I've already vaguely given you the answer

Algebra Level 5

Let $g(x)$ denote a quintic $\big(5^\text{th}$ -degree $\big)$ polynomial such that $g(n) = 2^n$ for $n = 1,2,3,4,5$ and $g(x)$ has a constant term of $5!$ .

Find $g(6)$ .

Bonus: Generalize this.

The answer is -56.

3 solutions

Otto Bretscher
Feb 27, 2016

We know that $\sum_{k=0}^n(-1)^k{n \choose k}g(k)=0$ for any polynomial $g(x)$ of degree $<n$ . For $n=6$ this gives $g(6)=-g(0)+\sum_{k=1}^{5}(-1)^{k+1}{6 \choose k}2^k=-5!+2^6=\boxed{-56}$

Moderator note:

Great explanation for how to easily determine the polynomial that approximates an exponential function on several initial values.

Otto, I believe your solution uses quite some tricks, which are not trivial. For example:

$\sum_{k=1}^{n}(-1)^{k+1}{n+1 \choose k}2^k = 2^{n+1} (-1)^{n+1} +1+(-1)^n$

or: $\sum_{k=0}^n(-1)^k{n \choose k}g(k)=0$ for any polynomial $g(x)$ of degree $\lt n$ , at least not for me, it took me a bit to actually prove this myself. Bringing first equation into a binomial form - it's inviting one to do so - works in fact quite cumbersomely.

Kalis To - 5 years, 3 months ago

You are right that one of the equations is non-trivial; it was discussed here . Look at @Alan Yan 's elegant solution!

The other equation is just playing around with the binomial expansion of $(2-1)^n$ .

Otto Bretscher - 5 years, 3 months ago

I think you meant to link this problem instead...

Pi Han Goh - 5 years, 3 months ago

Either works... the link I provided discusses the case $n=2016$ , but the two excellent proofs by @Aareyan Manzoor and @Alan Yan cover the general case.

I will post a similar problem that pushes this interesting problem a bit further.

Otto Bretscher - 5 years, 3 months ago

The generalization works here as $f(x)= 2^{x}[(1+x); at x= 1]$ but how does it hold true for any general polynomial of degree $<n$ ?

And how is first statement is arrived at is by taking the derivative of a $<n$ degree polynomial $n$ times, What does it mean to substitute $x= 6$ in the equation ?

Vishal Yadav - 4 years, 2 months ago

Zk Lin
Feb 29, 2016

This problem can be tackled by The Method of Differences since the " $n$ "s are consecutive. Here is the difference table for this question.

$n$	$g(n)$	$D_{1}(n)$	$D_{2}(n)$	$D_{3}(n)$	$D_{4}(n)$	$D_{5}(n)$
$0$	$120$	$-118$	$120$	$-118$	$120$	$-118$
$1$	$2$	$2$	$2$	$2$	$2$	$-118$
$2$	$4$	$4$	$4$	$4$	$-116$
$3$	$8$	$8$	$8$	$-112$
$4$	$16$	$16$	$-104$
$5$	$32$	$-88$
$6$	$\boxed{-56}$

Did the same way.

Anupam Nayak - 5 years, 3 months ago

Department 8
Feb 27, 2016

Consider a function $f(x)$ such that $f(x)=g(x)+2^x$ where $g(x)=a(x-1)(x-2)(x-3)(x-4)(x-5)$ where $a$ is a real number. First note that the constant term of $g(x)$ is -5! on multiplying but we are given that the constant term was 5! so $a$ must be equal to -1. So $g(x)=-(x-1)(x-2)(x-3)(x-4)(x-5)$ , now plugging $x=6$ in $f(x)$ we get $-5!+32=\boxed{-56}$ .

Thanks to Otto sir:

Here is a way to fix it: With your notation, make $f(x)=g(x)+\sum_{k=0}^{5}{x\choose k}$ $=a(x-1)(x-2)(x-3)(x-4)(x-5)$ , and choose $a$ so that $f(0)=5!$ as required.

@Pi Han Goh, Nice question!, @Otto Bretscher Sorry, if the solution is same as yours.

This is wrong. it is given that f(x) is a polynomial, but f(x) = g(x) + 2^x is not a polynomial

Pi Han Goh - 5 years, 3 months ago

Actually I wrote f(x) is a function

Department 8 - 5 years, 3 months ago

But adding 2^x to a polynomial doesn't give you a polynomial.

Amish Naidu - 5 years, 3 months ago

@Amish Naidu – Yes you are right as g(x) is a polynomial but when we add 2^x it becomes a function which is f(x). @Otto Bretscher is the solution right?

Department 8 - 5 years, 3 months ago

@Department 8 – Your f is actually the question's g, which is given to be a polynomial.

Amish Naidu - 5 years, 3 months ago

@Department 8 – I must confess that I don't understand your solution. As @Pi Han Goh and @Amish Naidu observe, you are finding $f(6)$ , but $f(x)$ is not a polynomial.

Otto Bretscher - 5 years, 3 months ago

@Otto Bretscher – I thought of changing the question so I just created a polynomial that is g(x). Now consider a function which satisfies the property as mentioned in the question that is f(x) then I think you understand what I mean the solution

Department 8 - 5 years, 3 months ago

@Department 8 – The key point is that the function you evaluate at 6 (whatever you name it) is not a polynomial as required in the problem. Also, its value at 0 isn't 5! as required in the problem.

Otto Bretscher - 5 years, 3 months ago

@Otto Bretscher – Oh now I understand what you mean to say thank you

Department 8 - 5 years, 3 months ago

@Department 8 – Here is a way to fix it: With your notation, make $f(x)=g(x)+\sum_{k=0}^{5}{x\choose k}$ $=a(x-1)(x-2)(x-3)(x-4)(x-5)$ , and choose $a$ so that $f(0)=5!$ as required.

Otto Bretscher - 5 years, 3 months ago

Refresh your memory about the definition of a Polynomial .

Calvin Lin Staff - 5 years, 3 months ago

Can you please explain the third and fourth line again. You have taken the constant term $-5!$ , but it is given as $5!$

Chirayu Bhardwaj - 5 years, 3 months ago

I've already vaguely given you the answer

The answer is -56.

3 solutions

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