Titles like these got way harder after we lost [sup].
Just write a programming function that does the same thing in the title!
What is a programming function?
[QUOTE=Krinkels;45892304]No, I said 'irregular', not 'ugly'.
Unless you come across a situation in the class where you'd absolutely need one (e.g. you need to do 13 iterations of Newton's Method, or they ask you how to graph something on a specific calculator), I wouldn't. Once you're in college, any situation where you would want or need a graphing calculator can be handled with a computer.[/QUOTE]
depends, sometimes the ti-84 is the only thing i have that can quickly do something, like 4 iterations of the SRK gas equation or solving matrixes for statics
[editline]8th September 2014[/editline]
also its damn useful for thermodynamics where my teacher insisted we use some abstract method for linear interpolation using matlab, i programmed a linear interpolation program on it in about 5 minutes, saved me like hours of work over the course of that class
Oh, I guess I'm wrong. Why would it take so much longer with matlab?
[QUOTE=Krinkels;45927706]Oh, I guess I'm wrong. Why would it take so much longer with matlab?[/QUOTE]
short answer, i press one button enter 5 values and i get a result, his way involved us entering the equation for interpolation each time, calling it as a function and then setting values to the variables
5 seconds of work vs 5 minutes of work, but he still insisted we use matlab over anything else, don't get me wrong its a powerful tool but you don't use a sledgehammer to put in a nail for a painting
I really suck at mathematics, but I need to find this out and I can't find anything useful elsewhere on the internet.
I got a bezier curve with 3 points [0, 0] [25, 90] [75, 100]
What I need is the Y coordinate of a point with X=25.
Can anyone help me out here?
Is it a quadratic Bezier curve?
-snip-
but the curve path doesn't touch that point ever, does it?
and yes, it's a quadratic Bezier curve
[editline]14th September 2014[/editline]
To make it more clear, I want to know the coordinates where the green line meets the red curve:
[url]http://jsfiddle.net/v6dLqoz0/1/[/url]
I've been wondering about this for a very long time, what are some of the applications of the infinite series convergence/divergence you learn in calc2 anyways?
[QUOTE=xmariusx;45974611]but the curve path doesn't touch that point ever, does it?
and yes, it's a quadratic Bezier curve
[editline]14th September 2014[/editline]
To make it more clear, I want to know the coordinates where the green line meets the red curve:
[url]http://jsfiddle.net/v6dLqoz0/1/[/url][/QUOTE]
Sorry this took so long.
The first given point is (0,0) so we don't have to include it in the equation. The x coordinate must be 25, so we need to find t in [0,1] such that:
50(1-t)t + 75(t^2) = 25
Only one of the two solutions lie in the given interval, namely sqrt(2)-1. We use that value as t in the next formula, whose coefficients are based on the y coordinates of the given points.
100(t^2) + 180t(1-t) = y
[url=http://www.wolframalpha.com/input/?i=100%28sqrt%282%29-1%29%5E2+%2B+180%28sqrt%282%29-1%29%281-%28sqrt%282%29-1%29%29]I threw this into WolframAlpha[/url] to get an answer around (25,60.8), which seems plausible.
took the ACT today and fuck that math section
hate it so much urhgurghrughruh
[QUOTE=Krinkels;45974848]Sorry this took so long.
The first given point is (0,0) so we don't have to include it in the equation. The x coordinate must be 25, so we need to find t in [0,1] such that:
50(1-t)t + 75(t^2) = 25
Only one of the two solutions lie in the given interval, namely sqrt(2)-1. We use that value as t in the next formula, whose coefficients are based on the y coordinates of the given points.
100(t^2) + 180t(1-t) = y
[url=http://www.wolframalpha.com/input/?i=100%28sqrt%282%29-1%29%5E2+%2B+180%28sqrt%282%29-1%29%281-%28sqrt%282%29-1%29%29]I threw this into WolframAlpha[/url] to get an answer around (25,60.8), which seems plausible.[/QUOTE]
Thank you so much!
Just a quick question, what if the first given point was not (0,0)?
[QUOTE=B!N4RY;45974755]I've been wondering about this for a very long time, what are some of the applications of the infinite series convergence/divergence you learn in calc2 anyways?[/QUOTE]
Shows up aaaallllll the time in physics. Like constantly 24/7.
Taylor expansion is ubiquitous in physics. If you have something small in some function, like perhaps a pendulum swinging at a small angle, Taylor expand and use the first few terms as an approximation. Of course, it's only the fact that we know the Taylor series converges to the true function we're approximation that allows us to do that!
(Almost) all of quantum mechanics is done in practice using some kind of perturbation theory, which amounts to calculating some value you want to know in terms of some value you [I]do[/I] know, using successively better approximations in the form of terms in an infinite series.
[QUOTE=xmariusx;45976840]Thank you so much!
Just a quick question, what if the first given point was not (0,0)?[/QUOTE]
[url=http://en.wikipedia.org/wiki/B%C3%A9zier_curve#Quadratic_B.C3.A9zier_curves]The solution wouldn't be much different.[/url] The part where I combine terms to get the quadratic into the form:
at^2 + bt + c = 0
Would be slightly longer because there's an extra term in there to begin with.
What's some of your guys favorite literature on tensor calculus?
[QUOTE=Falubii;45977387]What's some of your guys favorite literature on tensor calculus?[/QUOTE]
From a mathematician's viewpoint? If not necessarily, Sean Carroll's GR text does a good job, I think. Good balance of pedagogy and rigor. If you want a mathematician's viewpoint and don't mind a bit more difficulty, Lee's [I]Introduction to Smooth Manifolds[/I] is really the best thing for anything you want to be introduced to for differential geometry.
curious, what do you guys do? i'm guessing most are students but do some of you work as legit mathematicians?
Really having a tough time in Algebra II this year. My teacher is doing a very poor job at explaining things, and is pretty much repeating the problem and hoping I understand. Is there any way I can get online tutoring from someone in this thread? Would really help me long term, as I'm very stressed out right now over it.
[QUOTE=Chernobyl426;46009577]Really having a tough time in Algebra II this year. My teacher is doing a very poor job at explaining things, and is pretty much repeating the problem and hoping I understand. Is there any way I can get online tutoring from someone in this thread? Would really help me long term, as I'm very stressed out right now over it.[/QUOTE]
Have you tried [url=https://www.khanacademy.org/]Khan Academy[/url]? Also if you add me on steam and have any questions while I'm online I'll try to help, but Khan Academy is probably a better bet.
Given a differentiable real function f of a real variable t, is there a curve [b]r[/b](t) = <x(t),y(t),z(t)> such that the following conditions hold:
For all t, ||[b]r[/b](t)|| = f(t)
For all t where [b]r[/b]' is defined, [b]r[/b](t) is orthogonal to [b]r[/b]'(t)?
[QUOTE=Krinkels;46049805]Given a differentiable real function f of a real variable t, is there a curve [b]r[/b](t) = <x(t),y(t),z(t)> such that the following conditions hold:
For all t, ||[b]r[/b](t)|| = f(t)
For all t where [b]r[/b]' is defined, [b]r[/b](t) is orthogonal to [b]r[/b]'(t)?[/QUOTE]
I'm not sure if the problem is more advanced than I'm seeing it, but one example would be uniform circular motion. I think I've terribly misunderstood though.
calculus is fucking me backwards but i love it
finally figuring out how the product/quotient rules of derivatives work and i'm loving this feeling of accomplishment
[QUOTE=Krinkels;46049805]Given a differentiable real function f of a real variable t, is there a curve [b]r[/b](t) = <x(t),y(t),z(t)> such that the following conditions hold:
For all t, ||[b]r[/b](t)|| = f(t)
For all t where [b]r[/b]' is defined, [b]r[/b](t) is orthogonal to [b]r[/b]'(t)?[/QUOTE]
Not unless f is constant, the only solutions to the second condition are paths on a sphere:
|[b]r[/b]|^2 = x^2 + y^2 + z^2
f^2 = x^2 + y^2 + z^2
2f.f' = 2x.x' + 2y.y' + 2z.z'
f.f' = [b]r.r'[/b] = 0 (as [b]r[/b] and [b]r'[/b] are orthogonal)
=> f = 0 or f' = 0
=> f is constant
=> |[b]r[/b]| is constant
[IMG]http://latex.codecogs.com/png.latex?D%28x%29%20%3D%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%201%20%26%20x%20%5Ctext%7B%20is%20rational%3B%7D%5C%5C%200%20%26%20x%20%5Ctext%7B%20is%20irrational.%7D%20%5Cend%7Bmatrix%7D%5Cright.[/img]
I've shown that the Dirichlet function isn't continuous for x = 0 (or any other x for that matter) by the way of delta-epsilon argument, but I have no idea where to start on a rigorous proof of its non-differentiability.
Can anyone point me in the right direction? I'm also trying to do this for [B]D(x)·x[/B] and [B]D(x)·x²[/B]
[QUOTE=PederPauline;46065252][IMG]http://latex.codecogs.com/png.latex?D%28x%29%20%3D%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%201%20%26%20x%20%5Ctext%7B%20is%20rational%3B%7D%5C%5C%200%20%26%20x%20%5Ctext%7B%20is%20irrational.%7D%20%5Cend%7Bmatrix%7D%5Cright.[/img]
I've shown that the Dirichlet function isn't continuous for x = 0 (or any other x for that matter) by the way of delta-epsilon argument, but I have no idea where to start on a rigorous proof of its non-differentiability.[/B][/QUOTE]
You're basically already there, continuity is a necessary condition for differentiability.
If a function is differentiable at a point, it must be continuous. You've shown that it's not continuous anywhere, so it can't be differentiable anywhere.
If you haven't yet proved that differentiable implies continuous, then try something like this:
Fix an irrational number a.
Consider secant lines passing through (a,0) and (b_i,D(b)), where {b_i} is a sequence of real numbers. Have it so that lim b_i = a. Now choose {b_i} carefully so that the slopes of the tangents do not converge.
Then do something similar for a fixed rational number.
[IMG]http://latex.codecogs.com/gif.latex?%5Clim_%7Bx%5Crightarrow%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%7Bx%5E2+1%7D+%5Csqrt%7B1-x%7D%7D%7Bx+%5Csqrt%7B2-x%7D%7D[/IMG]
Find the limit
[IMG]http://latex.codecogs.com/gif.latex?%5Clim_%7Bx%5Crightarrow%200%7D%20%5Cfrac%7B2x%20-%20arcsin%28x%29%7D%7B3x+tan%282x%29%7D[/IMG]
Find the limit
[IMG]http://latex.codecogs.com/gif.latex?arccos%5Cleft%20%28%20%5Cfrac%7B1-x%5E2%7D%7B1+x%5E2%7D%20%5Cright%20%29%20%3D%20-2%20arctan%28x%29[/IMG]
For which real x is this true
[IMG]http://latex.codecogs.com/gif.latex?n%5E%7Bn+1%7D%20%3E%20%28n+1%29%5En%20%5C%3B%20%5C%3B%20%5C%3B%20%5C%3B%20%5C%3B%20n%20%5Cin%20%5Cmathbb%7BN%7D%2C%20n%20%5Cgeq%203[/IMG]
Prove by induction
They are from a test I did and I just want to see how you pros would go about it. (You don't have to do them all, one would suffice).
[QUOTE=JohanGS;46096284][IMG]http://latex.codecogs.com/gif.latex?%5Clim_%7Bx%5Crightarrow%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%7Bx%5E2+1%7D+%5Csqrt%7B1-x%7D%7D%7Bx+%5Csqrt%7B2-x%7D%7D[/IMG]
Find the limit
[IMG]http://latex.codecogs.com/gif.latex?%5Clim_%7Bx%5Crightarrow%200%7D%20%5Cfrac%7B2x%20-%20arcsin%28x%29%7D%7B3x+tan%282x%29%7D[/IMG]
Find the limit
[/QUOTE]
Whenever I think about limits I first look at the equation for the domain of x. Most of the time this will be somewhere in the denominator or in square roots. In the numerator we see that there is a sqrt(1 - x). This is an indicator that the maximum value of x is 1. Anything past that would be undefined. Since we know that, the limit as x approaches infinity will also be undefined.
To check this, think of it numerically. Look at the limit to the left and to the right of the x value you're looking for, in this case infinity. I know infinity isn't really a number, so let's say infinity is 99999.5. Put a value lower than that, 999998 into the equation, which would be the left side. What do you get? Undefined. Put 999999 in the equation, the right side. Also undefined. Therefore the limit as x approaches i finity is also undefined.
If you didn't understand the numerical check above, doing the second limit might help. So here there are limits for arcsin, the domain of ±π/2. But that doesn't help is because we're looking at the limit as approaches 0.
So let's put 0 in the equation and see what we get. Let's look at the denominator first. 2*0 + Tan(2*0) is 0, which would make the function undefined at x = 0 because you are dividing by 0. Does that mean that the limit is also undefined? Not neccesarily. Again let's test thus numerically. Plug in values that you know are defined and really close to 0, like ±0.00000001. You'll probably get something close to 0 the more precise your x value is. I would check but I don't have my calculator with me. But anyways, the f(x) values you get from looking at the ldft and right side of 0 should be the same. Since they are, that f(x) value, 0 is going to be our limit of f(x) as x approaches 0.
Graphically speaking this makes sense too. Imagine a continuous function that passes through the orgin. There would just be an open dot at the orgin, to signify that x cannot be 0.
I believe this should be right, but perhaps double check with someone else because not only am I typing on my phone but I am exhausted and should have slept ages ago.
The limits can be performed analytically using l'Hopital's rule.
1) I found it easiest to factor out x from the three sqrt terms (which will leave you with some imaginary terms when you evaluate the limit), decide that it's indeterminate, apply l'Hopital's rule, reevaluate the limit (imaginary stuff goes away), and you'll finally find the answer to be 1.
2) This one is really straightforward. Just apply l'Hopital's rule and reevaluate the limit.
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