MHB How Do You Calculate P(12) for a Polynomial Given Specific Conditions?

[Saitama]

Problem:
Let $P(x)$ be a polynomial of degree 11 such that

$$P(x)=\frac{1}{x+1}\,\,\,\text{for}\,\,\,x=0,1,2,\cdots 11$$
Then find the value of P(12).

Attempt:
I have done this kind of problem long before but I don't exactly remember the process.

I think it was something like this:

Define $g(x)=P(x)-\frac{1}{x+1}$, then
$$g(x)=P(x)-\frac{1}{x+1}=ax(x-1)(x-2)\cdots (x-11)$$
But I don't think the above is correct. I don't have the value of $a$.

Any help is appreciated. Thanks!

 

[chisigma]

Problem:
Let $P(x)$ be a polynomial of degree 11 such that

$$P(x)=\frac{1}{x+1}\,\,\,\text{for}\,\,\,x=0,1,2,\cdots 11$$
Then find the value of P(12).

Attempt:
I have done this kind of problem long before but I don't exactly remember the process.

I think it was something like this:

Define $g(x)=P(x)-\frac{1}{x+1}$, then
$$g(x)=P(x)-\frac{1}{x+1}=ax(x-1)(x-2)\cdots (x-11)$$
But I don't think the above is correct. I don't have the value of $a$.

Any help is appreciated. Thanks!

I'm afraid that You have to use the Lagrange Interpolating Polynomial...

$\displaystyle P(x) = \sum_{j=1}^{n}\ p_{j}(x)\ (1)$

... where...

$\displaystyle p_{j} (x) = y_{j}\ \prod_{k=1,\ k \ne j}^{n} \frac{x - x_{k}}{x_{j} - x_{k}}\ (2)$In Your case is n=12, $x_{k}=k - 1$, $y_{k} = \frac{1}{k}$...

Kind regards$\chi$ $\sigma$

 

[Opalg]

Problem:
Let $P(x)$ be a polynomial of degree 11 such that

$$P(x)=\frac{1}{x+1}\,\,\,\text{for}\,\,\,x=0,1,2,\cdots 11$$
Then find the value of P(12).

Attempt:
I have done this kind of problem long before but I don't exactly remember the process.

I think it was something like this:

Define $g(x)=P(x)-\frac{1}{x+1}$, then
$$g(x)=P(x)-\frac{1}{x+1}=ax(x-1)(x-2)\cdots (x-11)$$
But I don't think the above is correct. I don't have the value of $a$.

You are thinking along the right lines, but you would do better to define $g(x) = (x+1)P(x) - 1$. You would still have $g(x) = ax(x-1)(x-2)\cdots (x-11)$, but this way you can find $a$ by putting $x=-1$.

 

[Saitama]

You are thinking along the right lines, but you would do better to define $g(x) = (x+1)P(x) - 1$. You would still have $g(x) = ax(x-1)(x-2)\cdots (x-11)$, but this way you can find $a$ by putting $x=-1$.

Thank you Opalg! :)