Mathematics Thread

[ben1996123]

Isolate "y" in the following equation: 5y^2-(18/√5)y-7=0
The answer is √5, but I can't seem to figure out how to actually get there.

\( 5y^{2}-\frac{18y}{\sqrt5}-7=0 \)
\( y=\frac{\frac{-18}{\sqrt5}\pm\sqrt{(\frac{18}{\sqrt5})^{2}-4\times 5\times -7}}{2\times 5} \)
\( y=\frac{\frac{18}{\sqrt5}\pm\sqrt{204.8}}{10} \)
\( y=\frac{\frac{18\sqrt{5}}{5}\pm\frac{32\sqrt{5}}{5}}{10} \)

\( y=\frac{\frac{50\sqrt{5}}{5}}{10} \)
\( y=\frac{50\sqrt{5}}{50} \)
\( y=\sqrt{5} \)
Or:
\( y=\frac{\frac{-14\sqrt{5}}{5}}{10} \)
\( y=\frac{-14\sqrt{5}}{50} \)
\( y=-\frac{7\sqrt{5}}{25} \)

 

[ben1996123]

Thanks a lot Ben. Just one last question though, how did you get from that first step in the quote to the second one?

Quadratic formula

\( y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)

with a = 5, b = \( -\frac{18}{\sqrt{5}} \), c = -7


randomedit: \( 535.4916555^{i}\approx1 \)

 

[samehsameh]

Thanks a lot Ben. Just one last question though, how did you get from that first step in the quote to the second one?

\( \sqrt{204.8}=\sqrt{\frac{1024}{5}}=\frac{\sqrt{1024}}{\sqrt{5}}=\frac{32}{\sqrt{5}}=\frac{32 \sqrt{5}}{5} \)

 

[ben1996123]

\( \sqrt{204.8}=\sqrt{\frac{1024}{5}}=\frac{\sqrt{1024}}{\sqrt{5}}=\frac{32}{\sqrt{5}}=\frac{32 \sqrt{5}}{5} \)

That's exactly what I did

 

[qqwref]

has anyone here gone to IMO?

I went to MOPS once.

 

[ben1996123]

How do you isolate x?

\( y=x^{x} \)

failattempt:

Spoiler

\( y=x^{x} \)
\( ln(y)=xln(x) \)
\( \frac{1}{y}\frac{dy}{dx}=ln(x)+1 \)
\( \frac{dy}{dx}=x^{x}(ln(x)+1)=x^{x}ln(x)+x^{x} \)
\( \frac{dx}{dy}=\frac{1}{x^{x}ln(x)+x^{x}} \)
\( x=\int\frac{dy}{x^{x}ln(x)+x^{x}} \)
idk

 

[JonWhite]

I went to MOSP once.

Red/Blue/Black? What year?

 

[aronpm]

How do you isolate x?

\( y=x^{x} \)

The solution is x = ln(y)/W(ln(y)), where W(z) is the Lambert W function, which is defined as the inverse function of f(w) = we^w.

Yeah...

 

[qqwref]

Red/Blue/Black? What year?

Whoops, yeah, sorry about the typo. It was blue, in 2006. (Ancient history!)

 

[ressMox]

How do you isolate x?

\( y=x^{x} \)

failattempt:

Spoiler

\( y=x^{x} \)
\( ln(y)=xln(x) \)
\( \frac{1}{y}\frac{dy}{dx}=ln(x)+1 \)
\( \frac{dy}{dx}=x^{x}(ln(x)+1)=x^{x}ln(x)+x^{x} \)
\( \frac{dx}{dy}=\frac{1}{x^{x}ln(x)+x^{x}} \)
\( x=\int\frac{dy}{x^{x}ln(x)+x^{x}} \)
idk

I remember seeing that the answer could only be represented using "regular" (for lack of better word) functions and are instead defined with the use of analytically determined functions. (Similar to what aronpm said) The same way the integral of (e^x)/(x) is represented as Ei(x) where Ei(x) is defined as the exponential integral.

 

[tozies24]

I feel pretty stupid asking this, but where can I learn more about this kind of factoring?

It seems like I have forgotten my basic algebra skills.

You see that in the sum you have that (4x-5) and (x^2+3) to some power occur in both terms. You can factor out how ever many of those products you want and then see what remains.

Example: 3^7*2*5 +3*5*2^4= 3^5*2*4(3^2+2)

 

[tozies24]

Thanks for your reply, I understand that perfectly but the main problem I had was the coefficients, how do you know where they go from that image? I mean the "2x" and the "3", I always mess up with those when I'm doing questions.

So you see how you have an 8x out front of the first and a 12 out front of the second? The greatest common divisor between 8x and 12 is 4. So the writers of the book factored out a 4. Then you have 2x in front of the first term and 3 out front of the second term.

 

[vcuber13]

4*stuff*2x -> 8x*stuff
3*stuff*4 ->12*stuff

 

[ben1996123]

I'm bored. Someone give me a random derivative to do please. No weird functions like Li(x), x!, W(x).

 

[tozies24]

I'm bored. Someone give me a random derivative to do please. No weird functions like Li(x), x!, W(x).

x*sinx*e^x*lnx*cosx

 

[blah]

I love maths. I've been doing research on my own on various problems and theorems, even came up with something on my own.

I'd like to see different ways to solve this:

Proof that

It's pretty easy.

Spoiler

Spoiler

I had the same solution as y235 (ninja'd ), but just for lols I'll sub in an x somewhere else.

\( x=1+\frac{1}{2+{\frac{1}{1+x}}} \)

\( x=\frac{3x+4}{2x+3} \)

\( 2x^2+3x=3x+4 \)

\( x^2=2 \)

\( x={\sqrt2} \)
or
\( x=-{\sqrt2} \)

assumed convergence fail?

\( x = 1 - 1 + 1 - 1 + \cdots \Rightarrow x = 1 - x \Rightarrow x = \frac12 \Rightarrow \text{lololololol} \)

moral of the story: you can't just "set some infinite thingy equal to x" without first proving that such an x exists

 

[vcuber13]

how do you get that?

 

[AustinReed]

Hardest for me was basic algebra. Why? I had the teacher that couldn't teach. I suck at math because of him.

 

[ben1996123]

'Extended' product rule:

If \( y = f(x)g(x)h(x)\cdots \) with n functions, and where n(x) is the nth function, \( \frac{dy}{dx}=\sum_{r=1}^{n}\frac{r'(x)y}{r(x)} \). It's probably known, but I found it myself today so I thought I'd post it.

\( y=xsin(x)cos(x)ln(x)e^x \)

Spoiler

\( y=abcdf \) Not using e because \( e^x \)
\( a=x \)
\( b=sin(x) \)
\( c=e^x \)
\( d=ln(x) \)
\( f=cos(x) \)
\( \frac{dy}{dx}=abcd\frac{df}{dx}+abcf\frac{dd}{dx}+abdf\frac{dc}{dx}+acdf\frac{db}{dx}+bcdf\frac{da}{dx} \) extended product rule
\( \frac{da}{dx}=1 \)
\( \frac{db}{dx}=cos(x) \)
\( \frac{dc}{dx}=e^x \)
\( \frac{dd}{dx}=\frac{1}{x} \)
\( \frac{df}{dx}=-sin(x) \)
\( bcdf\frac{da}{dx}=sin(x)cos(x)ln(x)e^x \)
\( acdf\frac{db}{dx}=xcos^2(x)ln(x)e^x \)
\( abdf\frac{dc}{dx}=xsin(x)ln(x)cos(x)e^x \)
\( abcf\frac{dd}{dx}=\frac{xsin(x)cos(x)e^x}{x}=sin(x)cos(x)e^x \)
\( abcd\frac{df}{dx}=-xsin^2(x)ln(x)e^x \)
\( \frac{dy}{dx}=sin(x)cos(x)ln(x)e^x+xcos^2(x)ln(x)e^x \)\( +xsin(x)ln(x)cos(x)e^x+sin(x)cos(x)e^x-xsin^2(x)ln(x)e^x \)

\( \frac{dy}{dx}=e^x(sin(x)cos(x)ln(x)+xcos^2(x)ln(x)+xsin(x)ln(x)cos(x) \) \( +sin(x)cos(x)-xsin^2(x)ln(x)) \)

Here's a derivative for someone to do:

\( x^{sinh(x+y)}=\frac{2^{ln(xy)}}{y^{csc(x)}} \)

Hint: \( \frac{d}{dx}(sinh(x))=cosh(x), \frac{d}{dx}(cosh(x))=sinh(x) \)

 

[samehsameh]

Pretty good example ben its all simple calculus just lots and lots of it. Will post a solution tomorrow, its 2:54 am atm and ive got lectures tomorrow morning, been teaching some friends how to cube

Also someone can try \( \int_{-\infty}^{\infty} \! \frac{x\sin x}{x^2+a^2} \, \mathrm{d} x, a > 0 \) i.e. the area enclosed by \( y=\frac{x\sin x}{x^2+a^2} \) and \( y=0 \)