Q8Advanced Engineering Mathematics

The problem asks us to evaluate the forward difference operator $\Delta$ applied to the function $f(n)={\tan }^{-1}\left(\frac{n-1}{n}\right)$. The forward difference operator for a discrete function $f(n)$ with a unit step size ($h=1$) is fundamentally defined as $\Delta f(n)=f(n+1)-f(n)$. By applying this definition directly to our specific function, we obtain the initial expression:

Simplifying the arguments of the arctangent functions inside the expression yields:

Applying Trigonometric Identities

To simplify the difference of two inverse tangent functions, we rely on the standard inverse trigonometric identity for subtraction, which states that ${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)$. We assign $A=\frac{n}{n+1}$ and $B=\frac{n-1}{n}$. Substituting these terms into the identity formulation:

Now, we algebraically simplify the complex fraction by finding a common denominator for both the numerator and the denominator parts.

By multiplying the numerator and denominator by $n(n+1)$ to clear the fractions, we obtain:

Expanding the polynomial terms in the denominator: $n^2+n+n^2-n=2n^2$. Therefore, the final, fully simplified expression is exactly ${\tan }^{-1}\left(\frac{1}{2n^2}\right)$. This mathematically completes the required proof.