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Chapter 2: Problem 39

Expand (write out in full):\(\sum_{n=0}^{3} n ! x^{n^{2}}\)

Short Answer

Expert verified
1 + x + 2x^4 + 6x^9

Step by step solution

01

Understanding the Problem

The exercise asks us to expand the summation \( \sum_{n=0}^{3} n! x^{n^2} \). This means we need to calculate each term from \( n = 0 \) to \( n = 3 \) and write them out in full.
02

Calculate Each Term

We calculate the term for each \( n \) value from 0 to 3:- For \( n = 0 \), the term is \( 0! x^{0^2} = 1 \cdot x^0 = 1 \).- For \( n = 1 \), the term is \( 1! x^{1^2} = 1 \cdot x^1 = x \).- For \( n = 2 \), the term is \( 2! x^{2^2} = 2 \cdot x^4 = 2x^4 \).- For \( n = 3 \), the term is \( 3! x^{3^2} = 6 \cdot x^9 = 6x^9 \).
03

Write the Expanded Sum

We now write out the expanded form by summing all the terms we calculated:\[ \sum_{n=0}^{3} n! x^{n^2} = 1 + x + 2x^4 + 6x^9 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Summation Notation
Summation notation is a way to compactly express the sum of a sequence of terms. It's represented by the sigma symbol \( \Sigma \). The general structure is \( \sum_{n=a}^{b} f(n) \), where:
  • \(a\) is the lower bound, indicating the starting value for \(n\).
  • \(b\) is the upper bound, showing the ending value for \(n\).
  • \(f(n)\) is the expression we calculate and sum as \(n\) varies from \(a\) to \(b\).
For example, in \(\sum_{n=0}^{3} n! x^{n^2}\), you calculate the expression \(n! x^{n^2}\) for each \(n\) starting from 0 to 3.
This means you'll compute values separately for each \(n\) and then add them together. This provides a handy way to organize repetitive calculations efficiently.
Factorials
Factorials are a fundamental concept in mathematics that recur often in probability, combinatorics, and polynomial expansions. The factorial of a non-negative integer \(n\) is the product of all positive integers less than or equal to \(n\). It is represented as \(n!\).
Here鈥檚 a simple breakdown:
  • \(0! = 1\) by definition, and it's a cornerstone for many mathematical principles.
  • \(1! = 1\)
  • \(2! = 2 \times 1 = 2\)
  • \(3! = 3 \times 2 \times 1 = 6\)
  • And so on.
Factorials grow very rapidly with respect to \(n\), and you can observe this from how quickly the product escalates. In the exercise given, factorials are part of each term being expanded, helping determine the coefficient that is multiplied by each power of \(x\).
Exponents
Exponents are a short form to express repeated multiplication of the same base number. They have a structure \(b^n\) where:
  • \(b\) is the base - the number being multiplied.
  • \(n\) is the exponent - the number of times the base is used as a factor.
For example, in \(x^{n^2}\), \(x\) is the base and \(n^2\) is the power to which \(x\) is raised.
Consider some examples:
  • \(x^0 = 1\), because any non-zero number raised to zero is one.
  • \(x^1 = x\), just the base itself.
  • \(x^2 = x \times x\)
  • \(x^3 = x \times x \times x\)
In our problem, each term involves calculating \(x^{n^2}\), meaning the power depends on the square of \(n\). This is crucial in determining the degree of each term in the expanded polynomial.

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Most popular questions from this chapter

Make a table of \((x, y)\) values and sketch a fully labelled graph of the quadratic:\(y=-x^{2}-x+2\)

Solve the simultaneous equations:\(2 x-y=2, x^{2}-x y+y^{2}=1\)

Find \(x\) as a function of \(y=$$y=\frac{2 x+3}{3 x-2}\)

Factorize:\(x^{2}-8 x+15\)

Simplify if possible:\(\frac{2 x^{2}-3 x+1}{x^{2}-3 x+2}\)
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