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Question: If \({f_0}\left( x \right) = \dfrac{x}{{\left( {x + 1} \right)}}\,and\,{f_{n + 1}} = {f_0} \cdot {f_...

If f0(x)=x(x+1) and fn+1=f0⋅fn(x){f_0}\left( x \right) = \dfrac{x}{{\left( {x + 1} \right)}}\,and\,{f_{n + 1}} = {f_0} \cdot {f_n}\left( x \right) and =0, 1, 2,... then fn(x){f_n}\left( x \right) is

Explanation

Solution

Hint: Start by putting the value of n=0 in  fn+1=f0⋅fn(x)\,{f_{n + 1}} = {f_0} \cdot {f_n}\left( x \right), then put n=1 repeat the step 2 to 3 times observe the pattern of the answers and generalise while putting n=n.

Complete step-by-step answer:

Given, f0(x)=xx+1{f_0}\left( x \right) = \dfrac{x}{{x + 1}}
fn+1(x)=f0(x)⋅fn(x)→(1){f_{n + 1}}\left( x \right) = {f_0}\left( x \right) \cdot {f_n}\left( x \right) \to (1)
On putting the value of n = 0, we get,
⇒f1(x)=f0(x)⋅f0(x)\Rightarrow {f_1}\left( x \right) = {f_0}\left( x \right) \cdot {f_0}\left( x \right)
On putting the values of f0(x) {f_0}\left( x \right)\,, we get,
⇒f1(x)=(xx+1)2→(2)\Rightarrow {f_1}(x) = {\left( {\dfrac{x}{{x + 1}}} \right)^2} \to (2)
On putting the value of n = 1 in equation 1, we get,
⇒f2(x)=f0(x)⋅f1(x)\Rightarrow {f_2}\left( x \right) = {f_0}\left( x \right) \cdot {f_1}\left( x \right)
On putting the values off0(x) and f1(x){f_0}\left( x \right)\,and\,{f_1}\left( x \right), we get,
⇒f2(x)=(xx+1)3\Rightarrow {f_2}(x) = {\left( {\dfrac{x}{{x + 1}}} \right)^3}
On putting the value of n = 2 in equation (1) , we get,
⇒f3(x)=f0(x)⋅f2(x)\Rightarrow {f_3}\left( x \right) = {f_0}\left( x \right) \cdot {f_2}\left( x \right)
On putting the values of f0(x) and f2(x){f_0}\left( x \right)\,and\,{f_2}\left( x \right), we get,
⇒f3(x)=(xx+1)4\Rightarrow {f_3}(x) = {\left( {\dfrac{x}{{x + 1}}} \right)^4}
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So, from above equations, we can generalize the fn(x){f_n}\left( x \right)as
⇒fn(x)=(xx+1)n+1\Rightarrow {f_n}(x) = {\left( {\dfrac{x}{{x + 1}}} \right)^{n + 1}}.
Let us verify the equation by substituting the value of n as 1 i.e..,
⇒f1(x)=(xx+1)1+1=(xx+1)2→(3)\Rightarrow {f_1}(x) = {\left( {\dfrac{x}{{x + 1}}} \right)^{1 + 1}} = {\left( {\dfrac{x}{{x + 1}}} \right)^2} \to (3).
As we can see the value of f1(x){f_1}(x) is the same in both equations (2) and (3).
Hence, fn(x)=(xx+1)n+1{f_n}(x) = {\left( {\dfrac{x}{{x + 1}}} \right)^{n + 1}}is our requiredfn(x){f_n}\left( x \right).

Note: In these questions, the aim should be to get the pattern while equating the values which are known to us, once that is done a general equation can be formed such that different values can be put to get the desired result.