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June 9th, 2016, 01:51 AM   #1
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Polynomial Expansion

Find the coefficient of x^2 in the polynomial:

(1-x)(1+2x)(1-3x)....(1+14x)(1-15x)
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June 9th, 2016, 05:36 AM   #2
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Let's list out all the coefficients we need to add up. Every combination of two x-terms will be included in our sum, so we'll have 15C2 = 105 numbers to sum up.

If we multiply the coefficient -1 by other coefficients, we have:
-2, 3, -4, 5, -6, 7, -8, 9, -10, 11, -12, 13, -14, 15
There are 14 numbers and the sum is 7.

If we multiply the coefficient 2 by other coefficients, excluding the combination we've already counted, we have:
-6, 8, -10, 12, -14, 16, -18, 20, -22, 24, -26, 28, -30
There are 13 numbers and the sum is 12 plus a stray -30.

If we multiple the coefficient -3 by other coefficients, excluding the combinations we've covered, we have:
-12, 15, -18 ... -42, 45
There are 12 numbers, the sum being 18.

By the same vein, we also need to add:
10 / 2 * 4 - 15 * 4
10 / 2 * 5
8 / 2 * 6 - 15 * 6
8 / 2 * 7
6 / 2 * 8 - 15 * 8
6 / 2 * 9
4 / 2 * 10 - 15 * 10
4 / 2 * 11
2 / 2 * 12 - 15 * 12
2 / 2 * 13
0 / 2 * 14 - 15 * 14

Sum up all these numbers, and you've got the required answer -588.
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June 9th, 2016, 05:46 AM   #3
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The correct answer is one among the following:

a) -121
b) -191
c) -255
d) -291
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June 9th, 2016, 06:01 AM   #4
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The correct answer is one among the following:

a) -121
b) -191
c) -255
d) -291
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June 9th, 2016, 06:07 AM   #5
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Quote:
Originally Posted by Sashank View Post
The correct answer is one among the following:

a) -121
b) -191
c) -255
d) -291
Are you sure, or did you perhaps mean to write something else in the original question? I input the full expression into Wolfram Alpha and the coefficient of x^2 was also -588...
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June 9th, 2016, 07:25 AM   #6
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Good gracious me..this was a direct question from one of the entrance exams of a very reputed college. How could they make such a blunder...

Anyways thanks a lot
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June 9th, 2016, 07:33 AM   #7
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But I still ain't getting this part:

By the same vein, we also need to add:
10 / 2 * 4 - 15 * 4
10 / 2 * 5
8 / 2 * 6 - 15 * 6
8 / 2 * 7
6 / 2 * 8 - 15 * 8
6 / 2 * 9
4 / 2 * 10 - 15 * 10
4 / 2 * 11
2 / 2 * 12 - 15 * 12
2 / 2 * 13
0 / 2 * 14 - 15 * 14

Sum up all these numbers, and you've got the required answer -588.


Would you please explain
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June 9th, 2016, 07:45 AM   #8
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Quote:
Originally Posted by Sashank View Post
But I still ain't getting this part:

By the same vein, we also need to add:
10 / 2 * 4 - 15 * 4
10 / 2 * 5
8 / 2 * 6 - 15 * 6
8 / 2 * 7
6 / 2 * 8 - 15 * 8
6 / 2 * 9
4 / 2 * 10 - 15 * 10
4 / 2 * 11
2 / 2 * 12 - 15 * 12
2 / 2 * 13
0 / 2 * 14 - 15 * 14

Sum up all these numbers, and you've got the required answer -588.


Would you please explain
We can't simply add up all the numbers directly, since that would be too time-consuming, so we have to find patterns and to do it in a systematic manner. Let's call the coefficient created by multiplying two coefficients in the original expression x and y as $\displaystyle C_{x, y}$. We can deduce from the above examples that

$\displaystyle \sum_{i = x + 1}^{15} C_{x, i} = \begin{cases}
\frac{15 - |x|}{2} \times |x|, &\mbox{if } i \mbox{ is odd} \\
\frac{15 - |x| - 1}{2} \times |x| - 15 \times |x|, &\mbox{if } i \mbox{ is even} \end{cases}$

which allows us to deduce the rest.

Edit:
Or maybe, in a more informal manner:
When i is odd, count the number of positive-negative pairs, divide the number of pairs by two and multiply by x (since this is the difference between the elements of each pair)
When i is even, count the number of positive-negative pairs (excluding the last item), divide the number of pairs by two and multiply by x (again, the difference between the elements of each pair), and finally add in the final item times x.
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Last edited by 123qwerty; June 9th, 2016 at 08:11 AM.
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June 19th, 2016, 05:06 AM   #9
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Use Newton's identities to evaluate that elementary symmetric polynomial

$\displaystyle p_2=e_1p_1-2e_2,\sum ab=\frac{1}{2}(\sum a)^2-\frac{1}{2}\sum a^2$

$\displaystyle \sum a=\sum_{k=1}^{15} (-1)^k k=-1+\sum_{k=1}^7 [2k-(2k+1)]=-8$

$\displaystyle \sum a^2=\sum_{k=1}^{15} k^2=\sum_{k=1}^{15} (2C_k^2+C_k^1)=2C_{16}^3+C_{16}^2=1240$

$\displaystyle \sum ab=\frac{1}{2}(\sum a)^2-\frac{1}{2}\sum a^2=\frac{1}{2}(64-1240)=-588$
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