Why this equation cannot be solved in Mathematica?

I do not understand why the following equation cannot be solved in Mathematica.

``````Solve(Log((-Exp(x) + Sqrt(2 - 3*Exp(2*x)))/2) == c*x, x)
``````

The left hand side is nicely plotted in the third quadrant, and the right hand side is just a line crossing the origin. Therefore, there should be a solution for sure. However, Mathematica returns the following message.

“This system cannot be solved with the methods available to Solve”

contest math – Olympiad Inequality which can be solved using Cauchy–Schwarz inequality

$$x,y,z$$ are positive real numbers, such that $$x+y+z=3$$ . prove that :

$$sum_{mathrm{cyc}}frac{x}{x^3+y^2+z} leq 1$$

I tried many things , but I don’t think any of those are worth of mentioning.
However, I know problem can be solves using Cauchy–Schwarz inequality.

GSA SER Stuck ? (SOLVED) — GSA SEO Forum

Hello everyone, I have about 80 or so projects running on GSA right now creating tier 2 links for my web2.0 created with RankerX, I also have Xevil and CB running for captchas with proper setup, dedicated proxies are setup as well, everything was running great for 10 or 12 days, now looks like everything is “stuck” all I see on the console is:

name of project: loaded 2/16 URLs form site list

name of project: loaded 28/48 URLs form site list

name of project: loaded 0/98 URLs form site list

etc….

I use catchalls email as well, everything is green no error on the screen.

Any ideas ? Thank you.

(SOLVED) Never mind guys, the problem was with the CPU usage, for some stupid reason it was set as 20% instead of 60%.

np complete – Does padding with dummy bits allow an NP-problem to be solved in fast exponential time?

Here is a more extreme example:
$$mathrm{SAT_{PAD}} = {1^{2^n} 0 phi : text{phi is a satisfiable CNF on n variables}}.$$
This language is decidable in polynomial time.

What padding accomplishes is it constructs a language in a lower complexity class; it doesn’t actually help you solve the original problem. All it does is artificially modify the parameter $$n$$.

Running an algorithm for $$L$$ on the language $$L’$$ results in a running time which is $$O(1.01^n)$$, where $$n$$ is the size of the padded instance; the running time is still $$O(2^n)$$ in terms of the size of the original instance.

I don’t see any relevance to universal search. The main application of padding is to relate results about different complexity classes. For example, if P=NP then EXP=NEXP, since given any language in NEXP, you can pad it so that it is in NP, use the assumption to deduce that the padded language is in P, and then interpret the corresponding algorithm as proving that the original language is in EXP.

Is there a class of recurrence relations that can’t be solved using the substitution method?

Is there a class of recurrence relations that can’t be solved using the substitution method? Let me explain the motivation behind this question by an example.

Consider the recurrence relation $$T(n) = 2T(n/2) + n$$. It’s obvious that $$T$$ is in $$Theta(n log n)$$.

To verify that $$T(n)$$ is in $$O(nlog n)$$, one could use the substitution method and the inductive hypotheses $$T(n) ≤ cnlog n + d$$, where $$d$$ is to be replaced by term of lower order.

Inductive step:

$$T(n) = 2T(n/2) + n ≤ 2(cnlog n + d) = 2cnlog(n/2) + 2d + n = cnlog n + ell d + n,$$ where $$ell = 2$$.

As $$ell > 1$$ holds, one can choose $$(ell-1)d := -n$$. As a result, the last term $$n$$ vanishes in the next step:

$$cnlog n + ell d + n = cnlog n + d$$

Choosing $$c$$ properly for the the given $$d$$ should complete this proof. However, this approach sometimes fails if $$ell$$ equals 1.

This is illustrated by the following example, where the goal is to prove a lower bound:

$$T(n) = T(n – 2) + n^2$$

Hypothesis: $$T(n – 2) ≥ cn^3 + d$$ for all $$n’ ≤ n$$:

$$T(n + 1) = T(n – 1) + (n-1)^3 ≥ c(n-1)^3 + d + n^2$$

I would be very thankful for an explanation.

typescript – Having trouble with this Koan – solved it, but I don’t think what I wrote is correct in regards to Typing

I’m working through these Koans: https://github.com/paytonrules/typescript.koans#readme

I’m on file 3, collections, the very first function `foreach`:

``````export interface Dictionary<T> {
(index: string): T;
}

interface ArrayForEachIteratee<T> {
(value?: T, index?: number, collection?: Array<T>): any;
}

interface DictionaryForEachIteratee<T> {
(value?: T, key?: string, collection?: Dictionary<T>): any;
}

/**
* ### forEach
* should iterate over all items of array or all properties of an object
*
* ## Examples
*
*  let collection = ("first", "second", "third");
*  let result = ();
*  let iteratee = (value, index, collection) => result(index) = (index, value);
*
*  _.forEach(collection, iteratee); => result === ((0, 'first'), (1, 'second'), (2, 'thidrd'));
*
*  collection = {
*    "0": "first",
*    "1": "second",
*    "2": "third"
*  };
*  result = ();
*  iteratee = (value, index, collection) => result(index) = (index, value);
*
*  _.forEach(collection, iteratee); => result === (('0', 'first'), ('1', 'second'), ('2', 'thidrd'));
*
*/
export function forEach() {
}
``````

When I look at the test code, I can see that the first test sends it an array and a function, and the second test sends it a dictionary and a function:

``````describe("forEach", function () {
context("when collection is an array", function () {
it("should iterate over all items of array", function () {
const collection = ("first", "second", "third");
const iteratee = sinon.spy();
_.forEach(collection, iteratee);
sinon.assert.calledWithExactly(iteratee, "first", 0, collection);
sinon.assert.calledWithExactly(iteratee, "second", 1, collection);
sinon.assert.calledWithExactly(iteratee, "third", 2, collection);
});
});

context("when collection is an object", function () {
it("should iterate over all items of object", function () {
const collection: _.Dictionary<string> = {
"0": "first",
"1": "second",
"2": "third"
};
const iteratee = sinon.spy();
_.forEach(collection, iteratee);
sinon.assert.calledWithExactly(iteratee, "first", "0", collection);
sinon.assert.calledWithExactly(iteratee, "second", "1", collection);
sinon.assert.calledWithExactly(iteratee, "third", "2", collection);
});
});
});
``````

So I set up my code to accept an array or a dictionary, and the corresponding array / dictionary callback for the second argument:

``````export function forEach<T>(collection: Array<T> | Dictionary<T>, iteratee: ArrayForEachIteratee<T> | DictionaryForEachIteratee<T>): void {
if (Array.isArray(collection)) {
collection.forEach(iteratee);
} else {
Object.keys(collection)
.forEach(key => {
const value = collection(key);
iteratee(value, key, collection);
});
}
}
``````

My code passes the tests – I’m passing the right arguments to the right functions in the right ways – but I’m absolutely certain that I’m not doing it in the proper Typescript-y type-safe way. My editor is showing me lots of red.

Can I get some pointers on how an expert typescripter would gracefully handle this situation?

Google maps unavailable issue solved by playing Spotify or Amazon Music first

I am frequently facing this issue.

1. I would connect my Android 10 phone to our 2021 RAV4 Android Auto USB.

2. Then Google Maps would close up on my phone.

3. While that happens, the RAV4 console would notify blue tooth disconnected and Android Auto connected. Wait till these messages are completed.

4. But when I click on the RAV4 console Map button, the console would notify that no Map application was available or that Android Auto disconnected.

But fear not, I have a way to solve the Maps availability issue by taking these steps.

1. Connect my phone to the RAV4 Android Auto USB.
2. Ensure I do not press the Map button on the RAV4 console yet.
3. Play Spotify or Amazon Music on my phone first.
4. Then press the Map button on the RAV4 console.

Questions:

1. What is Google doing wrong that Spotify or Amazon Music is doing right?
2. What are the protocol disconnects between my phone and the RAV4 Android Auto that is causing these frequent disconnects. Feel free to refer me to other StackOverflow coding sites.

equation solving – Can’t this inequality really not be solved with Reduce?

Taking $$alpha$$ as given, I am trying to understand/know under which restrictions on $$n$$, the following holds:
$$0 leq n+alpha Wleft(-e^{frac{-n}{alpha}} right) leq 1.$$
i.e I want to solve for $$n$$ the following two inequalities: a) $$0 leq n+alpha Wleft(-e^{frac{-n}{alpha}} right)$$, and b) $$n+alpha Wleft(-e^{frac{-n}{alpha}} right) leq 1$$.
In addition, I know $$alpha >0$$, that both $$n$$ and $$alpha$$ are reals.

• I manage to get that a) is satisfied for $$ngeq alpha$$,
using: `Reduce(n+a*ProductLog(-E^(-n/a))>=0, n, Reals)` which returns `(a<0&&n==a)||(a>0&&n≥a)`.
• However I struggle to find any solution for the inequality b). I have tried all of the following:
``````(* for convenience I first define the function *)
f(n_, a_) =n+a*ProductLog(-E^(-n/a))

Reduce(f(n, a)<=1, n, Reals)
Reduce(f(n, a)<=1 && a>0, n, Reals)
Solve(f(n, a)<=1, n, Reals)
Solve(f(n, a)<=1 && a>0, n, Reals)
``````

Which all return:

“This system cannot be solved with the methods available to Solve/Reduce.”

Even though I specified the domain as suggested here.

Is the inequality in b) really not solvable for $$n$$? Could Anyone think of an alternative command or commands that would do the trick?

Thank you,

EDIT:
In my context, `f(n_, a_)`being strictly increasing in `n`, even the solution for the equality to 1 would suffice me.

``````Manipulate(Plot(f(n, a), {n, -1, 10}), {a, 0, 5, 0.05, Appearance -> "Labeled"})
``````

np complete – Can we solved Knapsack problem in polynomial time using algorithm A?

Suppose we have an algorithm called A.

A input:

1. $$list = {l_1,l_2, dots,l_n}$$ such that for every $$l_iin list$$ there is $$value_{l_i},weight_{l_i} in N$$
2. $$k in N$$
3. $$min_value in N$$
4. $$max_weight in N$$

A output:

Yes $$iff$$ there exists a subset $$S = {s_1,s_2, dots,s_t}$$ such that:

1. $$S subseteq list$$
2. $$|S|leq k$$
3. $$value_{s_1} + value_{s_2} + … + value_{s_t} geq min_value$$
4. $$weight_{s_1} + weight_{s_2} + … + weight_{s_t} leq max_weight$$

Algorithm A running time complexity is $$O(f(k)*n^{O(1)})$$.

Can we solved Knapsack problem in polynomial time using algorithm A?