Chapter 9: Problem 38
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 \(f(x)=(1+x)^{p}\).
Short Answer
Step by step solution
Convergence of the Series
Differentiate the Function
Verification When x = 0
Establish the Differential Equation
Solve the Differential Equation
Final Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ratio Test
- \( \, \sum_{n=0}^{\infty} a_n x^n \, \)
- \( \, \sum a_n \, \)
- \( \, \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \, \)
- \( \, a_n = \binom{p}{n} x^n \, \)
- \( \, \frac{a_{n+1}}{a_n} \, \).
This ratio simplifies to the absolute value of \( \, |x| \, \). So, for the series to converge,
- \( \, |x| < 1 \, \).
The Ratio Test ensures our power series becomes a valid representation of \( f(x) \) in this range.
Binomial Coefficient
- \( \, \binom{p}{n} \, \)
- \( \, (1 + x)^p = \sum_{n=0}^{p} \binom{p}{n} x^n \, \)
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
- \( \, n\binom{p}{n} = p\binom{p-1}{n-1} \, \),
These coefficients are vital when dealing with changes in index during calculations, as seen when manipulating series like in our differentiation process.
Differential Equation
- \( \, (1+x)f'(x) = pf(x) \, \),
Here, separating the terms yields
- \( \, \frac{f'(x)}{f(x)} = \frac{p}{1+x} \, \).
- \( \, \ln(f(x)) = p\ln(1 + x) + C \, \).
- \( \, f(x) = e^C (1+x)^p \, \),
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
- \( \, f(x) = 1 + \sum_{n=1}^{\infty} \binom{p}{n} x^n \, \)
The differentiation of each term yields
- \( \, n\binom{p}{n}x^{n-1} \, \),
- \( \, f'(x) = \sum_{n=1}^{\infty} n\binom{p}{n}x^{n-1} \, \).
- \( \, n\binom{p}{n} = p\binom{p-1}{n-1} \, \),
Ultimately redefining summation indices accordingly leads to solutions like
- \( \, p\sum_{m=0}^{\infty} \binom{p-1}{m}x^m = p\frac{f(x)}{1+x} \, \).