## calculus – Differentiability of a non-continuous function

So the question that comes to my mind is does differentiability require continuity? I have read it many a times but what I think of it like this, lets say a piecewise function as below is defined
$$f(x) =begin{cases} x^2 & x leq 0 \(2ex) x^2-1 & x gt mathbb 0 end{cases}$$

Now as $$xrightarrow0$$ the right hand derivative approaches zero and so does the left hand derivative.
So apparently the derivative should exit at all points, even though it isn’t continuous. Am I somehow wrong here? I recently completed my school so an answer at that level would be really helpful.

Posted on

## calculus and analysis – How does Mathematica obtain this result?

``````FullSimplify(
Sqrt(2 (Pi)) InverseFourierTransform(1/(x^2 - a^2), x, p),
Element(a, Reals))
``````

Gives the output

``````-(((Pi) Sign(p) Sin(a p))/a)
``````

But

$$int_{-infty }^{infty } frac{e^{-i k x}}{x^2-a^2} , dx$$ is not a defined integration. Mathematica also returns undefined as answer if you compute it.

So, I am trying to understand how does Mathematica calculates that Fourier transform of the non-integrable function.

Posted on

## calculus – Show that \$lim_{epsilonrightarrow0}y_epsilon(x)=delta(x)\$

For $$y_epsilon=frac{e^{frac{-x}{epsilon}}}{e^{frac{2}{epsilon}}}(e^{frac{2}{epsilon}}-e^{frac{2x}{epsilon}})$$ prove it converges to dirac’s delta as $${epsilonrightarrow0}$$.
Answer:
It is sufficient to prove $$frac{1}{epsilon}lim_{epsilonrightarrow0}int_0^x y_epsilonphi(x)dx=phi(0)$$, for any $$phi(x)in(0,1)rightarrowmathbb{R}, phiin C_c(mathbb{R})$$.
The $$frac{1}{epsilon}lim_{epsilonrightarrow0}int_0^x y_epsilon dx=1$$, after some calculations. Therefore I can take $$frac{1}{epsilon}lim_{epsilonrightarrow0}int_0^x y_epsilon(phi(x)-phi(0))dx$$.
But can I use continuity? How can I know that $$lvert x-0rvert?

Posted on

## calculus and analysis – Definite integral gets stuck in the calculation

$$f=frac{sinh ^{-1}left(e^{-2 k t} sinh (6 k)right)}{2 k}-2$$

for $$k=20$$ I have:

$$frac{df}{dt}=-frac{e^{-20 t} sinh (60)}{sqrt{e^{-40 t} sinh ^2(60)+1}}$$

``````f = -2 + ArcSinh(E^(-2 k t) Sinh(6 k))/(2 k)

Integrate(D(f, t), {t, 0, 10})
``````

And he just gets stuck and doesn’t move on. Then I tried to apply parallel computation, but got the error:

``````Needs("Parallel`Developer`")

f = -2 + ArcSinh(E^(-2 k t) Sinh(6 k))/(2 k)

Integrate(D(f, t), {t, 0, 10})

Parallelize::nopar1: !(*SubsuperscriptBox(((Integral)), (0), (10))(*SubscriptBox(((PartialD)), (t))f (DifferentialD)t)) cannot be parallelized; proceeding with sequential evaluation.
``````

I would be grateful for help in finding out the reason.

Processor AMD Ryzen 7 2700 Pro, 16 Gb RAM. When calculating this task, the processor is loaded by 8-10%.

Posted on

## multivariable calculus – Difference between a circle inside another given polar coordinates.

I have to calculate the area between a circle of radius $$r = a$$ inside another one of radius $$r = 2asin{theta}$$

Then, I just have to setup and calculate the integrals:

$$int_0^{2pi}int_0^{r}r,dr,dtheta – int_0^{2pi}int_0^rr,dr,dtheta,$$

The author also gives the following identity:
$$sin^2theta=frac{1}{2}(1-cos2theta)$$

The second integral is easy, since $$r = a$$:
$$int_0^{2pi}int_0^ar,dr,dtheta = pi a^2$$

The first integral I think it’s confusing, but it is done using $$r = 2asin{theta}$$:
$$int_0^{2pi}frac{r^2}{2}dtheta = int_0^{2pi}2a^2sin^2theta ,dtheta$$

Substituting using the identity:

$$int_0^{2pi}a^2(1-cos2theta) ,dtheta = 2pi a^2 + sin(4pi) = 2pi a^2$$

Finally subtracting:

$$2pi a^2 , – pi a^2 = pi a^2$$

I’ve done just a few exercises using polar coordinates so far, there’s something wrong in the process? Since the author didn’t give any specific value for $$a$$, I guess it’s fine to leave the answer as it is, or is there a constant $$a$$ that I’m missing?

Posted on

## multivariable calculus – Question about Divergence formula derivation posted

As a new user, I am not allowed to comment on someone else’s answer to a question. My only choice was to ask a new question about an old answer to a question.

User @Kcronix mentioned, in this question, that the divergence of an arbitrary vector field can be derived as the net flux through the boundary of a region $$R$$:
$$text{net flux}=oint_{partial text{Rect}} vec{V}cdot hat{n};mathbb{d}s=int_{R} vec{V}cdot hat{n};mathbb{d}s+int_{L} vec{V}cdot hat{n};mathbb{d}s+int_{T} vec{V}cdot hat{n};mathbb{d}s+int_{B} vec{V}cdot hat{n};mathbb{d}s$$

I have tried to define the same situation in a notebook and I figured that going counter-clockwise through the sides of the rectangle would be a good idea. So it would be simply rearranged in my case:
$$text{net flux}=oint_{partial text{Rect}} vec{V}cdot hat{n};mathbb{d}s=int_{B} vec{V}cdot hat{n};mathbb{d}s+int_{R} vec{V}cdot hat{n};mathbb{d}s+int_{T} vec{V}cdot hat{n};mathbb{d}s+int_{L} vec{V}cdot hat{n};mathbb{d}s$$

Thus, I defined my top flux as:
$$int_{T} vec{V}cdot hat{n};mathbb{d}s=int_{x+Delta x}^{x} left(M(X,y+Delta y)hat i+N(X,y+Delta y)hat jright)cdot hat j;mathbb{d}X=int_{x+Delta x}^{x}N(X,y+Delta y);mathbb{d}X$$

Please, notice how the integral goes from $$x+Delta x$$ to $$x$$, instead of going from $$x$$ to $$x+Delta x$$. This also happens to occur with the left side of the rectangle, going from $$y+Delta y$$ to $$y$$, instead of $$y$$ to $$y+Delta y$$.

This is my main question with this derivation. I know I can switch the boundaries of the integral by bringing a negative sign outside of it (from the Fundamental Theorem of Calculus), but still, I would get:
$$text{net flux}=int_{y}^{y+Delta y}M(x+Delta x,Y);mathbb{d}Y+int_{y}^{y+Delta y}M(x,Y);mathbb{d}Y\-int_{x}^{x+Delta x}N(X,y+Delta y);mathbb{d}X-int_{x}^{x+Delta x}N(X,y);mathbb{d}Xtag{1}$$
and I do not think I can reduce to the partial derivative definition with the above.

Is there anything I am missing?

Posted on

## calculus and analysis – Product and limit of many (>2) matrices

So I have a matrix like this

$$begin{bmatrix} 1+iA & iAe^{-2ian/N} \ -iAe^{2ian/N} & 1-iA end{bmatrix}$$

in mathematica form is like this

``````{{1 + I*A, (I*A)/E^(2*I*(na/N))}, {-(I*(A*E^(2*I*(na/N)))), 1 - I*A}}
``````

I want to multiply these matrices from n=1 to n=N, for example if I choose N=2 I would have

``````{{1 + I*A, (I*A)/E^(2*I*(a/2))}, {-(I*(A*E^(2*I*(a/2)))), 1 - I*A}} . {{1 + I*A, (I*A)/E^(2*I*(2a/2))}, {-(I*(A*E^(2*I*(2a/2)))), 1 - I*A}}
``````

Notice I put “.” not “* * ” because I want matrix product not some other thing.

After this accomplished I also would like to take the limit of this as well taking N to $$infty$$ for example. So I would have something like this, call above matrix $$A_{n,N}$$
$$prod_{n=1}^{N}A_{n,N}$$
where product is of course matrix product.
Thanks in advance for your answers.

Posted on

## multivariable calculus – Building an integral when there’s a trigonometric identity

Considering the following trigonometric equality

$$sin^2{theta} = frac{1}{2}(1-cos{2theta})$$

calculate the area of the R region that is outside the circle $$r = a$$ and inside the circle $$r = 2asin{theta}$$

How should I build this integral? Since I have a $$theta$$ angle and the radius is in function of $$theta$$. Should it be a double integral of $$theta$$ going from $$0$$ to $$2pi$$ and the second integral for $$r$$? What is the role of the trigonometric equality above?

Posted on

## calculus – Confusion in using double integrals & projection to calculate hemispherical surface area

As a preamble, if we have a surface given by $$z=z(x,y)$$ in $$mathbb{R}^3$$, i.e. $$z =z(x,y) = sqrt{a^2-x^2-y^2}$$ then $$z- z(x,y)=0$$ gives us a constant surface of a scalar field $$f$$, the $$grad$$ of which will tell us the normal vector to our surface: $$nabla f = mathbf{k} -frac{∂z}{∂x}mathbf{i}-frac{∂z}{∂y}mathbf{j}$$, the norm of which is $$sqrt{1+left(frac{partial z}{partial x}right)^2+left(frac{partial z}{partial y}right)^2}$$

A hemisphere of radius $$a$$ (with the $$z$$ axis as the axis of symmetry) is described by the equation $$x^2+y^2+z^2 (=r^2) = a^2$$ and has a surface area given by

$$iint dS = iint sqrt{1+left(frac{partial z}{partial x}right)^2+left(frac{partial z}{partial y}right)^2}dA$$

If you project onto the $$xy$$ plane.

This formula is obtained by considering an element of surface area $$mathbf{dS}$$ and its projected area $$mathbf{dA}$$, and relating the two by $$dA = cos(alpha) dS = mathbf{hat{n}}cdotmathbf{k}, dS rightarrow dS = dA/mathbf{hat{n}}cdotmathbf{k} = sqrt{1+left(frac{partial z}{partial x}right)^2+left(frac{partial z}{partial y}right)^2}dA$$ using $$mathbf{n}$$, the normal to the surface, from the preamble and $$mathbf{k}$$ the unit vector in the $$z$$ direction. For the above example we can use $$z =sqrt{a^2-x^2-y^2}$$ to find our partial derivatives, and then evaluate the integral using plane polar co-ordinates ($$0 leq theta leq 2pi, 0 leq r leq a$$) to reach an answer of $$2pi a^2$$ as expected.

However, if we use a normal vector to this surface expressed in 3D polar co-ordinates:
$$mathbf{hat{n}} = mathbf{r}/|mathbf{r}| = (xmathbf{i} + ymathbf{j} + zmathbf{r} )/ |mathbf{r}| = sinphi costheta mathbf{i}+sinphi sintheta mathbf{j}+cosphi mathbf{k}$$

our logic above the included diagram should still hold: $$dS = dA/mathbf{hat{n}}cdotmathbf{k}$$ (here projecting onto the $$xy$$ i.e. $$r,theta$$ plane).
This yields $$dS = dA/cosphi$$ for our 3d polar case, which has got me stuck on the next step of evaluating

$$iint dS = iint secphi dA = iint sec(phi) rdrdtheta$$

It is intuitively true that as the angle $$phi$$ increases, our surface area element on the hemisphere becomes more and more vertical w.r.t. the $$xy$$ plane and so this $$1/cosphi$$ makes our corresponding area element larger, but how should I actually evaluate the integral?

$$iint secphi dA = iint sec(phi) rdrdtheta$$ is certainly ringing alarm bells, which makes me think that projecting on the $$r,theta$$ plane in the first place was a mistake, which is a problem not encountered when working in 3D cartesians.

Is there such a thing as projecting onto the “$$theta,phi$$ plane” and then integrating over those two variables? – if this were to be a valid method I believe it could solve my problem here. Please do ask for further clarification if it is needed.

Posted on

## calculus and analysis – Extract real part of a complex function

I have a problem for extracting the real part of a complex number. The problem is the following: Suppose $$f(z)=frac{1}{z+frac{Delta_{1}}{z+frac{Delta_{2}}{z+frac{Delta_{3}}{z+…}}}}$$, in which z is a complex number $$(z= a+bi)$$ and $$Delta_{1}$$, $$Delta_{2}$$ … are number and $$f(z)$$ is a continued fraction.
And I want to compute $$phi(b)$$ since I would like to plot $$phi(b)$$ in the end

$$phi(b)=lim_{arightarrow0}Ref(a+bi)$$

But Mathematica doesn’t allow me to do that. I use “Limit” in Mathematica and set all $$Delta_{n} = 0 , n > 1$$ and $$Delta_{1}=1$$ , I get 0.
Even I let some $$Delta_{n}$$ to be nonzero, I still end up with 0.

Here is my code. I set $$Delta_{1}=1$$ and $$Delta_{2}=0.5$$, rest are zero.

``````Limit(ComplexExpand(Re(1/(a + b*I + (1/(a + b*I + (0.5/(a + b*I))))))), a -> 0)
``````

Can someone help me with this ? I am guessing the “Limit” function in Mathematica can’t not handle this problem.

Thank you very much !!!

Posted on