## Solve For the expression by substitution

If $$x=a(b-c),y=b(c-a),z=c(a-b)$$Find $$(frac{x}{a})^3+(frac{y}{b})^3+(frac{z}{c})$$

Options:
$$text{1: }frac{xyz}{abc}$$
$$text{2: }frac{xyz}{3abc}$$
$$text{3: }frac{abc}{xyz}$$
$$text{4: }frac{3abc}{xyz}$$

What I tried:

Try 1.
$$(frac{a(b-c)}{a})^3+(frac{b(c-a)}{b})^3+(frac{c(a-b)}{c}$$
But that leaves us with
$$(b-c)^3+(c-a)^3+(a-b)^3$$
That doesn’t match with any options given above.

Try 2

Since
$$frac{x}{(b-c)}=a,frac{y}{(c-a)})=b,frac{z}{(a-b)}=c$$
$$(frac{a(b-c)(b-c)}{x})^3+(frac{b(c-a)(c-a)}{y})^3+(frac{c(a-b)(a-b)}{z}$$

But that leaves us with:
$$(frac{acdot (b^2-2bc+c^2)}{x})^3+(frac{bcdot (c^2-2ac+a^2)}{y})^3+(frac{ccdot (a^2-2ab+b^2)}{z}$$

So is there any way where I can get the correct answer?

If I have missed something obvious, please be gentle.

## complex analysis – Computing a substitution with matrices

Consider the system
$$begin{pmatrix}frac{d}{dx} & -q\ -q & -frac{d}{dx}end{pmatrix}Psi=-ixiPsi.$$

Now, it is said that substituting $$Psi=e^{ixi x}Phi$$ with $$Phi=begin{pmatrix}Phi_1\Phi_2end{pmatrix}$$ results in
$$begin{pmatrix}frac{partial}{partial x} & -q\ -q & -frac{partial}{partial x}end{pmatrix}Phi=-2ixibegin{pmatrix}Phi_1\0end{pmatrix}.$$

I do not see it.

If I substitute, I get, on the left hand side,
$$begin{pmatrix}frac{d}{dx} & -q\ -q & -frac{d}{dx}end{pmatrix}begin{pmatrix}e^{ixi x}Phi_1\e^{ixi x}Phi_2end{pmatrix}=begin{pmatrix}frac{d}{dx}e^{ixi x}Phi_1-qe^{ixi x}Phi_2\ -qe^{ixi x}Phi_1-frac{d}{dx}e^{ixi x}Phi_2end{pmatrix}$$

## equation solving – Write expressions in simpler form with substitution for part of the epxression

I have faced this situation many times where I get values like this as solutions:

``````x = (a1 + a2 + a3 - a4 - a5*a6 + a7) * (a1 - a2 - a3 - a4 - a5*a6 + a7) * m
y = (a1 + a2 + a3 - a4 - a5*a6 + a7) * (a1 - a2 - a3 - a4 - a5*a6 - a7) * n
``````

Now, the value for `x` and `y` has `(a1 + a1 + a3 - a4 - a5*a6 + a7)` and `a1 - a2 - a3 - a4 - a5*a6` as common parts. Is there any built-in way to detect these common parts in Mathematica and write solutions like:

``````x = p * (q + a7) * m
y = p * (q - a7) * n

where p = (a1 + a2 + a3 - a4 - a5*a6 + a7) and q = a1 - a2 - a3 - a4 - a5*a6
``````

If there is no automatic way of doing this, can I tell Mathematica to substitute these values manually?

## induction – Substitution Method to Solve Recurrences

One approach to solve recurrences is the so called substitution method.

While practicing I encountered some recurrences, where non integer arguments can occur, e.g. T(n) = 2*T(n/2) + n, if n is not a power of 2.

My understanding of the substitution method, is that it only works, if all arguments are integers. Is this correct? Is is it reasonable to apply the substitution method to lower and upper bounds instead, e.g. T(n) ≤ T1(n), where T1(n) = 2*T1(ceil(n/2)) + n, in order to draw conclusions regarding T?

## undecidability – Is this string substitution problem decidable?

Take as input a finite set of string pairs. Each pair represents a substitution. Replace exactly one instance of the left with the right. A substitution can only be performed on x if the left is a substring of x. For example $$01rightarrow 10$$ means replace one 01 with 10, and can only be applied if 01 is in the string. The algorithm should decide if for the given set there exists a string such that applying a non – zero number of the substitutions yields the initial string.

I am wondering if this is a decidable task. It seems like it should be possible to establish an upper bound on the length of the string and number of substitutions, but I haven’t been able to. And from the other end I tried to build a test based on invariants, since invariants can be used to show a lot of sets can’t loop. For example

$${011rightarrow 101, 101rightarrow 110}$$

Can never produce a loop. We can show this since each rule decreases the average position of 1s in the string when used. Thus it can never produce a string with the same average position as the start thus it can never produce the start.

But there are some cases that I can’t think of a clever invariant. For example

$${1001rightarrow 0110}$$

Which clearly can’t form a loop, but I can’t think of a property which it always increases.

Is this problem decidable?

## Closure of context-sensitive languages under inverse language substitution

Fix a universal Turing machine $$T$$.

Let $$L_1$$ be the language of all sequences of configurations $$c_1 # c_2 # cdots # c_ell$$ which describe a valid computation of $$T$$, starting with an arbitrary initial configuration $$c_1 neq epsilon$$, and ending with a halting state; we do not allow any further configurations to appear. This language is clearly context-sensitive (it can be accepted in linear space even deterministically).

Let $$L_2$$ be the language of all sequences of configurations $$c_1 # c_1 # c_2 # cdots # c_ell$$ such that $$c_1 # c_2 # cdots # c_ell in L_1$$ (notice the initial configuration is repeated twice). This language is also context-sensitive.

Define a function $$f$$ from $$Sigma$$ to CSLs by $$f(sigma) = { sigma }$$ for $$sigma neq #$$ and $$f(#) = { #w : w in L_1 }$$.

Notice that $$f^{-1}(L_2)$$ consists of strings $$c_1#$$, where $$c_1$$ is an initial configuration of $$T$$ on which $$T$$ eventually halts. Thus $$f^{-1}(L_2)$$ is not computable.

## Context-Sensitive Grammars are closed under language substitution, and string homomorphism, but are they closed under inverse language substitution?

Wikipedia note on closure properties.

We define language substitution for a Context-Sensitive Language (CSL) $$S$$ over an alphabet $$Sigma$$ is a map from $$Sigma$$ into CSL’s, for example: $$f(abc) = L_1(a) L_2(b) L_3(c)$$ such that (I guess) the union of all $$L(s)$$ for $$s in f(S)$$ is defined to be $$L(f(S))$$ and $$L(f(S))$$ is known to be a CSL itself.

That is my interpretation of language substitution for CSL’s. Well languages are also closed under inverse of string homomorphisms, homomorphisms $$f$$ being a pecial case of language subtitution in which each $$a in Sigma$$ gets mapped to a singleton language $$L_1(a) = {f(a)}$$.

So my question is simple, yet probably hard or interesting to prove. That is, are CSL’s closed under inverse of language substitution?

Let $$f$$ be a language substitution taking $$S$$ to $$f(S)$$. Then $$f^{-1}(S) := bigcup f^{-1}(s)$$ I’m assuming. Is that a CSL?

## How to integrate without using trigonometric substitution: \$int{dfrac{x}{sqrt{1-x^4}}}dx\$?

How can I integrate the following without using Trigonometric substitution?
$$int{dfrac{x}{sqrt{1-x^4}}}dx$$
I tried substituting, $$t = 1 – x^4$$ but that didn’t work. The solution according to my book is
$$dfrac{1}{2}arcsin{left(x^2right)}+C$$

## natural language processing – Compute the edit distance between two words in which substitution is not allowed

This is a straighforward modification of the classical dynamic programming algorithm.

Let the first string be $$s = s_1 s_2 dots s_n$$ and the second string be $$t = t_1 t_2 dots t_m$$.
In the dynamic programming algorithm you define $$D(i)(j)$$ as the edit distance between $$s_1 s_2 dots s_i$$ and $$t_1 t_2 dots t_j$$.

The bases cases are $$D(0)(j)=j$$ and $$D(i)(0)=i$$ (for any $$i = 0, dots, n$$ and $$j=0, dots, m$$), and the recursive formula (for $$i>0$$ and $$j>0$$) is:

$$D(i)(j) = min begin{cases} 1 + D(i-1)(j) & mbox{ deletion of s_i};\ 1 + D(i)(j-1) & mbox{ insertion of t_j};\ 1 + D(i-1)(j-1) & mbox{ substitution of s_i with t_j}. end{cases}$$

You can simply modify the above formula by not contemplating the last case, i.e.:
$$D(i)(j) = min begin{cases} 1 + D(i-1)(j) & mbox{ deletion of s_i};\ 1 + D(i)(j-1) & mbox{ insertion of t_j}. end{cases}$$

By definition of $$D(cdot)(cdot)$$, the edit distance between $$s$$ and $$t$$ is $$D(n)(m)$$.

## calculus and analysis – Algorithmically imposing a substitution in a difficult integral

Consider the following integral:

$$I = frac{1}{pi c^2} intlimits_{r=0}^c 2 pi r e^{-frac{ left( sqrt{a^2 – r^2} -sqrt{b^2 – r^2} right)}{lambda}} dr$$

under the conditions $$a>b>c>0$$ and $$lambda > 0$$ are all in $$mathbb{R}$$.

This problem is too difficult for Mathematica (v. 11.3) to solve directly:

``````Assuming(a > b > c > 0 && (Lambda) > 0,
1/((Pi) c^2) Integrate(
2 (Pi) r Exp(- (Sqrt(a^2 - r^2) - Sqrt(b^2 - r^2))/(Lambda)),
{r, 0, c})
)
``````

However, if one makes the substitution $$k = sqrt{a^2 – r^2} – sqrt{b^2 – r^2}$$, then one gets the following integral:

$$frac{2}{c^2} intlimits_{k = a – b}^{sqrt{a^2 – c^2} – sqrt{b^2 – c^2}} left( frac{(a^2 – b^2)^2}{k^3} – k right) e^{-k/lambda} dk$$

This integral can be broken up and solved analytically, where Mathematica employs the

$$E_q (x) – intlimits_1^infty frac{e^{-x t}}{t^q} dt$$

which Mathematica implements as `ExpIntegralE(q,x)`.

I accept that finding this $$k$$ substitution requires “intelligence” that Mathematica does not yet have. But assume the user has this insight and wants to give it as a hint or condition to Mathematica. Hence the core of my question:

Question

In the integral for $$I$$, defined above, how would the user impose the $$k$$ substitution as a “hint” and have Mathematica perform all the substitutions (including differentials and limits) and produce an analytic solution for $$I$$?