Math and Logic Puzzles: Redux

[Ircher]

What exactly is a function in terms of a graph? As in, what inputs is it mapping? I may give this a try, but I would.need to know the previous.

[Ircher]

I would think you would have to give a bit more information to help us out on those puzzles; I don't see any patterns.

Not simple addition, subtraction, multiplication, division (try it and you'll see you get different answers for each)
Not modular arithmetic (i.e.: 18 mod X = 5 does not have the same value X as say if one tried 8 mod X = 8 and 24 mod X = 37)
Not non-decimal base system (Some of the sequences increase, others decrease, and in the first one, there is no change and this stuff implies the middle number would be different for each one)
Nothing to do with prime factorization
Doesn't seem to have anything to do with the digit sequence

Also, how can the same input give two different outputs??? (See puzzle B : Apparently, 2 gives both 17 and 31 as possible outputs after going through the middle)

[Ircher]

Prove that the derivative of an odd function is an even function and that the derivative of an even function is an odd function.

[Ircher]

Spoiler: Solution to vizIIsto problem

We know that xyz = 192 and z = 4, so xy = 48. In addition, p = (x + y) / 2. Since 48 is a small number, and we're dealing with integer factors, the best approach is to enumerate all the possible values of x and y snd calculate p for each case.

48 = 1 * 48 --> p = 24.5
48 = 2 * 24 --> p = 13
48 = 3 * 16 --> p = 9.5
48 = 4 * 12 --> p = 8
48 = 6 * 8 --> p = 7

Due to the commutative property of multiplication, we can ignore the remaining cases. Therefore, the answer must be b) 7. I haven't looked at the hint.

pedit: ninja'd

[Ircher]

1. Prove that % is distributed over addition, that is:
(x+y)%z=(x%z+y%z)%z

Spoiler: My Solution

1. Let x, y, z be arbitrary integers. Let a1 = x % z and a2 = y % z. Then by the definition of module, there exist integers q1 and q2 such that x = q1*z + a1 and y = q2*z + a2. Then, x + y = (q1 * z + a1) + (q2 * z + a2) = (q1 + q2)z + (a1 + a2) = (q1 + q2)z + (x % z + y % z). If 0 <= (x % z + y % z) < z then (x % z + y % z) is clearly the result of (x + y) % z. (Applying % z again to this quantity won't change its value.) If (x % z + y % z) >= z, then we can express (x + y) as (q1 + q2 + 1)z + (x % z + y % z - z). Note that in this case, 0 <= (x % z + y % z - z) < z. (Note that (x % z + y % z < 2z.) Otherwise, x % z + y % z < 0, so (x + y) = (q1 + q2 - 1)z + (x % z + y % z + z), and once more, 0 < (x % z + y % z + z) < z. Thus, (x % z + y % z) % z must be the result of (x + y) % z. Q.E.D.

I will note that #2 is much easier to do by directly applying the definitions of even and odd versus using the module operation.

[Ircher]

Let z₁ and z₂ be two complex numbers whose magnitudes are r₁ and r₂ respectively and whose phases are φ₁ and φ₂ respectively. Prove algebraically that z₁z₂ = r₁r₂ cis(φ₁ + φ₂), where cis(φ) = cos(φ) + i sin(φ).

Spoiler: My Attempt

Let z1 and z2 be complex numbers with magnitudes r1 and r2 and phases phi1 and phi2 respectively. Then (in polar form), z1 = r1(cos(phi1) + i * sin(phi1)) and z2 = r2(cos(phi2) + i * sin(phi2)). Then:

z1z2 = (r1*cos(phi1) + i * r1 * sin(phi1)) * (r2 * cos(phi2) + i * r2 * sin(phi2))
= r1 * r2 * cos(phi1) * cos(phi2) + i * r1 * r2 * cos(phi1) * sin(phi2) + i * r1 * r2 * sin(phi1) * cos(phi1) + i * i * r1 * r2 * sin(phi1) * sin(phi2)
= r1 * r2 * (cos(phi1) * cos(phi2) + i * cos(phi1) * sin(phi2) + i * cos(phi2) * sin(phi1) - sin(phi1) * sin(phi2)
= r1 * r2 * (cos(phi1 + phi2) + i * sin(phi1 + phi2))
= r1 * r2 * cis(phi1 + phi2).\box

Not too bad of a problem.

[Ircher]

Spoiler: Response to Ircher's attempt

= r1 * r2 * (cos(phi1) * cos(phi2) + i * cos(phi1) * sin(phi2) + i * cos(phi2) * sin(phi1) - sin(phi1) * sin(phi2)

You have an unmatched parenthesis on this line. Where should the other closing parenthesis go?

Spoiler:

Ah, it should go at the end of the line, so revised proof:

Let z1 and z2 be complex numbers with magnitudes r1 and r2 and phases phi1 and phi2 respectively. Then (in polar form), z1 = r1(cos(phi1) + i * sin(phi1)) and z2 = r2(cos(phi2) + i * sin(phi2)). Then:

z1z2 = (r1*cos(phi1) + i * r1 * sin(phi1)) * (r2 * cos(phi2) + i * r2 * sin(phi2))
= r1 * r2 * cos(phi1) * cos(phi2) + i * r1 * r2 * cos(phi1) * sin(phi2) + i * r1 * r2 * sin(phi1) * cos(phi1) + i * i * r1 * r2 * sin(phi1) * sin(phi2)
= r1 * r2 * (cos(phi1) * cos(phi2) + i * cos(phi1) * sin(phi2) + i * cos(phi2) * sin(phi1) - sin(phi1) * sin(phi2))
= r1 * r2 * (cos(phi1 + phi2) + i * sin(phi1 + phi2))
= r1 * r2 * cis(phi1 + phi2).\box

[Ircher]

Here is a fun integral to do from my probability and statistics course last semester:

The probability density function for the standard normal distribution is given by $f_Y(y)=ke^{-y^2/2}$ where $k=1 / \sqrt{2\pi}$. Show that the variance of the standard normal distribution is equal to 1 by computing the following integral:

$$\int_{-\infty}^{\infty}{y^2f_Y(y)dy}-\left(\int_{-\infty}^{\infty}{yf_Y(y)dy}\right)^2.$$

(You will probably want to copy the equations inside dollar signs into Desmos or something since the forums don't support rendering math really well at all.) You can show your work by just posting the latex code involved.

[Ircher]

You should prove the standard normal integral is 1 as well if you are going to use that argument. (Yes, it is a probability density function, so it equals 1, but the solution should be understandable by someone who has only taken math through multivariable calculus. This makes the problem more interesting.)

[Ircher]

6/9

[Ircher]

1/2
1/3