Question

Let the function $f:\left[ {0,1} \right] \to \mathbb{R}$ be defined by $f\left( x \right) = \dfrac{{{4^x}}}{{{4^x} + 2}}$. Then what is the value of $f\left( {\dfrac{1}{{40}}} \right) + f\left( {\dfrac{2}{{40}}} \right) + f\left( {\dfrac{3}{{40}}} \right) + ..... + f\left( {\dfrac{{39}}{{40}}} \right) - f\left( {\dfrac{1}{2}} \right)$?

Answer 1

Hint: A function $f\left( x \right)$ is given. Find the function $f\left( {1 - x} \right)$. Adding the functions, $f\left( x \right) + f\left( {1 - x} \right) = 1$. Omit $f\left( {\dfrac{1}{2}} \right)$ and arrange the given expression taking the first and last term within a bracket, then the second and the previous last within a bracket, same for other terms. There are a total of $19$ brackets. All the terms within brackets are of the form $f\left( x \right) + f\left( {1 - x} \right)$. Substitute $f\left( x \right) + f\left( {1 - x} \right) = 1$.

Formula Used:
${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$
Average of two numbers $a$ and $b$ is $\dfrac{1}{2}\left( {a + b} \right)$

Complete step by step solution:
The given function is $f\left( x \right) = \dfrac{{{4^x}}}{{{4^x} + 2}} - - - - - \left( i \right)$, where $f:\left[ {0,1} \right] \to \mathbb{R}$
If we replace $x$ by $\left( {1 - x} \right)$, then $f\left( {1 - x} \right) = \dfrac{{{4^{1 - x}}}}{{{4^{1 - x}} + 2}}$
Use the formula ${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$
$ \Rightarrow f\left( {1 - x} \right) = \dfrac{{\left( {\dfrac{4}{{{4^x}}}} \right)}}{{\left( {\dfrac{4}{{{4^x}}} + 2} \right)}}$
Simplify this expression.
$ \Rightarrow f\left( {1 - x} \right) = \dfrac{{\left( {\dfrac{4}{{{4^x}}}} \right)}}{{\left( {\dfrac{{4 + 2 \cdot {4^x}}}{{{4^x}}}} \right)}} = \dfrac{4}{{{4^x}}} \times \dfrac{{{4^x}}}{{4 + 2 \cdot {4^x}}} = \dfrac{4}{{4 + 2 \cdot {4^x}}}$
Take $2$ as common from numerator and denominator.
$ \Rightarrow f\left( {1 - x} \right) = \dfrac{4}{{2\left( {2 + {4^x}} \right)}} = \dfrac{2}{{2 + {4^x}}} - - - - - \left( {ii} \right)$
Adding $\left( i \right)$ and $\left( {ii} \right)$, we get
$f\left( x \right) + f\left( {1 - x} \right) = \dfrac{{{4^x}}}{{{4^x} + 2}} + \dfrac{2}{{2 + {4^x}}} = \dfrac{{2 + {4^x}}}{{2 + {4^x}}} = 1 - - - - - \left( {iii} \right)$
$\therefore $ The given expression is $f\left( {\dfrac{1}{{40}}} \right) + f\left( {\dfrac{2}{{40}}} \right) + f\left( {\dfrac{3}{{40}}} \right) + ..... + f\left( {\dfrac{{39}}{{40}}} \right) - f\left( {\dfrac{1}{2}} \right)$
Each term of the expression is a value of the function $f\left( x \right)$ at a number. The first term is at $x = \dfrac{1}{{40}}$ and the preceding term of the last term is at $x = \dfrac{{39}}{{40}}$. So, the middle term is at $\dfrac{1}{2}\left( {\dfrac{1}{{40}} + \dfrac{{39}}{{40}}} \right) = \dfrac{1}{2}\left( {\dfrac{{1 + 39}}{{40}}} \right) = \dfrac{1}{2}\left( {\dfrac{{40}}{{40}}} \right) = \dfrac{1}{2} \times 1 = \dfrac{1}{2}$
So, we can rewrite the given expression as
$ = f\left( {\dfrac{1}{{40}}} \right) + f\left( {\dfrac{2}{{40}}} \right) + ..... + f\left( {\dfrac{1}{2}} \right) + ..... + f\left( {\dfrac{{38}}{{40}}} \right) + f\left( {\dfrac{{39}}{{40}}} \right) - f\left( {\dfrac{1}{2}} \right)$
The middle term and the last term in the expression are equal. So, cancel the term $f\left( {\dfrac{1}{2}} \right)$.
$ = f\left( {\dfrac{1}{{40}}} \right) + f\left( {\dfrac{2}{{40}}} \right) + ..... + f\left( {\dfrac{{38}}{{40}}} \right) + f\left( {\dfrac{{39}}{{40}}} \right)$
Arrange the terms.
$ = \left\{ {f\left( {\dfrac{1}{{40}}} \right) + f\left( {\dfrac{{39}}{{40}}} \right)} \right\} + \left\{ {f\left( {\dfrac{2}{{40}}} \right) + f\left( {\dfrac{{38}}{{40}}} \right)} \right\} + ..... + \left\{ {f\left( {\dfrac{{19}}{{40}}} \right) + f\left( {\dfrac{{21}}{{40}}} \right)} \right\}$
From $\left( {iii} \right)$, we can conclude that the values of all the terms within the brackets are equal to $1$. There are a total of $19$ brackets.
So, the required value is sum of $19$ times $1$ i.e. $19$
Hence, the required value is $19$.

Note: Be careful while finding the middle term. If a series follows a pattern then the value of the middle term is the average of the ending values.