Is the function f(x) = cscx continuous if we restrict the domain to the closed interval [-3pi, 3pi] (in radians)? If not, where are the discontinuities?

Ok, so this is how I solve a problem like this:

cscx = 1/sinx

I am imagining a unit circle right now.

At x = 0 radians, f(x) is undefined (because 1/sin0 is the same as 1/0, which is undefined). So already we can answer the first question, and say that the function is not continuous over the set of real numbers contained within the restricted domain.

But to answer the second question:

The only discontinuities that occur are at the points where sinx = 0 (because 1/sinx is undefined when sinx = 0).

Within the closed interval [-3pi, 3pi], sinx = 0 when x = -3pi, -2pi, -pi, 0, pi, 2pi, and 3pi.

Therefore, the discontinuities occur at the points where x = -3pi, -2pi, -pi, 0, pi, 2pi, and 3pi.

I am also supposed to draw a sketch of the graph according to the question I'm working on, but there's no way I'm going to waste time explaining here how I figured out the right way to sketch the graph (it would take too much time to explain in "static" words). Suffice to say, according to Desmos, I got it right, so the answers above should be correct.

Another interesting thing about complex numbers is that you sometimes get extra solutions you wouldn't normally think of.

For example, consider

x^3 = 1

What's the solution?

x = 1

Obviously. But there's also two other solutions, if we allow x to be complex. They fall on the unit circle drawn in the complex plane around the origin, at angles 120 and 240 degrees from the positive real axis. Multiplying those numbers by themselves rotates them around the origin, so 3 * 120 = 360 and 3 * 240 = 720 leaving you back on the positive real axis at 1.

(March 8, 2018 at 3:49 am)robvalue Wrote: Very nice! Good work, and it shows what I was saying about how understanding helps the memory. It all follows, it's not just meaningless drivel to memorise.

Now do the formula for cubics Just kidding, I didn't see that done until university, and it hurt my poor brain.

Funnily enough, I remember hearing that there is also a formula for power 4 polynomials, but that there cannot be a general formula for power 5 or more. I wonder if my memory is correct there. There's probably some restriction about the kind of formula we're talking about.

Instead, we can use iterative methods and such, for approximate solutions.

Your memory is correct. The allowed operations are addition, subtraction, multiplication, division, and extraction of roots.

The quadratic formula shows it is possible to solve a general quadratic via these operations. And yes, there is a procedure for cubic and a quartic equations

it turns out that for every polynomial there is an object called a group that describes the symmetries between the roots of that polynomial. If that group has a certain property (called solvability) , then the polynomial is solvable via radicals. It turns out that the relevant group is NOT solvable for the general equation of degree 5 or higher.

A specific quintic that cannot be solved via radicals is x^5 -5x+5.

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(March 8, 2018 at 9:48 am)robvalue Wrote: Another interesting thing about complex numbers is that you sometimes get extra solutions you wouldn't normally think of.

For example, consider

x^3 = 1

What's the solution?

x = 1

Obviously. But there's also two other solutions, if we allow x to be complex. They fall on the unit circle drawn in the complex plane around the origin, at angles 120 and 240 degrees from the positive real axis. Multiplying those numbers by themselves rotates them around the origin, so 3 * 120 = 360 and 3 * 240 = 720 leaving you back on the positive real axis at 1.

The same thing happens for every polynomial equation of the form x^n =1. The result is a regular n-sided polygon in the complex plane.

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(March 8, 2018 at 3:53 am)Grandizer Wrote:

(March 8, 2018 at 3:49 am)robvalue Wrote: Very nice! Good work, and it shows what I was saying about how understanding helps the memory. It all follows, it's not just meaningless drivel to memorise.

Now do the formula for cubics Just kidding, I didn't see that done until university, and it hurt my poor brain.

Funnily enough, I remember hearing that there is also a formula for power 4 polynomials, but that there cannot be a general formula for power 5 or more. I wonder if my memory is correct there. There's probably some restriction about the kind of formula we're talking about.

lol, I don't even know what it's supposed to be like (the cubic formula). But I can imagine it being rather tedious to derive. Maybe a few years from now once I get into studying proofs for real.

The formula itself is nasty, but the procedure can be understood.

Given a cubic equation, a x^3 +b x^2 +c x +d=0, first divide by the leading coefficient to get

x^3 +B x^2 +C x +D=0

Then let  y=x-B/3. It turns out that when you do the algebra, this will eliminate the quadratic term, so you have
an equation of the form

y^3 + E y +F=0.

Now comes the trick. Multiplying out shows that (u+v)^3 -(3uv)(u+v) =u^3 + v^3. We try to force our equation into this form.

So, we put y=u+v, giving one more variable to play with and then require that -3uv=E, so v=-E/(3u),

Plugging in gives
0=y^3 +Ey +F=(u+v)^3 -(3uv)(u+v) +F =u^3 +v^3 +F =u^3 +(-E/(3u) )^3 +F.

Expand this and multiply through by 27u^3 gives

27 u^6 +27F u^3 -E^3 =0.

This is now a quadratic in u^3, which can be solved by the quadratic formula. This u^3, which gives u, which gives v, which gives y, which finally gives x.

In finding u from u^3, a cube root has to be taken and it is crucial to remember that there are three complex cube roots, which ultimately gives the three roots to the cubic.

Whew!

Say we have a function defined as follows:

f(x) = x^2 when x < 1
= 2x-1 when x >= 1

Is f(x) differentiable at point where x = 1?

One would think at first glance the answer is "no" because the two parts of the function are two different functions. However, the limit of the slope of the tangent line as it approaches the point of interest (where x = 1) from the left is the same as the limit of the slope as it approaches the same point of interest from the right. Using the power rule, we get the same thing. The derivative at point where x = 1 is consistently 2.

This means that f(x) is actually differentiable at point where x = 1. Did I get this correct? Or am I missing something here?

(March 15, 2018 at 1:49 am)Grandizer Wrote: Say we have a function defined as follows:

f(x) = x^2 when x < 1
     = 2x-1 when x >= 1

Is f(x) differentiable at point where x = 1?

One would think at first glance the answer is "no" because the two parts of the function are two different functions. However, the limit of the slope of the tangent line as it approaches the point of interest (where x = 1) from the left is the same as the limit of the slope as it approaches the same point of interest from the right. Using the power rule, we get the same thing. The derivative at point where x = 1 is consistently 2.

This means that f(x) is actually differentiable at point where x = 1. Did I get this correct? Or am I missing something here?

That is correct. if you do the *definition* of the derivative in terms of a limit for this function, you use the x^2 for the left limit and 2x-1 for the right limit. Since both limits in the definition are 2, f'(1)=2 and the function is differentiable there. It is even continuously differentiable (the derivative is a continuous function). But, it is not differentiable two times: the derivative of the derivative is not defined at x=1.

(March 18, 2018 at 10:43 pm)polymath257 Wrote:

(March 15, 2018 at 1:49 am)Grandizer Wrote: Say we have a function defined as follows:

f(x) = x^2 when x < 1
     = 2x-1 when x >= 1

Is f(x) differentiable at point where x = 1?

One would think at first glance the answer is "no" because the two parts of the function are two different functions. However, the limit of the slope of the tangent line as it approaches the point of interest (where x = 1) from the left is the same as the limit of the slope as it approaches the same point of interest from the right. Using the power rule, we get the same thing. The derivative at point where x = 1 is consistently 2.

This means that f(x) is actually differentiable at point where x = 1. Did I get this correct? Or am I missing something here?

That is correct. if you do the *definition* of the derivative in terms of a limit for this function, you use the x^2 for the left limit and 2x-1 for the right limit. Since both limits in the definition are 2, f'(1)=2 and the function is differentiable there. It is even continuously differentiable (the derivative is a continuous function). But, it is not differentiable two times: the derivative of the derivative is not defined at x=1.

I must confess it wasn't intuitive to me at the start (mainly because one part is a linear function, the other is quadratic), but after examining the curve closely in Desmos at the point of interest, I can see how it works.

(March 19, 2018 at 9:11 am)Grandizer Wrote:

(March 18, 2018 at 10:43 pm)polymath257 Wrote: That is correct. if you do the *definition* of the derivative in terms of a limit for this function, you use the x^2 for the left limit and 2x-1 for the right limit. Since both limits in the definition are 2, f'(1)=2 and the function is differentiable there. It is even continuously differentiable (the derivative is a continuous function). But, it is not differentiable two times: the derivative of the derivative is not defined at x=1.

I must confess it wasn't intuitive to me at the start (mainly because one part is a linear function, the other is quadratic), but after examining the curve closely in Desmos at the point of interest, I can see how it works.

A more interesting example is f(x)=x^2 sin(1/x) for x!=0 and f(0)=0.

Here, the derivative exists everywhere, and f'(0)=0 (use the limit definition), but f'(x) is badly discontinuous.

I was requested in another thread to talk a bit about continuous functions that are nowhere differentiable. This is not a simple topic, but it was an important one historically because it shows how our intuitions can diverge quite a bit from what is actually provable.

We first start with some definitions:

A real valued function f:R->R is said to be continuous if for each x in R and each e>0, there is a d>0 such that whenever y is in R with |x-y|<d, we get |f(x)-f(y)|<e.

In this, the d can depend on both x and e. This was discussed above and is a typical hurdle for math students.

Similarly, we say that f:R->R is differentiable at the point x if for there is a real number f'(x) such that for each e>0 there is a d>0 such that whenever |h|<d, we have
| ( f(x+h) -f(x))/h -f'(x) |<e

This is just a restatement of the limit definition of the derivative in terms of the epsilon-delta definition of limit. So, we have f'(x)=lim_{h->0} (f(x+h) -f(x) )/h. To say that f is differentiable at x is to say that this limit on h exists and is finite. To say that f is NOT differentiable means this limit does NOT exist.

Now, most 'typical' functions that are continuous are differentiable at 'most' points. Intuitively (danger!), a lack of differentiability at a point of continuity means there is a 'corner' at that point or a vertical tangent line. We are going to stretch that intuition to the breaking point.

We start out with a 'saw tooth' function g(x), which is defined to be periodic of period 4 (for convenience) and such that g(x)=x for 0<=x<=2 and g(x)=4-x for 2<=x<=4.

This function graphs something like /\/\/\/\/\/\/\/\/\/\ with a base for the slopes all 1 or -1 and the baseline of a triangle of length 4. This function fails to be differentiable (corner!) at every even integer.

Next, let g_n (x) =g( 4^n x)/4^n.

In this, the 4^n inside the parentheses make the function have a smaller period (it goes through things faster by a factor of 4^n) and the division by 4^n makes the total 'size' mush smaller. Again, though, g_n (x) is a sawtooth function /\/\/\/\/\/\, but now with a smaller height and a faster back-and-forth. A crucial aspect: the slopes of the lines re still 1 or -1.

Notice that all of these g_n are continuous.

Now let f(x) be the infinite sum of the g_n (x) (n from 1 to infinity).

A few crucial points:
1. This sum always converges. In fact, we can do a comparison test with with sum of 1/4^n which is a geometric series.
2. The function f(x) is still continuous. This is trickier. The point is that given an e>0, we can get some tail of the series to be smaller than e/2 since the sum of 1/4^n converges. Then, the finitely many g_n left over are all continuous, so we can pick the d>0 show that their total change is smaller than e/2 also. So, in the function f, we can get |f(x)-f(y)|<e by controlling finitely many of the g_n (the rest add up to be small automatically).

3. This function is differentiable nowhere.
And this is trickier still. The point is that for h=1/4^n, the difference quotient for the derivative adds up the slopes of the g's up to stage n. This is a sum of 1 or -1 for n terms. ALL the rest of the g's are periodic so contribute nothing to the difference quotient. In particular, for h=1/4^n, the difference quotient is even when n is even and odd when n is odd. But this means the limit as n->infinity cannot exist! Hence, the limit as h->0 does not exist!

The result of adding up all of these smaller and smaller, but faster, sawtooth functions is a limit that is continuous (since all the g's are and they are controlled nicely in size), but has 'corners everywhere'.

This, by the way, was NOT the first such continuous nowhere differentiable function. The first given was the sum of sin(3^n x)/2^n. The proof is harder for this example, though, but the idea is the same: control size by using a convergent sequence to keep heights low and make the slopes go wild by increasing periods. In the second case, the derivative 'should be' the sum of 3^n cos(3^n x)/2^n and the 3^n/2^n diverges badly when added up.

Even more dramatic, in the study of Brownian motion (molecules bouncing off of each other), the functions representing the motion of the particle are *usually* continuous and nowhere differentiable! The idea is that many small jolts make the velocity undefined.

(March 22, 2018 at 9:34 am)polymath257 Wrote: I was requested in another thread to talk a bit about continuous functions that are nowhere differentiable. This is not a simple topic, but it was an important one historically because it shows how our intuitions can diverge quite a bit from what is actually provable.

We first start with some definitions:

A real valued function f:R->R is said to be continuous if for each x in R and each e>0, there is a d>0 such that whenever y is in R with |x-y|<d, we get |f(x)-f(y)|<e.

In this, the d can depend on both x and e. This was discussed above and is a typical hurdle for math students.

Similarly, we say that f:R->R is differentiable at the point x if for there is a real number f'(x) such that for each e>0 there is a d>0 such that whenever |h|<d, we have
| ( f(x+h) -f(x))/h -f'(x) |<e

This is just a restatement of the limit definition of the derivative in terms of the epsilon-delta definition of limit. So, we have f'(x)=lim_{h->0} (f(x+h) -f(x) )/h. To say that f is differentiable at x is to say that this limit on h exists and is finite. To say that f is NOT differentiable means this limit does NOT exist.

Now, most 'typical' functions that are continuous are differentiable at 'most' points. Intuitively (danger!), a lack of differentiability at a point of continuity means there is a 'corner' at that point or a vertical tangent line. We are going to stretch that intuition to the breaking point.

We start out with a 'saw tooth' function g(x), which is defined to be periodic of period 4 (for convenience) and such that g(x)=x for 0<=x<=2 and g(x)=4-x for 2<=x<=4.

This function graphs something like /\/\/\/\/\/\/\/\/\/\ with a base for the slopes all 1 or -1 and the baseline of a triangle of length 4. This function fails to be differentiable (corner!) at every even integer.

Next, let g_n (x) =g( 4^n x)/4^n.

In this, the 4^n inside the parentheses make the function have a smaller period (it goes through things faster by a factor of 4^n) and the division by 4^n makes the total 'size' mush smaller. Again, though, g_n (x) is a sawtooth function /\/\/\/\/\/\, but now with a smaller height and a faster back-and-forth. A crucial aspect: the slopes of the lines re still 1 or -1.

Notice that all of these g_n are continuous.

Now let f(x) be the infinite sum of the g_n (x) (n from 1 to infinity).

A few crucial points:
1. This sum always converges. In fact, we can do a comparison test with with sum of 1/4^n which is a geometric series.
2. The function f(x) is still continuous. This is trickier. The point is that given an e>0, we can get some tail of the series to be smaller than e/2 since the sum of 1/4^n converges. Then, the finitely many g_n left over are all continuous, so we can pick the d>0 show that their total change is smaller than e/2 also. So, in the function f, we can get |f(x)-f(y)|<e by controlling finitely many of the g_n (the rest add up to be small automatically).

3. This function is differentiable nowhere.
And this is trickier still. The point is that for h=1/4^n, the difference quotient for the derivative adds up the slopes of the g's up to stage n. This is a sum of 1 or -1 for n terms. ALL the rest of the g's are periodic so contribute nothing to the difference quotient. In particular, for h=1/4^n, the difference quotient is even when n is even and odd when n is odd. But this means the limit as n->infinity cannot exist! Hence, the limit as h->0 does not exist!

The result of adding up all of these smaller and smaller, but faster, sawtooth functions is a limit that is continuous (since all the g's are and they are controlled nicely in size), but has 'corners everywhere'.

This, by the way, was NOT the first such continuous nowhere differentiable function. The first given was the sum of sin(3^n x)/2^n. The proof is harder for this example, though, but the idea is the same: control size by using a convergent sequence to keep heights low and make the slopes go wild by increasing periods. In the second case, the derivative 'should be' the sum of 3^n cos(3^n x)/2^n and the 3^n/2^n diverges badly when added up.

Even more dramatic, in the study of Brownian motion (molecules bouncing off of each other), the functions representing the motion of the particle are *usually* continuous and nowhere differentiable! The idea is that many small jolts make the velocity undefined.

Thanks for taking the time to answer this one for me. I'll give this a good read tomorrow morning when I'm all wide awake!

I sort of see it now. It seems it's all about continuing the pattern to infinity, starting from an intuitive sawtooth function.