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Pascal
Integration sans fundamental theorem of calculus?
Hope you all had a Merry Christmas yesterday. The challenge du jour:
Let n be an arbitrary fixed positive integer, and consider the right-hand Riemann sums of the function x^n obtained from a partition of [0, 1] into m equal intervals. These are:
[k=1, m]∑(1/m * (k/m)^n)
Using the fundamental theorem of calculus, it is of course trivial to prove that:
[m→∞]lim [k=1, m]∑(1/m * (k/m)^n) = [0, 1]∫x^n dx = x^(n+1)/(n+1) |[0, 1] = 1/(n+1).
Your mission, should you decide to accept it, is to prove the limit [m→∞]lim [k=1, m]∑(1/m * (k/m)^n) = 1/(n+1) _directly_, without using the fundamental theorem. Can you do it?
3 AnswersMathematics1 decade agoI have a slightly harder mathematical puzzle for you...?
Find constants a, b, c, and d such that:
4x³ - 3x + √3/2=a * (x-b) * (x-c) * (x-d)
Rules:
#1: Your factors must be exact. No numerical approximations allowed.
#2: Neither the imaginary constant, nor the square root of any negative number may appear in your answer.
#3: Your answer must be simple. Nothing more complicated than a trigonometric function may appear in your answer.
#4: Yes, this problem is solvable.
10 AnswersMathematics1 decade agoA mathematical experiment?
f(x) is a monic polynomial of degree 6. It has the following values:
f(0)=0
f(1)=1
f(2)=2
f(3)=3
f(4)=4
f(5)=5
Question 1: what is f(6)?
Question 2: do you think that other people are likely to get this question wrong?
5 AnswersMathematics1 decade agoWhat the heck does "pls" mean?
I see that word a lot after many of the questions in this section, and I'm wondering what it means. I know it doesn't mean please, because surely anyone who is going to ask people to take valuable time out of their day to do their homework for them would not be so ill-mannered as to mock all standards of politeness and decency by deliberately misspelling the word "please," so I assume that it must be some kind of internet abbreviation with which I am not familiar. So what does it mean?
17 AnswersMathematics1 decade agoWould Flatland be more widely remembered...?
Some of you may be familiar with the old classic "Flatland: A Romance of Many Dimensions" by Edwin Abbott (if you aren't, you can read it at http://xahlee.org/flatland/index.html ). It's really quite good, and an excellent introduction to higher-dimensional thinking, but a lot of people have never heard of it. So what I want to know is: do you think Flatland would be more widely remembered if it had been called "Shapes on a plane"?
3 AnswersOther - Entertainment1 decade agoDo people really believe this stuff?
And why? What possible mechanism do you imagine links the date of your birth to your fate in life?
22 AnswersHoroscopes1 decade agoI have a mathematical puzzle for you...?
Consider a ten-digit number with the following properties: each digit from 0-9 occurs in the number exactly once. The number itself is divisible by ten. If you remove the last digit, the nine-digit number so formed is divisible by nine. If you remove the last two digits, the eight-digit number so formed is divisible by eight. If you remove the last three digits, the seven digit number so formed is divisible by seven, and in general the n-digit number formed from the first n digits is divisible by n. Call any number with all of these properties Pascalian. Now the question:
Part A: List all the Pascalian numbers, if any exist.
Part B: Show, by human-verifiable proof, that such a list is exhaustive.
You may wish to make use of the following information in solving this problem: http://en.wikipedia.org/wiki/Divisibility_rule
Good luck.
3 AnswersMathematics2 decades ago