Let g(x) = log(f(x)) where f(x) is twice differentiable positive function on (0, ∞) such that f(x +

Q: Why do we take the logarithm of the functional equation? Can't we just differentiate $f(x+1) = x f(x)$ directly?

Great question! If you differentiate $f(x+1) = x f(x)$ directly, you get $f'(x+1) = f(x) + x f'(x)$. This introduces $f'(x)$ and $f(x)$ in a way that doesn't easily lead to the second derivative of $\log(f(x))$. By taking the log first, we transform the product $x f(x)$ into a sum $\log x + \log f(x)$, which turns the functional equation into a much simpler difference equation for $g(x)$. It's all about choosing the right tool for the job!

Q: How do we know it's a telescoping series? Is there a general rule?

A telescoping series occurs whenever you have a sum of terms of the form $a_{k+1} - a_k$. When you add these up, the positive part of one term cancels the negative part of the next. Whenever you see a difference equation like $g''(x+1) - g''(x) = \text{something}$, it is a massive hint that summing over a range will cause the intermediate values to vanish, leaving only the boundary terms. It's one of the most elegant techniques in calculus!

Q: Why does the summation go from $k=0$ to $N-1$ and not $N$?

Think about the boundary terms. We want the expression $g''(N + \frac{1}{2}) - g''(\frac{1}{2})$. If we sum from $k=0$ to $N-1$, the telescoping sum expands as $(g''(\frac{3}{2}) - g''(\frac{1}{2})) + (g''(\frac{5}{2}) - g''(\frac{3}{2})) + \dots + (g''(N + \frac{1}{2}) - g''(N - \frac{1}{2}))$. The terms $g''(\frac{3}{2}), g''(\frac{5}{2}), \dots$ all cancel out, leaving exactly the first and last terms we need. If we went to $N$, we would have an extra term $g''(N + 1 + \frac{1}{2})$ that we don't need.

Question

Let $g(x) = \log(f(x))$ where $f(x)$ is twice differentiable positive function on $(0, \infty)$ such that $f(x+1) = x f(x)$. Then, for $N = 1, 2, 3, \dots\dots\dots$ $g''\left(N + \frac{1}{2}\right) - g''\left(\frac{1}{2}\right) =$

Options:

(A) $-4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N-1)^2} \right\}$

(B) $4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N-1)^2} \right\}$

(C) $-4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N+1)^2} \right\}$

(D) $4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N+1)^2} \right\}$

Sigma Percentile · Accepted Answer

### The Beauty of Functional Equations

Welcome, fellow traveler in the world of mathematics! Today, we are going to unravel a problem that might look intimidating at first glance, but beneath its surface lies a beautiful, rhythmic structure.

We are dealing with a functional equation, $f(x+1) = x f(x)$, which is the defining property of the Gamma function. Our mission is to find the value of $g''(N + \frac{1}{2}) - g''(\frac{1}{2})$, where $g(x) = \log(f(x))$.

### Phase 1: The Logarithmic Bridge

We start with the given functional equation: $f(x+1) = x f(x)$. Since our target function $g(x)$ is defined as the natural logarithm of $f(x)$, we apply the logarithm to both sides of our equation.

Using the property $\log(ab) = \log a + \log b$, we transform the product on the right into a sum:

$$\log(f(x+1)) = \log x + \log(f(x))$$

Substituting $g(x) = \log(f(x))$, the equation becomes remarkably clean:

$$g(x+1) = \log x + g(x)$$

Rearranging this, we get a difference equation: $g(x+1) - g(x) = \log x$. This is the foundation of our entire solution.

### Phase 2: The Calculus Descent

Our goal is to find the second derivative, $g''(x)$. Let's differentiate our difference equation with respect to $x$. The first derivative gives us:

$$g'(x+1) - g'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}$$

We need the second derivative, so we differentiate once more with respect to $x$:

$$g''(x+1) - g''(x) = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}$$

This is the core relationship. It tells us exactly how the second derivative changes as we shift the argument by one.

### Phase 3: The Telescoping Symphony

We need to evaluate $g''(N + \frac{1}{2}) - g''(\frac{1}{2})$. This structure suggests a "telescoping series." To get from $\frac{1}{2}$ to $N + \frac{1}{2}$, we sum our core relationship $g''(x+1) - g''(x) = -\frac{1}{x^2}$ by substituting $x = k + \frac{1}{2}$ for $k = 0, 1, 2, \dots, N-1$.

Substituting $x = k + \frac{1}{2}$ into our relation, we get:

$$g''(k + 1 + \frac{1}{2}) - g''(k + \frac{1}{2}) = -\frac{1}{(k + \frac{1}{2})^2}$$

Simplifying the right-hand side, we note that $(k + \frac{1}{2})^2 = (\frac{2k+1}{2})^2 = \frac{(2k+1)^2}{4}$. Therefore, the term becomes:

$$-\frac{1}{(k + \frac{1}{2})^2} = -\frac{4}{(2k+1)^2}$$

Now, we sum both sides from $k=0$ to $N-1$:

$$\sum_{k=0}^{N-1} [g''(k + \frac{3}{2}) - g''(k + \frac{1}{2})] = \sum_{k=0}^{N-1} -\frac{4}{(2k+1)^2}$$

On the left side, the terms cancel out like falling dominoes:

$$(g''(\frac{3}{2}) - g''(\frac{1}{2})) + (g''(\frac{5}{2}) - g''(\frac{3}{2})) + \dots + (g''(N + \frac{1}{2}) - g''(N - \frac{1}{2}))$$

Everything in the middle vanishes, leaving us with $g''(N + \frac{1}{2}) - g''(\frac{1}{2})$.

### The Final Result

On the right side, we expand the sum to obtain the final expression:

$$g''(N + \frac{1}{2}) - g''(\frac{1}{2}) = -4 \sum_{k=0}^{N-1} \frac{1}{(2k+1)^2}$$

This expands to:

$$-4 \left\{ \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots + \frac{1}{(2N-1)^2} \right\}$$

We have navigated the functional equation, applied calculus, and utilized the power of telescoping series to arrive at the **final result**.

Animated Solution for Mathematics - Differentiation: Let $g (x) = lo g (f (x))$ where $f (x)$ is twice differentiable positive function on $(0, \infty)$ such that $f (x + 1) = x f (x)$ . Then, for $N = 1, 2, 3, \dots\dots\dots$ $g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =$

Select Answer:

Visualized Solution

Problem Setup

Logarithmic Transformation

Introducing $g (x)$

First Differentiation

Second Differentiation

Setting Up the Telescoping Series

Substituting $x = k + \frac{1}{2}$

Simplifying the RHS

Applying the Summation

Telescoping Cancellation

Evaluating the RHS Series

Final Result

The Sigma Insight: Higher Order Derivatives

The Beauty of Functional Equations

Phase 1

Phase 2

Phase 3

The Final Result

Similar Questions

Let $f : S \to S$ where $S = (0, \infty)$ be a twice differentiable function such that $f (x + 1) = x f (x)$ . If $g : S \to R$ be defined as $g (x) = lo g_{e} f (x)$ , then the value of $∣ g^{''} (5) - g^{''} (1) ∣$ is equal to :

There exist a function $f (x)$ , satisfying $f (0) = 1, f^{'} (0) = - 1, f (x) > 0$ for all $x$ , and

Let $f$ and $g$ be twice differentiable functions on $R$ such that $f^{''} (x) = g^{''} (x) + 6 x$ , $f^{'} (1) = 4 g^{'} (1) - 3 = 9$ , $f (2) = 3 g (2) = 12$ . Then which of the following is NOT true?

Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equal :

Let $f : R \to R$ be a twice differentiable function such that $(sin x cos y) (f (2 x + 2 y) - f (2 x - 2 y)) = (cos x sin y) (f (2 x + 2 y) + f (2 x - 2 y))$ for all $x, y \in R$ . If $f^{'} (0) = 1/2$ , then the value of $24 f^{''} (5 π /3)$ is:

If $f (x) = {x^{3} sin (\frac{1}{x}), 0, x \neq = 0 x = 0$ then

If $f (x) = x^{3} - x^{2} f^{'} (1) + x f^{''} (2) - f^{'''} (3), x \in R$ , then

If $y^{2} + lo g_{e} (cos^{2} x) = y, x \in (- \frac{π}{2}, \frac{π}{2})$ , then :

Let $y = lo g_{e} (\frac{1 - x ^{2}}{1 + x ^{2}}), - 1 < x < 1$ . Then at $x = \frac{1}{2}$ , the value of $225 (y^{'} - y^{''})$ is equal to

Let $f$ be a twice differentiable function such that $f^{''} (x) = - f (x)$ , and $f^{'} (x) = g (x), h (x) = [f (x)]^{2} + [g (x)]^{2}$ . Find $h (10)$ if $h (5) = 11$ .