Calculus for the mathmatically challenged

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I was never able to pay close attention to math in school. Now, I find myself fascinated by the potential inherent in quantum physics, I need to learn calculus and higher function mathmatics to properly grasp a lot of the work that is done in quantum physics. Can anyone recommend a good, free, online course to learn calculus and mathmatics? I'm 47 years old and working full time so returning to school for a hobby is out of the question. Thanks!

I can't think of any free, but may be able to send some powerpoint stuff with examples etc if thats any help? (just done engineering degree maths) I was rubbish at maths in school. pm if you'd like them.

https://class.coursera.org/calc1-001/class/index

Here you are. This is just calculus one though;however it's a good foundation. You may need to do alot of catch up though on the quizzes.

Try this if you haven't yet. It's a site at MIT that has on the web the course material they use to learn. Sometimes, the lecture notes are not avaialable.

http://ocw.mit.edu/index.htm

You can also search for this stuff at the Cambridge mathematics department. It's more likely that they have what you need.

http://www.maths.cam.ac.uk/studentreps/tripos.html

And the lecture notes section here http://www.archim.org.uk/

Thank you all! Looks like a bunch of resources so I will be checking them out. Thanks!

Go to Khan Academy. Lots of calculus videos there to watch and learn from.

There are some free on line calculators which will intergrate or differentiate for you. I have even found some which do laplace transformations. These are wounderful gadgets.

That would be helpful if I knew what that meant!
That's what I mean, I don't even have a clue what this all means, I barely passed High School algebra, but I want to learn!

Calc 1: finding slope of lines
Calc 2: Area under curves

Differentiation is all about maipulation of an equation to find another equation that will give you the slope of equation 1 at any instant. With respect to a certain variable..

Ie: f(x) = 4x^4

Differentiate using "the power rule"



What's happening there is you bring the exponent down in front and then subtract one from where it used to be//

So; f'(x) = 16x^3

and f''(x) = 48x^2

If f(x) was a position function and "x" variable was time, then f'(x) will give you velocity and f''(x) is acceleration.

Only if x is taken to mean time.

ruveyn

I would never forget that!

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Sorry, I try to break it down to one post and it's hard to use simple words for that..

All it means is: "use the rules to change one equation into another equation."

There's more to learn such as secant lines, tangent lines, limits, etc. Which they throw at you in the first few days, but the practical lesson of using the rules to change the equations around is all you need to pass the course.. and really you will pick it all up eventually anyhow.

The why it works part is interesting, but the first lesson I connected with was the one I posted above.

First lets explain in basic terms one half of calculus - differential calculus, or the portion of calculus dealing with derivatives.

From geometry, you should know that at any point on a circle there exists a tangent line that touches only that one point on the circle, and is also perpendicular to the radius line from the center of the circle to that same point on the circle:



Ok, well, other curves besides circles can have tangents too. For instance, the figure below illustrates the tangent line to a parabola at a given spot on the parabola:



Now, think back to the equation for a line, which is generally stated as y=mx+b. The important thing here is that m is the slope of the line. The slope of the line is expressed as the ratio of the rise of the line to the run of the line. For instance, a line with a slope of 5/2 rises 5 units for every 2 units moved to the right.

So a derivative of a function is another function that tells you the slope of the tangent line on the original function for any value of x.

For instance, y=x^2 is the equation of a simple parabola, as above.

The derivative of y=x^2 is y=2x (take my word for it, how I found that is unimportant for this discussion).

What this tells us is that when x is 0, for instance, the slope of the tangent line on the function y=x^2 is 0. This is easy enough to see - at x=0 the curve "peaks" and is neither rising or falling. At x=1, the slope is 2 - the tangent line at x=1 is rising 2 units for every unit moved to the right. At x=-2, the slope of the tangent on the parabola is -4 - for every 4 units the line rises, it moves backwards (to the left) one unit.

What the slope of the tangent line at a point on the parabola tells us, in a more practical sense, is how fast the parabola is rising or falling at that one particular point. The same is true for other types of curves and functions as well.

The reverse of differential calculus is integral calculus. Basically if function f(x) is the derivative of the function g(x), then the integral of f(x) is g(x) (not quite, but almost). Instead of telling us the slope of a tangent line, the derivative can be used to measure the area underneath a given curve.

The "power rule" that CornerPuzzlePieces posted earlier explains it well. The exponent 2 goes in front of the x and 1 is subtracted from the exponent. Anything (except for zero perhaps?) raised to the 1st power is itself so the exponent goes away.

Yes, and if you have f(x)= 5x + 6, that is the same as 5x^1 + 6x^0.

Bring the zero down in front and the whole term cancels. f'(x) = 5.

f'(x) = 5x^0 ((Notice this results in 0*5x^-1)) but it cancels because of 0.

f''(x) = 0

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IF you were going the opposite direction (integrating) you would not know the constant because it would be invisible!

So we get around this by adding a "C". And then substituting a known value into the equation.

Integrate by adding 1 to the power, then dividing the whole term by that new number.

Exact opposite of what we did above.. instead of bringing down and multiplying we put it up and divide.

(x^(n+1))
-----------
(n+1)

ie: f'(x) = 5x^0

f(x) = (5x^1)/1 = 5x, now we add a "C" since there may have been a constant term there. (Any number without an x. Aka x^0)

So

f(x) = 5x + C

To solve for C you are usually told the information..

That information are the initial or boundary conditions (in the case of a partial differential equations).

If the function is differentiable the derivative is uniquely determined. If it is anti-differentiable then it is determined only up to a constant. One needs initial conditions to nail down the anti-derivative uniquely.

ruveyn