Question

Let $f\left( t \right)={{e}^{\dfrac{-1}{t}}},t\text{ }>\text{ 0}$ for each positive integer n, ${{P}_{n}}\left( x \right)$ be the polynomial such that $\dfrac{{{d}^{n}}}{d{{t}^{n}}}f\left( t \right)={{P}_{n}}\left( \dfrac{1}{t} \right){{e}^{\dfrac{-1}{t}}}$ for all t>0. If ${{P}_{n+1}}\left( x \right)={{x}^{k}}\left[ {{P}_{n}}\left( x \right)-{{P}_{n}}'\left( x \right) \right]$ then k = ?

Answer 1

Here, we are given a function $f\left( t \right)={{e}^{\dfrac{-1}{t}}},t\text{ }>\text{ 0}$ and its ${{n}^{th}}$ derivative is also given in the form of ${{P}_{n}}\left( x \right)$ and we have to find value of k if ${{P}_{n+1}}\left( x \right)={{x}^{k}}\left[ {{P}_{n}}\left( x \right)-{{P}_{n}}'\left( x \right) \right]$. For this, we will first take $\dfrac{{{d}^{n}}}{d{{t}^{n}}}f\left( t \right)$ and find $\dfrac{{{d}^{n+1}}}{d{{t}^{n+1}}}f\left( t \right)$ by replacing n with n+1. After that, we will convert it into $\dfrac{d}{dt}\left( \dfrac{{{d}^{n}}}{d{{t}^{n}}}f\left( t \right) \right)$ form and put value of $\dfrac{{{d}^{n}}}{d{{t}^{n}}}f\left( t \right)$ and simplify. After simplifying we will compare the equation with the equation of ${{P}_{n+1}}\left( x \right)$ to find the value of k and conclude our result. We will use chain rule, product rule for performing differentiation. In the chain rule, we take the first term as it is and multiply it by the derivative of the first term. Both terms formed are added to find final differentiation. In the chain rule we first find derivative of ${{P}_{n}}\left( \dfrac{1}{x} \right)$ as ${{P}_{n}}'\left( \dfrac{1}{x} \right)$ and then multiply it by derivative of function inside the given function that is take derivative of $\dfrac{1}{x}$. We will also use the formula as $\dfrac{{{d}^{n+1}}}{d{{t}^{n+1}}}f\left( t \right)=\dfrac{d}{dt}\left( \dfrac{{{d}^{n}}}{d{{t}^{n}}}f\left( t \right) \right)$

Complete step-by-step answer:
We are given the equation as:
$\dfrac{{{d}^{n}}}{d{{t}^{n}}}f\left( t \right)={{P}_{n}}\left( \dfrac{1}{t} \right){{e}^{\dfrac{-1}{t}}}\cdots \cdots \cdots \cdots \left( 1 \right)$
Let us now change n to n+1, we get:
$\dfrac{{{d}^{n+1}}}{d{{t}^{n+1}}}f\left( t \right)={{P}_{n+1}}\left( \dfrac{1}{t} \right){{e}^{\dfrac{-1}{t}}}$
We can write as $\dfrac{{{d}^{n+1}}}{d{{t}^{n+1}}}f\left( t \right)$ as $\dfrac{d}{dt}\left( \dfrac{{{d}^{n}}}{d{{t}^{n}}}f\left( t \right) \right)$ thus we get:
$\dfrac{d}{dt}\left( \dfrac{{{d}^{n}}}{d{{t}^{n+1}}}f\left( t \right) \right)={{P}_{n+1}}\left( \dfrac{1}{t} \right){{e}^{\dfrac{-1}{t}}}$
Now let us solve this equation using the result from (1). Putting value from (1) in above equation we get:
$\dfrac{d}{dt}\left( {{P}_{n}}\left( \dfrac{1}{t} \right){{e}^{\dfrac{-1}{t}}} \right)={{P}_{n+1}}\left( \dfrac{1}{t} \right){{e}^{\dfrac{-1}{t}}}$
Using product rule on the left side of equation we get:
${{e}^{\dfrac{-1}{t}}}\cdot \dfrac{d}{dt}{{P}_{n}}\left( \dfrac{1}{t} \right)+{{P}_{n}}\left( \dfrac{1}{t} \right)\dfrac{d}{dt}\left( {{e}^{\dfrac{-1}{t}}} \right)={{P}_{n+1}}\left( \dfrac{1}{t} \right){{e}^{\dfrac{-1}{t}}}$
Using chain rule on left side we get:
${{e}^{\dfrac{-1}{t}}}\cdot {{P}_{n}}'\left( \dfrac{1}{t} \right)\cdot \left( \dfrac{-1}{{{t}^{2}}} \right)+{{P}_{n}}\left( \dfrac{1}{t} \right)\cdot {{e}^{\dfrac{-1}{t}}}\cdot \left( \dfrac{1}{{{t}^{2}}} \right)={{P}_{n+1}}\left( \dfrac{1}{t} \right){{e}^{\dfrac{-1}{t}}}$
Taking ${{e}^{\dfrac{-1}{t}}}$ common from both sides, we get:
${{P}_{n}}'\left( \dfrac{1}{t} \right)\cdot \left( \dfrac{-1}{{{t}^{2}}} \right)+{{P}_{n}}\left( \dfrac{1}{t} \right)\cdot \left( \dfrac{1}{{{t}^{2}}} \right)={{P}_{n+1}}\left( \dfrac{1}{t} \right)$
Taking $\dfrac{1}{{{t}^{2}}}$ common from left hand side we get:
$\dfrac{1}{{{t}^{2}}}\left( {{P}_{n}}\left( \dfrac{1}{t} \right)-{{P}_{n}}'\left( \dfrac{1}{t} \right) \right)={{P}_{n+1}}\left( \dfrac{1}{t} \right)$
Let us change $\dfrac{1}{t}$ by x, hence substituting $\dfrac{1}{t}=x$ in above equation we get:
${{x}^{2}}\left( {{P}_{n}}\left( x \right)-{{P}_{n}}'\left( x \right) \right)={{P}_{n+1}}\left( x \right)\cdots \cdots \cdots \cdots \left( 2 \right)$
We are given equation as:
${{x}^{k}}\left[ {{P}_{n}}\left( x \right)-{{P}_{n}}'\left( x \right) \right]={{P}_{n+1}}\left( x \right)\cdots \cdots \cdots \cdots \left( 3 \right)$
Comparing (2) and (3) we get:
k = 2
Hence, we have found that value of k is 2, if ${{P}_{n+1}}\left( x \right)={{x}^{k}}\left[ {{P}_{n}}\left( x \right)-{{P}_{n}}'\left( x \right) \right]$

Note: Students should be careful while applying chain rule. Don't forget to find derivatives of terms inside the function P(x). Here, P'(x) denotes $\dfrac{d}{dx}P\left( x \right)$. Product rule on two terms is applied as below:
$\dfrac{d}{dx}\left( u\cdot v \right)=v\cdot \dfrac{du}{dx}+u\cdot \dfrac{dv}{dx}$
Where, u and v are two functions of x. Students should carefully convert equations from n to n+1. While comparing make sure that all then rest terms are equal.