1/3x^ = 2x+1

- anonymous

1/3x^ = 2x+1

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- anonymous

What's your x (on the left-hand side) to the power of?

- anonymous

2

- anonymous

You need to remove any x's you have from the denominator. So, multiply everything through by 3x^2 (I said 3x^2 instead of just x^2 because we'd end up multiplying through by 3 anyway - may as well get it over and done with). So,

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## More answers

- anonymous

\[\frac{1}{3x^2}=2x+1 \rightarrow 1=3x^2(2x+1)=6x^3+3x^2\]

- anonymous

he may have meant (1/3)x^2...

- anonymous

Is that what you meant, mac?

- anonymous

that would sound more likely to be the homework assigned to an algebra class

- anonymous

Because solving 6x^3+3x^2=1 algebraically is not nice.

- anonymous

^not at all...

- anonymous

\[\frac{1}{3}x^2=2x+1 \rightarrow x^2=6x+3\] after multiplying through by 3. You have a quadratic to solve:\[x^2-6x-3=0\]

- anonymous

ergh... that still wont give integer roots

- anonymous

You can complete the square or use the quadratic formula.

- anonymous

Can't help that...

- anonymous

\[x^2-6x-3=0 \rightarrow x=3 \pm 2\sqrt{3}\]

- anonymous

as for the 6xcubed i would use long division... or rather synthetic... is that the best way?

- anonymous

Do you mean you would use long division or synthetic after finding a factor?

- anonymous

i would use them to check if i can guess a factor lol

- anonymous

Yes

- anonymous

i wonder where mac went...

- anonymous

The method for solving cubics is a pain, so I usually try to find one solution and, like you said, use long/synthetic division on the cubic and go from there.
Else, numerical solution, like Newton-Raphson method.

- anonymous

Mac's lurking...

- anonymous

newton-raphson method? whats that?

- anonymous

It's a method for finding solutions to equations when you can't do so algebraically (or can't be bothered). The only condition is that the function you're looking at has to be continuous (because you use calculus).

- anonymous

im curious, as the derivative cant tell you solutions... can it??

- anonymous

You're right, the derivative itself won't give you the solutions (unless you have a multiple root), but the method takes your guess, then uses calculus to find the equation of the tangent at the function value of your guess for x, then seeks out where the tangent cuts the x-axis. That point - where it cuts the x-axis - is closer to the root than your initial guess (usually...there are cases where it won't work). That new x-value is your new 'guess' which goes through the process again to yield a better 'guess' for x...you can keep going until you get to a level of accuracy you want.

- anonymous

... so this will only work with f(x) with a degree 3 or more so that f'(x) is at least quadratic... and you find the tangent, then find the zero of that equation, and then use that in the original and find the tangent and use the zero of that equation and so on?

- anonymous

The method will will work with any function that has at least one derivative. I'll try and find some information for you. It's a very useful tool to have, mainly because you end up having at your disposal a means to solve many practical problems that either don't have algebraic solutions, or where the solutions are a pain to get at and you don't need your answer to be exact.

- anonymous

but then you will never get the exact zero, you will only get closer and closer to it? as the zero of a tangent line can never be equal to the value put into the original function for that tangent line unless you have a multiple root case ( as you said above)

- anonymous

Check out the image - this is what's going on.
http://en.wikipedia.org/wiki/Newton's_method

- anonymous

thank you man

- anonymous

You're welcome ;)

- anonymous

what the heckkk that is freaking genius. damn newton -.-

- anonymous

Yeah, I think it's his best thing outside of the calculus and Laws of Motion.

- anonymous

when it comes to physics laws of that sort, id always put kepler over newton... idk why newton got all the fame for those, keplers were way better -.-

- anonymous

Try it with something like finding the square root of 2 - it will find it pretty damn quick.
\[x=\sqrt{2} \rightarrow f(x)=x^2-2=0 \rightarrow f'(x)=2x\]Then\[x_{n+1}=x_n-\frac{f(x)}{f'(x)} \rightarrow x_{n+1}=x_n-\frac{x^2_n-2}{2x_n}\]

- anonymous

i didnt understand the second half of that lol

- anonymous

I'm going to do an example.

- anonymous

You have to find something to start with, so we make an educated guess for sqrt(2).
Since \[1<2<4 \rightarrow \sqrt{1}<\sqrt{2}<\sqrt{4} \iff 1 <\sqrt{2}<2\] as sqrt function is monotonic increasing (i.e. the direction of the inequality is preserved). So we have for an initial guess for x,\[x \approx \frac{3}{2}\]Then, a better 'guess' for x will be,\[x_2 = \frac{3}{2}-\frac{\left( \frac{3}{2} \right)^2-2}{2\frac{3}{2}} = \frac{17}{12} \approx 1.416667\]and the square root of 2 is about \[1.414213562\]

- anonymous

You can see after one trial, it's pretty close.

- anonymous

The closer you are to your answer in the first place, the faster it will converge.
You could then plug 17/12 into the formula to get a better approximation, and continue.

- anonymous

Next one would be 577/408 ~= 1.414215686

- anonymous

dang, the formula, has you divide the tangent line function by the tangent line slope, why?

- anonymous

Well, going back to the sqrt of 2, when you start out, you don't know what the sqrt of 2 is, but you know it should be something, so call that something, x.
Then x =sqrt(2).
If I square both sides and subtract 2 from both sides, I get
x^2-2=0
So, finding x = sqrt(2) amounts to finding where on the x-axis
x^2-2
will cut it. So I plot my parabola, and imagine the setup that's in that Wikipedia pic. Since the method finds where the graph cuts the x-axis, it will find that x where x^2-2=0; that is, the x that came from x=sqrt(2).

- anonymous

yeah i get that, but i dont get how subtracting the tangent line formula divided by the slope of the tangent gets you the intersection of the tangent line?

- anonymous

The reason for why we divide the function by the derivative is due to the derivation. The derivation can be done in either a geometrical setting (using that picture, say) or with Taylor series expansion on an arbitrary function. I'm sure proofs would be online. It's a pretty old theorem ;)

- anonymous

oh wait i get it...

- anonymous

You know, you can derive it yourself by analyzing each step in that pic. and doing what the mathematics would dictate in that situation.

- anonymous

your taking y=mx+b and doing 0=mx+b so mx=-b so x=-b/m and you take that, which is the distance from the x value to the zero, and subtract it from your original point which will show you where on the x axis it is...

- anonymous

nope that made no sense... ill figure this out, gimmy a sec

- anonymous

You're half there.

- anonymous

It's difficult to explain online. They need to put a drawing feature in this thing.

- anonymous

lol they do, like tutorvista

- anonymous

I think mac must think his thread's been hijacked.

- anonymous

i think he left

- anonymous

so youre taking x and subtracting y/y'(x)

- anonymous

i still dont get why ><

- anonymous

Hang on, I'll see if I can put a small proof together.

- anonymous

how is the y value divided by the slope at that point, equal to the difference btween x and the intersection of the tangent line of x

- anonymous

no you dont need to make a proof of it i just need to understand how y/y' is the difference between x and the zero of y'

- anonymous

Wait, if I prove it using the geometrical route, you'll see...I'm just going to see if it's feasible (versus finding something online).

- anonymous

ok :/

- anonymous

Alright, writing it up isn't hard, but you'll see what's going on if you do a drawing like the one seen in Wikipedia.

- anonymous

And go along with it...

- anonymous

i can understand how it all works, but how does the drawing explain how the formula works?

- anonymous

(i already made a drawring btw lol)

- anonymous

Suppose x_1 is close to a root of the equation f(x)=0. Then the equation of the tangent at x=x_1 is\[y-y_1=m(x-x_1)\]that is,\[y-f(x_1)=f'(x_1)(x-x_1)\]This tangent crosses the x-axis at a point where x = x_2 and y = 0. Hence,\[0-f(x_1)=f'(x_1)(x_2-x_1) \rightarrow x_2=x_1-\frac{f(x_1)}{f'(x_1)}\]

- anonymous

Then repeat...

- anonymous

oh wow that makes sense...

- anonymous

Happy now?

- anonymous

yes :D thanks for spending time to teach me this stuff :O

- anonymous

No worries - it's good you want to know :)

- anonymous

its incredible that you can put up with teaching me like that, even alot of teachers would loose it pretty soon

- anonymous

i honestly am really curious as to where mac went...

- anonymous

he stopped observing a while ago even

- anonymous

You learnt quickly - it wasn't a problem. You would have learnt quicker if the interface was easier (but it's not the designers' fault - where's the holography already?).

- anonymous

Yeah, he probably got his answer and was all, "WTF is going on?!"

- anonymous

whats after differential equations?

- anonymous

Eh?

- anonymous

like course wise, you take calc1 calc2 calc3 diff eq then what?

- anonymous

Have no idea man, not American ;D

- anonymous

but you studied in europe, so the course order must be nearly identicle, especially in advanced mathematics, so instead i should probably ask, what did you do (courses) for you degree and btw, you have a masters? phD?

- anonymous

your degree*

- anonymous

Partial differential equations is usually end of the line for calculus. Then there's other stuff like Lebesgue integration (overtakes Riemann integral), and other stuff in algebra and geometry - the further you go, the more the areas start to link up. E.g. differential geometry, algebraic topology, etc. But for calculus, numerical and algebraic methods for solving partial differential equations is usually end of line.
If you know how to solve (methods to solve)
nth order, non-linear, inhomogeneous partial differential equations with non-constant coefficients,
you're done!

- anonymous

Since i did not understand any of
nth order, non-linear, inhomogeneous partial differential equations with non-constant coefficients
i guess im not done :D

- anonymous

lol, keep going, though!

- anonymous

do you know all that is taught of mathematics to the extent that research is your only option?

- anonymous

No, you can do whatever you want with it. I teach as well. You can use it in industry and government, as in modelling for things...haven't really sold it, have I?

- anonymous

well not in that sense, i meant like if you wanted to know more on math, your only option would be to study unsolved theorems or unproved research or do research on your own

- anonymous

anddd man i need help on something:
\[\int\limits_{}^{} (\sin x - \cos x) \div (\sin x)\]

- anonymous

nevermind i got it

- anonymous

Yes, you're right on your last point.

- anonymous

thats pretty crazy, that you know everything O.o

- anonymous

I don't know everything..!

- anonymous

you know everything in math...

- anonymous

No I don't ;)

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