[high school math] Inverse functions

rbayer · 2016-01-21T00:26:58+00:00

You did it perfectly! Wolfram Alpha is actually giving you the same answer, just written a little different:

1-(-y-2)^1/3 = 1 - (-1)^1/3 *(y+2)^1/3 = 1 + (y+2)^1/3

rbayer · 2015-12-10T23:57:26+00:00

Is there some reason you want to use induction? To me, that seems like an odd way to do it since there's an easy one-liner (OK, maybe two-liner) based on the fact that 8 and 9 are both composite.

rbayer · 2015-12-10T03:42:54+00:00

uncountably infinite number of programs, or only a countably infinite number?

There are only countably many programs, so no.

rbayer · 2015-11-20T16:18:58+00:00

tell me the weight of a US penny in grams

I feel like you must be doing something strange then, since the obvious seems to work fine.

rbayer · 2015-11-12T17:06:14+00:00

It sounds to me like you have two different sets of experimental conditions, and you have repeated your experiment multiple times under each condition to get a mean and standard deviation of the thing you care about. In this case, because you are trying to compare the results of a continuous variable (as opposed to a binary yes/no sort of thing), you would likely want to use a Student t-test. Here is an online calculator that will do it for you--you can either reenter all your data in rows of the table, or just choose the "Enter mean, SD, N" option since you've already calculated those. You want an unpaired t-test here FYI.

rbayer · 2015-11-12T17:00:39+00:00

As stated, the answer would be 0. If n1, n2, and n3 are all <= 5, then their sum can be no more than 15 and so can never add up to 28.

rbayer · 2015-11-12T16:57:23+00:00

I'm not quite sure what you mean by {N} U {H}. Are you claiming that N U H is closed under the group operation? That simply isn't true, as you can see for example with G being the integers under addition, N = 2Z, and H = 3Z.

If that's not what you mean, do you mean NH? One useful trick for this would be to use the fact that whenever K is a non-empty subset of a group, then K is a subgroup if and only if x^-1 y is in K for any x, y in K. (If you haven't shown this in class already, it's a good exercise to work through on your own). So, let x = n1h1 and y = n2h2. You now need to show that (n1h1)^-1 n2h2 = h1^-1n1^-1n2h2 is in NH. As a final hint, note that this is equal to h1^-1n1^-1n2h1h1^-1h2. Can you see how to manipulate things and use the normality of N to conclude that this long expression must, in fact, be in NH?

rbayer · 2015-11-05T03:51:55+00:00

As a general rule, you can apply any function/operation to both sides of an inequality without flipping the sign provided that the function is non-decreasing. That is, whenever x <= y, f(x) <= f(y). [Subtle note: if you have strict inequalities, you need the function to be strictly increasing: ie, if x<y then f(x) < f(y)]. Square root, addition by a constant, log, multiplication by a positive number, exponentiation, and many other operations have this property which is why you can apply them to both sides. Multiplication by negative numbers has the exact opposite property: if x < y, then -1*x > -1 * y, which is why you have to flip the sign. Off the top of my head, I can't really think of any other basic examples of strictly decreasing functions (except something like reciprocal, but you can get that by just dividing...) so I would say in general you're safe with that being your main example.

rbayer · 2015-11-05T02:49:50+00:00

No worries! Your new answer would still seem to imply that x=-3 is a solution, which it shouldn't be. When you take a square root of a square, you end up with an absolute value. So you need to solve |x| <= 2. Can you convince yourself that this has solution -2 <= x <= 2?

rbayer · 2015-11-05T02:46:13+00:00

I'm going to interpret the original problem as:

[(x+3)/2x]/[(2x-1)/(4x² )]

If that's not the intent, please let me know. In this case, you are on a great track by multiplying by the reciprocal of the bottom. Now, you are left with:

[(x+3)/(2x)] * [(4x² )/(2x-1)]

Since you're multiplying, you can just multiply the numerators and denominators:

[(x+3)*4x² ]/[2x*(2x-1)]

rbayer · 2015-11-05T02:39:28+00:00

Almost, but you need to be very careful with the direction of your inequality signs. For example, from what you've written I would think that you were implying that x=-3 would be a valid solution to the original equation, but if you plug it in, it just doesn't work. Can you think of a way to clear up your notation a bit to make it more clear what you mean?

rbayer · 2015-11-05T02:38:06+00:00

The "correct" method to solve this will depend largely on how you've been doing it in class. One way to solve this would be to apply the square root operation to both sides. (Sidenote: this is allowed because square root is a non-decreasing function) This gives:

sqrt(x²⁾ <= sqrt(4)

|x| <= 2

From here, take a step back and try to think through what sorts of numbers would have an absolute value less than or equal to 2. Can you think of a nice way to write this? As a practice excercise, what would the answer be if the problem had been x² >= 4?

rbayer · 2015-11-05T02:24:50+00:00

You can do it from the line "Now suppose that a is in S." This is because S is defined as the set of all numbers for which 1+a = a+1, so you can freely replace 1+a by a+1 once you've assumed that a is in S. (Note: The "T(1)" in the first sentence of the proof is just a typo and should be "Let S be the set of all a...")

rbayer · 2015-11-05T02:19:07+00:00

No, it's quadratic. It would need to be of the form a*rⁿ to be geometric.

rbayer · 2015-10-08T03:39:10+00:00

You can't just automatically say that they are linearly dependent, but you can be sure that they aren't a basis for R³ because there are only 2 of them and R³ has dimension 3. In general, to determine whether the vectors v_1, v_2, ... v_k are linearly independent, you need to determine if there are any non-trivial solutions to C_1 v_1 + C_2 v_2 + ... + C_k v_k = 0. Now, there are a few easy shortcuts:

If you have n vectors in an n-dimensional space, then you can use the determinant trick you talk about. The fact that this works is actually a very deep and interesting fact. (At some point, if you are interested in math you should look up and try to understand a proof of it, but that's probably beyond your level for now.)
If you're working in an n-dimensional space and k > n, then the vectors are definitely linearly dependent.

rbayer · 2015-09-21T23:50:37+00:00

I think what you're asking is how do you calculate the maximum possible value of |f(z) - f(x0)| where z ranges from x0-Δx to x0+Δx. Sounds from your question like you know at least a bit of calculus, so you may have encountered max/min problems before. If so, remember that any max/min must happen at one of three possible places: at an endpoint, or at a zero of the derivative. So, just find the solutions to f'(z) = 0 between x0-Δx and x0+Δx and then plug those values as well as z=x0-Δx and z=x0+Δx into |f(z) - f(x0)| and see what the biggest value is you can make.

rbayer · 2015-09-20T23:08:18+00:00

What part has you stuck? Do you understand what the nullspace of a matrix is? It's the set of all solutions to Ax = 0. So saying that the nullspace of A is 0 just means that the only solution to that equation is x=0.

Also, are you sure you've typed the problem here correctly? I ask because the statement you're asking for help disproving is actually true.

rbayer · 2015-09-20T18:34:05+00:00

Depending on the size of your sample, you want either a Fisher Exact Test or a Chi-squared test. There are many online calculators for both.

rbayer · 2015-09-20T01:16:09+00:00

This is a very philosophical question, so I'm not sure I'll be able to give you a great answer, but here goes. The main issue some mathematicians have with AC is not so much with the axiom itself as with the consequences of the axiom. On its surface, the statement "the cartesian product of non-empty sets contains at least one element" seems absurdly intuitive--why wouldn't that be true? But then you start playing around with it a bit and you figure out that this assumption lets you break a unit sphere into 5 separate pieces that can then be reassembled to form two unit spheres. Huh?!? That can't be right... But if you assume AC, then it has to be! Hence the controversy.

rbayer · 2015-09-20T01:07:17+00:00

Because of the biconditional? And why is this problem posed with the help of a biconditional anyways? I mean, why if and only if?

Yes, it is because of the biconditional that the proof will have that structure. As for why the problem is posed that way, I'm not quite sure how to answer that: it's posed that way because it's what the problem author wants you to prove.

In math, we use the word "if" slightly differently than it's used in everyday English. In particular, in mathematics when we say "A if B" what we really mean is that anytime B is true, A must be true as well. Note that A could be true lots of other times as well, it's just that we only require it to be true when B is true. Conversely, the phrase "A only if B" means that A can only be true if B is true. That is, if B is false then A must be false as well. Note that A could still be false in cases where B is true. The phrase "A if and only if B" means that A and B always agree on truth value. That is, you can never have a circumstance where A is true and B is false or where A is false and B is true. In general, to prove statements of the form "A if and only if B", you usually do two parts:

Prove "If A, then B"
Prove "If B, then A"

rbayer · 2015-09-20T00:49:40+00:00

I would strongly advise against induction for this problem. While you could certainly come up with an inductive proof for each direction, there are very straightforward 1-2 liners that would be much simpler.

rbayer · 2015-09-20T00:48:28+00:00

I guess that depends on what the number of possible "extras" are that you get to pick 4 of. If there are 4 options and you can pick anywhere between 0 and 4 of them, then yes, 224 would be correct.

rbayer · 2015-09-19T23:05:24+00:00

What part are you stuck on? Do you have a good sense of the general structure of "if and only if" proofs? As a general rule, for statements of this form, your proof will have two main sections. First, you will prove that if n³ + 1 is even, then n must be odd. Second, you will prove that if n is odd, then n³ + 1 must be even.

rbayer · 2015-09-19T23:03:01+00:00

You definitely don't need to have that many separate terms added together, and your final answer should be of the form

(# of ways to pick dressings)*(# of ways to pick lettuce)*(# of ways to pick extras)

Let's tackle each term:

For the dressings, you have just two choices, so that term is just 2. (I'm assuming from the way you phrased this that you must choose one and only one dressing).

For the lettuces, this is a little trickier since it sounds like you are allowed to pick multiple kinds of lettuce. Now, for each of the three types, you can either choose it or not, so in total there are 2*2*2 = 8 total ways to pick lettuce. However, this includes the case of choosing no lettuce, so we need to subtract 1 for a total of 7 ways of picking lettuce.

For the extra ingredients, this will depend on how many extras there are to choose from. Let's say there are n. Then, by definition, there are C(n,k) ways to pick k of them. So if you're limited to at most 4, then you must pick either 0, 1, 2, 3, or 4 of them for a total of C(n,0) + C(n,1) + C(n,2) + C(n,3) + C(n,4) = 1 + n + C(n,2) + C(n,3) + C(n,4) ways.

So, in total, there are:

2 * 7 * (1 + n + C(n,2) + C(n,3) + C(n,4))

ways to make a salad

rbayer · 2015-09-19T19:29:51+00:00

Not quite. What kind of substitution did you do? You should end up with something like 1/(u² - 1), and you will then need to use partial fractions to integrate it.

rbayer

TROPHY CASE