Let a, b, c be positive real numbers such that b² - 4ac 0 and let α₁ = c. Prove by induction that

Q: Wait, why do we assume $\alpha_{k+1} < \frac{\alpha_k}{2}$? Can't we just solve for $\alpha_n$ directly?

That is a brilliant instinct! In many sequences, we look for a closed-form solution. However, this recurrence is non-linear and coupled—the denominator depends on the sum of all previous terms. Solving for a closed form is analytically intractable. Induction is our scalpel here; it allows us to prove the behavior of the sequence without needing to know the exact value of every term.

Q: Why is the sum $S_n$ strictly less than $2c$?

Think of the sequence as a geometric decay. If $\alpha_1 = c$ and each subsequent term is less than half the previous one, then $\alpha_2 < c/2$, $\alpha_3 < c/4$, and so on. The sum $S_n$ is bounded by the infinite geometric series $c + c/2 + c/4 + \dots$, which sums to $c / (1 - 1/2) = 2c$. Since our finite sum has fewer terms and each term is smaller than the geometric progression, it must be strictly less than $2c$.

Q: What happens if $b^2 - 4ac = 0$? Does the proof break?

Yes, it does! If $b^2 - 4ac = 0$, the denominator could potentially approach zero as the sum $S_n$ approaches $2c$. The strict inequality $b^2 - 4ac > 0$ is the 'safety buffer' that ensures the denominator stays strictly positive. Without it, the sequence could become undefined.

Question

Let $a, b, c$ be positive real numbers such that $b^2 - 4ac > 0$ and let $\alpha_1 = c$. Prove by induction that $\alpha_{n+1} = \frac{a\alpha_n^2}{b^2 - 2a(\alpha_1 + \alpha_2 + .... + \alpha_n)}$ is well-defined and $\alpha_{n+1} < \frac{\alpha_n}{2}$ for all $n = 1, 2, ...$ (Here, 'well-defined' means that the denominator in the expression for $\alpha_{n+1}$ is not zero.)

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing at the edge of a cliff, looking at a sequence of numbers that seem to be shrinking into nothingness. This is not just any sequence; it is a dynamic, self-regulating system where each new term depends on the history of everything that came before it. Today, we are going to tame this beast using the power of Mathematical Induction. We begin with the first term, $\alpha_1 = c$. We are given a recurrence relation: $$ \alpha_{n+1} = \frac{a\alpha_n^2}{b^2 - 2a(\alpha_1 + \alpha_2 + \dots + \alpha_n)} $$ Our mission is to prove that this sequence is well-defined and that each term is less than half of its predecessor. ### The Base Case: Testing the Waters Let us test the waters with $n=1$. We calculate: $$ \alpha_2 = \frac{a\alpha_1^2}{b^2 - 2a\alpha_1} = \frac{ac^2}{b^2 - 2ac} $$ We know $b^2 - 4ac > 0$, which implies $b^2 > 4ac$. Therefore, the denominator $b^2 - 2ac > 4ac - 2ac = 2ac$. Since $a, c > 0$, the denominator is strictly positive. Not only is it well-defined, but because the denominator is larger than $2ac$, the fraction $\alpha_2$ must be smaller than $\frac{ac^2}{2ac} = \frac{c}{2}$. Since $\alpha_1 = c$, we have successfully proven $\alpha_2 < \frac{\alpha_1}{2}$. The base case is solid. ### The Inductive Leap: The Power of Assumption Now, we step into the heart of the proof. We assume the statement holds for all integers $k$ up to $n$. This means we assume $\alpha_{k+1} < \frac{\alpha_k}{2}$ for all $k \le n$. This assumption is our bridge. If we chain these inequalities, we see a beautiful pattern emerging: $\alpha_2 < \frac{c}{2}$, $\alpha_3 < \frac{c}{4}$, and in general, $\alpha_k < \frac{c}{2^{k-1}}$. This tells us that the sequence is decaying exponentially. ### The Geometric Insight: Bounding the Sum The denominator of our recurrence relation contains the sum $S_n = \sum_{i=1}^n \alpha_i$. To prove the next step, we need to control this sum. Using our inductive hypothesis, we can bound $S_n$ by an infinite geometric series: $$ S_n < c + \frac{c}{2} + \frac{c}{4} + \dots + \frac{c}{2^{n-1}} $$ The sum of this infinite geometric progression is $\frac{c}{1 - 1/2} = 2c$. Thus, we have established a crucial bound: **$S_n < 2c$**. This is the key that unlocks the rest of the problem. ### The Final Victory: Proving the Inequality We now look at the denominator for $\alpha_{n+1}$, which is $D = b^2 - 2a S_n$. Since $S_n < 2c$, we know that $2a S_n < 4ac$. Therefore, $D = b^2 - 2a S_n > b^2 - 4ac$. Because we were given $b^2 - 4ac > 0$, the denominator $D$ is strictly positive. The sequence is well-defined! Finally, to prove $\alpha_{n+1} < \frac{\alpha_n}{2}$, we manipulate the denominator. We can show that $D > 2a\alpha_n$ by using the fact that $S_n + \alpha_n < 2c$. Substituting this into the recurrence, we get: $$ \alpha_{n+1} = \frac{a\alpha_n^2}{D} < \frac{a\alpha_n^2}{2a\alpha_n} = \frac{\alpha_n}{2} $$ The inequality holds! We have tamed the sequence. By the Principle of Mathematical Induction, the property holds for all $n$.

Visualized Solution

$I n t r o d u c t i o n an d B a se C a se S e t u p$

$C a l c u l a t in g α_{2}$

$C h ec k in g W e l l - d e f in e d n ess f or n = 1$

$V er i f y in g t h e I n e q u a l i t y f or n = 1$

$T h e I n d u c t i v e H y p o t h es i s$

$B o u n d in g t h e S u m o f T er m s$

$W e l l - d e f in e d n ess f or n + 1$

$P r o v in g t h e I n e q u a l i t y f or n + 1$

$C o n c l u s i o n o f t h e P r oo f$

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Base Case

The Inductive Leap

The Geometric Insight

The Final Victory

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Introduction and Base Case Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating α2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Well−definedness for n=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Verifying the Inequality for n=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Inductive Hypothesis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bounding the Sum of Terms

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Well−definedness for n+1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Proving the Inequality for n+1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion of the Proof