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Chapter 19: Problem 25

Problem 18-43 presented a rather complicated proof that \(f=0\) if \(f^{\prime \prime}-f=0\) and \(f(0)=f^{\prime}(0)=0 .\) Give another proof, using Taylor's Theorem. (This problem is really a preliminary skirmish before doing battle with the general case in Problem 26, and is meant to convince you that Taylor's Theorem is a good tool for tackling such problems, even though tricks work out more neatly for special cases.)

Short Answer

Expert verified
Using Taylor's Theorem, it is proved that \( f = 0 \) given the conditions \( f''(0) = 0 \), \( f(0) = 0 \), and \( f'(0) = 0 \).

Step by step solution

01

Understand the given conditions

The differential equation given is: \[ f'' - f = 0 \]with the initial conditions: \[ f(0) = 0 \] and \[ f'(0) = 0 \].We need to prove that \( f = 0 \) using Taylor's Theorem.
02

Write the Taylor series for \( f(x) \)

According to Taylor's Theorem, the function \( f(x) \) can be expressed as a series: \[ f(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{\text{higher-order terms}}{n!} x^n \]Given the initial conditions \( f(0) = 0 \) and \( f'(0) = 0 \), we have: \[ f(x) = 0 + 0 \times x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{\text{higher-order terms}}{n!} x^n \]Thus, \[ f(x) = \frac{f''(0)}{2} x^2 + \frac{f'''(0)}{6} x^3 + \text{higher-order terms} \].
03

Substitute Taylor series into the differential equation

Express the Taylor series terms explicitly in the differential equation \( f'' - f = 0 \):\[ f(x) = \frac{f''(0)}{2} x^2 + \frac{f'''(0)}{6} x^3 + \text{higher-order terms} \].The second derivative \( f''(x) \) is:\[ f''(x) = f''(0) + \frac{f'''(0)}{2} x + \text{other terms} \].
04

Analyze the Taylor series for consistency with the differential equation

Rewrite the differential equation with the Taylor series:\[ f''(x) - f(x) = f''(0) + \frac{f'''(0)}{2} x + \text{other terms} -\bigg( \frac{f''(0)}{2} x^2 + \frac{f'''(0)}{6} x^3 + \text{other terms} \bigg) = 0 \].For this equation to hold for any \( x \), the coefficients of each power of \( x \) must be zero.
05

Equate coefficients and solve

From the equation:\[ f''(0) - \frac{f''(0)}{2} x^2 + \frac{f'''(0)}{2} x - \frac{f'''(0)}{6} x^3 + \text{other terms} = 0 \], we can see that each term coefficient must be zero:- The coefficient of the constant term: \( f''(0) = 0 \).- The coefficient of \( x \): \( \frac{f'''(0)}{2} = 0 \).
06

Conclude the solution

If \( f''(0) = 0 \), then the whole initial Taylor series for \( f(x) \): \[ f(x) = 0 + 0 \times x + 0 \times x^2 + \text{higher-order terms} \] simplifies to 0. Therefore, \( f(x) = 0 \) for all \( x \). This confirms that \( f = 0 \) under the given conditions.

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Key Concepts

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

Differential Equations
Differential equations involve equations that include derivatives of a function. In the given problem, our differential equation is \( f'' - f = 0 \). This tells us that the second derivative of the function \( f \) minus the function itself equals zero. Such equations help us describe various physical phenomena like motion, heat, and waves. Understanding these kinds of equations is key to solving problems involving rates of change and accumulated quantities. Here, we need to show that the function \( f \) is zero given the initial conditions.
Initial Conditions
Initial conditions specify the value of a function and its derivatives at a particular point. They are crucial in solving differential equations because they allow us to find a specific solution out of the many possible ones. In our exercise, we have two initial conditions: \( f(0) = 0 \) and \( f'(0) = 0 \). These conditions mean that at \( x = 0 \), the function \( f \) and its first derivative are both zero. These values narrow down the potential functions that satisfy our differential equation.
Series Expansion (Taylor's Theorem)
Taylor's Theorem provides a way to approximate functions using polynomials. The theorem states that a function can be expressed as an infinite sum of its derivatives evaluated at a specific point (usually zero). For our function \( f(x) \), we write the series expansion as:
  • \( f(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \text{higher-order terms} \)
Given our initial conditions, this simplifies to:
  • \( f(x) = 0 + 0 \times x + \frac{f''(0)}{2} x^2 + \frac{f'''(0)}{6} x^3 + \text{higher-order terms} \)
With \( f(0) = 0 \) and \( f'(0) = 0 \), the first two terms vanish. By substituting this series back into the original differential equation and equating the coefficients to zero, we prove that all terms must be zero, ultimately showing that \( f(x) = 0 \) for all \( x \).

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Most popular questions from this chapter

(a) Let \(f(x)=x^{4} \sin 1 / x^{2}\) for \(x \neq 0,\) and \(f(0)=0 .\) Show that \(f=0\) up to order 2 at \(0,\) even though \(f^{\prime \prime}(0)\) does not exist. This example is slightly more complex, but also slightly more impressive, than the example in the text, because both \(f^{\prime}(a)\) and \(f^{\prime \prime}(a)\) exist for \(a \neq 0 .\) Thus, for each number \(a\) there is another number \(m(a)\) such that $$(*) \quad f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{m(a)}{2}(x-a)^{2}+R_{a}(x),$$ $$\text { where } \lim _{x \rightarrow a} \frac{R_{a}(x)}{(x-a)^{2}}=0;$$ namely, \(m(a)=f^{\prime \prime}(a)\) for \(a \neq 0,\) and \(m(0)=0 .\) Notice that the function \(m\) defined in this way is not continuous. (b) Suppose that \(f\) is a differentiable function such that \((*)\) holds for all \(a\) with \(m(a)=0 .\) Use Problem 27 to show that \(f^{\prime \prime}(a)=m(a)=0\) for all \(a\). (c) Now suppose that \((*)\) holds for all \(a\), and that \(m\) is continuous. Prove that for all \(a\) the second derivative \(f^{\prime \prime}(a)\) exists and equals \(m(a)\).

Use the Taylor polynomial \(P_{1, a, f},\) together with the remainder, to prove a weak form of Theorem 2 of the Appendix to Chapter 11: If \(f^{\prime \prime}>0,\) then the graph of \(f\) always lies above the tangent line of \(f,\) except at the point of contact.

Suppose that \(a_{i}\) and \(b_{i}\) are the coefficients in the Taylor polynomials at \(a\) of \(f\) and \(g,\) respectively. In other words, \(a_{i}=f^{(i)}(a) / i !\) and \(b_{i}=g^{(i)}(a) / i ! .\) Find the cocficients \(c_{i}\) of the Taylor polynomials at \(a\) of the following functions, in terms of the \(a_{i}\) 's and \(b_{i}\) 's. (i) \(\quad f+g\). (ii) \(\quad f g\). (iii) \(\quad f^{\prime}\). (iv) \(\quad h(x)=\int_{a}^{x} f(t) d t\). (v) \(\quad k(x)=\int_{0}^{x} f(t) d t\).

Write each of the following polynomials in \(x\) as a polynomial in \((x-3)\) is only necessary to compute the Taylor polynomial at 3, of the same degree as the original polynomial. Why?) (i) \(x^{2}-4 x-9\). (ii) \(x^{4}-12 x^{3}+44 x^{2}+2 x+1\). (iii) \(\quad x^{5}\). (iv) \(a x^{2}+b x+c\).

(a) Prove that if \(f^{\prime \prime}(a)\) exists, then $$f^{\prime \prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)+f(a-h)-2 f(a)}{h^{2}}.$$ The limit on the right is called the Schwarz second derivative of \(f\) at a. Hint: Use the Taylor polynomial \(P_{2, a}(x)\) with \(x=a+h\) and with \(x=a-h\) (b) Let \(f(x)=x^{2}\) for \(x \geq 0,\) and \(-x^{2}\) for \(x \leq 0 .\) Show that $$\lim _{h \rightarrow 0} \frac{f(0+h)+f(0-h)-2 f(0)}{h^{2}}$$ exists, even though \(f^{\prime \prime}(0)\) does not. (c) Prove that if \(f\) has a local maximum at \(a,\) and the Schwarz second derivative of \(f\) at \(a\) exists, then it is \(\leq 0\). (d) Prove that if \(f^{\prime \prime \prime}(a)\) exists, then $$\frac{f^{\prime \prime \prime}(a)}{3}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a-h)-2 h f^{\prime}(a)}{h^{3}}.$$
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