Degree 99 Polynomial

It is given that for 100 values from 0 to 100 the equation $f(x) (x+1) = 1$ .

But it is a well known theorem that if a polynomial g(x), with $deg(g) = n$ , that satisfies g(x)=0 for n values $a_{1}, a_{2},..., a_{n}$ , then $g(x)=r(x-a_{1})(x-a_{2})...(x-a_{n})$ , for some real constant r .

Since $f(x)(x+1)-1=g(x)=0$ is satisfied for 100 values and $deg(g)=100$ , then

$f(x)(x+1)-1= \frac{rx(x-1)(x-2)...(x-100)}{x-a}$ , for the specified value a.

Now substituting $x=-1$ we get that $-1=r \times \frac{-101!}{-1-a}$ since there are and odd amount of negative terms in the numerator, and so $r=\frac{-(a+1)}{101!}$ .

Now we can use the given information that $f(101)=0$ , which implies that substituting $x=101$ will give

$-1=r \times \frac{(101-1)(101-2)...(101-100)}{101-a}$

$-1=\frac{-(a+1)}{101!} \times \frac{101!}{101-a}$

$-1=\frac{-(a+1)}{101-a}$

$a-101=-a-1$

$2a=100$ or $a=50$ .

We know that $f(x)=\frac{1}{x+1}$ for all integers ranging from $0$ to $100$ except $a$ . Thus, $(x+1)f(x)-1=0$ for all integers ranging from $0$ to $100$ except $a$ . This leads us to consider the 100 degree polynomial $g(x)=(x+1)f(x)-1$ , whose roots are the integers ranging from $0$ to $100$ except $a$ . (There are no more roots because $g$ is only of degree 100.) Therefore, we can write $g(x)=c\cdot \frac{x(x-1)(x-2)\cdots(x-100)}{x-a},$ for some nonzero constant $c$ . But we are also given that $f(101)=0$ , which implies $g(101)=-1$ . Plugging in $101$ to the above equation yields $c\cdot \frac{101!}{101-a}=-1\implies c=-\frac{101-a}{101!}$ Thus we can now write $g(x)$ as $g(x)=-\frac{101-a}{101!}\cdot \frac{x(x-1)(x-2)\cdots(x-100)}{x-a}$ Now notice that from our definition of $g(x)$ we have $g(x)+1=(x+1)f(x)$ , which means that $-1$ is a root of $g(x)+1$ . Thus plugging in $x=-1$ into $g(x)+1$ should give us $0$ : $-\frac{101-a}{101!}\cdot \frac{(-1)(-2)(-3)\cdots(-101)}{-1-a}+1=0,$
and after some cancellation and rearranging we get $a=\boxed{50}$ .

Consider $g(x) := (x + 1)f(x) - 1$ which is a polynomial with degree $100$ . Note that the roots of $g(x)$ are the $100$ integers (From $f(x) = \frac{1}{x + 1}$ for these integers). And also note that $g(101) = g(-1) = -1$ , as $f(101) = 0$ . Let the roots of $g(x)$ be $a_{1}, a_{2}, ... , a_{100}$ (which are the $100$ integers.). Let $a$ be the 'leftover' integer. Let $g(x) = A(x - a_{1})(x - a_{2})...(x - a_{100})$ . Note that $g(101) = (101 - a_{1})(101 - a_{2})...(101 - a_{100})$ $= \frac{A(101!)}{101 - a}$ (Since $a_{i}$ ranges through $0$ to $100$ , except $a$ , and thus $100 - a_{i}$ ranges through $1$ to $101$ , except $101 - a$ ) and similarly $g(-1) = A(-1 - a_{1})(-1 - a_{2})...(-1 - a_{100})$ $= (-1)^{100}(1 + a_{1})(1 + a_{2})...(1 + a_{100})$ $= A(1 + a_{1})(1 + a_{2})...(1 + a_{100}) = \frac{A(101!)}{a + 1}$ . Since $g(101) = g(-1) = -1$ , $101 - a = a + 1 \Rightarrow a = \boxed{50}$ .

Let $g$ be a polynomial such that $g(x)=(x+1)f(x)-1$ . The $100$ values that satisfy $f(x)=\frac{1}{x+1}$ are clearly the only zeros of $g$ , since $g$ is a polynomial of degree $100$ . Observe that $g(-1)=g(101)=-1$ . Now, let those 100 values that satisfy $f(x)=\frac{1}{x+1}$ be $x_1, x_2,\ldots x_{100}$ . $g(x)$ can be written as $A(x-x_1)(x-x_2)\ldots (x-x_{100})$ , where $A$ is a nonzero constant. Then $g(101)=A(101-x_1)(101-x_2)\ldots (101-x_{100})=\frac {101!A}{101-a}$ and also $g(1)=A(-1-x_1)(-1-x_2)\ldots (-1-x_{100})=A(1+x_1)(1+x_2)\ldots (1+x_{100})=\frac{101!A}{1+a}$ . So we have $g(-1)=g(101) \Rightarrow \frac {101!A}{101-a}=\frac{101!A}{1+a} \Rightarrow 101-a=1+a$ since $A$ is not zero. Solving this gives us $a=50$ .

Let $g(x)=(x+1).f(x)-1$ . As f(x) is a polynomial of degree 99, g(x) is a polynomial of degree 100.

However, as $f(x) = \frac {1}{x+1}$ for exactly 100 values of x from ${0;1; \ldots ;100}$ we have that g(x)=0 for all such values.

Let the values be $x_1; x_2; \ldots ; x_{100}$ . Thus, $g(x)= k. (x- x_1).(x- x_2) \ldots (x- x_{100})$ for some real constant k.

We have that g(-1)=-1 since g(-1)=(-1+1).f(-1)-1=-1 and g(101)=102.f(101)-1=-1 since f(101)=0.

Thus: $k . (-1-x_1).(-1-x_2) \ldots (-1-x_{100})$ $= k.(101-x_1).(101-x_2) \ldots (101-x_{100}).$

However k cannot equal 0 because f(-1) is different from 0.

Thus, $(-1-x_1).(-1-x_2) \ldots (-1-x_{100})=$

$(101- x_1).(101- x_2) \ldots (101-x_{100})$

Let the value a for which $f(a) \neq \frac {1}{a+1}$ be $x_{101}$

Then, $(-1-x_1).(-1-x_2) \ldots (-1-x_{100})$ $=(1+x_1).(1+x_2) \ldots (1+x_{100})$

$= \frac {(1+0).(1+1) \ldots (1+100)}{1+x_{101}}$

and $(101-x_1).(101-x_2) \ldots (101-x_{100})$

$= \frac {(101-0).(101-1) \ldots (101-100)}{101-x_{101}}$ .

Since both the numerator are equal and non-zero, we must have that: $1+x_{101}=101-x_{101}$ , thus $x_{101}=50$ .

Thus, the only value of a, $0 \leq a \leq 100$ such that $f(a) \neq \frac {1}{a+1}$ is 50.

Let $g(x) = f(x)(x+1)-1$ . The problem implies that $g(x) = 0$ for 100 integer values from 0 to 100 except for $a$ .

Also $g(-1) = -1$ and $g(101) = f(101) \cdot 102 -1 = -1$

Now, since $g(x)$ is a polynomial of degree 100, then $g(x) = A(x-x_1) \cdots (x-x_{100}) = A \frac{\displaystyle\prod_{i=0}^{100}(x-i)}{x-a}$

$g(-1) = 1$ implies $1 = A\frac{\displaystyle\prod_{i=0}^{100}(-1-i)}{-1-a} = \frac{A \cdot 101!}{a+1}$

$g(101) = 1$ implies $1 = A\frac{\displaystyle\prod_{i=0}^{100}(101-i)}{101-a} = \frac{A \cdot 101!}{101-a}$

From both of these equations we obtain $a+1 = 101-a \Rightarrow a = 50$ as desired.

Consider f(x)=N/(x+1), where N is divisible by x+1. Since N=1 for x= all integers between 0 and 100 inclusive except one value (which we shall assume as a), N must be of the form: 1+k x(x-1)(x-2)...(x-100)/(x-a) Note that we define the value of f(x) at x=a or x= -1 to be their limiting values so that they are continuous (the reason for which shall become obvious as we proceed). We first determine k in terms of a so that N is divisible by (x+1) Put x=-1 and solve for N=0, for which we get, k=-(1+a)/{q(101)}, where q(n)=factorial of n (n belongs to N) Next we put f(101)=0: N=0; 1+k {q(101)}/(101-a) Putting above value of k, we get a=50 We can easily see that this is the odd one out among our set of natural numbers from 0 to 100 that would yield f(x)=1/(x+1) since (x-50) no longer exists in the numerator we constructed. Fortunately, neither does x=50 coincidentally yield 1/51; hence is the required answer.

Let $q(x)= (1+x)f(x) -1$ Then except for one $k$ in $S=\{0,1,2,...,100 \}$ , every integer in $S$ is a root of $q$ . Thus there exists $\alpha$ in $\mathbb{R}$ such that q(x)=\alpha \prod_{i=0}_{i\neq k}^{100} (x- i) Using the fact that $q(101)=-1$ yields $\alpha = \frac{101-k}{101!}$ Finally, noticing that $q(-1)=-1$ , and \prod_{i=0}_{i\neq k}^{100} (-1- i) = (-1)^{100}\frac{101!}{1+k} we get $\frac{101-k}{101!} \frac{101!}{1+k} = -1 \iff 101-k=-1-k \iff k = 50$

Consider the polynomial $g(x)$ such that $g(x)=(x+1)f(x)-1$ . Note that if $f(x)=\frac{1}{x+1}$ , $g(x)=0$ . Hence, each of the $100$ values satisfying $f(x)=\frac{1}{x+1}$ is a root of $g(x)$ . We then can write $g(x)$ as $g(x)=\displaystyle\frac{c(x)(x-1)(...)(x-100)}{x-a}$ , where $a$ is the desired missing value.

Since $f(101)=0$ , $g(101)=102f(101)-1=\displaystyle\frac{c(101)(101-1)(...)(101-100)}{101-a}$ so $-1=\displaystyle\frac{c*101!}{101-a}$ .

Also, $g(-1)=(-1+1)f(-1)-1=\displaystyle\frac{c(-1)(-1-1)(...)(-1-100)}{-1-a}$ so $-1=\displaystyle\frac{-c*101!}{-1-a}$ .

Equating our two equations, we find that $a+1=101-a$ , so $a=\boxed{50}$ .

f(x)= 1/x+1 for 100 integers between 0 to 100 we can write (x+1)f(x)-1 = p(x-a {1})(x-a {2})..............(x-a {100}) .....eqn 1 where a {1}, a {2}...... a {100} are integers belonging to [0,1,2,3.....100] Since the degree of f(x) is 99, therefore f(-1)=0 It is given that f(101)=0 On applying the above 2 conditions on equation 1, we get that a_{i} does not contain 50

Let P(x)=(x+1)f(x)-1, then P(x)=kx(x-1)...(x-100)/(x-a) P(101)=102f(101)-1=0.=> k.101!+101-a=0(I) f(x)=[kx(x-1)...(x-100)+(x-a)]/[(x-a)(x+1)] Q(x)=kx(x-1)...(x-100)+(x-a).Q(-1)=0.=>-101!k-a-1=0.(II) (I)+(II)=> a=50

Since $101$ is a root of polynomial , we have $f(x)=(x-101)\cdot p(x)$ where degree of $p(x)$ is $98$ .

Now since $f(x) = \frac{1}{x+1}$ for all values between $0$ and $100$ , we have $p(x)=\frac{1}{(x+1)(x-101)}$ Now its easy to see that , if $x=a$ satisfies then $x= 100-a$ also satisfies the condition . now all the values except $50$ work in pairs ( because $50 = 100-50$ .

$\therefore$ we have $\boxed{a=50}$

Let $S$ be the set of all integers between $0$ and $100$ for which $f(x) = \frac{1}{x+1}$ for all $x \in S$ . From the problem statement, we have $|S| = 100$ .

Since $101$ is a root of $f(x)$ , we can rewrite $f$ as $f(x) = (101-x)Q(x)$ , where $Q(x)$ is a polynomial of degree $98$ . Hence, $Q(x) = \frac{f(x)}{101-x} = \frac{1}{(x+1)(101-x)}$ for all $x \in S$ . Define $P(x) = \frac{1}{(x+1)(101-x)}$ . Then, substituting $100-x$ into the equation, we get $P(x) = P(101-x)$ . So $x \in S$ if and only if $100-x \in S$ (*).

Let $a$ be an integer from $0$ to $100$ which does not belong to $S$ . From (*), we conclude that both $a$ and $100-a$ do not belong to $S$ . But, there can be at most one such $a$ . So, we must have $a = 100-a \Leftrightarrow a = 50$ .

Let $f'(x)$ be defined in the same way as $f$ above, except

that $f'(a) = \frac{1}{a+1}$ .

Since $f'(101) = 0$ , this means $(x - 101)$ is a factor of

$f'(x)$ . Let $\begin{aligned} h'(x) &= \frac{f'(x)}{x-101} \\ &= \frac{1}{(x+1)(x-101)} \mbox{with } 0 \le x \le 100 \end{aligned}$ Similarly let $h(x)$ be $\frac{f(x)}{x-101}$ .

$h'(x)$ is even; specifically, $h'(x) = h'(100 - x)$ .

This means that if $h'(a) \neq h(a)$ then $h'(100-a) \neq h(100-a)$ .

But it was specified that only one value of $f$ differs from

$f'$ (on the domain we are looking at), therefore only one

value of $h(x)$ can differ from $h'(x)$ . So we we must have

$a = 100-a$ , i.e. $a = \boxed{50}$ .