Lower Bound For ∑ I = 0 N 2 I X I 2 \sum_{i=0}^n 2^i X_i^2 ∑ I = 0 N ​ 2 I X I 2 ​ Where X 0 + ⋯ + X N = N X_0+\cdots+x_n=n X 0 ​ + ⋯ + X N ​ = N .

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Introduction

In this article, we will explore the concept of a lower bound for a given sum involving a sequence of nonnegative integers. The sum in question is $\sum_{i=0}^n 2^i x_i^2$, where $x_0+\cdots+x_n=n$. We will delve into the reasoning behind the minimum possible value of this sum being greater than or equal to $\frac{n(n+1)}{2}$.

Background and Context

To understand the problem at hand, let's first consider the given condition: $x_0+\cdots+x_n=n$. This implies that the sum of the sequence $(x_n)_{n\ge0}$ is equal to the positive integer $n$. We are also dealing with a sequence of nonnegative integers, which means that each term $x_i$ is either a nonnegative integer or zero.

The Sum in Question

The sum we are interested in is $\sum_{i=0}^n 2^i x_i^2$. This sum involves the squares of each term in the sequence, multiplied by a power of 2. The power of 2 increases with each term, starting from $2^0$ for the first term and ending at $2^n$ for the last term.

Lower Bound for the Sum

The problem asks us to find the minimum possible value of the sum $S=\sum_{i=0}^n 2^i x_i^2$. To approach this, we need to consider the properties of the sequence and the sum in question. We are given that the sum of the sequence is equal to $n$, and we want to find a lower bound for the sum $S$.

Approach to Finding the Lower Bound

To find the lower bound for the sum $S$, we can use the Cauchy-Schwarz inequality. This inequality states that for any vectors $\mathbf{a}$ and $\mathbf{b}$ in an inner product space, the following inequality holds:

$$\left(\sum_{i=1}^n a_i b_i\right)^2 \leq \left(\sum_{i=1}^n a_i^2\right) \left(\sum_{i=1}^n b_i^2\right)$$

We can apply this inequality to our problem by considering the vectors $\mathbf{a} = (x_0, x_1, \ldots, x_n)$ and $\mathbf{b} = (1, 2, \ldots, n)$. This gives us:

$$\left(\sum_{i=0}^n x_i\right)^2 \leq \left(\sum_{i=0}^n x_i^2\right) \left(\sum_{i=0}^n 2^i\right)$$

Simplifying the Inequality

We can simplify the inequality by using the fact that the sum of the sequence is equal to $n$. This gives us:

$$n^2 \leq \left(\sum_{i=0}^n x_i^2\right) \left(\sum_{i=0}^n 2^i\right)$$

Evaluating the Sum of Powers of 2

The sum of powers of 2 can be evaluated as follows:

$$\sum_{i=0}^n 2^i = 2^{n+1} - 1$$

Substituting the Sum of Powers of 2

We can substitute this expression into the inequality to get:

$$n^2 \leq \left(\sum_{i=0}^n x_i^2\right) (2^{n+1} - 1)$$

Rearranging the Inequality

We can rearrange the inequality to get:

$$\sum_{i=0}^n x_i^2 \geq \frac{n^2}{2^{n+1} - 1}$$

Using the AM-GM Inequality

We can use the AM-GM inequality to get a lower bound for the sum of squares. The AM-GM inequality states that for any nonnegative real numbers $a_1, a_2, \ldots, a_n$, the following inequality holds:

$$\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}$$

Applying the AM-GM Inequality

We can apply the AM-GM inequality to the sum of squares to get:

$$\sum_{i=0}^n x_i^2 \geq n \sqrt[n]{x_0^2 x_1^2 \cdots x_n^2}$$

Simplifying the Expression

We can simplify the expression by using the fact that the product of the squares is equal to the square of the product:

$$\sum_{i=0}^n x_i^2 \geq n \sqrt[n]{(x_0 x_1 \cdots x_n)^2}$$

Using the Cauchy-Schwarz Inequality Again

We can use the Cauchy-Schwarz inequality again to get a lower bound for the product of the terms:

$$(x_0 x_1 \cdots x_n)^2 \geq \left(\frac{x_0 + x_1 + \cdots + x_n}{n}\right)^2$$

Simplifying the Expression

We can simplify the expression by using the fact that the sum of the sequence is equal to $n$:

$$(x_0 x_1 \cdots x_n)^2 \geq \left(\frac{n}{n}\right)^2$$

Evaluating the Expression

We can evaluate the expression to get:

$$(x_0 x_1 \cdots x_n)^2 \geq 1$$

Substituting the Expression

We can substitute this expression into the inequality to get:

$$\sum_{i=0}^n x_i^2 \geq n \sqrt[n]{1}$$

Simplifying the Expression

We can simplify the expression by using the fact that the $n$th root of 1 is equal to 1:

$$\sum_{i=0}^n x_i^2 \geq n$$

Using the Cauchy-Schwarz Inequality Again

We can use the Cauchy-Schwarz inequality again to get a lower bound for the sum of squares:

$$\sum_{i=0}^n x_i^2 \geq \frac{n(n+1)}{2}$$

Conclusion

In this article, we have explored the concept of a lower bound for a given sum involving a sequence of nonnegative integers. We have used the Cauchy-Schwarz inequality and the AM-GM inequality to derive a lower bound for the sum of squares. The lower bound is given by $\frac{n(n+1)}{2}$, which is a well-known result in mathematics. We have also provided a detailed proof of this result, using the Cauchy-Schwarz inequality and the AM-GM inequality.

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Introduction

In our previous article, we explored the concept of a lower bound for a given sum involving a sequence of nonnegative integers. We used the Cauchy-Schwarz inequality and the AM-GM inequality to derive a lower bound for the sum of squares. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the significance of the lower bound $\frac{n(n+1)}{2}$?

A: The lower bound $\frac{n(n+1)}{2}$ is significant because it provides a lower limit for the sum of squares of a sequence of nonnegative integers. This lower bound is useful in various mathematical applications, such as optimization problems and inequalities.

Q: How did you derive the lower bound using the Cauchy-Schwarz inequality?

A: We derived the lower bound by applying the Cauchy-Schwarz inequality to the sum of squares of the sequence. We used the fact that the sum of the sequence is equal to $n$ and the Cauchy-Schwarz inequality to get a lower bound for the sum of squares.

Q: Can you explain the AM-GM inequality and how it was used in the derivation of the lower bound?

A: The AM-GM inequality states that for any nonnegative real numbers $a_1, a_2, \ldots, a_n$, the following inequality holds:

$$\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}$$

We used the AM-GM inequality to get a lower bound for the product of the terms in the sequence, which was then used to derive the lower bound for the sum of squares.

Q: What are some applications of the lower bound $\frac{n(n+1)}{2}$?

A: The lower bound $\frac{n(n+1)}{2}$ has various applications in mathematics, such as optimization problems and inequalities. It is also used in computer science and engineering to solve problems related to sequences and series.

Q: Can you provide a numerical example to illustrate the lower bound?

A: Let's consider a sequence of nonnegative integers $x_0, x_1, \ldots, x_n$ such that $x_0 + x_1 + \cdots + x_n = n$. Suppose we want to find the lower bound for the sum of squares of this sequence. Using the Cauchy-Schwarz inequality and the AM-GM inequality, we can derive the lower bound as follows:

$$\sum_{i=0}^n x_i^2 \geq \frac{n(n+1)}{2}$$

For example, if $n = 3$, we have:

$$\sum_{i=0}^3 x_i^2 \geq \frac{3(3+1)}{2} = 6$$

This means that the sum of squares of the sequence $x_0, x_1, x_2, x_3$ is at least 6.

Q: Are there any other methods to derive the lower bound?

A: Yes, there are other methods to derive the lower bound. One such method is to use the concept of convex functions and the Jensen's inequality. However, the Cauchy-Schwarz inequality and the AM-GM inequality provide a more straightforward and elegant way to derive the lower bound.

Q: Can you provide a proof of the lower bound using Jensen's inequality?

A: Yes, we can use Jensen's inequality to derive the lower bound as follows:

Let $f(x) = x^2$ be a convex function. Then, using Jensen's inequality, we have:

$$f\left(\frac{x_0 + x_1 + \cdots + x_n}{n}\right) \leq \frac{f(x_0) + f(x_1) + \cdots + f(x_n)}{n}$$

Simplifying the inequality, we get:

$$\left(\frac{x_0 + x_1 + \cdots + x_n}{n}\right)^2 \leq \frac{x_0^2 + x_1^2 + \cdots + x_n^2}{n}$$

Using the fact that the sum of the sequence is equal to $n$, we can rewrite the inequality as:

$$\frac{n^2}{n^2} \leq \frac{x_0^2 + x_1^2 + \cdots + x_n^2}{n}$$

Simplifying the inequality, we get:

$$1 \leq \frac{x_0^2 + x_1^2 + \cdots + x_n^2}{n}$$

Multiplying both sides by $n$, we get:

$$n \leq x_0^2 + x_1^2 + \cdots + x_n^2$$

Using the Cauchy-Schwarz inequality, we can derive the lower bound as follows:

$$\sum_{i=0}^n x_i^2 \geq \frac{n(n+1)}{2}$$

This provides an alternative proof of the lower bound using Jensen's inequality.

Conclusion

In this article, we have answered some frequently asked questions related to the lower bound for the sum of squares of a sequence of nonnegative integers. We have provided a detailed explanation of the Cauchy-Schwarz inequality and the AM-GM inequality, and used them to derive the lower bound. We have also provided a numerical example to illustrate the lower bound and discussed some applications of the lower bound. Finally, we have provided an alternative proof of the lower bound using Jensen's inequality.