# Download PDF by Julian Lowell Coolidge: An introduction to mathematical probability

By Julian Lowell Coolidge

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**Example text**

Xk } be a uniform partition of the interval, with each subinterval having length x and with k = 2 being even. The integral b f (x) dx a 26 2. THE RIEMANN INTEGRAL is approximated by the expression x {f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + . . 2f (xk−2 ) + 4f (xk−1 ) + f (xk )} . 3 It can be determined that, if |f (iv) (x)| ≤ M, then the error in applying Simpson’s method is not greater than M(b − a)5 . 13 with the same partition {0, 1/4, 1/2, 3/4, 1}, using Simpson’s Rule. It is straightforward to calculate, since f (x) = 1/(1 + x 2 ), that f (iv) (x) = 120x 4 − 240x 2 + 24 .

Let h(x) = x · sin(1/x) 0 x =0 x=0 if if on the interval [−1, 1]. Is h Riemann integrable? Why or why not? 5. The integral ∞ 0 sin x dx x is what is called an improper Riemann integral. But we may make sense of it by evaluating N lim N →+∞ 1/N sin x dx . x Do so. 6. Provide the details of the assertion that, if f is Riemann integrable on the interval [a, b], then for any > 0 there is a δ > 0 such that if P is a partition of mesh less than δ then sup f − inf f j Ij Ij j < . [Hint: Follow this scheme: Given > 0, choose δ > 0 as in the definition of the integral.

THE LEBESGUE SPACES Lp 51 (b) If neither f Lp = 0 nor g L1 = 0 and instead either f Lp = ∞ or g Lq = ∞, then the right-hand side of Hölder’s inequality is infinity. So the inequality is trivially satisfied. (c) If p = 1, q = ∞ or p = ∞, q = 1, then the inequality is trivially confirmed as follows (in the first instance): R f (x) · g(x) dx ≤ ≤ R R |f (x)| · |g(x)| dx |f (x)| · g = g L∞ · = g L∞ · f R L∞ dx |f (x)| dx L1 . Thus, we may consider 1 < p < ∞, 1 < q < ∞, and we may assume that f Lp and g Lq are both finite.

### An introduction to mathematical probability by Julian Lowell Coolidge

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