$$ 1 imes 3 imes 5 imes .... imes (2x-1)

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$$ 1	imes 3	imes 5	imes ....	imes (2x-1)

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The inequality you've presented involves the product of the first $$ x $$ odd integers compared to $$ x^x $$: $$ 1 \times 3 \times 5 \times \dots \times (2x - 1) < x^x $$ Let's analyze and understand this inequality step by step. ## Understanding the Components 1. **Product of Odd Integers:** The left-hand side (LHS) of the inequality is the product of the first $$ x $$ odd integers: $$ 1 \times 3 \times 5 \times \dots \times (2x - 1) $$ This product is known as the **double factorial** of $$ (2x - 1) $$, denoted as $$ (2x - 1)!! $$. The double factorial for an odd number is defined as: $$ (2x - 1)!! = 1 \times 3 \times 5 \times \dots \times (2x - 1) $$ 2. **Right-Hand Side $$ x^x $$:** The right-hand side (RHS) is simply $$ x $$ raised to the power of $$ x $$. ## Objective We want to determine for which values of $$ x $$ the following inequality holds: $$ (2x - 1)!! < x^x $$ ## Analyzing the Inequality ### 1. **Base Cases:** Let's test the inequality for small positive integer values of $$ x $$: - **For $$ x = 1 $$:** $$ (2(1) - 1)!! = 1 < 1^1 = 1 \quad (\text{Equality holds}) $$ - **For $$ x = 2 $$:** $$ 1 \times 3 = 3 < 2^2 = 4 \quad (\text{Inequality holds}) $$ - **For $$ x = 3 $$:** $$ 1 \times 3 \times 5 = 15 < 3^3 = 27 \quad (\text{Inequality holds}) $$ - **For $$ x = 4 $$:** $$ 1 \times 3 \times 5 \times 7 = 105 < 4^4 = 256 \quad (\text{Inequality holds}) $$ - **For $$ x = 5 $$:** $$ 1 \times 3 \times 5 \times 7 \times 9 = 945 < 5^5 = 3125 \quad (\text{Inequality holds}) $$ These base cases suggest that the inequality holds for $$ x \geq 2 $$. ### 2. **Asymptotic Analysis:** To understand whether the inequality holds for larger values of $$ x $$, let's perform an asymptotic analysis using Stirling's approximation. #### Stirling's Approximation for Factorials: Stirling's formula approximates the factorial of a large number $$ n $$ as: $$ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n $$ #### Approximating the Double Factorial: The double factorial $$ (2x - 1)!! $$ can be expressed in terms of the regular factorial: $$ (2x - 1)!! = \frac{(2x)!}{2^x x!} $$ Using Stirling's approximation for both $$ (2x)! $$ and $$ x! $$: $$ (2x)! \approx \sqrt{4\pi x} \left(\frac{2x}{e}\right)^{2x} $$ $$ x! \approx \sqrt{2\pi x} \left(\frac{x}{e}\right)^x $$ Plugging these into the expression for $$ (2x - 1)!! $$: $$ (2x - 1)!! \approx \frac{\sqrt{4\pi x} \left(\frac{2x}{e}\right)^{2x}}{2^x \sqrt{2\pi x} \left(\frac{x}{e}\right)^x} = \sqrt{2} \cdot \frac{(2x)^{2x}}{e^{2x} \cdot 2^x \cdot x^x \cdot e^{-x}} = \sqrt{2} \cdot 2^x \cdot x^x \cdot e^{-x} $$ So, $$ (2x - 1)!! \approx \sqrt{2} \cdot 2^x \cdot x^x \cdot e^{-x} $$ #### Comparing with $$ x^x $$: The inequality $$ (2x - 1)!! < x^x $$ becomes: $$ \sqrt{2} \cdot 2^x \cdot x^x \cdot e^{-x} < x^x $$ Dividing both sides by $$ x^x $$ (which is positive for $$ x > 0 $$): $$ \sqrt{2} \cdot 2^x \cdot e^{-x} < 1 $$ Taking the natural logarithm of both sides: $$ \ln(\sqrt{2}) + x \ln(2) - x < 0 $$ $$ \frac{1}{2}\ln(2) + x(\ln(2) - 1) < 0 $$ Let $$ \ln(2) \approx 0.6931 $$, so: $$ \frac{1}{2} \times 0.6931 + x(0.6931 - 1) < 0 $$ $$ 0.3466 - 0.3069x < 0 $$ $$ 0.3069x > 0.3466 $$ $$ x > \frac{0.3466}{0.3069} \approx 1.127 $$ ### 3. **Conclusion from Asymptotic Analysis:** The analysis shows that for $$ x > 1.127 $$, the inequality $$ \sqrt{2} \cdot 2^x \cdot e^{-x} < 1 $$ holds, implying: $$ (2x - 1)!! < x^x \quad \text{for} \quad x > 1.127 $$ Since $$ x $$ is typically considered an integer in this context, the inequality holds for all integers $$ x \geq 2 $$. ## Summary - **For $$ x = 1 $$:** $$ 1 = 1 $$ (equality holds) - **For $$ x \geq 2 $$:** $$ (2x - 1)!! < x^x $$ holds true Therefore, the inequality $$ 1 \times 3 \times 5 \times \dots \times (2x - 1) < x^x $$ is **true for all integers $$ x \geq 2 $$**. The inequality $$1 \times 3 \times 5 \times \dots \times (2x - 1) < x^x$$ holds true for all integers $$x \geq 2$$.

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