Equations

[sleek wyvern]

Can someone give me equations like 5a=3b-1 or somethinganf i try to solve it

[pine iron]

You mean linear equations like of the form: $4x=3x-5$?

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

Yes

Would the answer be 5?

For that example

[pine iron]

\begin{enumerate}
\item $5x=3x-2$
\item $2x=5x+3$
\item $69x=420x-96$
\item $8x=3x+8$
\item $7x+5=4x+3$
\item $4x-2=10x+4$
\item $ax=5x+3$
\item Find a number $x$ such that number plus it's double equal to that number plus 4.
\item $ax=4-2x$
\end{enumerate}

Like that?

[sleek wyvern]

Number 3 and 5 look too advanced for me

Lol

Im going to try number 1

[pine iron]

I added a few more

They're challenge ones (7-9)

[sleek wyvern]

Which one do you recommend me to do

[pine iron]

well all of them but which do you have issues with?

[sleek wyvern]

And I woud like to add that im not in any course about number theory and stuff like that i just like learning more

[pine iron]

do you want me to get rid of the fractions?

[pine iron]

well number 8 is doable with just your knowledge

[sleek wyvern]

So you want me to solve for x or a in number i

[clear needle]

XxMrFancyu2xX

[pine iron]

solve for x in all of them

treating all other variables (a) as constants (1, 2, 3, 4, etc...)

[sleek wyvern]

Yeah 8 is kind of hard but I got x=4 not sure if its right though

[pine iron]

nope

So take look: "Find a number x such that number plus its double"

so x+2x

"is equal to that number plus 4"

so =x+4

[sleek wyvern]

Instead of these can you give me simple diophantine equations

Like maybe something like x+2y=3 or something

[pine iron]

alright

[sleek wyvern]

You dont have to give me a list give me one one by one

[pine iron]

very well

I realize with diophantine equations, you may want integer solutions 😅

[sleek wyvern]

To me it doesnt really matter if itd an integer or not as long as the equations true

[pine iron]

smthn like $3x+4y=5$?

[sleek wyvern]

Im just trying to test myself

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

Did I already do this one?

Im going to try to solve it eitherway

[pine iron]

idk

[sleek wyvern]

I couldnt solve this one unfortunately.

If I remember you would use either a formula or the euclidean algorithm

[pine iron]

Ok so, for LDE we have to convert to an HLDE

We have to find the gcf of both coefficients

gcf(3,4)=1 which is divisible by 4 (obviously)

so we're all good to proceed

[sleek wyvern]

Is the answer for x=-5/7 and y=5/7?

[pine iron]

I don't think so....

but generally LDE require solutions over the integers

so now we find a case where the equation works

which is $x=-1$ and $y=2$, we get $3(-1)+4(2)=-3+8=$\fbox{5}

[clear needle]

XxMrFancyu2xX

[pine iron]

Ok so now we let values $u=x+1$ and $v=y-2$

[clear needle]

XxMrFancyu2xX

[pine iron]

then we get $3u+4v=3(x+1)+4(y-2)$ do you understand how I got this?

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

No i dont understand

[pine iron]

question, what math class are you in?

[sleek wyvern]

Like I said im not experienced. Im in precalc but ive been seeing videos on diophantine equations and understand a little

[pine iron]

ah ok so this is an independent study, good learning math is great, so what part are you not getting?

[sleek wyvern]

Ok so I get the gcf part but dont understand how you got x=-1 and y=-2

[pine iron]

it's just one solution

to the equation $3x+4y=5$

[clear needle]

XxMrFancyu2xX

[pine iron]

there are infinitely many it's just best to chose one with the littlest numbers possible

to make your life easier

[sleek wyvern]

I used a different technique to solve this one

[pine iron]

How did you do it?

oh well that has only one variable

3x+4y=5 has two

[sleek wyvern]

Ok so you know how an even integer is 2k and odd integer is 2k+1

[pine iron]

mhm

[sleek wyvern]

So what i did was substitute the values in corresponding to the coefficents so it would be 4(2k)-3(2k+1)=-5 and solve for k

Because 4 is even and 3 is odd

And then once you solved for k you plug it back in

So x=2k and y=2k+1

[pine iron]

There are few issues: in the equation: $4x=3x-5$ you are giving the variable $x$ two different forms.\~\

Also you are assuming that $2k$ and $2k-1$ are going to produce two different results when it's just going to produce the $k$th even and odd number, 7 and 8 or 9 and 10

[clear needle]

XxMrFancyu2xX

[pine iron]

Ok, wait hang on

your trying to solve for y in an equations with only x's?

[sleek wyvern]

No you solve for k and whatever number you get plug it back in to the substitution

If you want i can show you with a picture in my whiteboard

[pine iron]

ok?

@sleek wyvern ?

[sleek wyvern]

I know the picture is aligned weird my bad

Im just experimenting with things for the fun of it

[pine iron]

also maybe put the original equation as 4x-3y=5 instead of 4x-3x=5

but I think what you're doing is just finding one solution of the original diophantine equation

I haven't seen such a method before but hey there very well could be

[sleek wyvern]

Sorry for wasting your time i just wanted to see if my method works

[pine iron]

I am letting $2m$ and $2m+1$ be coefficients of a diophantine equation and $c$ the constant on the end.\~\

$(2m)(2k+1)+(2m+1)(2k)=c$\
$4km+2m+4km+2k=c$\
$8km+2m+2k=c$\~\

Huh, tbh idk what to do with that, but that's pretty cool, it might be a way of generating a solution

[clear needle]

XxMrFancyu2xX

[pine iron]

no no, no time was wasted

[sleek wyvern]

This might be too advanced for me but i wouldnt mind doing basic calculus problems

[pine iron]

$\int_{-\infty}^{\infty}{e^{-x^{2}}}dx$

[clear needle]

XxMrFancyu2xX

[pine iron]

lmao "basic"

[sleek wyvern]

I memorized this one its square root of pi

Lol

[pine iron]

prove it 😳

[sleek wyvern]

Thats...too hard for me right now

[pine iron]

\begin{center}
We begin with the integral over the entire bell curve, and multiply to get a double integral over the real plane using \textit{Fubini's Theorem}.
\end{center}
$$I=\int^{\infty}{-\infty}{e^{-x^{2}}}dx\Rightarrow I^2=\int^{\infty}{-\infty}{e^{-x^{2}}}dx\cdot\int^{\infty}{-\infty}{e^{-y^{2}}}dy\Rightarrow I^2=\iint{\mathbb{R}^{2}}{e^{-(x^2+y^2)}}dydx$$

\begin{center}
Next, we transform with 3d polar coordinates using the Jacobean. Notably, $x=\rho\cos\theta$ and $y=\rho\sin\theta$, therefore $x^2+y^2=\rho^2$. So out task is computing $dydx$ and to transform into $d\rho d\theta$.
\end{center}
$$J_{\rho\theta}=\frac{d(x,y)}{d(\rho,\theta)}=\begin{vmatrix}\frac{\partial x}{\partial\rho} & \frac{\partial x}{\partial\theta}\ \frac{\partial y}{\partial\rho} & \frac{\partial y}{\partial\theta}\end{vmatrix}$$
$$\Rightarrow\begin{vmatrix}\frac{\partial}{\partial\rho}\left[\rho\cos\theta\right] & \frac{\partial}{\partial\theta}\left[\rho\cos\theta\right]\ \frac{\partial}{\partial\rho}\left[\rho\sin\theta\right] & \frac{\partial}{\partial\theta}\left[\rho\sin\theta\right]\end{vmatrix}=\begin{vmatrix}\cos\theta & -\rho\sin\theta \ \sin\theta & \rho\cos\theta\end{vmatrix}$$

$$\Rightarrow -\rho\cos^2\theta-(\rho\sin^2\theta)=-\rho(\cos^2\theta+\sin^2\theta)=-\rho$$
$$\frac{d(x,y)}{d(\rho,\theta)}=-\rho\Rightarrow d(x,y)=-\rho\cdot d(\rho,\theta)$$

\begin{center}
Now we must find our bounds. It trivial that $\theta\in[0,2\pi)$ and $\rho\in[0,\infty)$ so those must be our bounds, so we have:
\end{center}
$$I^2=\int^{2\pi}{0}{\int^{\infty}{0}{-\rho e^{-\rho^{2}}}d\rho}d\theta\Rightarrow I^2=2\pi\int^{\infty}{0}{-\rho e^{-\rho^{2}}}d\rho\Rightarrow\pi\left(e^{-\rho^{2}}\right|^{\infty}{0}$$
$$\Rightarrow\pi(-e^{-\infty}+e^{-0})\Rightarrow I^2=\pi\Rightarrow I=\sqrt{\pi}=\int^{\infty}_{-\infty}{e^{-x^{2}}}dx$$

\begin{center}
q.e.d.
\end{center}

[clear needle]

XxMrFancyu2xX

[pine iron]

ok anyways, you really don't want to do calculus until you have strong foundation in trig, alg, and precalc

specifically the rules of Algebra

[sleek wyvern]

When i mean basic calculus i mean like u substitution

[pine iron]

do you know any calc?

[sleek wyvern]

A little yes

[pine iron]

do you understand limits conceptually and computationally?

[sleek wyvern]

I wouldnt say conceptually but maybe computationally

Like the derivative is just the limit approaching 0

[pine iron]

ok

[pine iron]

the limit of $\frac{f(x+h)-f(x)}{h}$ as $h$ approaches 0 or $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

[sleek wyvern]

yes

[clear needle]

XxMrFancyu2xX

[pine iron]

Ok, how is your power rule?

[sleek wyvern]

Decent

[pine iron]

So if I gave you the problem $\frac{d}{dx}\left[6x^{79}\right]$ you would be able to do it?

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

Yes

[pine iron]

so then what is $\frac{dy}{dx}$

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

Is it 474x^78 ?

[pine iron]

✅

Ok so challenge, what about: $\frac{d}{dx}\left[\sqrt{x}\right]$ knowing that $\sqrt{x}=x^{\frac{1}{2}}$

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

1/2(squareroot of x)

[pine iron]

nice!

[sleek wyvern]

I already memorized that one because ive done it a million times

[pine iron]

ah

what about $\frac{d}{dx}[\sqrt[n]{x}]$

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

1/n(x)^1-n/n ?

This is ones tricky

[pine iron]

Ok, so we have $\frac{d}{dx}[\sqrt[n]{x}]\Rightarrow\frac{d}{dx}[x^{\frac{1}{n}}]$, yes?

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

Yes

Bring power to front

And then subtract the power

[pine iron]

Now we drop the exponent down in front and subtract from it: $\frac{1}{n}(x^{\frac{1}{n}-1})$

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

Yes

[pine iron]

simplify the exponent and we get $\frac{1}{n}(x^{\frac{1-n}{n}})\Rightarrow\frac{\sqrt[n]{x^{1-n}}}{n}$

[sleek wyvern]

How do i know the exponent is positive?

1-n/n

[clear needle]

XxMrFancyu2xX

[pine iron]

I messed it up 💀

but that should be it

[sleek wyvern]

Can you give me a limit problem

[pine iron]

I'll start easy: $\lim_{x\to3}\frac{x^2-9}{x-3}$

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

6

[pine iron]

nice, quick and easy

$\lim_{x\to n}\frac{x^2-n^2}{x-n}$

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

2n?

[pine iron]

✅

Ok so you got that down, $\lim_{x\to0}xe^{-x}$

[sleek wyvern]

Would it be just x?

Wait

[pine iron]

oh wait, hang on I think I did this wrong... 💀

[sleek wyvern]

No worries

[pine iron]

$\lim_{x\to\infty}xe^{-x}$

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

Would it be 0?

[pine iron]

yes

how did you get that?

[sleek wyvern]

Bring down e^x to the denominator

And then use lhopitals rule

[pine iron]

Can you think about it conceptually without using l'Hôspital's Rule?

[sleek wyvern]

Well I know that e^x is growing really fast

Because its exponential

[sleek wyvern]

Im going to say probably not

[pine iron]

So we have the fraction $\frac{x}{e^x}$

[clear needle]

XxMrFancyu2xX

[pine iron]

yes?

[sleek wyvern]

Yes

[cobalt gull]

Just knowing e^-x shrinks faster than x grows is all you need to know

[pine iron]

pretty much that...

for sufficiently large x

e^x will dominate the linear x

[cobalt gull]

Or any polynomial x^n for that matter

[sleek wyvern]

Gotcha

[pine iron]

So then what about: $\lim_{x\to\infty}\frac{e^x}{x!}$?

[clear needle]

XxMrFancyu2xX

[cobalt gull]

[sleek wyvern]

Thats too hard for me atm

[pine iron]

think about it, factorials always grow faster than any exponential for sufficiently large x

[cobalt gull]

This isn't obvious for most new calculus students

[sleek wyvern]

Isnt n factorial just n(n-1)(n-2)...

[pine iron]

fair enough

well then you take that problem as a thought (i forgor correct word) ig, if you have any other question or problems let me know otherwise, is that it?

[sleek wyvern]

Can we stick with the basic limits

[pine iron]

Ok, how about $\lim_{x\to3}\frac{x^4-81}{x^2-9}$?

[clear needle]

XxMrFancyu2xX

[cobalt gull]

[pine iron]

[cobalt gull]

$\lim_{x\to0} x\sin(\frac{1}{x})$

[clear needle]

SWR

[pine iron]

isn't it: $\lim_{x\to0}\frac{\sin(x)}{x}$?

[clear needle]

XxMrFancyu2xX

[pine iron]

or are you considering a different limit?

[sleek wyvern]

Too hard

[pine iron]

think of x^4-81 as (x^2)^2-9^2

[sleek wyvern]

I did do that

[pine iron]

$\frac{x^{2^2}-9^2}{x^2-9}=\frac{(x^2-9)(x^2+9)}{x^2-9}$

[clear needle]

XxMrFancyu2xX

[sleek wyvern]

Oh im dumb lol i factored the denominator

[pine iron]

ah

well that could work just be more steps

so then what would be the limit?

[cobalt gull]

A different limit

[sleek wyvern]

18?

So 1-10 what are my limit skills at I would say im at a 4

[pine iron]

$\lim_{0\to\infty}\frac{\ln x}{x}$

[clear needle]

XxMrFancyu2xX

[pine iron]

bruh wow im tired

i really just said limit 0 to infity 🤦‍♂️

you get the idea x to infinity

[sleek wyvern]

0

[pine iron]

how did you get that?

[sleek wyvern]

Just used lhopitals again as the derivative of lnx is 1/x and derivative of x is 1

[pine iron]

ok good

[ocean elk]

how on earth did you go from asking about 5a-3b=1 to integrals

[pine iron]

dw honestly got carried away with it, math does that ig

[sleek wyvern]

Too advanced for me

I think basic calculus is more up my ally

[pine iron]

ok @sleek wyvern I am seriously sleep deprived so I'm gonna go to sleep you can close the thingy if you want by typing .solved or .close I assume you know that but if you want maybe you can get someone else

[pine iron]

https://math.libretexts.org/Courses/Mount_Royal_University/MATH_2150%3A_Higher_Arithmetic/5%3A_Diophantine_Equations/5.1%3A_Linear_Diophantine_Equations

[ocean elk]

I can’t tell if you’re joking

[sleek wyvern]

Im not

[ocean elk]

5a-3b= 1 is algebra

[pine iron]

no we we're trying to do Diophantine equations

and he wasn't getting it

[pine iron]

this may help @sleek wyvern

ok gn ppl 👋

[sleek wyvern]

.close