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This might be a stretch, but the only way that I can see it working is if the middle and youngest child were born within 1 year of each other. Here are the equations I came up with:
Middle + 1 = 2 * (Youngest - 1) = 2 * Youngest - 2
Oldest - Youngest = Middle + 1
These give us
Middle = 2 * Youngest - 3
Oldest = 3 * Youngest - 2
Now, we're subject to the fact that the oldest has never been twice the age of the younger two children. We're also subject to the fact that the middle child has to be older than the youngest child, obvioiusly. Here are some values based on the equations above:
Youngest 0 1 2 3 4 5
Middle -3 -1 1 3 5 8
Oldest -2 1 4 7 10 13
Clearly the first 3 columns are impossible. The last 2 are also ruled out since the oldest child is or would at some point have been twice the age of the middle child. This remains true as the ages increase, so we're only left with ages of 3, 3, and 7. This of course depends on the youngest and middle child being born within one year of each other. It's not common, certainly, but it is physically possible. This would make the father 2 * (3 + 3 + 7) = 26 years old.
Edit: Whoops, that's not right.
I have a question for you Stumpe. What do you believe 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... is equal to?
I'm not completely sure, but I think they want you to separate those 4 terms into 2 groups and factorize those groups to find a common factor, like this:
You should be able to simplify it from there into the form (a + b)(c + d).
Sorry, I made some notational errors in my last post that I've edited, I don't know if they'll help or not. In case they don't I'll try to explain it again.
Im not sure I understand. if b_{k} diverges, how does it imply that a_{k} cannot converge? Isnt this implying that if a series diverges absolutely then it diverges conditionally? (Which is wrong).
Again, you're confusing sequences and series. I don't have a proof of this, but my belief is that if the sequence b_{k} diverges (that is to say, a_{k} diverges absolutely) then a_{k} will diverge.
This is, however, NOT true for series, which you showed.
However, in order for a series to converge it's terms must converge to 0. This can easily be seen by considering the definitions of series and limits. A series has a limit L if and only if for every real number e > 0 there exists a natural number N such that
for every n > N.Now, remember that
.Now, let's assume that a_{k} does not converge to 0. If it does not then we can always find an e > 0 such that
, which means that the series cannot have a limit.To reiterate, don't confuse series with sequences. A series can diverge absolutely and converge conditionally, but a sequence can't (as far as I know). In order for a series to converge the sequence of it's individual terms must converge to 0.
You're confusing series and sequences.
and are sequences, is a series. Specifically, is the sequence of individual terms in the series, and is the absolute value of the individual terms in the series.Using this distinction you can see that if
then the series cannot converge.To see this, consider a sequence
. This is called a sequence of partial sums (I think). The limit as this new sequence approaches infinity will be the same as the infinite series we're working with. Note that we can rewrite this new sequence recursively like this: . From a high level we can see that if diverges, then will diverge as well. This is by no means rigorous, but it should give you something to start with.After thinking about it, I'm definitely wrong. My method basically ignores every other term in the series, which isn't going to give an accurate picture of an alternating series. Thinking about it, every other term is more than 1 away from the term before it, so the series can't possibly converge.
Well, something's screwy here. mathsy's and my alterations seem to be equivalent, yet we're drawing different conclusions from them. For example, in mathsy's series at n = 6, the sum is -37/60. Similarly, in my series at k = 2 (which is the same point as n = 6 in mathsy's series) the sum is again -37/60, so I'm confident that our series are the same.
However, our evaluation of them is completely different. Mine is bounded by
, which is convergent. However, I can't find a single weakness in mathsy's method either. I'm confused.First of all, you'd better not start that sum at k=0 or you'll be in trouble.
Secondly, I'm not that familiar with these tests you used, but I'm pretty sure that your line of reasoning is wrong.
may diverge, but that doesn't guarantee that your sum will diverge too, unless I'm mistaken.Assuming I'm right and that you can't evaluate the limit that way, my suggestion would be to combine every other term in the series and then evaluate that series. Let k be an odd number, then we'd get this:
This gives us the difference between 2 consecutive terms in the series, so we won't have to deal with the -1^k any more. Rewriting this into a form we can use with summation notation we get
You have to change the terms a little bit since we're counting by 2 instead of by 1. Hopefully this will be a little easier to work with.
Edit: If all of my tinkering around was correct then the sum will, in fact converge. This can be see by comparing it to the sum of the reciprocals of the squares: 1 + 1/4 + 1/9 + 1/16 ... = pi^2/6.
Ricky wrote:i is the square root of -1, an "imaginary" number (although it really exists).
So, if I'm doing this correctly... would this be the solution for the exponential term?
I'm assuming that the reason why it comes out to 1 is that you ignore all of the subsequent terms becuase they would all be imaginary.
Not quite jtiger. You can't just ignore all of the terms with i in them, they must be added up as well. It just so happens that they all cancel out. The i needs to be squared / cubed / etc. as well. That makes the series
Note that since i is the square root of -1, i squared is -1, which makes the third term of the series, (i^2 pi^2)/2! become (-pi^2)/2, which is a real number. By extention, i^3 = -i, and i^4 = 1. This means that every other term in the series, starting with the first, will be a real number, and every other term starting with the second will be imaginary. We can separate the Taylor series into real and imaginary parts:
If you are so inclined you can try to work out the sums. Just know that the left sum must equal -1 and the right sum must equal 0.
Ignore Lilian's and Judy's responses altogether. David and Margaret disagreed on when they met, which means one of them must be lying about that. This in turn means that one of them is telling the truth about who stole the purse, narrowing it down to Judy or Theo. Theo says that he is innocent and that Margaret took the purse, one of which must be false (it is logically impossible for his third statement to be false). However, his second statement must be false because we've already narrowed it down to Judy and Theo, so this must mean that he's telling the truth about his innocence, which means Judy is the theif.
2. I'm not going to give a rigorous proof for this, but I will tell you that it converges. This is easy to see if you define the sequence recursively:
As you can see, after n = 2 every successive value will be equal to the previous value multiplied by a number whose absolute value is less than 1. The value of the sequence will bounce between positive and negative numbers, but it will converge to 0. I leave the proof to you.
First she takes the cat across the river. Then she heads back and picks up the mouse and takes it across the river. When she gets to the other side she puts the mouse down and picks the cat back up. She goes back to the other side and leaves the cat while picking up the cat food. She then takes the cat food to the other side of the river and leaves it with the mouse. Finally, she goes back to pick up the cat and then takes it to the side with the mouse and cat food.
Let's start with the sequence a(n). Notice that for any n, a(n) will be either (4/2)^(-n) = 2^(-n) or (6/2)^(-n) = 3^(-n). This means that for any n,
Now, note that starting at n = 2, 3^n > 2^(n + 1). This means that in cases where the numerator of r(n) is 3^(-(n + 1)) r(n) is less than 1, whereas when the numerator is 2^(-(n + 1)) r(n) is greater than 1. This all means that r(n) has no limit for the sequence a(n), so Mathematica was right.
As for p(n), I think you switched your a(n) and your b(n). The limit of p(n) using the series a(n) once again doesn't exist. It will continually flip between 1/2 and 1/3. This must have been what Mathematica was trying to tell you with the interval.
Use a similar trick when working on b(n). For any n, b(n) = 2^n or b(n) = 2^(-n). You can see from this that the function r(n) will alternate between approaching 0 or approaching infinity, depending on which version of b(n) is in the numerator, and you can also see that p(n) will alernate between 1/2 and 2.
These proofs are far from rigorous so you still have quite a bit of proving to do, but this should get you started.
First transform all of the sec's and tan's into their sin and cos equivalents:
From here you can factor the numerator and match the original expression to one of your choices.
Does it end at ...979899, or is there some new pattern after that?
Well, as stated this is trivial to prove. For any positive integer n add the numbers -(n - 1) + -(n - 2) + ... + -1 + 0 + 1 + ... + (n - 2) + (n - 1) + n. Then every positive number less than n is cancelled by it's negative counterpart and the sum of the series is n. This works for any positive n, even the powers of two, and can be easily reversed for negative values of n.
Assuming that you meant a sum of consecutive positive integers, consider the series n - k, n - (k - 1), n - (k - 2), ..., n - 1, n, n + 1, ..., n + (k - 1), n + k. Adding these together cancels out all of the values except the n's, of which there are 2k + 1. The sum of those numbers will be n(2k + 1) = S, where n and k can be any positive integers. If S is a composite number that is not a power of 2, then let 2k + 1 be one of it's odd prime factors and n represent the rest of its prime factors. If S is prime then let n = 1 and 2k + 1 = S.
In cases where 2k + 1 > 2n the series will stretch into the negatives, but since those negatives will cancel with their positive counterparts we can still construct a series of positive integers only that will add up to S. For example, for S = 5 we would have n = 1 and k = 2, and our series would be -1 + 0 + 1 + 2 + 3. However, the -1 and 1 cancel out, and the 0 can be ignored, so our series will be 2 + 3. This also shows us why we can't find any such series for a power of 2: 2k + 1 is always odd, and powers of 2 have no odd prime factors.
I'd be wary of using that one line solution, since it can be interpretted two different ways by the compiler. If it evaluates right to left then it will set x = y first, then subtract that from x + y, which is now y + y, and you'll end up with y = y and x = y. Of course, I realize that you aren't suggesting actually using such code, but I thought that should be pointed out.
You're very close, just leave out the 1 and use the ceiling function (I can't tell from your post if you were using the ceiling or floor function):
int x = a;
int y = b;
x = x + y;
y = x - y;
x = x - y;
You assumed that the sets that you were working with were sets of numbers, but that is never explicitly stated in the question. They could be sets of words, or computer brands, or people. Therefore you can't use Cantor's Diagonal Argument, because it only works with sets of numbers.
Consider how the expression looks for a given n. In expanded form it will look like this:
Notice that the every fraction is less than or equal to 1. This tells us that for a given n, we know that n!/n^n <= 1/n, because 1/n is a member of the expanded expression and it is being multiplied by other fractions that are less than 1. Then you point out that 1/n tends to 0 as n tends to infinity, and qed.
Combine the logarithms:
ln(a) - ln(b) = ln(a / b)
So in this case you'll get
ln(5x+4) - ln(4x+5) = ln( (5x+4) / (4x+5) )
Now you can take the limit as x approaches infinity and get a meaningful answer.