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All I have done is noted that
by the domain-splitting property of the Riemann integral. But since I've defined F(y) as
I can write your original integral as
.Differentiating and using the chain rule, we get (can't write apostrophes in LaTeX):
G'(x) = -sin(x)F'(cos(x)) - cos(x)F'(sin(x))
But what are F'(sin(x)) and F'(cos(x))?
First, put
so that we can write
Now, how might we find G'(x)?
I'm getting pi/8 without having to take any limits.
You might like to call the sum S and consider the integral
where Re(a) and Re(b) > 0.
He is trying to get you to notice that
for a certain f(x).
Bumping on an old topic is called necroposting, and there are good things and bad things there. It's good to necropost if the topic is relevant, e.g. a topic about a mathematician. Or irrelevant topics such as a topic in "Help Me!" that has been answered already (you already know the answer, right? So why need more answers?)
Can't quite see how such posts are necessarily irrelevant -- very often guests (like myself) are searching for solutions across the internet hoping to stumble upon a thread on some forum where someone is having the same problem with a question, hoping that someone has answered it. If it has already been answered, sometimes it is good to get another perspective on things; perhaps see an alternative solution that unpicks some deeper insight into the problem that we'd never thought of before. I believe someone 'bumped' my old "Laplace Transform of tan(t)" thread a few years later, both here and on another forum, to add something that I found interesting to read about.
Did they disappear, or were they always missing?
This thread is a bit of a mess: http://mathisfunforum.com/viewtopic.php?id=14826
[sup]The sup[/sup] [sub]and sub[/sub] tags don't work.
The fancy apostrophes are missing from older posts, and the text 'near' LaTeX code is affected. See here: http://mathisfunforum.com/viewtopic.php?id=14950
Apostrophes no longer work in LaTeX, and presumably true for other characters:
I have also been noticing the missing character issues lately -- especially the 'fancier' (i.e. curled) apostrophes that have suddenly disappeared from everyone's posts. The sup and sub tags no longer work (which makes some older posts very tedious to read!). Some characters no longer work in LaTeX, and any (unrendered) sentence written on the same line as LaTeX text ends up changing its font style. Not to mention that in the hide tags, a lot more characters don't work -- even ones like simple exponents fail to show.
3 years late, but might be useful to people looking up the same thing.
First, note that your second partition is a refinement of the first with one extra point -- namely, y. I prove the general case for any two partitions P and P* where P* is a refinement of P, so that your result follows as an immediate corollary.
Let P be a partition of [a,b], with U(f,P) and L(f,P) being the upper and lower Darboux sums of f with respect to the partition P. Write the original partition P as
and the partition P* as
where x* is the extra point. Then:
But note that
and
substituting, we see that the RHS of the inequality is greater than or equal to 0, thus
.The inequality
is done similarly. Thus, by repeated application of this proof, this holds true for any refinement of P. Using this result, let
be a common refinement of P and Q. Then:
and
but, obviously
hence
QED.
Hi;
This is what I get plugging into that formula:
This is the correct answer. However, for the benefit of future viewers, I demonstrate a quick way to tackle such problems (without the need for memorising formulae). Recall the following theorem:
Let f be locally Riemann integrable over I, and let F: I -> R be an indefinite integral of f. Then:
(i) F is continuous on I;
(ii) F is differentiable at each interior point c ∈ I at which f is continuous, and satisfies F'(c) = f(c);
(iii) If f is continuous on I, then F is a primitive of f.
This theorem allows us to perform the following manipulations on Anakin's problem.
We let
, and put. Then:So by the chain rule:
.hi Anakin,
In other words you can substitute into the integral process any alternative variable.
Not strictly true -- you'd need the 't' to act as the dummy variable.
I think it should be OK, the period looked a little odd, anyway -- thanks for the offer, though.
Do you think it would be better for continuity purposes?
The real reason was that I forgot to put it on because I'm used to it being filled in automatically. It doesn't matter though.
Probably less so, but I'm not bothered if they do. Let them find it.
I am amazed that they actually would look here. If they find this then the secret is out.
If you google "FP1 linear interpolation formula" -- which a lot of A-level students would do -- the thread comes up as the second result. It was just one person who mentioned it, and he wasn't sure if zetafunc actually went to UCL, just that we did the same A-levels. The reason is that there are only 50 people in the country who did one of the subjects I mentioned, and even fewer with my particular subject combination. I think my 'secret' is fairly safe, however -- don't think it'll compromise anything.
I've split my post into two parts. Feel free to read the more elaborate explanation if you'd like to learn a little more about indefinite integrals.
Quick reason: The constant is needed because it is still technically a possibility for the function. For instance, consider the integral:
.Clearly, one solution is x². Another is x² + 2. Another is x² - 5. They all differentiate to form 2x. Hence, we add a 'constant of integration' to account for this.
More elaborate explanation:
To add to ShivamS' answer, a slightly more precise way of defining an indefinite integral is as follows. Let f be locally Riemann integrable over I. Then, an indefinite integral of f is a function F: I -> R defined by
for some a ∈ I. From the domain-splitting property for integrals, it follows that two indefinite integrals differ by a constant. Further, it follows that:
(i) F is continuous on I;
(ii) F is differentiable at each interior point c ∈ I at which f is continuous, and satisfies F'(c) = f(c);
(iii) If f is continuous on I, then F is clearly a primitive of f.
Such a primitive is not unique, because one can always add a constant to F.
(As an aside, you might like to know that (ii) allows one to differentiate integrals!)
In real life?!
Yes -- I denied it, but someone from uni asked about what I did at A-level. My answer was pretty distinctive and they matched what I posted in that thread, which led them to think I was zetafunc. But I'm not too bothered by it, not the end of the world if they know. It's a year with 200+ mathematicians, a lot of whom regularly look at maths forums. It's mostly The Student Room that they use, but I guess 1 person does look at this forum on occasion.
My username did have a trailing decimal point, but I decided to be super-adventurous and live dangerously by omitting it!
It is nice for a while, but then when you're doing absolutely nothing at all, just sitting in the kitchen drinking booze during the small hours, you feel so... filthy. Actually, that happens during term time too. There's a word we have for that over here -- it's a play on the word 'procrastination' but it's a bit too rude to post here. Yep, I go to one -- you might be able to find out which one if you looked through the FP1 thread, not sure if I should mention it though seeing as I've already had one person come up to me and ask me if I'm zetafunc!
Not sure if I have a favourite colour -- but let's go with green (as in, basil green).
Yes, just finished my first year exams, got a month left living with my flatmates in halls to chill and do sweet sod all.
Effects of Row/Column Operations on Determinants
Let the matrix B be obtained from the matrix A by applying the row operation e. Then:
.These results hold true for the corresponding column operations.
Triangular Matrices
Let A be an n x n matrix. Then, A is lower triangular if a_ij = 0 whenever i < j. Similarly, A is upper triangular if a_ij = 0 whenever i > j. The determinant of a triangular matrix is given by the product of the diagonal entries, i.e.
.Hi everyone, thanks for the welcome.
I was going to continue posting as a guest, but apparently guests can't post anymore... and for a while I couldn't register either!
Determinant of an n x n Matrix
Let A be an n x n matrix with entries (a_ij). Then, the determinant of A is given by
where S_n is the group of permutations of {1, 2, ... , n}.