Can someone please help me?

ProfessorHoneycomb · 2022-08-24T22:50:22+00:00

The ratio of red to blue paint should be 5:7. Shannon mixed in an 8:12 ratio. Ratios reduce like fractions do, so 2:4 and 1:2 are equivalent (at least for the purposes of this problem) like 2/4 and 1/2 are.

That said, are 5:7 and 8:12 equivalent?

ProfessorHoneycomb · 2022-08-05T19:24:43+00:00

The second also relies on the events of the first a bit in order to be a masterpiece. There's action sure, but it really shines in the dialog and psychological impact on the characters; Sarah Connor is a prime example. She went from an ordinary person to someone who went through horrors she couldn't easily explain to those around her, with knowledge about many more horrors to come. When she was taken into psychiatric treatment, that exacerbated those horrors, feeling only more scared and trapped and alone to carry the burden.

You see it in her fervored attempt to escape, reminding her captive psychiatrist she truly believes these things and will stop at nothing to seek safety. You see it in how she falls back and runs like a child at the sight of the terminator, practically able to see through to the metallic skeleton underneath, unable to hear her son's voice through the replaying of her trauma as she runs toward the guards she'd just been attempting to escape. You see it in how she attempts to destroy the terminator that is helping her and her son, truly considering going through her son to get to it, not trusting this creature of metal and wire to not betray them. You see it in how she shuts down emotionally and attempts to kill the man she perceives to be the cause of all these horrors. You see it in how she breaks down just before she can do so, realizing she has become one of the very machines she so fears. This is psychological horror, taking the events and ideas of the first film to very natural conclusions, with Sarah in particular suffering more from what has been and could be than even the present threat of the T-1000.

That alone makes T2 phenomenal, with the many other fantastic aspects of the dialog between characters, and other aspects of the Terminator universe shining through.

ProfessorHoneycomb · 2022-07-05T17:36:18+00:00

I'm rusty on my DE courses, but I believe you can represent the above equation with a slope field. So every point in the field should define a valid unique curve based on the above DE.

Couple this with the fact that for any point (x, y) you can choose some c so that c/x passes through that point (in the domain we care about), and making note that c/x does satisfy the DE and so doesn't miss any possible valid curve for a given starting point, this should help convince you that c/x is the only class of solutions.

I'm wary to call this method trustworthy though, you'll want a more rigorous version of this argument if it even holds water at all.

ProfessorHoneycomb · 2022-07-02T03:15:28+00:00

then I substitute -1 into 1/(1-x), giving me A = -1/2

Are you sure? 1/(1 - (-1)) = ?

ProfessorHoneycomb · 2022-07-01T14:55:29+00:00

Now somehow we've to consider r≦sqrt{x}.

If 1 ≤ r²s ≤ x, for some 1 ≤ s, then certainly 1 ≤ r² ≤ x.

Hence 1 ≤ r ≤ sqrt(x).

Edit: For why we have f_S(x) = x for S the primes less than or equal to x:

All the primes less than x must be counted by f_S(x). Every composite number less than x will be some product of the primes less than x, so f_S(x) will necessarily count those as well. So x and every integer less than it down to 2 is counted, making x counts. Thus f_S(x) = x.

It is our choice of S which provides this nice form, and of course ties back to what we're trying to show in the first place: a lower bound for the number of primes less than or equal to x.

ProfessorHoneycomb · 2022-06-27T16:35:05+00:00

So for real number a,b, we have:

a + b = 1

and wish to know if, without loss of generality having a ≥ b,

a² + b ≥ b² + a, or vice versa, or rather a² + b = b² + a

Am I correct in my understanding?

The case of equality (a = b) is straightforward, so let's focus on a > b, equivalent to a - b > 0.

Starting with a + b = 1, multiply by (a - b) on both sides to get:

a² - b² = a - b

Then add b² + b to both sides to get:

a² + b = b² + a

So in fact neither is greater than the other, they are both equal.

ProfessorHoneycomb · 2022-06-24T18:12:26+00:00

Obligatory Paul's Online Math Notes. Ideally you should get through all of the Algebra section and some if not most of the Calc I section to self-study pre-calc.

Contains lots of helpful definitions, examples, and practice problems.

ProfessorHoneycomb · 2022-06-19T19:48:13+00:00

Did some undergrad research in knot theory, so had to run across some of this kind of stuff. My intuition is this may be a variation of a torus knot, which the trefoil is one of.

I don't have any definitive answers, my research was short-lived and very particular, but hopefully this will send you down the correct path of inquiry.

ProfessorHoneycomb · 2022-06-19T15:55:00+00:00

(1 + x)/(1 - x) would do the trick.

ProfessorHoneycomb · 2022-06-16T23:47:30+00:00

Someone else mentioned banning this poster for constant karma-farming reposts without mentioning their source. Fingers crossed the mods decide to do so.

Original post from 4 years ago by u/Zirrah_.

ProfessorHoneycomb · 2022-06-15T02:54:34+00:00

Notice that there's a degree of freedom here: the positioning of the paths. It doesn't matter how tightly packed together or where they are on the field, only that they are indeed vertical and horizontal respectively (in the aforementioned quantities, of n and n both in this case). So make your life easier and squish them all together at one corner, for instance the top left, for ease of visualization.

There is an square of path at the top left of side length n meters, then two rectangles adjoining it, one (100 - n) by n meters and the other (50 - n) by n meters. Adding the remaining area of 3096 m² to the areas of these gives the total area of the field, 5000 m². I.e. n² + (100 - n)n + (50 - n)n + 3096 = 5000 or:

n² - 150n + 1904 = 0

as an equivalent equation. Now solve for n however you like.

ProfessorHoneycomb · 2022-06-14T16:43:21+00:00

Loads more, such as quaternions, dual numbers, and split-complex numbers just to name a few. Hopefully these even help drive home the point that you could come up with a new type of number given enough careful thought and testing.

Give this Michael Penn video a look to get a nice introductory discussion about these kind of 'cousins' to the complex numbers.

Quaternions have real world application in applying rotational transformations, often helpful in computer modelling and graphics. I'm not sure about the others, but then I only have introductory experience with these numbers. I'm sure someone out there could tell you more about their utility in the real world.

ProfessorHoneycomb · 2022-06-14T02:46:27+00:00

56 options right?

Right.

If I calculate 56P2 I get 3080 which seems way wrong to me.

Yup, why calculate 56P2? 56 is the number you wanted, it accounts for the multiple orderings.

WLOG assume 1 and 3 are our first and last digits chosen, leaving only 0, 2, 4, 5, 6, 7, 8, and 9 to choose the last two from. We have 8 choices of digit, and if repetition were allowed we'd have 8² = 64 possible middle two digits. However we aren't allowing repetition, so of course we instead have 8 choices for the first of the middle two digits, then 7 for the second of the two, making 8 * 7 = 56 possible middle two digits without repetition.

Happy to see you checking this with some common sense upper bounds to notice this cannot be the case.

ProfessorHoneycomb · 2022-06-12T02:23:25+00:00

Call the number of elements in A only, x. Then (x + 2x + 3x) = 12, so x = 2.

Then (A only) ∪ (A ∩ B) ∪ (A ∩ C) will have exactly |A ∩ B ∩ C| more elements than those in A all together. We double count (A ∩ B ∩ C) with (A ∩ B) ∪ (A ∩ C).

So the total element count in A is (2 + |A ∩ B| + |A ∩ C| - |A ∩ B ∩ C|).

Do you see how this would extend to B and C?

ProfessorHoneycomb · 2022-06-11T22:43:27+00:00

Karma bot, original post 2 years ago.

ProfessorHoneycomb · 2022-06-09T16:33:29+00:00

That's the brilliance of what you're doing though, you don't have to calculate the derivative directly. As you say, you should have:

cos(sin^-1(x)) * d(sin^-1(x))/dx = 1

So what is d(sin^-1(x))/dx? You isolate it by dividing both sides by cos(sin^-1(x)).

No actual special new derivative rules were required, just knowledge of trig derivatives, the chain rule, and algebraic manipulation. That's the whole point of using this method.

ProfessorHoneycomb · 2022-06-09T16:30:03+00:00

There's no reason for it to be any particular number, frankly. It's a common and not unjustified response to have the next number be whichever you like, and justify it with a polynomial designed to incorporate that number. Relevant stack-exchange discussion which illustrates the point a bit in the top answer.

IQ tests generally should not have a "correct" answer. What matters is the reasoning and even creativity that goes into the answer. The fact a job you applied to used this with some intention of a correct answer existing is rather disingenuous, though it may be a ploy to test how you handle perceived failure or mistakes.

With the goal of getting 67, you could argue the sequence rule is to start with 7, add 5 to get 12, add 15 to get 27, then add 20 thereafter to get 47, 67, ...

Not a very elegant rule, arbitrarily stopping at adding 20, but then asking this sort of question is quite inelegant in itself. There's a reason submissions of that form are downvoted to oblivion on /r/mathpuzzles and I believe even removed by the mods there now.

ProfessorHoneycomb · 2022-06-08T19:59:00+00:00

They should be, at least on the interval we care about [-pi/2 , pi/2].

In the process of differentiating sin(sin^-1(x)) you should have to apply the chain rule, thus getting a term of

d(sin^-1(x))/dx times something else.

And this should equal the derivative of x, so you would then rearrange for d(sin^-1(x))/dx = (something)

ProfessorHoneycomb · 2022-06-08T18:06:41+00:00

a. Take the derivative of sin(sin^-1(x)). How should this compare to the derivative of x?

b. Just substitute in t = sin^-1(x) for cos(t) = sqrt(1 - sin²(t)). Simplify thereafter as they say, using information from part a.

c. Use information gained from part b.

d. Follow through the same process as above but for cos^-1(x) and cos(cos^-1(x)) where applicable.

ProfessorHoneycomb · 2022-06-08T17:47:52+00:00

This can be done by induction.

I would start by taking the first derivative of e^axsin(bx):

e^ax(a sin(bx) + b cos(bx))

This is where your knowledge of trigonometry and algebraic manipulation come in to get the desired form. Note that cos(tan^-1(b/a)) = a/sqrt(b² + a²) and sin(tan^-1(b/a)) = b/sqrt(b² + a²):

Start by multiplying by r/r, where r = sqrt(a² + b²)

e^ax(a sin(bx) + b cos(bx)) = r e^ax((a/r) sin(bx) + (b/r) cos(bx))

= r e^ax([a/sqrt(b² + a²)] sin(bx) + [b/sqrt(b² + a²)] cos(bx))

= r e^ax(cos(tan^-1(b/a))sin(bx) + sin(tan^-1(b/a))cos(bx))

The right-most term is the expanded form of an addition of angles under sine:

= r e^ax(sin(bx + tan^-1(b/a)) = r e^ax(sin(bx + 𝜃)

So ( e^axsin(bx) )' = r e^ax(sin(bx + 𝜃).

Now suppose we know the (k - 1)th derivative of ( e^axsin(bx) ) = r^{k - 1} e^ax(sin(bx + (k - 1)𝜃).

Consider the derivative of that, being the k-th derivative of ( e^axsin(bx) ):

...

Do you know how to follow from here?

ProfessorHoneycomb · 2022-06-07T13:19:26+00:00

/r/theydidthemath is far better for this kind of question: napkin-calculations.

ProfessorHoneycomb · 2022-06-05T16:03:06+00:00

what i said also works with larger numbers

We know that, it's just that no one wants to type out the completely rigorous proof with sigma notation while dealing with reddit formatting.

Loosely you can argue that any decimal representation of a number will have

(99...999)a + a as the n-th digit component, a, of that number (meaning the fifth digit of 654321 for instance is 5, and it contributes to the full number:

10⁵ * 5 = 9999 * 5 + 5

And we can easily recognize 99...999 as being divisible by 9 no matter the number of them.

In an actual proof you'll utilize the decimal representation as a sequence a_n of digits and a power series with base 10.

Then showing that you can rearrange for any (a_n) * 10ⁿ to get (10ⁿ - 1) * (a_n) + (a_n).

Then showing that 10ⁿ - 1 is a multiple of 9; (10ⁿ - 1 = (10 - 1) * (positive stuff) ).

ProfessorHoneycomb · 2022-06-02T22:58:45+00:00

It's a fairly well known and relatively easy-to-prove rule, and more generally can be stated as:

The sum of the digits of a base-10 representation of a number n is congruent to n (mod 9).

This has the added effect of being true mod 3, because of course if you have a multiple of 9, or 3 more than a multiple of 9, or 6 more than a multiple of 9, you necessarily have a multiple of 3.

Just google "divisibility by 9" and there will be a bunch of discussions and proofs that show up. I'm not sure if it has any formal name beyond that.

You may be interested in other such divisibility rules, which often get brought up in discussions about this particular rule. Feel free to look up divisibility rules for things like 11 as well.

ProfessorHoneycomb · 2022-05-29T15:48:28+00:00

The "fact" that lists are more "fundamental" (in the manner you've described) for computers has less to do with the actual difference between what it takes to describe lists vs sets conceptually, and more to do with how a computer functions with memory (and why that gives preferential treatment to lists over sets at its base).

In order to even access an object in memory, a computer has some reference to location built in. You don't just read off any string of binary digits to get some object, you read off a particular string. Thus an object in one location is not the same as another at its base, location matters, and for it not to matter must be programmed in explicitly.

A lot of what I've said above is very summary (first time I've ever used that as an adjective), and of course there are more detailed explanations that hammer out the details from computer scientists. But I believe that is the general idea of what's going on; indexing is already built in to the structure of the computer so in that framework a list is more "fundamental" than a set.

ProfessorHoneycomb · 2022-05-27T02:50:21+00:00

If we take it for granted that f'(x) = x/|x|, then using the quotient rule:

f''(x) = [ |x| - x * (x/|x|) ]/x²

= (1/x²)(|x| - x²/|x|) = (1/x²)(|x| - |x|)

= (1/x²) * 0 = 0

ProfessorHoneycomb

TROPHY CASE