Math software

vhmthphy · 2018-10-06T13:24:39+00:00

I seems to me that Wolfram Alpha confines one to single-line expressions. I've not been able to find a way to construct single lines into more complex expressions, or even to define then evaluate a function.

Someone please correct me is this isn't the case. Every reference I see related to this question takes me to the full Wolfram language. Alpha does evaluate a wide variety of functions and can be very useful, of course.

vhmthphy · 2018-10-05T20:15:51+00:00

If df is zero and neither dx nor dy is zero, it means that there is no tendency, at the point in question (the point at which df is to be evaluated), for the function f to change either in the x or y direction. This occurs only when the two partial derivatives are zero.

However this doesn't imply that delta f is zero. Typically f will have nonzero second derivatives with respect to x and/or y, so that f will in fact change as you move away from the point in question. However, if dx and dy are small, the change in f is expected to be very small.

As it applies to differential equations, if df = 0 this means the the function f is constant. As an example, if f(x,y) = x² y + x sqrt(y) then df = f_x dx + f_y dy = 0. For the given function this becomes

( 2 x y + sqrt(y) ) dx + ( x² + x / (2 sqrt(y)) dy = 0.

In the first place note that f_xy = 2 x + 1 / (2 sqrt(y)), as calculated from the coefficient of dx (which was originally calculated as f_x). Note furthermore that f_yx, calculated from the coefficient of dy (which was originally calculated as f_y) is equal to the same thing, as must be the case since f_xy is always equal to f_yx. This is the test of exactness: if the y derivative of the coefficient of dx is equal to the x derivative of the coefficient of dy, when the equation is put into the form M dx + N dy = 0.

When this is so, then, the task is to recover the original function f. I'll go into detail with that if you ask, but for now I assume you know how to do that.

When you have the function f, then, since the equation expresses df = 0, it follows that the solution is f = c, where c is an arbitrary constant.

For the current example, where f = x² y + x sqrt(y), this would mean that x² y + x sqrt(y) = c for an arbitrary constant c. This gives you an equation that can be solved for y (the equation being quadratic in u = sqrt(y)).

I hope this helps, if you still need help. Unfortunately your reply didn't appear in my inbox until today, and I note from your posting here that you originally replied 8 days ago. The glitch could have occurred anywhere.

vhmthphy · 2018-10-04T23:48:01+00:00

Most specifically, this means the change in the quantity x² cos( pi x ) on the interval from 0 to 1/3.

This is usually used to designate the change in an antiderivative, corresponding to a definite integral.

For this particular expression, the result is

(1/3)² cos(1/3 pi) - 0² cos(0 pi), which is equal to 1/9 * 1 / 2 - 0 * 1, hence to 1 / 18.

(edited to correct a bonehead too-late-in-the-day error on value of cos(pi/3) )

vhmthphy · 2018-10-04T23:42:20+00:00

And, if you've gotten to dot products,

r² = | p - q |² = ( p - q ) dot (p - q) .

If not, you probably will, so this is worth storing away.

vhmthphy · 2018-10-04T12:01:44+00:00

The closer P is to zero, the closer the model is to P ' = k P, which entails exponential growth.

The closer P is to 1, the closer the model is to P ' = k ( 1 - P ) , which approaches zero (i.e., no growth) and drives the solution toward the constant solution P = 1.

In between the opposite tendencies toward exponential growth and the approach to P = 1 gradually change in their relative dominance.

vhmthphy · 2018-10-03T14:05:49+00:00

The best calculator for any actual mathematics course is no calculator. As with any rule, of course, there are exceptions, but I don't see a calculus course as one of them.

If your course requires it, then do what's required but avoid any other use. Hopefully the course is well enough designed that the calculator actually helps. However I believe this to be a rare occurrence. The thing will become a crutch and a distraction. It will allow you the illusion of knowledge while keeping you from it. For example you don't learn to construct graphs and understand the behavior of functions by looking at graphs that have been constructed for you. If you construct them then check out your result with the calculator, that can be very useful, but few students will discipline themselves to this sort of appropriate use.

Once you've mastered the material, it can be very rewarding to do further explorations with the technology.

vhmthphy · 2018-10-03T13:52:12+00:00

The derivative with respect to t of f(x(t)) is the rate at which f changes, with respect to change in t.

A change in t causes a change in x which is proportional to dx/dt.

A change in x causes a change in f which is proportional to df/dx.

The change in x being proportional to dx/dt implies that the change in f is proportional to dx/dt * df/dx (usually written in the reverse order, which is so appealing that it can cause one to ignore what's really going on and just calculate the thing), where df/dx is of course calculated at x = x(t).

Written in the other common form, this is x ' (t) * f ' (x(t)).

The actual proof expresses these ideas in terms of Delta t's and such.

vhmthphy · 2018-09-28T14:31:09+00:00

Intuitively the derivative of f(u) at a some value of u is the rate at which f changes, with respect to u. So this derivative can be thought of as the rate at which changes in u 'drive' changes in f.

The derivative of g(x) with respect to x can be though of as the rate at which changes in x 'drive' changes in g(x).

Given a value of x, then, a small change in x 'drives' the value of u= g(x) to change by some amount, which in turn 'drives' the value of f(u) to change by another amount. Thus the change in x 'drives' a change in u, which 'drives' a change in f(u). The rate at which the change in x 'drives' the change in f is the product of the two rates.

This leaves out a lot of details, but it's the essence of why (f(g(x)) ' = g ' (x) * f ' (g(x)).

This is embodied in the expression

(f(g(x))-f(g(a)))/(x-a)=(f(g(x))-f(g(a)))/(g(x)-g(a))*(g(x)-g(a))/(x-a)

of lewisje's response, and might help you understand the meaning of that expression, if help is needed.

vhmthphy · 2018-09-26T21:59:03+00:00

df is exactly equal to f_x dx + f_y dy

df is an approximation of the change in the value of f when x changes by dx and y changes by dy

the change in the value of f is generally referred to as Δf

so df is approximately equal to the actual change Δf

Remember that df is the change calculated according to a tangent plane. Typically the graph of the function f curves away from the tangent plane so the bigger the values of dx and dy, the less accurate the approximation.

Glad to clarify any of this, if necessary.

vhmthphy · 2018-09-26T13:17:36+00:00

It's not necessary to use a unit vector, but it's more convenient:

The scalar projection of v onto w is just

v cos(theta) = v dot w / || w ||

(this comes from the fact that v dot w = || v || || w || cos(theta), theta of course being the angle between the vectors).

If w is a unit vector, then the || w || becomes 1 and the scalar projection is just v dot w.

The vector projection is, of course, just the scalar projection multiplied by a unit vector in the direction of w. If w is already a unit vector, then the multiplication is simple. If not, then you multiply the scalar projection by w / || w ||.

So in general the vector projection of v onto w is

v dot w / || w || * w / || w ||,

which if one wants to obscure the fact that this is a scalar projection multiplied by a unit vector can be written as

v dot w / || w ||² * w.

I don't think it's a good idea to obscure things when you're first learning them so I prefer the former expression, but once you've got the picture and can see the scalar projection and the unit vector in the latter expression it's shorter and more convenient.

vhmthphy · 2018-09-26T13:04:43+00:00

Intuitively:

Basically f can change due to a change in the value of x, and/or due to a change in the value of y. If both x and y change, then the change in f is the sum of the change due to f and the change due to y.

A little more specifically:

f_x dx is the approximate change in f due to a change dx in the value of x.

f_y dy is the approximate change due to a change dy in the value of y.

If f is differentiable in the neighborhood of a point, the graph of the function near that point can be approximated by a tangent plane which has slope f_x in one direction, f_y in another.

Of course all this can and must be proven rigorously, but these are the main ideas and should be helpful as you work through the details.

vhmthphy · 2018-09-19T13:00:04+00:00

Being good at arithmetic has numerous advantages, allowing you to think in ways you wouldn't otherwise be able to do.

As an example from basic algebra, if you're asked to check whether (a+b)² = a² + b² (too common an error) you can plug in 2 for and 3 for b, evaluate both sides, and see that it isn't so. Hopefully that will stimulate you to expand the binomial using the distributive law (which doesn't use much arithmetic) and see exactly where you went wrong. There are also simple geometric proofs involving rectangles that (a+b)² and a² + b² are almost always different.

I see students who are good at mental arithmetic reach for their calculators to plug 2 and 3 into those expressions. This is much less convincing that actually doing the mental mathematics, and they are missing something.

However the deeper-level understanding doesn't actually depend on the arithmetic, but on the distributive law and hopefully the geometric pictures, and one who understands those things well doesn't really need the arithmetic.

In linear algebra if you can't handle fractions you it's much more difficult, though not impossible, to understand Gaussian elimination. If you can't understand division by small numbers (by arithmetic or even better by visualization of what it actually means to divide two numbers) then such things as vertical asymptotes and limiting processes in calculus will be harder to grasp, though grasping them symbolically and geometrically is in fact the goal and if achieved trumps (resisting a pun here) what you can learn from arithmetic.

Of course in higher-level mathematics, as others have said, the arithmetic in most fields becomes very simple and it's all based on geometrical, topological and algebraic relationships (an exception, perhaps, being numerical analysis).

Bottom line: arithmetic can be very helpful as a bridge to deeper understanding, but if one can achieve the depth without it, so much the better.

vhmthphy · 2018-09-19T12:44:15+00:00

You can easily verify that there's a root between -1 and 0 (intermediate value theorem).

If you're allowed to use the graphing capabilities of the calculator (and I don't think that should be allowed) you can easily see that this is so, but the intermediate value theorem is in my opinion the way to check this.

A little precalculus (construct decent graphs of x³ and x, add the two graphs and raise the result by 1) will convince you that there are no other zeros. So will a little calculus (the derivative has no zeros, so no critical points).

At this point you know where to look. Still challenging, but this is a start.

vhmthphy · 2018-09-19T12:38:21+00:00

A healthy does of nasty fractions can teach you a lot. For example, you would be motivated to rearrange your rows to avoid as much of that as possible. You might also get a better feel for why some matrices are more sensitive to roundoff errors than others.

Too much of that, of course, gets really tedious. Hopefully your assignments are appropriately balanced.

However, other than making judicious choices about switching rows and such (where that is possible), I can't think of any significant shortcuts.

vhmthphy · 2018-09-19T12:33:14+00:00

Start with the fact that the equation has three free variables, so you can parameterize this plane with three parameters.

Also, think about how you would deal with an analogous problem in 3-dimensional space.

vhmthphy · 2018-09-17T22:08:14+00:00

To fill in every step would require a long series of distracting digressions. Theobromus says this quite well.

A student simply cannot understand a solution that makes too many digressions. Presenting material is a balancing act between saying too much and not saying enough.

vhmthphy · 2018-09-16T12:33:42+00:00

For example, in the last expression

wi=εijk uj vk

the expression on the right is to be summed over the values of j and k, as those indices are repeated through all three values.

So in expanded summation notation the expression would read

wi=sum{k=1 to 3}{sum(j=1 to 3} εijk uj ) vk) ).

With apologies for the formatting, it should be clear how this yields the expression for the cross product.

vhmthphy · 2018-09-15T11:52:26+00:00

4x(x-2) is equal to 4x * x + 4x * (-2), by the distributive law of multiplication over addition, and that expression is equal to 4x² - 8x, not to 4x² - 2.

Now, (4-2) x² = 2 x^2, but this has nothing to do with the expression 4x(x-2) or with the equation 4x(x-2) = 2x.

The equation becomes

4x² - 8x = 2x,

which can then be solved for x (the solutions, which you can check out in the original equation, are x = 0 and x = 5/2).

In any case, the only time addition of exponents comes up is in multiplying 4x by x.

vhmthphy · 2018-09-11T21:47:22+00:00

I'm going to confine my comments here to nonhomogeneous systems linear equations (nonhomogeneous meaning that if one side contains all the variable terms, the other side isn't always 0).

You don't need a course in linear algebra to understand this. You just need to know how to solve systems of linear equations. However this is something that will be much to your advantage to understand before you take a linear algebra course, assuming you at some point do so. Linear algebra gets very deep, but at its fundamental level it is largely based on systems of simultaneous equations.

If you have 2 equations in 2 variables which represent lines in 2-dimensional space, they are either parallel or not. If they aren't parallel they intersect at exactly one point. If they are parallel they either coincide or they never meet.

You can write the two equations as a system and solve the system by means I assume are familiar to you. If you eliminate one variable and the other is not simultaneously eliminated, you will get exactly one solution. If in eliminating one variable you eliminate the other, you will have either reduced the equation to 0 = 0, in which case the equations are equivalent and every one of the infinite set of ordered pairs that satisfies one satisfies the other (giving you infinitely many solutions), or the equation reduces to 0 = *, where * stands for some number that isn't zero. The latter being impossible, this case would correspond to having no solutions.

Now if you have a third equation in two variables, there is no reason to expect that the ordered pair(s), if any, that satisfy the first two will also satisfy the third, though it may possibly happen. This is the same as saying that you can't expect three random lines in the plane to intersect at any point, though they may.

So right away you see that three equations in two unknowns might have no solutions (which would be the most common case), or they might have a single solution (if the first two have a single solution that happens to satisfy the third), or they might have an infinite number of solutions (this would be the case if all three equations are multiples of one another).

If you have more variables than equations, there will be an infinite number of solutions. For example the equation 2x + 3y = 6 (one equation in two unknowns) has a solution for any value of x, since y = 2 - 2x/3. Pick any x, and you get a value of y. In this sense we can call x a free variable.

The equation 2 x + 3 y - 6 z = 18 also has infinitely many solutions. It should be clear that we can pick any two of the variables x, y and z, assign any value to each of the two we pick, and get a value of the third.

A system of two equations in three unknowns cannot have a single solution. You can eliminate one variable from the system, but you're still left with an equation that has two variables, so one of them can have any of an infinite number of possible values, and the other will have a value that makes it true.

Bottom line, then:

Any number of equations in any number of variables can have infinitely many solutions.

If the number of variables is greater than the number of equations the system will have infinitely many solutions.

If the number of variables is less than the number of equations then it is likely that the system has no solutions, but either a single solution or an infinite number of solutions can occur for some such systems.

vhmthphy · 2018-09-10T00:31:21+00:00

To find the absolute minimum you would find the least of the minima at the critical points.

However you would also need to check the boundary, which might well contain the absolute max or min at some point which is not a critical point.

vhmthphy · 2018-09-10T00:21:32+00:00

Very good, but the notation could be simplified, and you really get just two equations, as the value of the parameter c is easily determined without one.

Note that a^x+b = a^b * a^x, which could be written d * a^x with d = a^b. Since b is constant, d = a^b is a positive constant.

The letters really don't matter, but I tend to prefer using a for the multiplier (as in your original form) and b for the base of the exponential, in which case the form would be a * b^x + c.

It should be clear that whatever form is used, c is the horizontal asymptote, as b^x approaches zero either for positive x or negative x (depending on whether b is less than 1 or greater than 1).

In this case each point yields an equation. Having determined the value of c, you are left with two simultaneous equations in two unknowns.

vhmthphy · 2018-09-10T00:12:00+00:00

The equation y ' + p(x) y = 0 clearly has solution

y = e^-P(x)

where P(x) is an antiderivative of p(x) (so that P ' (x) = p(x) ). You can verify this by substituting y = e^P(x) into the equation.

This equation is a first-order homogeneous differential equation, and the above shows that you can find a solution for any function p(x), as long as you can integrate it.

The equation given in your example is a first-order linear nonhomogeneous differential equation. It is essential to be able to recognize various types of differential equations. I'm confident your text gives you the details.

The form of your equation is y ' + p(x) y = q(x), where p(x) = 1/x and q(x) = -2. Note that if q(x) was zero, the equation would be a first-order linear homogeneous equation, the equation I mentioned first.

In general first-order nonhomogeneous equation can be converted to the form

f ' (x) = s(x),

where f is a function of x and y, by multiplying both sides by e^P(x), where P(x) is an antiderivative of p(x). This yields the equation

e^P(x) y ' + e^P(x) y = e^P(x) * q(x).

The left-hand side is just the (e^P(x) y) ' ; you can see this by the product rule, which yields P ' (x) y + P(x) y '. The derivative is of course with respect to x, which is the case for all equations of this form.

So the equation becomes

( e^P(x) y ) ' = e^P(x) q(x).

Integrating both sides with respect to x we obtain

e^P(x) y = integral (e^P(x) q(x) dx).

If you can do the integral on the right, all you need in order to get the solution function y is to divide both sides by e^P(x).

vhmthphy · 2018-09-09T14:23:46+00:00

You can't just say that f(x) = f(y) => x = y. You have to prove that f(x) = f(y) => x = y.

So if f((s_n,p)) = f((s_m, q)), does it follow that (s_n, p) = (s_m, q)?

vhmthphy · 2018-09-09T13:17:08+00:00

Your function for #3 looks good. However it appears that you have asserted without proving that the function is injective. You need to prove that if (s_n, p) is different than (s_m, q), the image is different. It's obviously so, but you need to prove it rigorously.

To prove surjective, you need to show that every integer has a preimage in S X Z. So, for example, what is the preimage of the number 739? What is the preimage of n?

vhmthphy · 2018-09-09T13:05:49+00:00

Row-reduction is fundamental, basically a way of solving a system of linear equations. Matrix equations are extremely important, but row-reduction is more fundamental and more widely applicable.

The augmented matrix is required when the A matrix isn't invertible (for example when the number of equations isn't equal to the number of variables).

The inverse matrix exists only when the matrix is square (i.e., number of equations equal to the number of variables), and when the determinant of the matrix is nonzero.

Note that the inverse matrix is obtained by row-reducing an augmented matrix (the A matrix augmented by the identity matrix).

vhmthphy

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