Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove Theorem D as follows: Let $$ f(x)=1+\sum_{n=1}^{\infty}\left(\begin{array}{l} p \\ n \end{array}\right) x^{n} $$ (a) Show that the series converges for \(|x|<1\). (b) Show that \((1+x) f^{\prime}(x)=p f(x)\) and \(f(0)=1\). (c) Solve this differential equation to get \(f(x)=(1+x)^{p}\).

Step-by-Step Solution

Answer

The series converges for \(|x|<1\); differentiating and solving gives \(f(x)=(1+x)^p\).

Step 1: Convergence of the Series

The power series for the function \( f(x) = 1 + \sum_{n=1}^{\infty} \binom{p}{n} x^n \) is examined. To show convergence for \(|x|<1\), we use the Ratio Test. Consider the general term \(a_n = \binom{p}{n} x^n\). The ratio \(\frac{a_{n+1}}{a_n}\) is:\\[\frac{\binom{p}{n+1} x^{n+1}}{\binom{p}{n} x^n} = \frac{\frac{p!}{(n+1)!(p-n-1)!}x}{\frac{p!}{n!(p-n)!}}\]Simplifying gives:\[\frac{p-n}{n+1}x\]As \(n\) approaches infinity, the absolute value of the expression tends to \(|x|\). For the series to converge, we need \(|x| < 1\). Thus, the series converges for \(|x|<1\).

Step 2: Differentiate the Function

Differentiate the function \(f(x)\) with respect to \(x\) to find \(f'(x)\). The differentiation of each term results in:\[f'(x) = \sum_{n=1}^{\infty} n\binom{p}{n}x^{n-1}\]Rewriting using the binomial coefficient identity \(n\binom{p}{n} = p\binom{p-1}{n-1}\), we have:\[f'(x) = \sum_{n=1}^{\infty} p\binom{p-1}{n-1}x^{n-1}\]which shifts the summation index:\[f'(x) = p\sum_{m=0}^{\infty} \binom{p-1}{m}x^m = p\frac{f(x)}{1+x}\], as shown further in the next steps.

Step 3: Verification When x = 0

To confirm \(f(0)=1\), substitute \(x=0\) into the series. The only non-zero term is the constant term:\\[f(0) = 1 + \sum_{n=1}^{\infty} \binom{p}{n} \cdot 0^n = 1\]Hence, \(f(0)=1\).

Step 4: Establish the Differential Equation

Substitute and rearrange the terms from the derivative to establish the differential equation:\[f'(x) = p(1+x)^{p-1}\]Thus from the binomial function we recognize:\[(1+x)f'(x) = pf(x)\]As verified by the differentiated term identity \(p(1+x)^{p-1}\) from Step 2 and 3.

Step 5: Solve the Differential Equation

The differential equation \((1+x)f'(x) = pf(x)\) is separable. Rewrite it as:\[\frac{f'(x)}{f(x)} = \frac{p}{1+x}\]Integrate both sides:\[\int \frac{f'(x)}{f(x)} dx = \int \frac{p}{1+x} dx\]This results in:\[\ln(f(x)) = p\ln(1+x) + C\]Taking exponentials of both sides gives:\[f(x) = e^C(1+x)^p\], where \(e^C=1\) since \(f(0)=1\). Finally, we have:\[f(x) = (1+x)^p\]

Step 6: Final Conclusion

We have shown that \(f(x) = (1+x)^p\) solves the differential equation and satisfies the initial condition. Therefore, the original function and theorem have been proven as required.

Key Concepts

Ratio Test

To determine if the power series converges, we use the Ratio Test, which is a tool for analyzing the convergence of an infinite series. A power series has the form

• \( \, \sum_{n=0}^{\infty} a_n x^n \, \)

where each \( a_n \) is a constant. The Ratio Test specifically examines the ratio of successive terms in a series. This test states that a series

• \( \, \sum a_n \, \)

converges absolutely if the limit

• \( \, \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \, \)

By applying the Ratio Test to our series, we consider the terms

• \( \, a_n = \binom{p}{n} x^n \, \)

The next task is to calculate the ratio of

• \( \, \frac{a_{n+1}}{a_n} \, \).

After simplifying \( \, \frac{\binom{p}{n+1} x^{n+1}}{\binom{p}{n} x^n} = \frac{p-n}{n+1}x \, \)
This ratio simplifies to the absolute value of \( \, |x| \, \). So, for the series to converge,

• \( \, |x| < 1 \, \).

This means the series for \( f(x) \) converges when \( |x| < 1 \).
The Ratio Test ensures our power series becomes a valid representation of \( f(x) \) in this range.

Binomial Coefficient

The binomial coefficient plays a crucial role in expanding expressions and analyzing power series. Generally written as

• \( \, \binom{p}{n} \, \)

it quantifies the number of ways to select \( n \) items from \( p \) without considering the order. This feature often appears in the binomial expansion

• \( \, (1 + x)^p = \sum_{n=0}^{p} \binom{p}{n} x^n \, \)

although in infinite series contexts such as with our function \( f(x) \), it extends indefinitely.
In the presented problem, the binomial coefficient component indicates how each power of \( x \) contributes differing amounts to the formation of \( f(x) \).
Recognizing terms such as

• \( \, n\binom{p}{n} = p\binom{p-1}{n-1} \, \),

we respect its mathematical importance during differentiation and series manipulations.
These coefficients are vital when dealing with changes in index during calculations, as seen when manipulating series like in our differentiation process.

Differential Equation

Understanding how to establish and solve a differential equation is pivotal in connecting the derivation process of a series with finished functional expressions. To solve

• \( \, (1+x)f'(x) = pf(x) \, \),

we first acknowledge its separability. Such equations can often be split to allow integration of each variable independently, leading to a comprehensive solution.
Here, separating the terms yields

• \( \, \frac{f'(x)}{f(x)} = \frac{p}{1+x} \, \).

Integration ensues, constructing a logarithmic relationship

• \( \, \ln(f(x)) = p\ln(1 + x) + C \, \).

Upon exponentiating both sides, this resolves as

• \( \, f(x) = e^C (1+x)^p \, \),

where \( e^C = 1 \) according to initial condition \( f(0) = 1 \).
This relationship showcases how establishing a differential equation can unify series representation with recognizable algebraic forms like powers of \( (1+x) \). Developing such equations provides a bridge from theoretical derivation to practical application.

Function Differentiation

Differentiation helps in understanding how functions change, particularly through the lens of rate and slope analysis. Differentiating

• \( \, f(x) = 1 + \sum_{n=1}^{\infty} \binom{p}{n} x^n \, \)

involves evaluating the rate of change of each term.
The differentiation of each term yields

• \( \, n\binom{p}{n}x^{n-1} \, \),

which transforms the entire function into

• \( \, f'(x) = \sum_{n=1}^{\infty} n\binom{p}{n}x^{n-1} \, \).

Using identity manipulations, such as

• \( \, n\binom{p}{n} = p\binom{p-1}{n-1} \, \),

aids in recognizing more manageable expressions.
Ultimately redefining summation indices accordingly leads to solutions like

• \( \, p\sum_{m=0}^{\infty} \binom{p-1}{m}x^m = p\frac{f(x)}{1+x} \, \).

Such processes underpin how differentiation of series assists in forming integral connections, further permitting conclusions about the behavior of \( f \) through differential equations.