Proof that the Sum of Products is Zero
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This is part of my answer to a question of a contest. But I ask about a really specific part of my answer and I can show the behaviour, but somehow don't know how to proof it generally. This is probably because of my missing experience with the summation symbol. The proof should be fairly simple, since it's right there but I don't know how to handle it.
There is a row of b Zeros for Example: $ Z_q = 0; Z_{q+1} = 0 ... Z_{q + b - 1} = 0$
$q+b-1 = p quad S(p) = 0 quad if quad p < 2b $
$$S(n) = sum_{m=0}^n Z_m * Z_{n - m}$$
When I write down the Sums, it seems rather obvious that this must be true:
$S(p) = Z_{0}*Z_{p} + Z_{1}*Z_{p-1} + Z_{2}*Z_{p-2} ... + Z_{p} * Z_{p - p}$ $For;example:;S(5) = Z_0 * Z_5 + Z_1 * Z_4 + Z_2 * Z_3 + Z_3*Z_2 + Z_4* Z_1 + Z_5*Z_0$
$If ; b = 2 ; then ; 5 > 2*b,; but ; if ; b ge 3 ; then ,5 < 2 * b land S(5) = 0 qquadqquadqquadqquadqquadqquad$
But I somehow fail to proof it generally.
proof-verification summation proof-writing
asked Jan 22 at 13:56
KarlTheGreatKarlTheGreat
12
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add a comment |
$begingroup$
This is part of my answer to a question of a contest. But I ask about a really specific part of my answer and I can show the behaviour, but somehow don't know how to proof it generally. This is probably because of my missing experience with the summation symbol. The proof should be fairly simple, since it's right there but I don't know how to handle it.
There is a row of b Zeros for Example: $ Z_q = 0; Z_{q+1} = 0 ... Z_{q + b - 1} = 0$
$q+b-1 = p quad S(p) = 0 quad if quad p < 2b $
$$S(n) = sum_{m=0}^n Z_m * Z_{n - m}$$
When I write down the Sums, it seems rather obvious that this must be true:
$S(p) = Z_{0}*Z_{p} + Z_{1}*Z_{p-1} + Z_{2}*Z_{p-2} ... + Z_{p} * Z_{p - p}$ $For;example:;S(5) = Z_0 * Z_5 + Z_1 * Z_4 + Z_2 * Z_3 + Z_3*Z_2 + Z_4* Z_1 + Z_5*Z_0$
$If ; b = 2 ; then ; 5 > 2*b,; but ; if ; b ge 3 ; then ,5 < 2 * b land S(5) = 0 qquadqquadqquadqquadqquadqquad$
But I somehow fail to proof it generally.
proof-verification summation proof-writing
asked Jan 22 at 13:56
KarlTheGreatKarlTheGreat
12
$endgroup$
add a comment |
$begingroup$
This is part of my answer to a question of a contest. But I ask about a really specific part of my answer and I can show the behaviour, but somehow don't know how to proof it generally. This is probably because of my missing experience with the summation symbol. The proof should be fairly simple, since it's right there but I don't know how to handle it.
There is a row of b Zeros for Example: $ Z_q = 0; Z_{q+1} = 0 ... Z_{q + b - 1} = 0$
$q+b-1 = p quad S(p) = 0 quad if quad p < 2b $
$$S(n) = sum_{m=0}^n Z_m * Z_{n - m}$$
When I write down the Sums, it seems rather obvious that this must be true:
$S(p) = Z_{0}*Z_{p} + Z_{1}*Z_{p-1} + Z_{2}*Z_{p-2} ... + Z_{p} * Z_{p - p}$ $For;example:;S(5) = Z_0 * Z_5 + Z_1 * Z_4 + Z_2 * Z_3 + Z_3*Z_2 + Z_4* Z_1 + Z_5*Z_0$
$If ; b = 2 ; then ; 5 > 2*b,; but ; if ; b ge 3 ; then ,5 < 2 * b land S(5) = 0 qquadqquadqquadqquadqquadqquad$
But I somehow fail to proof it generally.
proof-verification summation proof-writing
asked Jan 22 at 13:56
KarlTheGreatKarlTheGreat
12
$endgroup$
This is part of my answer to a question of a contest. But I ask about a really specific part of my answer and I can show the behaviour, but somehow don't know how to proof it generally. This is probably because of my missing experience with the summation symbol. The proof should be fairly simple, since it's right there but I don't know how to handle it.
There is a row of b Zeros for Example: $ Z_q = 0; Z_{q+1} = 0 ... Z_{q + b - 1} = 0$
$q+b-1 = p quad S(p) = 0 quad if quad p < 2b $
$$S(n) = sum_{m=0}^n Z_m * Z_{n - m}$$
When I write down the Sums, it seems rather obvious that this must be true:
$S(p) = Z_{0}*Z_{p} + Z_{1}*Z_{p-1} + Z_{2}*Z_{p-2} ... + Z_{p} * Z_{p - p}$ $For;example:;S(5) = Z_0 * Z_5 + Z_1 * Z_4 + Z_2 * Z_3 + Z_3*Z_2 + Z_4* Z_1 + Z_5*Z_0$
$If ; b = 2 ; then ; 5 > 2*b,; but ; if ; b ge 3 ; then ,5 < 2 * b land S(5) = 0 qquadqquadqquadqquadqquadqquad$
But I somehow fail to proof it generally.
proof-verification summation proof-writing
proof-verification summation proof-writing
asked Jan 22 at 13:56
KarlTheGreatKarlTheGreat
12
asked Jan 22 at 13:56
KarlTheGreatKarlTheGreat
12
edited Jan 22 at 14:45
KarlTheGreat
asked Jan 22 at 13:56
KarlTheGreatKarlTheGreat
12
asked Jan 22 at 13:56
KarlTheGreatKarlTheGreat
12
asked Jan 22 at 13:56
KarlTheGreatKarlTheGreat
12
12
add a comment |
add a comment |
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S(p) is the sum of p + 1 products. Since the first b products are equal to zero and the last b products are equal to zero, all products are equal to zero if $(p+1) - 2b le 0 to p + 1 le 2b to p < 2b$
answered Jan 22 at 15:18
KarlTheGreatKarlTheGreat
12
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$begingroup$
S(p) is the sum of p + 1 products. Since the first b products are equal to zero and the last b products are equal to zero, all products are equal to zero if $(p+1) - 2b le 0 to p + 1 le 2b to p < 2b$
answered Jan 22 at 15:18
KarlTheGreatKarlTheGreat
12
$endgroup$
add a comment |
$begingroup$
S(p) is the sum of p + 1 products. Since the first b products are equal to zero and the last b products are equal to zero, all products are equal to zero if $(p+1) - 2b le 0 to p + 1 le 2b to p < 2b$
answered Jan 22 at 15:18
KarlTheGreatKarlTheGreat
12
$endgroup$
add a comment |
$begingroup$
S(p) is the sum of p + 1 products. Since the first b products are equal to zero and the last b products are equal to zero, all products are equal to zero if $(p+1) - 2b le 0 to p + 1 le 2b to p < 2b$
answered Jan 22 at 15:18
KarlTheGreatKarlTheGreat
12
$endgroup$
S(p) is the sum of p + 1 products. Since the first b products are equal to zero and the last b products are equal to zero, all products are equal to zero if $(p+1) - 2b le 0 to p + 1 le 2b to p < 2b$
answered Jan 22 at 15:18
KarlTheGreatKarlTheGreat
12
answered Jan 22 at 15:18
KarlTheGreatKarlTheGreat
12
answered Jan 22 at 15:18
KarlTheGreatKarlTheGreat
12
answered Jan 22 at 15:18
KarlTheGreatKarlTheGreat
12
12
add a comment |
add a comment |
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