## automatons – find derivative tree for CFG

I need to draw the derived tree for 1-2-3 (3-4) * 5 * 6 of the grammar
How many possibilities are there trees?

Vn = {expr, term, factor, number}

Vt = {(,), -,, 0 … 9}
P = {
expr-> expr-expr | term
term-> term
factor | factor
factor-> number | (expr)
number-> 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
}
S = expr

The possibilities that I can find are: (1-2) – (((3-4) * 5) * 6), 1- (2- (((3-4) * 5) * 6))

Is there another possibility?

## Calculation and analysis. The integral of the derivative does not work.

In `11.3.0 for Microsoft Windows (64 bits) (March 7, 2018)` writing:

``````To integrate[D[ArcTan[Sqrt[1 + x^5]], X], X]
``````

I get:

ArcTan[Sqrt[1 + x^5]]

but if I write:

``````To integrate[D[ArcTan[x Sqrt[1 + x^5]], X], X]
``````

Mathematics The integral can no longer be solved. Does something escape me?

## Real analysis – Explicit expression for derivative.

Leave $$f:[-pi,pi]^ 2 rightarrow [0,1]$$ be a smooth function such that $$f (0) = 0$$, $$0$$ it is the only critical point of $$f$$ and the sets of levels $$A_t: = left {z in [pi,pi]^ 2; f (z) = t right }$$ They are connected and homeomorphic to a circle to $$t> 0.$$

An example of such $$f$$ would be $$f (z) = frac { Vert z Vert ^ 2} { Vert ( pi, pi) Vert ^ 2}.$$

I would like to know: Can you get an explicit expression for the derivative of line integrals?

$$F (t): = int_ {A_t} g (z) dz$$

where $$g in C ^ { infty} ([-pi,pi]^ 2).$$

It seems that there is no standard expression for the derivative that immediately gives the derivative of this function $$F$$.

## Multivariable calculation: what is the derivative function?

I would like to illustrate my confusion on this issue by constructing the theme from more or less first principles. Leave $$U subseteq mathbb {R} ^ n$$ be an open subset and let $$f: U to mathbb {R} ^ m$$. We say that $$f$$ it is totally differentiable in $$a in U$$ If there is a linear function. $$D_ {a} f: mathbb {R} ^ n a mathbb {R} ^ m$$ such that the following holds.

$$lim_ {x to a} frac {|| f (x) -f (a) -D_ {a} f (x-a) ||} {|| x-a ||} = 0$$

We called $$D_ {a} f$$ the total derivative of $$f$$ to $$a$$. My question concerns how we define the derived function. For the family case in which $$n = m = 1$$, this is simple since the total derivative is reduced to only

$$(D_ {a} f) (b) = big ( frac {df} {dx} big { rvert} _a big) b$$

for all $$b in mathbb {R}$$. Here, $$frac {df} {dx} big { rvert} _a$$ Denotes the usual definition of the derivative of $$f$$ to $$a$$. Then we define the derived function. $$f & # 39 ;: U to mathbb {R}$$ as

$$f & # 39; (y): = frac {df} {dx} big { rvert} _y$$

for all $$y in U$$. Now consider the case when $$m = 1$$. Representing the action of $$D_ {a} f$$ through the Jacobian matrix, we have

$$(D_ {a} f) (b) = begin {bmatrix} frac { partial {f}} { partial {x_1}} big { rvert} _a & cdots & frac { partial {f}} { partial {x_n}} big { rvert} _a end {bmatrix} begin {bmatrix} b_1 \ vdots \ b_n end {bmatrix}$$

for all $$b = (b_1, cdots, b_n) in mathbb {R} ^ n$$. This suggests that we define the derived function. $$Df: U to mathbb {R} ^ n$$ as

$$(Df) (y): = big ( frac { partial {f}} { partial {x_1}} big { rvert} _y, cdots, frac { partial {f}} { partial {x_n}} big { rvert} _y big)$$

for all $$y in U$$. Maybe we could make a similar argument for the case in which $$n = 1$$. How does this extend to cases where the Jacobian matrix is ​​not a simple column or row vector? What definition of the derived function is more natural in those cases? By natural, I mean that the definition should allow important identities such as the product rule and the chain rule to retain their obvious forms.

## tracing: How to manipulate a partial derivative diagram of self-defined bivariate functions?

I have a linear function by parts (spline) that describes a progressive income tax program, with higher marginal rates for higher and higher incomes.

I want to understand the implications of a rule that some tax-exempt income could still increase your average rate in the taxable part. Basically, with `X` being taxable income and `Y` being exempt income, your taxes will be `f (x + y) / (x + y) * x`, where `F` It is the non-linear scheme that would apply to income without exemptions. I can define `F` as a linear function by parts, see above.

To calculate the marginal tax rates on taxable and exempt income, I would simply need the partial derivatives of this expression, with the particular function `F` plugged in, and this I can plot against the two variables. Or trace against one and use the other as a parameter to `Handle`.

The derivative looks good in Mathematica, although it is a bit difficult to verify with so many cases.

However, I have empty plots, which, of course, do not make sense to manipulate. Where is this going wrong?

``````Handle[Plot[D[Tax[x+y]/ (x + y) * x, y], {and, 0,200000}], {x, 0,200000}]
``````

Think about `Tax` in this way:

``````Tax [z_]: = By parts[{
{0,0    <=z< 11265},
{0+ 0.0856(z-11265),11265<=z}
}]
``````

## gm. General Mathematics: first derivative with respect to x and the Pell equation.

When you graph the Pell equation:
x = $$sqrt (Dy ^ 2 + 1)$$and the first derivative of y with respect to x of the Pell equation, d / dx $$sqrt ((x ^ 2 – 1) / D)$$ for D> 1, you will see them intersect at x = 1.618. Is there any evidence that shows that the point of intersection is exactly equal to $$phi$$, where $$phi$$ = ($$sqrt5 + 1) / 2$$? If not, can someone prove it (or refute it)?

## calculation and analysis – Find the derivative of a loss function with respect to the output of a neural network

I have a loss function $$L$$, specifically a `CrossEntropyLayer["Probabilities"]`. I also have a trained neural network $$text {net}$$ that outputs a softmax layer, mapping $$out = text {net} (entry)$$. I'm experimenting with backward propagation – all I'm trying to find is $$frac {dL (out)} {d (out)} | _ {input = bf {x}}$$, and I can not figure out how to do it.

Of course, that derivative is an abbreviated form of a list of partial derivatives for each (4315) of the probabilities returned from the `SoftmaxLayer`.

I have seen `NetPortGradient`, but it only seems to be able to calculate the gradient of the output with respect to something, instead of the gradient of the loss with respect to the output.

## numerics – Error with NSum: returns NSum :: nsnum: Summand (or its derivative) f[n] it is not numeric at the point n = 17

Consider the following example (I had a lot of trouble finding an example of minimal work, I think it's already compact enough).

``````Omega0 = 1.
t = 2
nAvg = 10

Omegan[n_] : = Omega0 * Sqrt[n + 1]

F[n_] : = By parts[{{Cos[Omegan[{{Cos[Omegan[{{Cos[Omegan[{{Cos[Omegan[n]* t / 2]^ 2 *
ABS[Exp[-nAvg/2]* Sqrt[nAvg]^ n / Sqrt[Factorial[Factorial[Factorial[Factorial[n]]],
0 <= n <= 20}}, 0]NSum[F[F[F[f[n], {n, 1, 100}]
``````

If you run this short script, you should return it:

NSum :: nsnum: Summand (or its derivative) […] great message […] It is not
numerical at point n = 16

This problem that I am facing occurs only with some specific function. It happens with this complicated looking function that I gave you, but if you try a simpler one, the script can work correctly.

My questions :

First: I would like to understand why I have this error

Second: How to solve it?

Extra question

Is it really more efficient to use NSum?[] what N[Sum[Sum[Suma[Sum[]]. Because I've read (I do not remember where) that when Mathematica sees N[Sum[Sum[Suma[Sum[]]understands that the sum must be done numerically (instead of trying the symbolic method, THEN numerically approximates).

Extra infos:

I've already seen Is this an NSum error?

With some functions it solves the problem to add NSumTerms-> number, with others not. The thing is that I would like to be able to face this problem "in general", so I need to understand what is happening (I read the documentation and I do not).

In summary: how to make numerical sums in general with mathematics? In my specific case I have functions that can be defined in parts. In general, my function can be a product / sum of functions by parts, so it is not obvious at first glance to know the limit of the sum without looking more closely, which I would like to avoid.

## External derivative

I have a 3-manifold with Riemannian metric $$g = omega_1 ^ 2 + omega_2 ^ 2 + omega_3 ^ 2$$, where $$omega_i$$ They are 1-forms (coframe fields).

I am in this situation:
$$-du wedge omega_2 wedge omega_3 (u ^ {- 1} p + frac {3} {2} u ^ {- 1}) + omega_2 wedge d omega_3 = d omega_2 wedge omega_3 PS$$, which should be possible to write as:
$$-du wedge omega_2 wedge omega_3 (u ^ {- 1} p + frac {3} {2} u ^ {- 1}) = d ( omega_2 wedge omega_3)$$.

How do I get it $$omega_2$$ Y $$omega_3$$?

## indexed variable – Derivative that is not evaluated when the function is defined

I have the following code

``````ExD = Sum[D[y[n, t], t], {n, 0, 1}];
ExS = Sum[y[n, t], {n, 0, 1}];
[Sigma][0]    = 1;
[Sigma]& # 39;[0] = 1;
Y[0, t_] = [Sigma][0]    + t [Sigma]& # 39;[0]
ExD
ExS
``````

Note that when it is executed, ExD is still general for D[y[0,t], t], while ExS prints and[0,t] how it was defined It seems I can not understand why the first derivative with respect to the second variable of y[0,t] does not evaluate even though[0,t] It is clearly defined and printed correctly. I need to use Evaluate[] here?