(3) If F ( X ) = A X A X + A ; ( A \textgreater 0 , A ≠ 1 F(x)=\frac{a^x}{a^x+\sqrt{a}} \, ; \, (a\ \textgreater \ 0, A \neq 1 F ( X ) = A X + A A X ; ( A \textgreater 0 , A  = 1 ], Find F\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+\cdots+f\left(\frac{99}{100}\right ].

Mar 9, 2025 by ADMIN 287 views

**Solving a Complex Summation Problem in Mathematics**

Introduction

In mathematics, solving summation problems can be a challenging task, especially when dealing with complex functions. In this article, we will explore a specific problem involving a function $f(x)$ and its summation over a range of values. The problem is as follows:

If $f(x)=\frac{a^x}{a^x+\sqrt{a}} \, ; \, (a\ \textgreater \ 0, a \neq 1$ ], find $f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+\cdots+f\left(\frac{99}{100}\right$ ].

Understanding the Function

Before we dive into solving the summation problem, let's take a closer look at the function $f(x)$ . The function is defined as:

f(x)=\frac{a^x}{a^x+\sqrt{a}}

where $a$ is a positive real number greater than 0 and not equal to 1.

Properties of the Function

To better understand the behavior of the function, let's examine its properties. One of the key properties of the function is that it is a rational function, meaning it is the ratio of two polynomials. Specifically, the numerator is a power of $a$ raised to the power of $x$ , while the denominator is the sum of a power of $a$ raised to the power of $x$ and the square root of $a$ .

Simplifying the Function

To simplify the function, we can rewrite it as:

f(x)=\frac{a^x}{a^x+\sqrt{a}}=\frac{a^x}{a^x(1+\frac{1}{a^{x-\frac{1}{2}}})}=\frac{1}{1+\frac{1}{a^{x-\frac{1}{2}}}}

This simplified form of the function will be useful in solving the summation problem.

Solving the Summation Problem

Now that we have a simplified form of the function, let's tackle the summation problem. We want to find the sum of the function evaluated at $x=\frac{1}{100}, \frac{2}{100}, \ldots, \frac{99}{100}$ .

Using the simplified form of the function, we can rewrite the summation as:

f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+\cdots+f\left(\frac{99}{100}\right)=\sum_{k=1}^{99}\frac{1}{1+\frac{1}{a^{\frac{k}{100}-\frac{1}{2}}}}

Evaluating the Summation

To evaluate the summation, we can use the formula for the sum of a geometric series. Specifically, we can rewrite the summation as:

\sum_{k=1}^{99}\frac{1}{1+\frac{1}{a^{\frac{k}{100}-\frac{1}{2}}}}=\sum_{k=1}^{99}\frac{a^{\frac{k}{100}-\frac{1}{2}}}{a^{\frac{k}{100}-\frac{1}{2}}+1}

Using the formula for the sum of a geometric series, we can rewrite the summation as:

\sum_{k=1}^{99}\frac{a^{\frac{k}{100}-\frac{1}{2}}}{a^{\frac{k}{100}-\frac{1}{2}}+1}=\frac{1}{2}\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}-\frac{1}{2}}}{a^{\frac{k}{100}-\frac{1}{2}}+1}-\frac{1}{a^{\frac{k}{100}-\frac{1}{2}}+1}\right)

Simplifying the Summation

To simplify the summation, we can use the formula for the sum of a geometric series. Specifically, we can rewrite the summation as:

\frac{1}{2}\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}-\frac{1}{2}}}{a^{\frac{k}{100}-\frac{1}{2}}+1}-\frac{1}{a^{\frac{k}{100}-\frac{1}{2}}+1}\right)=\frac{1}{2}\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}-\frac{1}{2}}-1}{a^{\frac{k}{100}-\frac{1}{2}}+1}\right)

Using the formula for the sum of a geometric series, we can rewrite the summation as:

\frac{1}{2}\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}-\frac{1}{2}}-1}{a^{\frac{k}{100}-\frac{1}{2}}+1}\right)=\frac{1}{2}\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}}-a^{-\frac{1}{2}}}{a^{\frac{k}{100}}+a^{-\frac{1}{2}}}\right)

Evaluating the Final Summation

To evaluate the final summation, we can use the formula for the sum of a geometric series. Specifically, we can rewrite the summation as:

\frac{1}{2}\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}}-a^{-\frac{1}{2}}}{a^{\frac{k}{100}}+a^{-\frac{1}{2}}}\right)=\frac{1}{2}\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}}-a^{-\frac{1}{2}}}{a^{\frac{k}{100}}+a^{-\frac{1}{2}}}\right)=\frac{1}{2}\left(\frac{a^{\frac{1}{100}}-a^{-\frac{1}{2}}}{a^{\frac{1}{100}}+a^{-\frac{1}{2}}}\right)\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}}}{a^{\frac{k}{100}}+a^{-\frac{1}{2}}}\right)

Using the formula for the sum of a geometric series, we can rewrite the summation as:

\frac{1}{2}\left(\frac{a^{\frac{1}{100}}-a^{-\frac{1}{2}}}{a^{\frac{1}{100}}+a^{-\frac{1}{2}}}\right)\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}}}{a^{\frac{k}{100}}+a^{-\frac{1}{2}}}\right)=\frac{1}{2}\left(\frac{a^{\frac{1}{100}}-a^{-\frac{1}{2}}}{a^{\frac{1}{100}}+a^{-\frac{1}{2}}}\right)\left(\frac{a^{\frac{1}{100}}}{a^{\frac{1}{100}}+a^{-\frac{1}{2}}}\right)\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}}}{a^{\frac{k}{100}}+a^{-\frac{1}{2}}}\right)

Final Answer

Using the formula for the sum of a geometric series, we can rewrite the summation as:

\frac{1}{2}\left(\frac{a^{\frac{1}{100}}-a^{-\frac{1}{2}}}{a^{\frac{1}{100}}+a^{-\frac{1}{2}}}\right)\left(\frac{a^{\frac{1}{100}}}{a^{\frac{1}{100}}+a^{-\frac{1}{2}}}\right)\sum_{k=1}^{99}\left(\frac{a^{\frac{k}{100}}}{a^{\frac{k}{100}}+a^{-\frac{1}{2}}}\right)=\frac{1}{2}\left(\frac{a^{\frac{1}{100}}-a^{-\frac{1}{2}}}{a^{\frac{1}{100}}+a^{-\frac{1}{2}}}\right)\left(\frac{a^{\frac{1}{100}}}{a^{\frac{1}{100}}+a^{-\frac{1}{2}}}\right)\left(\frac{a^{\frac{99}{100}}-a^{-\frac{1}{2}}}{a^{\frac{99}{100}}+a^{-\frac{1}{2}}}\right)

Simplifying the expression, we get:

\frac{1}{2}\left(\frac{a^{\frac{1}{100}}-a^{-\frac{1}{2}}}{a^{\<br/> **Solving a Complex Summation Problem in Mathematics: Q&A** ===========================================================

Introduction

In our previous article, we explored a complex summation problem involving a function $f(x)$ and its summation over a range of values. In this article, we will provide a Q&A section to help clarify any doubts and provide additional insights into the problem.

Q: What is the function $f(x)$ and how is it defined?

A: The function $f(x)$ is defined as:

Introduction

Understanding the Function

Properties of the Function

Simplifying the Function

Solving the Summation Problem

Evaluating the Summation

Simplifying the Summation

Evaluating the Final Summation

Final Answer

Introduction

Q: What is the function f(x)f(x)f(x) and how is it defined?

Q: What is the function $f(x)$ and how is it defined?