Anonymous

'A Lazy Cloud'.

Anonymous

Clouds at ground level.

Anonymous

Taking this back to in-topic, fog is a Mathematical representation of a function of a function. Usually written with a circle (not an o), so it looks like this: (f ○ g)(x) It is the same as: f(g(x)) So if you have a value for "x", put it into the g(x) function to get an output value, then take that value and put it into the f(x) function as an input to get the final result. An example of this would be: f(x) = x² and g(x) = x + 2 Find (f ○ g)(3) We solve for g(3) first: g(3) = 3 + 2 g(3) = 5 Then use that as an input into the other function, so solve for: f(5) = 5² f(5) = 25 So your answer is: (f ○ g)(3) = 25

Anonymous

In mathematics, it is a composition of functions (a new function that is composed by causing one function to act with the other). "o" is not the letter "o". It is an "operator", a symbol (just like x, +, ÷) that tells you to perform an operation on two "things". A function is a set of instructions that tell you what to do with a variable (or many variables) given as input, in order to find the image (the output) of the instructions acting on the variable(s). Sounds complicated, but it is not. However, functions do sometimes act strangely. f(x) = 2x^2 - 4x + 6 is a simple example of a function. Think of it as a recipe. "f" is the name of the recipe (it can be any letter, it can be names or even symbols) x is the "secret ingredient" (the variable) . What follows the equal sign are the instructions as to what you must do with the secret ingredient once you are told what it is. f(3) would mean: Use function "f", with value 3 for x. So you pick out function f: f(x) = 2x^2 - 4x + 6 and you use "3" wherever the recipe shows an "x": f(3) = 2(3^2) - 4(3) + 6 = 18 - 12 + 6 = 12 f(3) = 12 You can have anything (almost) as the "secret ingredient" including another function. Just like the recipe for Eggs Benedict will tell you to prepare a "Hollandaise Sauce" (which is another recipe), forcing you to look it up somewhere else in the recipe book. But the rule is simple: whatever you are given as the "secret ingredient" (the value of x), is what you blindly put in the function. f(0) = 2(0^2) - 4(0) + 6 = 6 f(w-5) = 2(w-5)^2 - 4(w-5) + 6 = 2(w^2 - 10w + 25) - 4w + 20 + 6 = ... f(cat) = 2(cat)^2 - 4(cat) + 6 OK; that last one makes no sense. But it does show how you apply a function. When a function acts on another function ( let's say g(x) = 8x - 3 ) then you work it the same way f(g(x)) = 2[ g(x) ]^2 - 4[ g(x) ] + 6 f(g(x)) = 2(8x - 3)^2 - 4(8x - 3) + 6 f(g(x)) = 2(64x^2 - 48x + 9) - 32x + 12 + 6 f(g(x)) = 128x^2 - 96x + 18 - 32x + 18 f(g(x)) = 128x^2 - 128x + 36 If you are only doing this once or twice, you stop there and you are happy. However, if you are going to use this more often, you create a new function (a composition) that combines both recipes into a new, single recipe. THAT is the "fog" (however, when we can, we use a smaller "∘" that is raised a bit from the line, just to show clearly that it is not the letter "o"). (f∘g)(x) = 128x^2 - 128x + 36 (but many people still pronounce it "fog") This is a new recipe (its name is f∘g) and it is the composition of f(x) = 2x^2 - 4x + 6 g(x) = 8x - 3 In most cases, there is no difference between (f∘g)(x) and f(g(x)) For a given value of x, they will spit out the same output; but there are cases where there are subtle differences (for example, a different domain or range). ------- f(x) = 1/x g(x) = 1/(x+1) f is not defined when x=0 (division by 0) g is not defined when x = -1 (division by zero). f(g(x)) = 1/[1/(x+1)] is not defined when x = -1, because the "secret ingredient" g(x) does not exist at x=-1 However, the composition (f∘g)(x) = x+1 is defined, even when x = 0 and when x = -1 (f∘g)(0) = 0+1 = 1 (f∘g)(-1) = -1 + 1 = 0 That's because (f∘g) is a new function, with its own restrictions which may be different than the individual restrictions on f and/or on g.

Anonymous

Example: f(x) = ax + b g(x) = cx + d (fog)(x) = f[g(x)] → you substitute x in f(x) by g(x) f(x) = ax + b ← you substitute x by g(x) (fog)(x) = a[g(x)] + b (fog)(x) = a.[cx + d] + b (fog)(x) = acx + ad + b (gof)(x) = g[f(x)] → you substitute x in g(x) by f(x) g(x) = cx + d ← you substitute x by f(x) (gof)(x) = c[f(x)] + d (gof)(x) = c.[ax + b] + d (gof)(x) = acx + bc + d …and you can observe that: (fog)(x) ≠ (gof)(x)

Anonymous

Fog is a cloud of water droplets that is near ground level and sufficiently dense to reduce visibility to less than 1000 m.

Anonymous

A cloud on the ground

Anonymous

When visibility is low

Anonymous

Fog is cloud formation at the ground level. Normally water vapour condenses around the pollutant, like smoke. This happens in the cold countries are in Delhi in winter.

Anonymous

it's a fog of war

Anonymous

minute water droplets suspended in the air. smog is that mixed with smoke.

Anonymous

Clouds

Anonymous

Fog is a visible aerosol consisting of minute water droplets or ice crystals suspended in the air at or near the Earth's surface.

Anonymous

A fog is a weather phenomenon while a f circle g is the "composition of functions" where you take g and plug it as an argument into f.