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Since
we know that
Now the function f(n) = 2f(n-1) is equivalent to f(n) = 2^n, and f(n) = 3f(n-1) ==> f(n) = 3^n. So we know that
. After that I'm stuck, sorry.You have it backwards. You're solving x/10 = 25, when it's asking for 10/x = 25.
You're given 3 points on this line, (0,0) (1,0.9) (10,0). You can use those to determine a and b.
I made a spreadsheet to visualize the values and found this:
To determine if a specific n satisfies the conditions for a given x, find the unique prime factors of n+1 and call them p_1, p_2, ..., p_i. If x+1 mod p_1, x+1 mod p_2, ..., x+1 mod p_i all do not equal 0 then n satisfies the conditions for x.
I'm not sure if this is helpful and I haven't worked a proof for it (although I can see how a proof could be done), but maybe it'll lead you in the right direction.
Edit: A more useful way of stating the above would be this:
Given x, let the set A contain all prime numbers p_1, p_2, ..., p_i which are not factors of x+1. Every n that satisfies the condition has the property that every prime factor of n+1 belongs to A.
You know how to use the quadratic formula right? h is your variable, so a = 3, b = -4R, and c = R^2. If you solve for this you should get h = 1/3 R. Although this particular equation is easy to factor too: (3h - R)(h - R) = 0.
You can get the correct answer by solving for the height, you just aren't being careful with what you're solving for. When you solved for
you got 2/3 R, which is the radius. However, if you solved you probably got 1/3 R, right? But remember that this answer is in terms of the height of the cylinder. Remembering that b + h = R, you get b = R - 1/3 R = 2/3 R, which is what you got when you solved for the radius directly.Thanks for the help, but I don't really understand how
.
This is given to us. The question is to prove that A(x) converges whenever
and since
we know that S(x) converges, which means A(x) also converges since they're equal.mathsy, regarding question 2 that makes sense for |x| < 1/2, but I don't see a restriction in the question that x must be chosen such that A(x) converges. My understanding (based on a quick look on Wikipedia) is that a Cauchy product can converge even if one of series being multiplied diverges. Although now that I've taken a second look I can see that this is never stated anywhere in the article, so perhaps both series do have to converge for the Cauchy product to converge?
I believe #1 can be proven with the ratio test. I'm probably going to butcher the terminology but this should be understandable.
The function A is equivalent to the series
Define a new function S
So A(x) = S(x). Now the ratio test says that if
then S(x) will converge. So what is
? Well, it'sSo we know that S(x) converges, which of course means A(x) does as well since they're equal.
I read #2 differently. I think it's trying to say
That would make it far, far more difficult than #1 though, so I'm not sure. In any case it's well beyond me so I'm afraid I can't help on that one.
For the first induction you need to prove 2 base cases, n = 1 and n = 2. From there you can use the following induction.
Treat this formula as a function of n:
a, c, and d are constants that we need to solve for. We are given 4 points on this function, though we only need 3 to solve it. Using the given information we know that
You can see from here that a = 2, d = 2, and c = 1, so we're left with
To test this, try n = 4:
which is what we expected. From here you can find any term you need.
This is kind of annoying since you can't prove a base case until n=5, but the induction proof is pretty simple. Assuming
we want to prove that
You can see that the inequalities of the terms outside of the brackets are already known to be true, so if we can show that the terms inside of the brackets follow the same inequalities then our induction is complete.
which is true for all n>=0.
You need to check 2 more cases.
CASE-IV: 9(11a+b) = 10x9
11a+b = 10 ⇒ a=1 and b=-1 Not Possible
CASE-V: 9(11a+b) = 19x18
11a+b = 38 ⇒ a=3 and b=5
a+b+c = 19
c = 19 - (a+b) = 19 - (3+5) = 11 Not Possible
If you're interested in a more curvy increase you can modify the equation to
So your example becomes
You then add x*1^2, x*2^2, x*3^2, etc. for each successive term. So the series for you example is 1000, 1000 + 27.2727... = 1027.2727..., 1000 + 4*27.2727... = 1109.0909..., 1000 + 9*27.2727... = 1245.4545..., 1436.3636..., 1681.8181...
If you're ok with a straight line it's pretty easy. Let S be your starting value, T be your cumulative total, c be the number of additional values you're going to add, and x is the increment size. This will give you the equation
For your example we have S = 1000, T = 7500, and c = 5, so
Earlier i was wondering if big perfect square xxxxxxxxxab exists such that xxxxxxxxxba is also perfect square (a,b≠0). Now i see that such a number can't be any greater than 7744
No, there are infinitely many such squares. Specifically, the square of 100z + {12, 38, 62, 88} for any non-negative integer z will end in xxxxxxxxx44. However, there are only 2 examples of perfect squares with this attribute that don't end in either xxxxxxx00 or xxxxxxx44 - 169 and 196.
Well, you can narrow your search fairly easily. There are two ways for your attribute to hold true: the two squares must have a difference of less than 100 or the unit's and ten's digits are equal. For the first method there are only a finite number of possibilities since the distance between consecutive squares strictly increases. Specifically, only the squares of 4 through 50 are eligible, and it's a pretty simple matter to brute force. As it happens there is exactly one such pair of squares - 13 & 14.
For the second method you should note that in multiplication the final two digits of the product are decided solely by the final two digits of the multiplicands. Again you can simply brute force numbers from 4 to 99 and check by hand. For squares found using this method you can add any multiple of 100 and it will still work. Aside from the obvious multiples of 10, the squares that fall under this category are 12, 38, 62, and 88.
In short, the numbers you are looking for are the squares of 10, 12, 13, 14, 20, 30, 38, 40, 50, 60, 62, 70, 80, 88, and 90, and any multiple of 100 plus any of those numbers except 13 and 14.
Poker movies, the worst. Every bet made in every movie is an illegal bet. They string bet, they call bet, all illegal. No poker dealer or player would ever allow another player to do that. No technical advisers?
Could you expand? I play poker sometimes with my friends, and have played once or twice at a casino table so I know we aren't playing by horribly inaccurate rules, but I've never heard of string bets or call bets.
Also, I think you guys might like http://tvtropes.org/pmwiki/pmwiki.php/Main/HomePage
Integration by parts isn't my strong suit, but I think I can help. We have
Let's rewrite A(b) as follows:
Now use integration by parts:
I'm not sure if it can be further simplified from there.
Your problem is missing something. If you use x<-1 and an even value for n the inequality breaks down.
How much have you figured out so far? Conceptually, the first derivative of f(t) gives you the rate at which the oxygen level is increasing at time t, and the second derivative gives you the rate of the rate of increase. So you're looking for the maximum of f'(t), which can be determined by finding a critical point in f''(t).
What you're really interested in is the length of a side of the pyramid at height z, since A(z) is just that number squared. Let's call this length L(z), and note that it is simply a linear function. You have L(0) = 0 and L(h) = w, so L(z) = zw/h, giving us A(z) = (zw/h)^2. Integrate this over 0 to h and you get your answer.
It's been stipulated that M < 0, so