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anonymous
 5 years ago
1/3x^ = 2x+1
anonymous
 5 years ago
1/3x^ = 2x+1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What's your x (on the lefthand side) to the power of?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3x^2}=2x+1 \rightarrow 1=3x^2(2x+1)=6x^3+3x^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he may have meant (1/3)x^2...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that what you meant, mac?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that would sound more likely to be the homework assigned to an algebra class

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because solving 6x^3+3x^2=1 algebraically is not nice.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3}x^2=2x+1 \rightarrow x^2=6x+3\] after multiplying through by 3. You have a quadratic to solve:\[x^26x3=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ergh... that still wont give integer roots

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can complete the square or use the quadratic formula.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^26x3=0 \rightarrow x=3 \pm 2\sqrt{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as for the 6xcubed i would use long division... or rather synthetic... is that the best way?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you mean you would use long division or synthetic after finding a factor?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i would use them to check if i can guess a factor lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i wonder where mac went...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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 NewtonRaphson method.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0newtonraphson method? whats that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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
 5 years ago
Best ResponseYou've already chosen the best response.0im curious, as the derivative cant tell you solutions... can it??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You'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 xaxis. That point  where it cuts the xaxis  is closer to the root than your initial guess (usually...there are cases where it won't work). That new xvalue 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
 5 years ago
Best ResponseYou've already chosen the best response.0... 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
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0Check out the image  this is what's going on. http://en.wikipedia.org/wiki/Newton's_method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what the heckkk that is freaking genius. damn newton .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I think it's his best thing outside of the calculus and Laws of Motion.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when 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
 5 years ago
Best ResponseYou've already chosen the best response.0Try it with something like finding the square root of 2  it will find it pretty damn quick. \[x=\sqrt{2} \rightarrow f(x)=x^22=0 \rightarrow f'(x)=2x\]Then\[x_{n+1}=x_n\frac{f(x)}{f'(x)} \rightarrow x_{n+1}=x_n\frac{x^2_n2}{2x_n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i didnt understand the second half of that lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm going to do an example.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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,\[x_2 = \frac{3}{2}\frac{\left( \frac{3}{2} \right)^22}{2\frac{3}{2}} = \frac{17}{12} \approx 1.416667\]and the square root of 2 is about \[1.414213562\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can see after one trial, it's pretty close.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0Next one would be 577/408 ~= 1.414215686

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dang, the formula, has you divide the tangent line function by the tangent line slope, why?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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^22=0 So, finding x = sqrt(2) amounts to finding where on the xaxis x^22 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 xaxis, it will find that x where x^22=0; that is, the x that came from x=sqrt(2).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah 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
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0You know, you can derive it yourself by analyzing each step in that pic. and doing what the mathematics would dictate in that situation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your 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
 5 years ago
Best ResponseYou've already chosen the best response.0nope that made no sense... ill figure this out, gimmy a sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's difficult to explain online. They need to put a drawing feature in this thing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol they do, like tutorvista

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think mac must think his thread's been hijacked.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so youre taking x and subtracting y/y'(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i still dont get why ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hang on, I'll see if I can put a small proof together.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how 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
 5 years ago
Best ResponseYou've already chosen the best response.0no 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
 5 years ago
Best ResponseYou've already chosen the best response.0Wait, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, 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
 5 years ago
Best ResponseYou've already chosen the best response.0And go along with it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can understand how it all works, but how does the drawing explain how the formula works?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(i already made a drawring btw lol)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Suppose x_1 is close to a root of the equation f(x)=0. Then the equation of the tangent at x=x_1 is\[yy_1=m(xx_1)\]that is,\[yf(x_1)=f'(x_1)(xx_1)\]This tangent crosses the xaxis at a point where x = x_2 and y = 0. Hence,\[0f(x_1)=f'(x_1)(x_2x_1) \rightarrow x_2=x_1\frac{f(x_1)}{f'(x_1)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wow that makes sense...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes :D thanks for spending time to teach me this stuff :O

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No worries  it's good you want to know :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its incredible that you can put up with teaching me like that, even alot of teachers would loose it pretty soon

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i honestly am really curious as to where mac went...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he stopped observing a while ago even

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, he probably got his answer and was all, "WTF is going on?!"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats after differential equations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like course wise, you take calc1 calc2 calc3 diff eq then what?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have no idea man, not American ;D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0Partial 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, nonlinear, inhomogeneous partial differential equations with nonconstant coefficients, you're done!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since i did not understand any of nth order, nonlinear, inhomogeneous partial differential equations with nonconstant coefficients i guess im not done :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, keep going, though!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know all that is taught of mathematics to the extent that research is your only option?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
 5 years ago
Best ResponseYou've already chosen the best response.0well 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
 5 years ago
Best ResponseYou've already chosen the best response.0anddd man i need help on something: \[\int\limits_{}^{} (\sin x  \cos x) \div (\sin x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, you're right on your last point.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats pretty crazy, that you know everything O.o

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know everything..!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you know everything in math...
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