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Old 2016-07-28, 02:12   #1
jvang
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Default Polynomial Problem

I'm in an Algebra II class, and the current topic is polynomials. I was working on a problem, but I have no idea where to begin.

Suppose \(f\) is a polynomial such that \(f(0)=47, f(1)=32, f(2)=-13,\) and \(f(3)=16.\) What is the sum of the coefficients of \(f\)?

Clearly, to find the sum of the coefficients, I need to find the function \(f\). I'm not sure how to find \(f\) given a set of points. One potential option I saw on the Internet was polynomial interpolation, but the Wikipedia article was very complicated. Is there a simpler explanation of this technique, if it is the right approach? If not, what is the right approach to this problem?

Thanks!
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Old 2016-07-28, 04:14   #2
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f(x)=a*x^4+b*x^3+c*x^2+d*x+47, since there are 4 roots assuming a single variable.

Substitute the root given for x in four separate equations then solve for the coefficients.
Check all roots via substitution once the function is obtained then sum your coefficients.

This is how I would start.

Last fiddled with by jwaltos on 2016-07-28 at 04:21
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Old 2016-07-28, 04:21   #3
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Quote:
Originally Posted by jwaltos View Post
f(x)=a*x^4+b*x^3+c*x^2+d*x+47, since there are 4 roots assuming a single variable.
Off by 1, I think.
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Old 2016-07-28, 04:24   #4
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Quote:
Originally Posted by jwaltos View Post
f(x)=a*x^4+b*x^3+c*x^2+d*x+47, since there are 4 roots assuming a single variable.
This is how I would start.
Why are there four roots? I see evidence for two. With 4 ordered pairs given, create a polynomial with four unknown coefficients, including the constant term. Since f(0) = 47, 47 is the constant term. That leaves three unknown coeffs in a cubic (degree 3).
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Old 2016-07-28, 04:24   #5
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Probably, Jack Daniels, Jim Beam and Johnny Walker were all giving advice.
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Old 2016-07-28, 04:43   #6
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Surely f(1) = 32 = the sum of the coefficients of f(x) for a polynomial of any degree, as all xn = 1 so we are reduced to an*1 + an-1*1 + an-2*1 + ........ + a0 = 32
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Old 2016-07-28, 05:27   #7
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Quote:
Originally Posted by Antonio View Post
Surely f(1) = 32 = the sum of the coefficients of f(x) for a polynomial of any degree, as all xn = 1 so we are reduced to an*1 + an-1*1 + an-2*1 + ........ + a0 = 32
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Old 2016-07-28, 13:44   #8
jvang
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I think I get the idea. So the polynomial must be of the form \(ax^3+bx^2+cx+d,\) since there are 4 given points, and one of them is the \(y\) intercept.

So if \(x=1,\) all of the \(x\) values stay the same since \(1\) to any power is \(1\). However, if \(f(1)=32,\) then that is \(a+b+c+47,\) so I would need to subtract \(47\) from \(a+b+c\) to get \(-7\) as the answer.

Either that or I overthought this problem.

Last fiddled with by jvang on 2016-07-28 at 13:47
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Old 2016-07-28, 17:08   #9
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Whatever the polynomial, f(1) is the sum of its coefficients. You weren't given the degree of the polynomial, because you don't need it.

(because if f(x) = sum a_i x^i, then f(1) = sum a_i because all the powers of 1 are 1)

Equally f(-1) is the alternating sum of its coefficients, so f(-1)+f(1)/2 is the sum of the even-numbered coefficients.

This turns into a root into Fourier analysis: think about $f(\omega)$ where $\omega$ is a root of unity.

Last fiddled with by fivemack on 2016-07-28 at 17:09
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Old 2016-07-29, 03:35   #10
LaurV
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@OP: Are you sure the problem asks for the coefficient's sum? As the problem is given, it is either tricky (because you only need to know f(1) and nothing else, no information about the degree is necessary, and the other 3 values given are futile) or either it wanted to ask for the sum of the roots. In this case, you need all 4 values, make s system with 4 equations and 4 variables, check if they are not linear (i.e. the system may not be independent, in such case a lower-degree poly will fit) and/or use Vieta's formulas to get the sum.

Last fiddled with by LaurV on 2016-07-29 at 03:36 Reason: link
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Old 2016-07-29, 14:46   #11
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Quote:
Originally Posted by LaurV View Post
@OP: Are you sure the problem asks for the coefficient's sum?
I'm pretty sure that's what it's asking. Otherwise, how do you know it's a quartic f(x) and not a quintic, say f(x) * (x - 7)?
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