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First you should simplify the expression. Notice that you can remove one n from the expression and get n/(n-1)! as your series. I would then separate the second n from the expression by making another faction that is less than 1:
The second fraction will be less than 1 for all n > 3, so it can be dropped for the purpose of determining convergence. We're now left with the more simple summation of 1/(n-3)!. Since we're testing for convergence all that matters is how the sum behaves as n approaches infinity, so in order to simplify the expression further we can remove the -3 portion in the denominator to leave us with the following expression:
From here do a comparison test. Find a series that you know converges and show that it is greater than your series. This will prove that your series converges. My advice would be to go with 1/2^n.
Which part of Jane's method is confusing you? The rearranging of the function or determining the range from the new expression? If it's the latter then the explanation is simple. The expression
can take any value except 0. This can be rather easily shown by noting that the graph is continuous everywhere except at x = -3/2 and finding the limits as x approaches -infinity, -3/2 from the left, -3/2 from the right, and +infinity. Once you've convinced yourself of this fact you can see that the added 1/2 makes the only unreachable value 1/2.
Diane is completely wrong. Karen is kind of right; she can get all 6 colors with just 6 pennies, it's just not likely. In order to guarantee that you'd get one of every color you would need (5/6)*n + 1 pennies, where n is the total number of gumballs in the machine.
In other words the gradient is positive?
Technically, that is sufficient but not necessary for a function to be increasing, since not all increasing functions are differentiable. However, if the function is differentiable and its gradient is positive you can be sure that it is increasing.
It depends on how you look at it. Cash-wise he's down 30$ from where he started, but he's also down a cap. If you count the cap as a loss, which I would, he's down $50 total, which makes sense. He sold a $20 cap for $20, so he broke even on that exchange, but then the neighbor demanded his $50, so that's a $50 loss.
mathsy is right, it doesn't matter how fast the plane is moving relative to the ground, it needs to move relative to the air in order to produce lift. The only way a plane in such a scenario could take off would be if the engines were angled downward, like a Harrier jet. If the engines are aligned horizontally like normal then the plane can't possibly take off, no matter how fast it moves relative to the runway.
First of all, Joseph, we're working with volume here. In that case you would say that the first container is 27 cubic inches, not 27 inches to the third power. If it's a unit of measurement that's being raised to a power then it represents another unit, like square inches represent area, cubic inches represent volume, etc.
Secondly, I would just use ratios for this. Your 27 cubic inch container holds 16 oz. of peanut butter, and you want to know how much peanut butter can be held in a 67.5 cubic inch container. To do that, set these two fractions equal to each other:
To set this up you put the container volumes in each numerator, and the corresponding denominator is the amount of peanut butter each container holds. In the first case that would be 16 oz., in the second case it's x. Now just solve for x using basic algebra:
Set u = 1 - x^2, then you get
It should be straightforward from there.
First we'll remove all of the constant terms from the limit:
Here we'll use a trick that will let us use L'Hopital's Rule. Notice that lim(n approaches infinity) n = infinity, while lim(n approaches infinity) 1/n = 0. It's also trivial to show that lim(n approaches infinity) sin(2pi/n) = 0. So we transform the expression like this:
Now we have a limit of a fraction whose numerator and denominator both approach 0 as n approaches infinity. From here you can use L'Hopital's Rule.
Ok, I'm getting a better idea of what you're looking for. I definitely underestimated how much work you already put into this. I have a few more questions now. What are you trying to maximize? Is it the total amount of talent you can recruit with your budget? How do you account for the different positions? Do you value a good fullback as much as you value a good quarterback? How do you penalize yourself for landing fewer recruits than you target for?
Also, I have a question on how your recruiting works. Does each person determine how many trips they're going to take for each player, input it all at once, and see what happens? Or is it more incremental? For example, do you choose on the first day that you want to do a head coach visit, then the next day you decide to bring him in for a campus visit? If so, it sounds like you could really benefit from an online algorithm, which is an algorithm that accounts for unpredicability in the future. The best example would be the stock market. You don't know how stocks are going to perform in the future, but with the right algorithm you can make more money by buying and selling stocks at certain intervals under certain conditions, rather than just buying a stock and holding it for a set length of time before selling it. Using something like this you could determine when it's best to back off a recruit and spend that money elsewhere.
If you want to do this mathematically I think you need to estimate your chance of getting a recruit depending on how much you spend. It doesn't matter if the average recruit has a talent/cost rating of 5 and you find one with a rating of 20, if you only have a 5% chance to get him you're wasting your money. In the best case you would come up with an equation to determine your (estimated) percent chance of landing a recruit after spending x dollars recruiting him. If that's not possible, then at least come up with dollar amounts that you think will give you a 99%, 75%, 50%, and 25% chance of landing him.
Basically, I think you're looking to maximize the expected value of your recruits. For example, let's say that you want 2 offensive linemen, and there are 5 that meet your bare minimum to consider recruiting: A, B, C, D, and E. They have rankings of:
A - 8.5
B - 8.2
C - 8
D - 7
E - 6.5
Then let's say that this table represents how much you have to spend to get them:
99% 75% 50% 25%
A 1000 800 650 450
B 900 725 600 400
C 850 700 500 300
D 700 500 400 250
E 600 450 300 200
Let's also say that your offensive lineman recruiting budget is $1500. This would be enough to guarantee that you can get recruit A and give you either a 75% chance with D, 50% chance with C, or a 25% chance with B. Multiply those probabilities by their respective talent amounts to find your expected total talent for a particular recruiting strategy. Also keep in mind that in this scenario you're only targetting 2 linemen. You'll need to factor in a "punishment" factor if you miss out on one of your targets and finish with just 1 lineman.
You'll end up with an equation that looks something like this:
If you want to maximize your Talent/Cost ratio for a given dollar amount try a couple strategies by hand. You'll make life much easier on yourself if you can find an equation between dollars spent and the probability of landing a recruit. That way you can get one equation to work with. Otherwise, you just have to brute force a couple calculations by hand (or write a computer program to do them for you) of a couple different recruiting strategies, such as targetting 2 or 3 or 4 players, and how much to spend on each of them.
Going back to our example, let's assume you came up with the following equations instead:
A - ChanceToGet = (DollarsSpent / 100)^2
B - ChanceToGet = 10 + (DollarsSpent / 100)^2
C - ChanceToGet = 15 + (DollarsSpent / 100)^2
D - ChanceToGet = 15 + (DollarsSpent / 100)^2.5
E - ChanceToGet = 25 + (DollarsSpent / 100) ^ 2.5
Plug these into the Recruit#_probability in the equation above and you'll get a single equation that encompasses how much total talent you can expect. Divide the whole thing by the amount of money that you spend and you can then maximize your expected talent / dollars spent.
If you do decide to go through all of this work, you'll end up with an equation with many variables. More precisely, it will have a number of variables equal to the number of recruits in your equation plus 1. If I remember my calculus correctly you can use the gradient and an optimization algorithm called gradient descent to find your maximum value. If this is the calculus thing you were talking about, the gradient, let us know and someone here can help you out.
Is your estimated total cost an estimate of what it will take to recruit a given player with 100% accuracy? Meaning, for a given player do you think that if you spend your estimated total cost on him you will get him 100% of the time?
There are a number of factors you need to consider. Too many to translate into a reliable mathematical model, in my opinion.
First of all, you need to consider how valuable the recruit is to your team. Let's say you have 2 quarterbacks on your team. If they are both sophomores, or a sophomore and a freshman, you could look for more of a project at quarterback (if you have a statistic for "upside"), or even afford not to recruit a quarterback at all. On the other hand, if your quarterbacks are a junior and a senior you'll need to spend more money to get a quarterback who will be ready to go quickly.
You also need to consider what it could mean to get a big time player on your team. Maybe you already have 3 or 4 running backs, but there's a recruit out there with 99 speed, 99 break tackle, etc. who would single-handedly impact your team. It might be worth your while to over-spend on recruiting such a player, if only to ensure that your opponent doesn't get him.
Speaking of other players, you should know them as well as you know yourself. Study their rosters and their tendencies so you'll know what kind of players they're likely to go after. If they tend to stay close to home to get "value" out of their recruiting budgets, you can afford to underspend on players close to you because you'll have little competition. Then you'll have some extra cash to outspend your opponents for a select handful of highly ranked recruits in their home states.
On the other hand, if your opponents tend to go for the top players regardless of how far away they are they'll spread their recruiting budgets thin. Overspend a little bit on top tier recruits in your own state, while targetting some solid second tier recruits from other states. Even though you won't be able to spend much, you'll have little competition because your opponents will be spending all of their money on just the top tier recruits.
Also, keep in mind what kind of players your're looking for. Do you use a straight ahead running game? Skimp on recruiting receivers and small, quick running backs. Focus on beefy, strong offensive linemen and powerful running backs. If you prefer a wide open passing attack go nuts on quarterbacks and receivers, but settle on lower rated running backs. In particular, look for running backs who have good pass blocking skills but lower than average running skills, these guys will probably not be targetted by other players.
This goes for defense as well. If you prefer to blitz a lot and play man coverage try to get bigger corners who can jam receivers at the line and smaller, faster linebackers who can rush through gaps in the blocking. If you prefer to sit back in zone coverage and let your defensive line rush the passer by themselves get faster corners who can play better in coverage.
Even if it's outside your comfort zone you should try to do the opposite of what everyone else is doing. Most players are probably going to build passing oriented offenses, which will leave most of the good running backs for you. They'll also probably go for the fastest defenders they can get in order to stop the passing game, which will make it easier for you to run all over them. Go for big, strong corners who can jam receivers at the line and blitz with your linebackers and safeties.
As far as how to recruit your players, give yourself a certain amount of money to recruit each position. Then determine the probability that you can get a certain player by spending a certain amount on him. This will help you decide between targetting 3 or 4 offensive linement when you need 2. There's no good mathematical way to do this, just make as good of a guess as you can as to how many players you can get. In some cases you might be better off spending $1200 each on 4 recruits to get 2. Other times you'll be better off spending $1600 each on 3. This is where knowing your opponents' tendencies will help you the most. If you know your opponent will spend whatever it takes to get his top 2 choices at offensive lineman, let him have them and target the next 3 on your list. That way you can get them for cheap and use the leftover cash to snatch his #1 choice at quarterback.
The first 2 of these uses the rule
1. Multiply out the left side to get
2. Factor the left side to get
3. Remember that tan(x) = sin(x) / cos(x) and sec(x) = 1 / cos(x).
George, you don't understand what infinity is. Your last post proves this:
Do you call
{2,3,4,5,...1001}
and
{2,3,4,5,...1000}
the same?The same concept applies to this situation also. The only difference is that this is an infinite case. And in order to compare, I applied the larger concept.
You're right in that the only difference between this example and
{1, 2, 3, 4, 5, ...} vs. {2, 3, 4, 5, 6, ...} is infinity. You're wrong, however, in that the same concept does not apply. Infinity is not a normal number. Indeed, it is not a number at all. You can't treat it like a number. Infinity + 1 = Infinity doesn't make sense, but neither does Infinity + 1 != Infinity, because you're treating infinity like a number when it isn't one. Until you accept the nature of infinity, discussion of this topic is pointless.
I'm not sure if this is rigorous enough for you, but here's my shot at it.
Our first observation is that we can't get stuck in a cycle (unlike the Collatz conjecture). This is due to the fact that our denominator n is always decreasing, as it is always set to either m (which is less than n) or to n - m, which is also less than n. We also know that m never increases (as it is either set to m or n - m, whichever is less) and must stay less than n, so we know that over (finite) time it will decrease.
Since we can never get stuck in a repeating cycle the only way that we won't reach 1/2 is if m' = 0 or m' = n'. Note that in order for m' to reach 0 we would need m = n in the previous step, so we need only consider this mode of failure.
Now, in order to reach m' = n', we need 2m = n in the previous step. This is equivalent to saying that our previous fraction was m/2m. This fact is easy to show, as no matter which direction we go from the previous step, in order for m' = n' we find that n - m = m -> n = 2m.
Now, it gets a little messy from here. What I'm going to try to show is that if an' = bm', a<b and a and b are positive coprime integers, then there exist positive coprime integers c and d, c<d, where cn = dm. If this is the case, then n and m cannot be coprime, and we'll never reach a point by going backwards where n and m can be coprime. Since they must be coprime when we start we know that we can never reach the point where m = n.
This all tells us that to get to an' = bm' we need either an = (a+b)m or bn = (a+b)m. Since both a and b are coprime to and less than a+b, let c = a or b and d = a+b, and we have our two new coprime multiples.
Finally, we need to show that if a<b are coprime integers and an = bm implies that m and n are not coprime (likely, there is already a named theorem that shows this, but I don't know of one so I'll prove it for the sake of completeness). From an = bm we find that a = m * (b/n). Since a and b are coprime, every prime factor of b must be present in n to be eliminated. If n has more prime factors than are required to eliminate b, then they must also be present in m, since a is an integer. This would mean that m and n are not coprime. However, if n has only the exact same prime factors as b, then n = b. This would further imply that m = a. However, this is no possible, as m<n but a<b. This means that m and n must share prime factors.
Finally, as stated above, no matter how far back we trace we can never find an instance where m and n are not coprime. Thus, every fraction where m and n are coprime will eventually reach 1/2. The only way this isn't true is if there is a failure mechanism other than a repeating cycle or m = n, which have been shown to be impossible.
I think the point that is confusing you is how the first pick alters the odds after the first door is opened.
The reason that the odds defy common sense is that the first door that they open and show is is not chosen randomly. They never show you the winning door and then ask if you want to stick with the door you chose first or to change (because what fun would that be?). They will always show you a losing door and then ask you if you want to keep your original choice or change.
Now, consider what would happen if they chose the door to show you completely randomly, meaning that they might show you the winning door. Let's say for the sake of argument that door A is the winner (the probabilities are the same if door B or door C is the winner). What are the odds that they'll show you the winning door before you even get to decide to keep your door or change? There is a 0% chance of this if you choose door A, but a 50% chance if you choose door B or door C, for a total of a 33% chance (0 * 1/3 + 2 * 1/3 * 1/2). That's a 33% chance of losing before you even get to decide to keep your door or switch.
Now, that means that you have a 66% chance of having a meaningful decision after the first door is opened. You now have a 50-50 shot of winning, whether you keep your door or change, giving you a 33% chance to win. Note that this is the same chance you have of winning if you keep your door in the original game, but this time you have no chance of improving your odds.
Since you know something about how the door that they open is chosen, you can improve your odds. Basically, that 33% chance to lose right away (if they show you the winning door before asking you to switch) is added to the door that you did not choose. Intuitively it's confusing, but it's what actually happens.
Finally, to address this:
One thing logically that confuses me is my original statement that if someone else in the audience said, "I think the car is behind door A" after the first goat is shown to you (and let's assume that he was in the bathroom when you picked your original door and the first goat was shown, so he didn't see any of this information). He has a 50% chance of being correct, right? If that's the case, how can I have a 33% chance with door A and he has a 50% chance.
There are two points that I want to make here. The first is that while you have a 33% chance of being right with door A while he has a 50% chance, he also has only a 50% chance of being right if he chooses door B (or door C, whichever one wasn't shown to you) while you have a 66% chance. You have the advantage. The second point I want to make is that he has a 50-50 chance of being right because he's choosing between two doors. You have a 33-66 chance of being right because you're choosing between three doors. Even though one of the doors was shown to you, you're still choosing between all 3 doors because your original choice affects your odds.
I hope that helps.
1. Let's call the speed of the car C and the speed of the train T. The train is 20 kph faster than the car, so T = C + 20. We also know that in 20 hours the train will travel as far as the car can in 30 hours, so that gives us 20T = 30C. This gives us a system of 2 simultaneous equations. To solve it we'll use substition.
From the first equation we know what T equals in terms of C, namely T = C + 20. We'll substitute this value in for T in the second equation to give us 20(C + 20) = 30C. From here we get 20C + 400 = 30C -> 10C = 400 -> C = 40. Plug this value of C into the first equation to give us T = 40 + 20 = 60 kph. To check this, the train will travel 60 * 20 = 120 km in 20 hours, while the car will travel 40 * 30 = 120 km in 30 hours, so we have the right answer.
2. Since A finishes the job in 6 hours, after 1 hour he's finished 1/6 of the job, meaning that there is still 5/6 of the job left to go. After 2 more hours A will do another 2/6 of the job for a total of 3/6, or 1/2 of a full job. That means that J did 1/2 of a job in 2 hours, so multiply that by 2 and we find that J can do a whole job in 4 hours.
3. This one's simple if you have a calculator. Just plug 66 in for H to give you 66 = 1.2L + 27.8 -> 1.2L = 66 - 27.8 = 38.2 -> L = 38.2 / 1.2 = 31.833333.... inches = 31 5/6 inches.
When dealing with decimals, don't let them scare you. Just treat them like normal numbers until you have to either add, subtract, mulitply, or divide them. Then just break out a calculator and keep going.
4. We know that 5 workers build 9 chairs in 4 days. That means that 5 workers build 9/4 chairs in 1 day, which we can further simplify to discover that 1 worker builds 9/20 of a chair in 1 day. Now multiply that value by 8 since we have 8 workers and we see that 8 workers build 18/5 chairs in 1 day. From here, divide the number of chairs they need to build by the number of chairs they build per day for your answer. In this case we have to build 18 chairs, at 18/5 chairs per day gives us 18 / (18/5) = 5 days.
If any of these explanations don't make sense just let me know and I or someone else can explain in more detail.
What are these percentages of? Total pay, hours worked, units sold?
I got the same thing. I can't manage to come up with a way to prove it mathematically, but I feel like it has to be the answer.