Problem of the Day The graph of a function f is shown above.Which of the following statements about f is false? A) f is continuous at x = a B) f has a relative maximum at x = a C) x = a is in the domain of f D) lim f(x) is equal to lim f(x) E) lim f(x) exists x a + x a - x a a

Problem of the Day The graph of a function f is shown above.Which of the following statements about f is false? A) f is continuous at x = a B) f has a relative maximum at x = a C) x = a is in the domain of f D) lim f(x) is equal to lim f(x) E) lim f(x) exists x a + x a - x a a

You have 1 minute to sum all the integers from 1 to 100.

Mathematicians needed a concise way for writing sums Sigma notation does this It uses the uppercase Greek letter sigma Σ

Sigma notation does this The sum of n terms a 1, a 2, a 3,..., a n equals Σ a i and means a 1 + a a n i = 1 n index of summation (i, j, k most of the time) upper bound of summation lower bound of summation (any interger < upper bound) Σ j 3 = j = 2 5

Properties of Summation Σ Ka i = i = 1 n K Σ a i i = 1 n Σ (a i ± b i ) = i = 1 n Σ a i + i = 1 n Σ b i i = 1 n

Summation Formulas (page 260) Σ c = i = 1 n cn Σ i = i = 1 n n(n + 1) 2 Σ i 2 = i = 1 n n(n + 1)(2n + 1) 6 Σ i 3 = i = 1 n n 2 (n + 1) 2 4

Evaluating a Sum Σ i + 1 n 2 i = 1 n Evaluate for n = 10, 100, 1000 Σ i + 1 i = 1 n 1 n 2 Factor out constant (property 1)

Evaluating a Sum Σ i + 1 n 2 i = 1 n Evaluate for n = 10, 100, 1000 Σ i + 1 i = 1 n 1 n 2 Factor out constant (property 1) Split apart the sum (property 2) Σ i i = 1 n 1 n 2 Σ 1 i = 1 n + ( (

Evaluating a Sum Σ i + 1 n 2 i = 1 n Evaluate for n = 10, 100, 1000 Σ i + 1 i = 1 n 1 n 2 1 n 2 n(n + 1) + n 2 ( ( Factor out constant (property 1) Split apart the sum (property 2) Summation formulas n + 3 2n Simplify Σ i i = 1 n 1 n 2 Σ 1 i = 1 n + ( (

n + 3 2n as n approaches ∞ what does the sum approach? what do you get when n = 10, 100, 1000?

n + 3 2n as n approaches ∞ what does the sum approach? lim n ∞ n + 3 = 1 2n 2 what do you get when n = 10, 100, 1000?

You have 1 minute to sum all the integers from 1 to x 101 = (because you did each number twice) Σ i = (100)(101) = i = 1 n

In truth, Carl Friedrich Gauss ( ) was asked by his teacher to sum the integers from 1 to 100. When he had the correct answer in only a few moments, the teacher was astonished. He developed the method mentioned on the previous screen.

Problems 1& 3 on your homework are to be written out (do not use properties and formulas)

Attachments