Problem 22
Question
(a) Expand \(f(x)=(a+x)^{4}\) by multiplying out or by using Pascal's triangle. (b) Rewrite \(f(x)\) as \(\left[a\left(1+\frac{x}{a}\right)\right]^{4}=a^{4}\left(1+\frac{x}{a}\right)^{4}\). Use the binomial series to expand \(\left(1+\frac{x}{a}\right)^{4}\), multiply by \(a^{4}\), and demonstrate that the result is the same as in part (a).
Step-by-Step Solution
Verified Answer
Following both methods of expanding \(f(x) = (a+x)^4\), we obtain the same result showing that the binomial theorem is correctly applied.
1Step 1: Expand Directly
In the first part, expand \(f(x) = (a+x)^4\) directly using the binomial theorem, which is \((p+q)^n = \sum_{k=0}^{n} {n \choose k} p^{n-k} q^k\). This results in \((a+x)^{4}={4 \choose 0}a^{4}x^{0}+{4 \choose 1}a^{3}x^{1}+{4 \choose 2}a^{2}x^{2}+{4 \choose 3}a^{1}x^{3}+{4 \choose 4}a^{0}x^{4}=a^{4}+4a^{3}x+6a^{2}x^{2}+4ax^{3}+x^{4}\).
2Step 2: Expand Using Binomial Series
For the second part, rewrite \(f(x)\) as \([a(1+\frac{x}{a})]^{4} = a^{4}(1+\frac{x}{a})^{4}\). Using the binomial series to expand \((1+\frac{x}{a})^{4}\) gives \((1+\frac{x}{a})^{4}={4 \choose 0}(1)^{4}(\frac{x}{a})^{0}+{4 \choose 1}(1)^{3}(\frac{x}{a})^{1}+{4 \choose 2}(1)^{2}(\frac{x}{a})^{2}+{4 \choose 3}(1)(\frac{x}{a})^{3}+{4 \choose 4}(\frac{x}{a})^{4}=1+4\frac{x}{a}+6\frac{x^{2}}{a^{2}}+4\frac{x^{3}}{a^{3}}+\frac{x^{4}}{a^{4}}\)
3Step 3: Multiply by \(a^{4}\)
Multiply the expansion from the previous step by \(a^{4}\) to achieve the final expanded form. It is then that we obtain the same result as in part (a): \(a^{4}+4a^{3}x+6a^{2}x^{2}+4ax^{3}+x^{4}\).
4Step 4: Compare Results
Comparing the results of step 1 and step 3, you can affirm they are identical, thus verifying the expansion was accurately performed in both cases.
Key Concepts
Pascal's TrianglePolynomial ExpansionBinomial Series
Pascal's Triangle
Think of Pascal's Triangle as a magical pyramid of numbers. It's a tool that helps you easily find the coefficients you need when expanding binomials using the Binomial Theorem. If you look at Pascal's Triangle, the numbers are arranged in rows. Each row corresponds to the powers of a binomial.
- The top of the triangle starts with the number 1.
- Every row begins and ends with the number 1.
- Each number inside the triangle is the sum of the two numbers directly above it.
Polynomial Expansion
Polynomial expansion is the process of opening up an expression. It involves writing it out in its expanded form, using the Binomial Theorem. For example, expanding \((a + x)^4 \)takes the form of the sum of terms that include powers of \(a\) and \(x\). Each term in the expanded form is determined by:
- The coefficient from Pascal's Triangle.
- Powers of \(a\) that decrease from 4 to 0.
- Powers of \(x\) that increase from 0 to 4.
Binomial Series
The Binomial Series lets you rewrite and expand expressions of the form \( (1 + u)^n \) for any given \( n \). In the problem, it shows how \[ \( a^4(1 + \frac{x}{a})^4 \) \] can be expanded using this series. The Binomial Series works like a charm by employing this expansion:\[(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots\]By using this series, you can set \( u = \frac{x}{a} \). Calculate each term multiplying it with \( a^4 \) and notice the results mirror the expanded form from earlier calculations. This highlights why the Binomial Series is a cornerstone in polynomial mathematics.
- Helps in efficiently handling fractional powers.
- Keeps calculations tidier by converting complex expressions into a series.
- Often leads to recognizable patterns in solutions.
Other exercises in this chapter
Problem 22
Use your knowledge of improper integrals to give an upper and lower bound for \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}\)
View solution Problem 22
Show that \(\sum_{k=0}^{\infty} \frac{(2 x)^{x}}{k !}\) is a solution to the differential equation \(f^{\prime}(x)=2 f(x) .\) What familiar function does this s
View solution Problem 23
Let \(\sum_{k=1}^{\infty} a_{k}\) be a series and \(S_{k}=a_{1}+a_{2}+\cdots+a_{k}\) its \(k\) th partial sum, where \(k=1,2,3, \ldots\) Let \(L\) be a constant
View solution Problem 23
Show that if \(f(x)=\sum_{k=0}^{\infty} a_{k} x^{k}\) is a power series solution to \(f^{\prime}(x)=-f(x)\), then \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{k}
View solution