Talking With Jon - Episode 16 Clip - You choose the new band member
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3 years ago
Talking With Jon - Episode 16 Clip - You choose the new band member
For exponent 1, (x+y)1 is x+y. For exponent 2, (x+y)2 is (x+y)(x+y), which forms terms as follows. The first factor supplies either an x or a y; likewise for the second factor. Thus to form x2, the only possibility is to choose x from both factors; likewise for y2. However, the xy term can be formed by x from the first and y from the second factor, or y from the first and x from the second factor; thus it acquires a coefficient of 2. Proceeding to exponent 3, (x+y)3 reduces to (x+y)2(x+y), where we already know that (x+y)2= x2+2xy+y2. Again the extremes, x3 and y3 arise in a unique way. However, the term x2y is either 2xy times x or x2 times y, for a coefficient of 3; likewise xy2 arises in two ways, summing the coefficients 1 and 2 to give 3.
This suggests an induction. Thus for exponent n, each term has total degree (sum of exponents) n, with n−k factors of x and k factors of y. If k is 0 or n, the term arises in only one way, and we get the terms xn and yn. If k is neither 0 nor n, then the term arises in two ways, from xn-k-1yk × x and from xn-kyk-1 × y. For example, x2y2 is both xy2 times x and x2y times y, thus its coefficient is 3 (the coefficient of xy2) + 3 (the coefficient of x2y). This is the origin of Pascal's triangle, discussed below.Talking With Jon - Episode 16 Clip - You choose the new band member
For exponent 1, (x+y)1 is x+y. For exponent 2, (x+y)2 is (x+y)(x+y), ...all »Talking With Jon - Episode 16 Clip - You choose the new band member
For exponent 1, (x+y)1 is x+y. For exponent 2, (x+y)2 is (x+y)(x+y), which forms terms as follows. The first factor supplies either an x or a y; likewise for the second factor. Thus to form x2, the only possibility is to choose x from both factors; likewise for y2. However, the xy term can be formed by x from the first and y from the second factor, or y from the first and x from the second factor; thus it acquires a coefficient of 2. Proceeding to exponent 3, (x+y)3 reduces to (x+y)2(x+y), where we already know that (x+y)2= x2+2xy+y2. Again the extremes, x3 and y3 arise in a unique way. However, the term x2y is either 2xy times x or x2 times y, for a coefficient of 3; likewise xy2 arises in two ways, summing the coefficients 1 and 2 to give 3.
This suggests an induction. Thus for exponent n, each term has total degree (sum of exponents) n, with n−k factors of x and k factors of y. If k is 0 or n, the term arises in only one way, and we get the terms xn and yn. If k is neither 0 nor n, then the term arises in two ways, from xn-k-1yk × x and from xn-kyk-1 × y. For example, x2y2 is both xy2 times x and x2y times y, thus its coefficient is 3 (the coefficient of xy2) + 3 (the coefficient of x2y). This is the origin of Pascal's triangle, discussed below.«
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