying

But since there's no x term, it still boils down to a quartic in y

[thin sigil]

How would I go about solving that nicely?

Note : This system has a nice solution

x = 9, y = 8

[limber oasis]

Not sure you can do better than reducing to a quartic in general

[thin sigil]

So there's no generalized method for solving systems of higher order like these through matrices or some Gaussian algorithm hack?

[limber oasis]

That just made stumble upon a math paper whose intro starts with "since the beginning of time"
https://www.sciencedirect.com/science/article/pii/S2590037423000146

[thin sigil]

"Since the beginning of time, determining the roots of polynomials has been a significant problem"

As if mankind had nothing better to do at the beginning

[thin sigil]

Well, I guess there's not much to be done here, I'll poke around and see if I can come up with something, tyvm for the help though

.close

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[lean herald]

Trying to help my daughter with test studying. Teacher provided the answer but we have no idea what to do to get there. Here is the problem:

The function f is given by f(x) = sin(2.25x + 0.2) . The function g is given by g(x) = f(x) + 0.5 . What are the zeros of g on the interval 0 ≤ x ≤ π ?

[keen rain]

Sure what have you tried so far to approach finding the zeros?
& What's the answer the teacher provided? :]

[lean herald]

Answer is 1.540 and 2.471 and we plugged in the formula so we start with sin(2.25x + 0.2) + 0.5 but she doesn’t know what to do next and I know even less. We tried isolating x but that gave us weird decimals

[keen rain]

Yes your goal is to solve
sin(2.25x + 0.2) + 0.5 = 0

as we want the roots of g(x)

(roots = zeros)

[lean herald]

Ok so do we need to isolate the x?

[keen rain]

So you can start by getting rid of everything on the left slowly, starting with - 0.5 on both sides

to get
sin(2.25x + 0.2) = -0.5

[lean herald]

0.2 right? Ok with you so far

[keen rain]

oops yes

[lean herald]

Then we do sin -1 on both sides?

[keen rain]

yup, inverse of sine

[lean herald]

Ok so we have 2.25x + 0.2 = sin-1 -0.5?

[keen rain]

yes, but exactly here came the tricky part, if you now solve for x, meaning -0.2 on both sides and then dividing by 2.25 on both sides

[lean herald]

That’s what we did first time

[keen rain]

You'll get a negative value: -0.332...

that's because the sine function is periodic, so we actually found a valid solution for x, but we want to find the solutions within 0≤x≤π

as the task stated

note that if
sin(x) = A
where A is some value

then sin(x+2π) = A as well

since the sine function is 2π-periodic

in fact sin(x+2πn) = A for any whole number n

[regal herald]

remember to change the range youre looking in

[keen rain]

And secondly, when sin(x) = A
then sin(π-x) = A as well

and that is the trick to get the values you seek :)

so instead of solving
sin(2.25x + 0.2) = -0.5

you actually solve two equations:

sin(2.25x + 0.2 + 2πn) = -0.5

sin(π - (2.25x + 0.2) + 2πn) = -0.5

And here you can do the same manipulations as before

so using arcsin

and moving everything over to solve for x

both equations, namely

[lean herald]

Is this part supposed to be known by her? She’s asking why we are adding 2 pi we got lost sorry!

[keen rain]

Ah np, usually this property should be taught in the class, where when we look at the behavior of the sine and cosine functions

we can observe that they contain a pattern which repeats every 2π

(as it's a curve with period 2π)

[lean herald]

Ok you just refreshed her memory she was thinking 1pi not 2

[keen rain]

ah kk np

[lean herald]

So let us try it one sec

[keen rain]

sure

It can be a little difficult getting used to solving equations with sine and cosine since we essentially solve for an infinite amount of solutions

which is why they restricted the solutions to be within the range from 0 to π

[lean herald]

Can you explain how to use arcsin? Trying to solve first one

sin(2.25x + 0.5 + 2πn) = -0.5

[keen rain]

Sure, using arcsin on both sides we get:
2.25x + 0.2 + 2πn = arcsin(-0.5)
like before

oops I copied 0.5 above again, it should be 0.2 :)

[lean herald]

N is the whole number right?

[keen rain]

yes

N is a new variable, it can be any whole number

[lean herald]

Ok so next we would subtract 2piN from both sides?

[keen rain]

Yup, you can subtract 0.2 and 2πn:
2.25x = arcsin(-0.5) - 0.2 - 2πn

[lean herald]

Then divide everything by 2.25 on both sides

[keen rain]

exactly

[lean herald]

What do we put for N now?

In the calculator

[keen rain]

Yes the primitive approach would be to test some values

so you could try -1, 0, 1 for instance

and see which values for x you get

and you'll notice that for N=0, the result is negative, but for N=-1, the result is between 0 and π! 🦩

And for N=-2 it's too big again, greater than π

which means for this first equation, the only solution we can get is for N=-1

[lean herald]

Oh ok so essentially you would try values until finding one that fits in between 0 and pi

[keen rain]

in theory you'd solve explicitly for which N the equation holds, but since your daughter was probably recently introduced to sine I think the teacher intended her to test values for N yes

[keen rain]

At best you also try the values around it

because we know for N=-2 it's too big

and for N=0 it's too small

which means only N=-1 could work

Try the same approach for the second equation and you'll find the second desired value :]

sry gtg but wish thee gl with the rest of the task!

[lean herald]

Ok thank you, appreciate it!

Will keep at it have a good one

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[elder mantle]

I think this is wrong because $|\grad T|$ = 6 not 3

[jolly parrot]

Hamdy Hisham

[elder mantle]

right?

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[elder mantle]

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[fading crescent]

I am currently doing conic sections right now, but I think the hardest part for me is understanding what is given. How do I write the standard form equation fo the circle given the ends of a diameter?

[fading crescent]

so far

I found the center by using the midpt formula

center = (7, -5)

then to find the radius i plugged in one of the points into x and y in the standard form equation

my final answer that I got was:

(x-7)^2 + (y+5)^2 = 61

are my steps correct?

<@&286206848099549185>

[pearl pond]

@fading crescent Has your question been resolved?

[covert lotus]

@fading crescent

ok

,w (x-7)^2 + (y+5)^2 = 61

[jolly parrot]

Results provided by WolframAlpha

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

Requested by fwgx

[covert lotus]

interesting

ok so we have 12,1 and 2,-11 as end point of diameter of a circle

to find the centre

we need to find the x and y axis

so it would be 12,-11

the diamter is 20

and the radius is 10

so we have

(x-12)^2+(y+11)^2=10^2

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[covert lotus]

uh

idk

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[waxen smelt]

Can I please get help with this

[waxen smelt]

I thought we can write a as:

del z/del w * del w/del x

but

if i get the situation right

then

is this whats hapepning

they are are solving for z from the constraint and placing it into w

and then doing implicit differentiation?

i dont get it

they just did this

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.reopen

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✅

@waxen smelt Has your question been resolved?

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@waxen smelt Has your question been resolved?

[pearl pond]

@waxen smelt Has your question been resolved?

[pearl pond]

@waxen smelt Has your question been resolved?

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@waxen smelt Has your question been resolved?

[smoky silo]

can anyone help me with this one? I got 0.354 using u-sub, but I use direct integration on calculator, the answer is 1.152.

[pearl pond]

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[errant vortex]

$\operatorname{proj}_{\mathbf{b}}\left(\mathbf{v}+\left( \right)\right) = + = $.

[errant vortex]

.close

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[leaden schooner]

hi

[leaden schooner]

whast does this mean?

[sharp vigil]

there are many different equations that can be made describing the same plane

[woeful stump]

trivially (0, 1)a + (1, 0)b and (0, 2)a + (2, 0)b form the same plane

[leaden schooner]

ohhh

what does the second part mean

[sharp vigil]

so any point can be represented as a vector from the origin.

1. if you have a point on the plane, then you can get a vector pointing from that point to any other point on the plane by adding specific scalar multiples of any two vectors in the plane that aren't in the same direction
2. adding that to a vector pointing from the origin to the starting point gives you a vector pointing from the origin to the point
3. you now have an equation for any point on a plane in terms of two vectors in that plane, and any point on that plane

[leaden schooner]

ah that makes sense

so say for any point

u just change the scalar multiples

so that it satisties whatever vector ur trying to find?

[sharp vigil]

yes

[leaden schooner]

mmm okay okay

thanks

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[somber snow]

I'm trying to solve this differential equation using powerseries but I got stuck.

[cursive wolf]

you don't want the x in the coefficients

[somber snow]

I figured, how do I get rid of it?

[cursive wolf]

bring the x into the x^n, making it x^(n+1), and then adjust the summation indices to get it back to x^n

you probably also want some initial condition so you have something to start your coefficient recursion

[somber snow]

The exercise doesn't provide me with one

[cursive wolf]

ah ok then you're probably just gonna say a_0 is some fixed value and determine all over coefficients from that

[somber snow]

Is this want you ment?

[cursive wolf]

yeah!

then you can isolate the n = 0 term from the left sum and then compare the rest of the two sums

[somber snow]

But is it possible to combine the series now that they have different starting indexes?

[cursive wolf]

nah, you need to remove it by writing $\sum_{n=0}^\infty (...) = 2 a_2 + \sum_{n=1}^\infty (...)$

[jolly parrot]

Lartomato

[cursive wolf]

and then go do your usual stuff

[somber snow]

Okay, give me a few minutes to try this

So like this? And then I start isolating for n=0,1 etc

[cursive wolf]

yea!

i think you can already use that a_2 = 0 b/c it's the "constant" term

[pearl pond]

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[sweet quest]

is there a name for this derivative rule?

[hollow cobalt]

Sum rule or derivative of sum is sum of derivatives

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[midnight haven]

PLs can someone solve this for me I did the question but my answer doesn't tally with my friends answers

[pearl pond]

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[elder flower]

if i have the expression of a function on the interval R+, how do i get its expression on R- , and R ?

[elder flower]

<@&286206848099549185>

[cursive wraith]

in most cases you can't

especially for piecewise functions

[midnight haven]

i dont get this question

[cursive wraith]

unless you give us the full context we can't help further

[midnight haven]

if a function is defined on R+ ofc not

[elder flower]

i stumbled across exercices where i have the expression of the function when x ∈ R-

and they tell me to give the expression of the function on R+ or R

with the absolute value ig

<@&286206848099549185>

[cursive wraith]

send a picture or something

give examples on what they exactly want

[polar hill]

Send it through

[elder flower]

if f is an odd function defined on R, so for every x ∈ R+ we have f(x) = x^2 + 3x

Calculate f(x) if x ∈ R-

and if x ∈ R

[cursive wraith]

there we go

f is an odd function

that's why you always need to share all details

[elder flower]

SO f(-x) = -f(x)

but what does it change?

@cursive wraith

[cursive wraith]

well

if x is negative

then -x is positive

and you know f(-x)

so f(x) = ....

[elder flower]

f(-x) is -f(x) right?

[cursive wraith]

no that's not what I was hinting at when I said this

[midnight haven]

what

thats just odd function criterion

[cursive wraith]

f(x) = x^2 + 3x when x is non-negative

so when x is negative, f(x) =....

[elder flower]

wait

x is negative

but -x is positive so i have it ( cuzz x ∈ R+)

and f is odd

so f(-x) = -f(x) so f(x) = -f(-x)

and i have to calculate -f(-x) because -x is positive right @cursive wraith ,

?

[cursive wraith]

yes

[pearl pond]

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[unkempt rapids]

can soemone explain this to me why this happens

[unkempt rapids]

what if the absolute value is x-3 and x > -2

[heady finch]

If x approaches -2 you will have |(-2)-3|

Which is |-5| = 5

[unkempt rapids]

but why does it end up -x+5 here if x<5 @heady finch

[heady finch]

Basically absolute value just makes any negative values positive

The moment the value enclosed in the function || becomes negative the function will make the value positive

So if x is below 5 (x-5) will be negative so the function will make it positive by taking the negative values of the negative value hence making it positive

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[storm sage]

Is this correct

[storm sage]

Solving for d) 1

Or proving

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[unkempt rapids]

Im confused why is the answer -9

[unkempt rapids]

did the teacher make a mistake cuz i got -4

[sharp smelt]

let that answer should be -4, yes

I think

,w lim x--> -2 (x^2|x-3|)/(x-3)

[jolly parrot]

Results provided by WolframAlpha

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Requested by physicsrocks

[sharp smelt]

yes

-4

[unkempt rapids]

oh okay thank you

is weird that the teacher did a mistake

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[cyan fractal]

I'm not fully sure how to figure this out? I mostly guessed on my answer but would appreciate if someone could explain

[west vault]

you've ever heard of area probabilty?

[midnight haven]

think about their areas

[cyan fractal]

not very much it wasnt well explained

[balmy helm]

first, calculate the area of the circle, and then calculate the area of the square, and then area of square/area of circle = probability

since the probability that a randomly chosen point on the circle lies inside the square is the ratio of the area of the square to the area of the circle

[balmy helm]

i obtain the answer 2/pi which approximates to the answer chosen here

[cyan fractal]

ohh okay thank you

[cyan fractal]

when the circle is inscribed in the square, do i do the opposite? like area of the circle/area of the square= probability

[balmy helm]

yes

[balmy helm]

interestingly enough, the fourth option here is the answer to your most recent question "when the circle is inscribed in the square"

pi/4

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[pearl pond]

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[junior pawn]

is there a way to transform the product a(a+1)...(a+n-1)/n! into something with factorial?

[junior pawn]

(a is a real number)

[trail shoal]

(a+n-1)!/(a-1)! maybe?

[junior pawn]

it would work if a€N

but a€R

[trail shoal]

Uhhh then i think you're outta luck

[junior pawn]

i think there's somethink to do if we multiply by a^n and devide by a^n
now we can work with N

[limber oasis]

If you could it would be known

[junior pawn]

i know this is possible with a = 1/2 so i thought it'd be possible with anything

[worldly glacier]

Yes very easily

You can do it two ways

Write out the infinite sum of 1/(1-x)

[junior pawn]

it's the sum of x^n

[worldly glacier]

Write it out please

[junior pawn]

1 + x + x^2 etc

[worldly glacier]

Mhm and you times it a times

Then you realize the coef is the same thing as the amount of the non negative solutions to x_1 + … + x_a = n

[junior pawn]

i do not really get it, what is the second way?

[worldly glacier]

You should understand both ways

[junior pawn]

i don't like mhm so i would be glad to avoid it x)

[worldly glacier]

Other way is really easy

Too easy for my liking

But fine

Re index the summation with n=0

(Change n = 0 and get rid of the 1)

[junior pawn]

ye we just 1 + the sum

[worldly glacier]

Just get rid of the 1 please with reindexing

And send a picture

[junior pawn]

you are asking this, right

[worldly glacier]

Yes but get rid of the 1

By setting n = 0

[junior pawn]

then we just start at 0 the sum and no more 1 because x^0 = 1

[worldly glacier]

Then do this

And you’re done

@junior pawn

[junior pawn]

wait n€N

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[brittle grotto]

functional analysis: I have to rpove that, if A,B:X->X are both bounded operators acting on a banach space X then the spectrums of AB and BA coincide. I know that the spectrum of an operator coincides with the spectrum of its adjoint. So .. \sigma(AB)=\sigma((AB)')=\sigma(B'A')... the A and B are in order but its their adjoint what else do I do? My other idea was to show that the regular points are the same for both, so the spectrum (its complement) is the same

[brittle grotto]

show if lamba is such that I-lambda AB is invertible so is I-lambda BA

where I is the identity operator

[merry carbon]

(you might wanna ask this in #advanced-analysis as well )

[pearl pond]

@brittle grotto Has your question been resolved?

[merry carbon]

(also I may be tripping, but does "the spectrum of an operator [coincide] with the spectrum of its adjoint"?)

[pearl pond]

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[sweet quest]

so which common denominator form should be used?

which form is the convention?

I think the rightmost form is not good since there's a negative exponent on top

so im debating between these two

which one should I use?

which one is preferred by mathematicians/math majors?

[inland ivy]

It really doesnt matter

They are all equivalent

There’s no preference usually, though I’d go with the last one

Just because it has the lowest power in denominator

[sweet quest]

yes it has the lowest power in denominator but it is also the only one with a negative power on the numerator

[inland ivy]

Oh yeah didn’t notice that

The middle one then I suppose

[inland ivy]

To reiterate

[pearl pond]

@sweet quest Has your question been resolved?

[sweet quest]

i see

[sweet quest]

also photomath gave me a different answer too

I guess it just doesn't really matter?

[inland ivy]

Indeed

As I said

[sweet quest]

personally though, since the original form has a whole number power on the denominator, I think the answer should also be a whole number power

so personally I don't really like the photomath's answer

personally i like this one the most

but yeah anyways thanks for helping me! @inland ivy

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[low ravine]

i cant solve this https://cdn.discordapp.com/attachments/1018703621841494076/1235274009109594222/IMG_3594.jpg?ex=6633c612&is=66327492&hm=ef6bdf6887c8012592319b54189945b1e0efd19b1aeceb738b1ec88b60c12f09&

[low ravine]

its supposed to result 2

[pearl pond]

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[cerulean smelt]

isn't this a reflection in y axis

[pearl pond]

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[midnight haven]

Any ideas ?

[feral sedge]

Lhopitals allowed?

[midnight haven]

No

Only factorize

[tight moss]

toughhhh

[midnight haven]

we can substitute sqrt(x)=y^60

[feral sedge]

You could try multiply both sides by the conjugate of the denominator, then do that again

[tight moss]

real analysis was a pain in the ass for me

[feral sedge]

y^8-1 / y^15-1

there's a common factor on both sides now, look for a root

[midnight haven]

how did it become y^8-1 / y^15-1

[feral sedge]

factored out y^12 from the top

as y->1 that becomes 1

[midnight haven]

solved

ty

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[ruby tinsel]

Any ideas ?

[ruby tinsel]

Im so bad at that

[worthy lance]

Did you try anything?

[ruby tinsel]

I try this

Wait

Bruh ...

@worthy lance

<@&286206848099549185>

[worthy lance]

I cant really read anything there tbh

[ruby tinsel]

Wait ill rewrite them @worthy lance

[worthy lance]

Ok so

x-2 goes to - infinity right?

And sqrt(3-x) goes to positive infinity

So you have -infinity + infinity

What grows faster, x or sqrtx?

@ruby tinsel

[ruby tinsel]

X i guess

I didnt understand what do u mean by growz

[worthy lance]

Well, how fast it goes to infinity it means

For example x^2 grows faster than x

[ruby tinsel]

Yeah but we didnt study this yet even if its logically right

[worthy lance]

So because x grows faster than sqrt(x) that means the limit tend to where x tends

This is basic

You must have studied it

[ruby tinsel]

So i need a way to prove it

[ruby tinsel]

Can u write it if u have any paper cuz i really understand if i see this

Itsokay if u cant u help me enough

[compact veldt]

,rotate

[jolly parrot]

Couldn't find an attached image in the last 10 messages.

May 1, 2024, 10:21 PM

[worthy lance]

Sure

You can compare the growth rate like this

In the numerator we will put 2sqrt(x)

In the denominator we will put x

2sqrt(x)/x

Have you ever rationalized?

Now you are gonna do the opposite

You will get 2/sqrtx

Because x goes to infinity

We get 2/inf = 0

This proves that x grows faster than sqrtx

Hence

x+2sqrt(-x) = -infinity

[ruby tinsel]

Im really sorry i understand that but can u do it in a paper or something like that

Im so bad at English cuz i study math in french

[worthy lance]

For what? If you understand what is the problem?

[ruby tinsel]

And ididnt manage to understand to many word

Like rationalized

[worthy lance]

$\frac{2\sqrt{u}}{u} = 2\frac{\sqrt{u}}{u} = 2\frac{1}{\sqrt{u}}$

[jolly parrot]

Samuel

[worthy lance]

$\lim_{u \to \infty} \frac{2\sqrt{u}}{u} = 0$

[jolly parrot]

Samuel

[worthy lance]

$\lim_{x \to -\infty} (x - 2 + 2\sqrt{3-x}) = -\infty$

[jolly parrot]

Samuel

[worthy lance]

I used u instead of x so you dont mix

But you can replace with x

The above with u is to prove that x grows faster than sqrt(x)

[ruby tinsel]

I still didnt understand can u explain in a random exercice

Or a similar one

[worthy lance]

Imagine you have the limit when x goes to infinity of x^2-x

Ok?

[ruby tinsel]

Okay

[worthy lance]

And you dont know for some reason which one grows faster

What u do is divide one with the other

[ruby tinsel]

x^2

[worthy lance]

x^2/x

[ruby tinsel]

Equal x

[worthy lance]

If you get 0, then x grows faster

If u get infinity, then x^2 grows faster

In this case x^2/x = x = infinity

In our case we have

Sqrt(x)/x

This is equal to 1/sqrt(x)

1/infinity = 0

[worthy lance]

This means denominator grows faster

[ruby tinsel]

Let me try this

[worthy lance]

In our case x

[ruby tinsel]

@worthy lance tysmmmmmm

I figure it out

That was helpful

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[frigid raft]

can someone explain this question

[white juniper]

do you know how to find f(x+2)?

[frigid raft]

(x+2)squared +2(x+2)+1

[white juniper]

yeah

so they ask you to calculate (x+2)^2+2(x+2)+1 - (x^2+2x+1)

if you get 4x+8 your work is done

[frigid raft]

ohh

thx

take care!

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[midnight haven]

could this be solved using the divergence theorem

if so why isn't this working

[pearl pond]

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[pearl pond]

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✅

[formal tiger]

how do i get an explict formula for this type of sequence? i wasnt there in class for the formula

[cursive wraith]

an+1 = can + b

check what recursive relation un = an - b/(1-c) verifies

[formal tiger]

what does that mean 😭😭

[cursive wraith]

that means*

express un+1 in terms of un

or un in terms of un-1

[midnight haven]

hello

this is my channel

?

[cursive wraith]

oh sorry

[midnight haven]

wtf are you doing

[formal tiger]

it said channel closed

i thought

[formal tiger]

^

[gloomy scroll]

#❓how-to-get-help :)

[midnight haven]

i reopened it and deleted the messages to avoid clutter

[gloomy scroll]

It‘s closed but not ready for another question

[midnight haven]

no

i reopened it

i still need help

[formal tiger]

oh my bad doorbell camera

[midnight haven]

np

[formal tiger]

sorry

[midnight haven]

it's ok

lemme jjust close this channel tbh

.close

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[warped ocean]

how do i evaluate $\lim_{x\rightarrow 0}\frac{5^x-3^x}{x}$ without l'hopital's

[jolly parrot]

kaiyan

[cursive wraith]

$\lim_{x\to 0}\frac{f(x)-f(0)}{x} = ...?$

[jolly parrot]

rafilou2003

[warped ocean]

i'm not sure

[cursive wraith]

well that's the definition of something you should be familiar with

(f(x+h)-f(x))/h...

[warped ocean]

oh the definition of a derivative

where does it come in here?

[cursive wraith]

well

[cursive wraith]

find the appropriate f

[cursive wraith]

that makes this match

[gloomy scroll]

(imo, factoring a 3^x helps)

[warped ocean]

how do i factor it out here?

[gloomy scroll]

I‘m not sure I get you.

You factor like you would factor any expression

[warped ocean]

i kinda forgot how to factor out exponential functions

[gloomy scroll]

Oh. Well it would give you something like $3^x (\frac{3^x}{5^x} -1)$

[jolly parrot]

ℑμΤ𝛄𝛗θ

[gloomy scroll]

But the left bit simplifies some more

[warped ocean]

how do I get there though?

[worthy lance]

You can rewrite the expression as ((5^x-1)-(3^x-1))/x

Split in two fractions

[warped ocean]

what then

[worthy lance]

Remember that the lim x->0 (a^x-1)/x =lna

[warped ocean]

got it

thanks

.close

[pearl pond]

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[latent crystal]

yow

[latent crystal]

i understand composite and inverse function but i cant understand this 💀

help

<@&286206848099549185>

<@&286206848099549185>

[weak lintel]

find it's higher and lower range where -4<=x<=4

[latent crystal]

-4 ≤ 2 - x ≤ 4?

[latent crystal]

p

help what is this

[weak lintel]

it is the range of f(x) not the domain

[latent crystal]

ok

hmm

how do i do that

[weak lintel]

sketch the graph first and you will understand

,w plot |2-x| where -4<=x<=4

[jolly parrot]

Results provided by WolframAlpha

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

Requested by pubuyun

[weak lintel]

the maximum value is 6, and the minimum is 0

[latent crystal]

yeah

so 0 <= x <= 6?

or what….

[weak lintel]

actually, 0<=f(x)<=6

[latent crystal]

o

so is that the andwr

@weak lintel

[weak lintel]

yes

[latent crystal]

how do i calculate without sketching a graph

@weak lintel

[weak lintel]

first we know |a|>=0, it equals 0 when a=0

[latent crystal]

what is a

[weak lintel]

any number

[latent crystal]

tes

yes

[weak lintel]

so when 2-x=0, it meets the minimum 0

[latent crystal]

omg yes

how did we find 6

as the maximum

[weak lintel]

compare f(x) where x=-4 and x=4

[latent crystal]

whar

[weak lintel]

you will find that when x=-4 it's bigger

[latent crystal]

hmm wai

oh

ohhhh

tys,

i got it now

[pearl pond]

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[hard crystal]

If we are given 2 points in a plane and a vector lying in the plane can i write the eqn of plane?

[marsh tangle]

You know the direction ratios, you know the two points so yes

[hard crystal]

How?

What's the topic name ?

[marsh tangle]

3d geometry, equation of a plane

[midnight haven]

Isn’t it just a(x-x0) +b(y-y0) +c(z-z0) = 0

Or am I wrong

[balmy helm]

yes i agree

[marsh tangle]

$(x_1, y_1, z_1)$ \newline.
$(x_2, y_2, z_2)$
Direction ratios of the plane = $(x_2-x_1, y_2-y_1, z_2-z_1)$

[jolly parrot]

penguin

[midnight haven]

Phew

I have a calc final next week wanted to make sure I got that right

[marsh tangle]

a, b, c = direction ratios which you know from the given two points so go on

[marsh tangle]

Glgl

[midnight haven]

And you can just pick either of the two points?

Or do you have to pick the initial

[hard crystal]

Also suppose A and B are two points and making an angleθ with the given vector then i want to find direcn perp to AB lying in that plane how do i find this

[marsh tangle]

What, no. You have two points, just take the direction ratio

[hard crystal]

Direcn ratio mean direcn vector?

[marsh tangle]

Yeah, which is (∆x,∆y,∆z)

[marsh tangle]

the angle isn't going to cause a huge problem as you can easily change from parametric points to polar to x,y,z easily

Do you have a specific problem which you couldn't solve based on that?

[hard crystal]

Yeah but it is related to physics

[marsh tangle]

Share it here if you don't mind

[hard crystal]

Ok

[marsh tangle]

,rotate

[hard crystal]

I can find θ using dot product
θ is angle b/w AB and P

[hard crystal]

Then i can find E along equator but i have trouble finding E axis i know the magnitude but how do i write the direction?

[marsh tangle]

Yeah you can find the angle between the line AB and P which is sin^-1(n1.n2/|n1||n2|)

[hard crystal]

cos^-1?

[marsh tangle]

For a line and a plane, the angle will be the 90-theta so that's why

[hard crystal]

There are 2lines only?

[marsh tangle]

Line and a plane

[hard crystal]

how?

AB and P are vectors so why 90-θ?

[marsh tangle]

Depends on what you are considering theta tho, I am considering between the normals

[hard crystal]

Oh got it i am considering θ is angle b/w AB and P

Not b/w their normals

[marsh tangle]

Makes sense

[hard crystal]

.

[marsh tangle]

The equator should have the same direction ratios as that of the plane

Direction ratios are the vector directions in 3D terms, maybe you're not introduced to this topic yet

[hard crystal]

That will be (x2-x1)î+(y2-y1)j+(z2-z1)k divided by its magnitude ?

[marsh tangle]

indeed (that's your unit vector right there)

[hard crystal]

So i multiply unit vector with the magnitude of E along equator i geth E equator

But how do i write unit vector of E axis?

[marsh tangle]

Dir ratios in i, j, k (whole divided by the magnitude)

[hard crystal]

There isn't any direcn ratios given perpendicular to AB ?

[marsh tangle]

There are but you can only come up with one equation from this question. Although since you're talking about the equator I'd assume that the perpendicular to the axis and equator meet at the middle point of the plane

That one equation being
Product of direction ratios = 0

[hard crystal]

Ok let us assume (e,f,g) are direcn ratios of axis so (x2-x1)e+(y2-y1)f+(z2-z1)g=0?

[marsh tangle]

correct

[hard crystal]

So the qstn has to give coordinate in some sort of parameter right because from one eqn we can't get values of e f and g?

[marsh tangle]

they should

[hard crystal]

Ok thanks.

.close

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[pearl pond]

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[paper ice]

why is the answer b?

[dense plank]

Im guessing you have to use $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

[jolly parrot]

Renz

[dense plank]

or just $\frac{y_2-y_1}{x_2-x_1}$

[jolly parrot]

Renz

[dense plank]

ye just use this

,,\frac{5.224-5.221}{3.0001-2.9999}

[jolly parrot]

Renz

[paper ice]

bluh

oh, ok

how bout this?

why is it c?

[dense plank]

well if you look at first 2 points

their difference is approximately 1.5

while on the last 2 points their difference is approximately 1.2~

and we can see that those differences are getting smaller and smaller

so its not linear

[paper ice]

h, ok

.close

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[dense plank]

and also we have C and D to choose from right

option D states it appears like a curve

well that is vague because theres other curves other than quadratics

[paper ice]

that wouldn't be enough

[paper ice]

mhm

[pearl pond]

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[muted wigeon]

Hello!
I need help with ODE's. I have to show that this ODE is exact (?, sorry my native tongue is German and I don't know how to say certain things in English ;)). And afterwards I have to calculate the general solution. I also uploaded everything I did and my German script (formulas stay the same I suppose).
I struggle calculating the potential line but maybe I already made a mistake earlier or there is a different way. Help is very much appreciated 🙂

[pearl pond]

@muted wigeon Has your question been resolved?

[muted wigeon]

<@&286206848099549185>

[valid slate]

Sorry @muted wigeon

Can not understand German algebra problem

[muted wigeon]

How would you solve the ODE?

Maybe there is an alternative way to what I did in university. But thanks for looking at it regardless!

[pearl pond]

@muted wigeon Has your question been resolved?

[muted wigeon]

<@&286206848099549185> last try I guess...

[pearl pond]

@muted wigeon Has your question been resolved?

[pearl pond]

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[elder flower]

french helpers? <@&286206848099549185>

[pearl pond]

@elder flower Has your question been resolved?

[pearl pond]

@elder flower Has your question been resolved?

[civic lagoon]

can you provide translation

[pearl pond]

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[elder flower]

for 4 question is the variation table of x on R

[elder flower]

and the question is to prove that f(1/2) < f(x) <f(-1) if x ∈ [ -1 ; 1/2 ]

the expression of f(x)= − x^2− 2+ 1

[pearl pond]

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[glossy lily]

what rule is used here? Why doesn't it just become 9 - e^1/2 ?

[trail shoal]

e^(ln9-½t) = e^ln9 * e^-½t

=9e^-½t

[glossy lily]

you're a life saver bog

you've helped me so much today

[trail shoal]

Just keep shooting

[glossy lily]

I probably will lol 🥲

.close

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[midnight haven]

I need help solving an inequality
2x²+6x-20≤0

[midnight haven]

i divided both with 2 to simplify it which becomes x²+3x-10≤0, but idk what to do from here

[severe quarry]

How did you get <= 10?

[midnight haven]

oops, i meant 0

spelling mistake

[severe quarry]

Yep.

Do you see how you could factor the LHS?

[midnight haven]

whats LHS?

[severe quarry]

left hand side

[midnight haven]

uh not really

[severe quarry]

Ok. Well, let's find the zeros first, then

Set $x^2 + 3x - 10 = 0$.

[jolly parrot]

Kepe

[severe quarry]

Can you solve that for x?

[midnight haven]

with the quadratic formula?

[severe quarry]

Sure

[midnight haven]

i got -5 and 2

[severe quarry]

Yep

[severe quarry]

Now, left and right and between these zeros, x^2 + 3x - 10 is either > 0 or < 0, right?

[midnight haven]

yeah

[severe quarry]

Yeah. So we just need to check what it is in each interval

First check the interval (-infty, -5)

Then (-5, 2)

Then (2, infty)

[midnight haven]

what do you mean by checking intervals?

[severe quarry]

Now there are two options you could take to do this

One is quicker, the other a bit slower.

The slower option would be to plug in any value into x^2 + 3x - 10 from the intervals, and see if it gives you something positive or negative.

So take e.g. -6, that surely lies in the interval (-infty, -5)

(-6)^2 + 3(-6) - 10 = 36 - 18 - 10 = 36 - 28 = 8 > 0

So we know that x^2 + 3x - 10 is positive in the interval (-infty, -5)

Does that make sense?

[midnight haven]

Yeah i got it

[severe quarry]

Yeah, do that with the other two intervals too

[midnight haven]

like this

[severe quarry]

Yeah, a table like that should have the same idea

[midnight haven]

alright ill try that

Alright betseen -5 and 2 its negative

and 5, -> is positive

[severe quarry]

Yeah

[severe quarry]

Now, there is a quicker way too

[midnight haven]

how so

[severe quarry]

x^2 + 3x - 10 describes a parabola, right?

[midnight haven]

yeah

[severe quarry]

And that parabola is opened upwards, right?