Polynomial Reciprocal on Integers

Consider $g(x)= xf(x)-1$ . We know that for all $n \in \{1, 2, 3, 4, 5\}$ , $f(n)= \frac{1}{n}$ , so for all $n \in \{1, 2, 3, 4, 5\}$ , $nf(n)-1= 0 \implies g(n)=0$ Note that the degree of $g(x)$ is $5$ . We have shown that $1, 2, 3, 4, 5$ are roots of $g(x)$ , so they must be the only roots of $g(x)$ . We then conclude $g(x)= c(x-1)(x-2)(x-3)(x-4)(x-5)$ for some non-zero constant $c$ . This implies $xf(x)= c(x-1)(x-2)(x-3)(x-4)(x-5)+1$ Notice that $x$ divides the L.H.S, so $0$ must be a root of the polynomial in the R.H.S. So, $c(0-1)(0-2)(0-3)(0-4)(0-5) + 1=0$ $\implies -5! \times c + 1= 0$ $\implies c= \frac{1}{5!}$

Thus, we obtain $xf(x)= \frac{1}{5!}(x-1)(x-2)(x-3)(x-4)(x-5)+1$ Since $0$ is the root of the polynomial in the R.H.S, we know its constant term is $0$ . Let $c_1, c_2, c_3, c_4, c_5$ be the coefficients of $x^1, x^2, x^3, x^4, x^5$ of $\frac{1}{5!}(x-1)(x-2)(x-3)(x-4)(x-5)+1$ . Then, we have $f(x)= c_5x^4 + c_4x^3 + c_3x^2 + c_2x + c_1$ This implies $f(0)=c_1$ By Vieta's formula, we obtain $c_1= \frac{1 \times 2 \times 3 + 1 \times 2 \times 4 + 1 \times 3 \times 4 + 2 \times 3 \times 4}{120}$ $\implies f(0)= \frac{137}{60}$ Our desired answer is $137+60= \boxed{197}$ .

Consider the polynomial xf(x)-1 As f(x) is of degree 4, xf(x)-1 must be of degree 5 Using the given condition, we know that 1,2,3,4,5 are roots of xf(x)-1 by substituting f(x)=1/x Therefore,$$xf(x)-1 \equiv A(x-1)(x-2)(x-3)(x-4)(x-5) \text{for some constant }A$$ By substituting 0 into both polynomials, we get $$-1 = A(-1)(-2)(-3)(-4)(-5)=-120A \Rightarrow A = 1/120$$ By expanding and comparing the 'x' terms of the polynomials, we get $$xf(0) = A((-1)(-2)(-3)(-4)+(-1)(-2)(-3)(-5)+(-1)(-2)(-4)(-5)+(-1)(-3)(-4)(-5)+(-2)(-3)(-4)(-5))x = 274x/120 = 137x/60$$ Since 137 and 60 are coprime, the answer is 137+60=197

Lets consider the polynomial $Q(n)=nf(n)-1$ which has degree 5. We note that $Q(n)=0$ for $n=1,2,3,4,5$ So by factor theorem; $Q(n)=A(n-1)(n-2)(n-3)(n-4)(n-5)$ for some real constant A. If n=0, $Q(0)=(0)(f(0))-1=-1$ , thus $A(-1)(-2)(-3)(-4)(-5)=-1$ . Hence we have $A=1/(5!)=1/120$ Hence we will get $nf(n)-1=(1/120)(n-1)(n-2)(n-3)(n-4)(n-5)$ . So, $f(n)=\frac {(n-1)(n-2)(n-3)(n-4)(n-5)+120} {120n}$ . The linear term in the numerator will become the constant in $f(n)$ . We can derive the expansion's coefficient's by using Vieta's theorem. So we can get $f(0)=\frac {(1\times 2\times 3\times 4+1\times 2\times 3\times 5+...+2\times 3\times 4\times 5)} {120}$ $=1/5+1/4+1/3+1/2+1=\frac {137}{60}$ Hence the require value is $137+60=197$

Let the function be F(x)=p x^4+q x^3+r x^2+s x^1+t, F(1)=p+q+r+s+t =1 , F(2)=16p+8q+4r+2s+t =1/2 , F(3)=81p+27q+9r+3s+t =1/3 , F(4)=256p+64q+16r+4s+t =1/4,
F(5)=625p+125q+25r+5s+t =1/5,
F(2)-F(1)=15p+7q+3r+s =-1/2 -----(1), F(3)-F(2)=65p+19q+5r+s=-1/6 ----------(2), F(4)-F(3)=175p+37q+7r+s=-1/12 -------------(3), F(5)-F(4)=369p+61q+9r+s=-1/20 ------------(4), (2)-(1)=50p+12q+2r=1/3 -----------(5), (3)-(2)=110p+18q+2r=1/12 ---------(6), (4)-(3)=194p+24q+2r=1/30 ----------(7), (6)-(5)=60p+6q=-1/4 -------------(8), (7)-(6)=84p+6q=-1/20 -----------(9), From (8) and(9) ,p=1/120 ,q=-1/8 , substituting the values of p and q in (5) we get ,r=17/24, substituting the values of p,q and r in (1) we get ,s=-15/8, we have ,p+q+r+s+t=1, t=137/60, F(0)=t=137/60=a/b, a+b=137+60=197

The general form of f(x) is as follows (a,b,c,d, and e are constants):

f(x) = a x^4 + b x^3 + c x^2 + d x + e

But we know that f(1) = 1, f(2) = 1/2, f(3) = 1/3, f(4) = 1/4, and f(5) = 1/5, using the fact that f(n) = 1/n for n = 1,2,3,4, and 5. So, we may write:

f(1) = a + b + c + d + e = 1

f(2) = 16a + 8b + 4c + 2d + e = 1/2

f(3) = 81a + 27b + 9c + 3d + e = 1/3

f(4) = 256a + 64b + 16c + 4d + e = 1/4

f(5) = 625a + 125b + 25c + 5d + e = 1/5

This is a system of five equations in five unkowns. It is straightforward to solve for a,b,c,d, and e. All we really need to know, however, is e, because

f(0) = e

Solving the above system gives e = 137/60

So, 137 + 60 = 197 is the answer.

a+b+c+d+e=1 16a+8b+4c+2d+e=1/2 81a+27b+9c+3d+e=1/3 256a+64b+16c+4d+e=1/4 625a+125b+25c+5d+e=1/5 get: a= 0.0083 b=-0.1250 c=0.7083 d= -1.8750 e= 2.2833=137/60

Let $f(x)=A(x-2)(x-3)(x-4)(x-5)+B(x-1)(x-3)(x-4)(x-5)+C(x-1)(x-2)(x-4)(x-5)+D(x-1)(x-2)(x-3)(x-5)+E(x-1)(x-2)(x-3)(x-4)$ . As $f(1)=1$ , we get, $A=\frac{1}{24}$ . As $f(2)=\frac{1}{2}$ , we get, $B=-\frac{1}{12}$ . As $f(3)=\frac{1}{3}$ , we get, $C=\frac{1}{12}$ . As $f(4)=\frac{1}{4}$ , we get, $D=-\frac{1}{24}$ . As $f(5)=\frac{1}{5}$ , we get, $E=\frac{1}{120}$ . Now, $f(0)=120A+60B+40C+30D+24E=5-5+\frac{10}{3}-\frac{5}{4}+\frac{1}{5}=\frac{137}{60}$ . So, $a+b=197$

We have $f(n) = \frac{1}{n}$ for $n \epsilon {1,2,3,4,5}$

So $nf(n)-1=0$ . Let $g(x)= xf(x)-1$ . And also $g(x)=0$ for $x\epsilon {1,2,3,4,5}$ and so they are the roots of $g(x)$

We can clearly see that, since $f(x)$ is of degree 4 then $g(x)$ will be of degree 5

Now, we have enough information about $g(x)$ , so we can state -

$g(x) = xf(x)-1 = a(x-1)(x-2)(x-3)(x-4)(x-5)$

Now we have $a$ as unknown. We can simply find out its value by setting $x=0$

So, we get - $0\times f(0) - 1 = a\times(-1)(-2)(-3)(-4)(-5) = -5!\times a$

$\Rightarrow a= \frac{1}{5!}$

Therefore, we get $f(x) = \frac{\frac{1}{5!}(x-1)\ldots (x-5)+1}{x}$

Since, we don't know if $x$ cancels out or not, we see that $f(0) = undefined$ .

Therefore, we proceed further as follows -

Let $f(x) = \frac{1}{5!}(ax^4+bx^3+cx^2+dx+e)$ . Hence $f(0)=\frac{e}{5!}$

Now, again we have another unknown namely $e$

Now we see that $g(x) = xf(x)-1 = x\times\frac{1}{5!}(ax^4+bx^3+cx^2+dx+e) -1 = \frac{1}{5!}(ax^5+bx^4+cx^3+dx^2+ex) -1$

Since, $1,2,3,4,5$ are roots of $g(x)$ , we can easily get the value of $e$ by Vieta's Relations -

$e = 1.2.3.4 + 1.2.3.5 + 1.3.4.5 + 1.2.4.5 + 2.3.4.5 = 274$

Therefore, $f(0) = \frac{274}{120} = \frac{137}{60} = \frac{a}{b}$

Hence, $a+b = \boxed{197}$

a+b+c+d+e=1 16a+8b+4c+2d+e=1/2 81a+27b+9c+3d+e=1/3 256a+64b+16c+4d+e=1/4 625a+125b+25c+5d+e=1/5

Let f(x) = ax^4 + bx^3 + cx^2 + dx + e. Form simultaneous equations using results for F(1) etc and solve.

$n\cdot f(n)-1=0$

Let $g(n)$ be a polynomial of degree $5$ , such that $g(n)=n \cdot f(n)-1$ . The roots of $g(n)$ are $1,2,3,4,5$ .

Write $g(n)=k(x-1)(x-2)(x-3)(x-4)(x-5)$ Put $n=0$ : $g(0)=-1=k \cdot -1 \cdot 5!$

$k=\frac{1}{5!}$

We have: $n \cdot f(n)=g(n)+1$

Observe that $g(n)$ is a a polynomial with degree $5$ , and it has no constant term, so we can divide both sides by $n$ to determine $f(n)$ .

Thus, $f(n)=\frac{{\frac{1}{5!}(n-1)(n-2)..(n-5)+1}}{n}$ The expansion of the above terms are ugly, but we do know that the $n$ in the denominator will cancel out. So write $f(n)=\frac{1}{5!}(ax^4+bx^3+cx^2+dx+e)$ . The answer to the question is $\frac{e}{5!}$ .

Now, consider the expansion of $(n-1)(n-2)..(n-5)$ . To evaluate $f(0)$ we only require to know the co-efficient of the linear term( the term $e$ ). We can find the linear term by vieta's relations of a degree $5$ polynomial:

$e=1\cdot 2\cdot3\cdot4+1\cdot2\cdot3\cdot5+1\cdot2\cdot4\cdot5+1\cdot3\cdot4\cdot5+2\cdot3\cdot4\cdot5=274$

Thus $f(0)=\frac{274}{120}=\frac{137}{60}$

Our answer is : $\boxed{197}$ .

We get that $n f(n) - 1 = c(n - 1) \cdots (n - 5)$ Using $n = 0$ , we get that $c = \dfrac{1}{120}$

Then the constant term in $\dfrac{(n - 1) \cdots (n - 5) + 120}{120 n}$ is $\dfrac{120 / 1 + 120 / 2 + 120 / 3 + 120 / 4 + 120 / 5}{120} = \dfrac{137}{60}$

Whenever one thinks of polynomials one thinks of its roots and consequently working with them, right? Now here we're given that $f(x)$ has degree 4, so not much info is given about its roots. Now look at the second condition $f(n) = \frac{1} {n}$ , which on modification gives $nf(n)-1=0$ . Now clearly, the above is a polynomial, call it $h(n)$ . Note that $h(n)=0$ for $n=1,2,3,4,5$ . ( This is the reason for writing $f \times n -1$ so that the roots can be obtained) . Since nothing is known about the leading coefficient of f and consequently h, let $h(n) = t (x-1)(x-2)(x-3)(x-4)(x-5)$ for some real number t. Note that h(n) is written like this because if a polynomial has roots $p_{i} ,i=1(1)n$ , then that polynomial can be written as $q \times (x-p_{1})(x-p_{2})....(x-p_{i}).....(x-p_{n})$ where q is real, the leading co-efficient of the polynomial.

Now h( nf(n)-1 ) has been written in a 'good form', but what about 't'? What is its value? . Since it's an identity in n, everything can be plugged in for n. Since the value of f isn't known at any point plugging in n=0 would be great!

So $(0\times f(0))-1 = t \times (-1)^{5} 120$ giving $t= \frac{1}{120}$ .

Now for the next step write $f(x) = ax^{4}+bx^{3}+cx^{2}+dx+e$ , where $a,b,c,d,e$ are assumed to be reals. So $f(0) = e$ , the target variable whose value is to be determined.

So finally writing the entire equation:

$xf(x)-1 = x(ax^{4}+bx^{3}+cx^{2}+dx+e)-1 =ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex-1$ So finally,

$ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex-1 = \frac{1}{120} \times(x-1)(x-2)(x-3)(x-4)(x-5)$

As mentioned earlier the target variable is $e$ , the co-efficient of $x$ on both sides of the above equation. So equating the co-efficient of $x$ ( this is only simple algebraic manipulation, only co-efficient of x is to be collected from both sides and nothing else!! It may seem long and tiresome but it's not that much hectic, trust me. ),

$e= \frac {274}{120} = \frac{137}{60}, so, a=137, b=60, a+b=\boxed{197}$ .

Man! A long solution!!

using all given values of f(x), and as degree is 4, we get4th derivative as a constant. solving, we get f(0) as 137/60. a=137, b=60. hence a+b=197

Let $g(x)=f(x) - \frac {1}{x}$ . Then, $g(1)=g(2)=g(3)=g(4)=g(5)=0$ . So $(x-1),(x-2),(x-3),(x-4),(x-5)$ are factors of $g(x)$ by the Remainder-Factor Theorem. But $g(x)$ will then be a degree 5 polynomial. Thus, $g(x)=\frac{(x-1)(x-2)(x-3)(x-4)(x-5)}{Ax+B}$ , where $A,B$ are real numbers. $f(x)=\frac{(x-1)(x-2)(x-3)(x-4)(x-5)}{Ax+B}+\frac 1x$ =\frac{x^6-15x^5+85x^4-225x^3+274x^2-120x+Ax+B}{Ax^2+Bx}). For $f(x)$ to be a polynomial, its least exponent must be 2. So $-120x+Ax+B=x(-120+A)+B=0$ , which gives $A=120$ , $B=0$ . $f(x)=\frac{x^4-15x^3+85x^2-225x+274}{120}$ . Thus, $f(0)=\frac{137}{60}$ , and $137+60=197$ .

We first obtain the polynomia $f(x)$ . For $x=1,2,3,4,5$ , we have

$xf(x)-1=C(x-1)(x-2)(x-3)(x-4)(x-5)$ for some real constant $C$ .. At $x=0$ , one can obtain $C$ as $C=\frac{1}{120}$ . It follows that the required polynomial is

$f(x) = \frac{120+(x-1)(x-2)(x-3)(x-4)(x-5)}{120x}$

Also, observe that (\f(x)) has the form

$f(x) =\frac{1}{120} (x^{4}-Ax^{3}+Bx^{2}-Cx+D)$

From Viete's formulas applied to $\frac{(x-1)(x-2)(x-3)(x-4)(x-5)}{x}$ gives

$D = (2)(3)(4)(5)+(1)(3)(4)(5)+(2)(1)(4)(5)+(2)(3)(1)(5)+(2)(3)(4)(1)=274$

Hence,

$f(x) =\frac{1}{120} (x^{4}-Ax^{3}+Bx^{2}-Cx+274)$

Now, by setting $x=0$ , we obtain

$f(0) =\frac{1}{120} (274)=\frac{137}{60}$

From which we see that $a+b=\boxed{197}$

Because $f(n)=1/n$ for $n=1,2,3,4,5$ we see that $nf(n)-1=0$ for those same values. Let $nf(n)=q(n)$ , we see that because $f(n)$ is a degree 4 polynomial, q(n) is a degree 5 polynomial. Furthermore, $q(0)=0*f(0)=0$ .

The rest of this problem is a tedious computation involving polynomial interpolation. If you haven't seen it, wiki does a passable job at explaining it here: http://en.wikipedia.org/wiki/Lagrange_polynomial.

Because we have 6 points of data for a degree 5 polynomial we can pin it down explicitly, divide by n and then evaluate the resulting degree 4 polynomial $f(n)$ at 0 yielding $f(0)=\frac{137}{60}$ . So the answer is $137+60=\boxed{197}$

Here is the computation:

$q(n)=\frac{(n)(n-1)(n-2)(n-3)(n-4)}{120} +\frac{(n)(n-1)(n-2)(n-3)(n-5)}{-24} + \frac{(n)(n-1)(n-2)(n-4)(n-5)}{12} + \frac{(n)(n-1)(n-3)(n-4)(n-5)}{-12} +\frac{(n)(n-2)(n-3)(n-4)(n-5)}{24}$

Where the coefficient of the term without $n-a$ (for $a=1,2,3,4,5)$ is computed by plugging $a$ into the numerator and then dividing by the result, so that when we plug $a$ into the modified term we get 1 as we must. Likewise because $f(0)=0$ the coefficient of the term without $n$ in the numerator is 0.

Now we evaluate $\frac{q(n)}{n}$ at 0: $=\frac{1}{5}-\frac{5}{4}+\frac{10}{3}-5+5 = \frac{137}{60}$

If f(x)=ax⁴+bx³+cx²+dx+e, to x belongs to {1,2,3,4,5}, we have the equality ax⁴+bx³+cx²+dx+e=1/x. In other words, ax⁵+bx⁴+cx³+dx²+ex-1=0. The roots of this equation are 1,2,3,4,5. By Girard, we now that: -(-1)/a=5.4.3.2.1=120 => a=1/120 e/a=2.3.4.5+1.3.4.5+1.2.4.5+1.2.3.5+1.2.3.4=274 => e=274a=274/120=137/60 But we are looking for f(0)=e, so the answear is 137+60=197 !!

$xf(x)-1=a(x-1)(x-2)...(x-5)$ ...(1)

So that

$a=\frac{1}{5!}$

from ...(1)

We get constant of $x$ is

$\frac{137}{60}$

So, $a+b=197$

f(n)=1/n for n=1, 2, 3, 4, 5

Let us assume a function g(x)=x f(x) -1= a (x-1) (x-2)(x-3)(x-4)(x-5).

Clearly, g(x)=0 for n=1, 2, 3, 4, 5.

Equation constant terms in LHS and RHS of x f(x) -1= a (x-1) (x-2)(x-3)(x-4)(x-5).

We get a=1/120.

f(0)= lt_{x-> 0} \frac{(x-1)(x-2)(x-3)(x-4)(x-5)+120}{120x}=\frac{137}{60}

Notice that nf(n)-1 is a quintic polynomial which is zero at n=1,2,3,4,5, and therefore must be of the form cg(n) where c is constant and $g(n) = \prod_{i=1}^5 (n-i)$ . Substituting n=0 gives c=-1/g(0), and thus f(n) = (g(0)-g(n))/ng(0). Since f is continuous, $f(0) = \lim_{n\to 0} f(n) = -g'(0)/g(0)$ by the definition of the derivative of g. -g'(n)/g(n) is the (negative) logarithmic derivative of g, which is $-\sum_{i=1}^5 1/(n-i)$ . At n=0 this evaluates to $1+1/2+1/3+1/4+1/5 = 137/60$ .

Let $g(x)=x\cdot f(x)-1$ . Thus $g(x)$ is a degrees 5 polynomial.

and we have $g(1)=g(2)=g(3)=g(4)=g(5)=0$ so

$g(x)=x\cdot f(x)-1=c_0(x-1)(x-2)(x-3)(x-4)(x-5)$

$c_0$ have to equal $\frac{1}{5!}$ . when we divide both sides by $x$ we will get the constant in right hand side is

$\frac{1\cdot2\cdot3\cdot4+1\cdot2\cdot3\cdot5+1\cdot2\cdot4\cdot5+1\cdot3\cdot4\cdot5+2\cdot3\cdot4\cdot5}{5!}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{137}{60}$

therefore the answer is $137+60=197$