Question

Suppose that a function \(f\) has derivatives of all orders near \(x=0 .\) By the Fundamental Theorem of Calculus, \(f(x)-f(0)=\int_{0}^{x} f^{\prime}(t) d t\) a. Evaluate the integral using integration by parts to show that $$f(x)=f(0)+x f^{\prime}(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) d t.$$ b. Show (by observing a pattern or using induction) that integrating by parts \(n\) times gives $$\begin{aligned} f(x)=& f(0)+x f^{\prime}(0)+\frac{1}{2 !} x^{2} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} x^{n} f^{(n)}(0) \\ &+\frac{1}{n !} \int_{0}^{x} f^{(n+1)}(t)(x-t)^{n} d t+\cdots \end{aligned}$$ This expression is called the Taylor series for \(f\) at \(x=0\).

Short answer

Question: Evaluate the given integral using integration by parts and derive the identity for the Taylor series of a function. Solution: We have shown that the integral can be evaluated using integration by parts as follows: $$f(x) = f(0) + x f^{\prime}(x) + \int_{0}^{x} f^{\prime \prime}(t)(x-t) dt$$ Moreover, we have used induction to prove that by integrating by parts n times, we can obtain the Taylor series formula: $$f(x)=f(0)+xf^'(0)+\frac{1}{2!}x^2f''(0)+\cdots+\frac{1}{n!}x^n f^{(n)}(0)+\frac{1}{n!}\int_0^x f^{(n+1)}(t)(x-t)^n dt+\cdots$$

Step-by-step solution

Part a: Integration by Parts

To show that \(f(x)=f(0)+xf'(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) dt\), we will first apply integration by parts on the integral \(\int_{0}^{x} f^{\prime}(t) dt\). Let \(u = f'(t)\) and \(dv = dt\). Then, \(du = f''(t)dt\) and \(v = t\). Integration by parts states that \(\int u dv = uv - \int v du\). Therefore: $$\int_{0}^{x} f^{\prime}(t) dt = \left[t f^{\prime}(t)\right|_{0}^{x} - \int_{0}^{x} t f^{\prime \prime}(t) dt.$$ Now, evaluate the term within the bracket: $$\left[t f^{\prime}(t)\right|_{0}^{x} = x f^{\prime}(x) - 0 f^{\prime}(0) = x f^{\prime}(x).$$ So, we get the following: $$\int_{0}^{x} f^{\prime}(t) d t = xf^{\prime}(x) - \int_{0}^{x} t f^{\prime \prime}(t) dt$$ Therefore, according to the Fundamental Theorem of Calculus: $$f(x)-f(0) = x f^{\prime}(x) - \int_{0}^{x} t f^{\prime \prime}(t) dt$$ Rearrange the equation to find \(f(x)\): $$f(x) = f(0) + x f^{\prime}(x) + \int_{0}^{x} f^{\prime \prime}(t)(x-t) dt$$

Part b: Finding the Taylor Series Formula

Let's use induction to prove that, after integrating by parts n times: $$\begin{aligned} f(x)&=f(0)+xf^'(0)+\frac{1}{2!}x^2f''(0)+\cdots+\frac{1}{n!}x^nf^{(n)}(0)\\ &+\frac{1}{n!}\int_0^x f^{(n+1)}(t)(x-t)^n dt+\cdots \end{aligned}$$ Base case (n=0): We already proved part a, which is the base case for n=0. Inductive step: Assume the identity is true for n=k, that is: $$\fbox{1} \begin{aligned} f(x)&=f(0)+xf^'(0)+\frac{1}{2!}x^2f''(0)+\cdots+\frac{1}{k!}x^kf^{(k)}(0)\\ &+\frac{1}{k!}\int_0^x f^{(k+1)}(t)(x-t)^k dt+\cdots \end{aligned}$$ We need to prove that the identity is also true for n=k+1: $$\fbox{2} \begin{aligned} f(x)&=f(0)+xf^'(0)+\frac{1}{2!}x^2f''(0)+\cdots+\frac{1}{(k+1)!}x^{k+1}f^{(k+1)}(0)\\ &+\frac{1}{(k+1)!}\int_0^x f^{(k+2)}(t)(x-t)^{k+1} dt+\cdots \end{aligned}$$ Integrate the k-th term of \fbox{1} by parts: Let \(u=f^{(k+1)}(t)\) and \(dv=(x-t)^kdt\). Then \(du=f^{(k+2)}(t)dt\) and \(v=-\frac{1}{k+1}(x-t)^{k+1}\). Applying integration by parts: $$\int_0^x f^{(k+1)}(t)(x-t)^k dt = \left[-\frac{1}{k+1}f^{(k+1)}(t)(x-t)^{k+1}\right]_0^x + \frac{1}{k+1}\int_0^x f^{(k+2)}(t)(x-t)^{k+1} dt$$ By substituting this last equation into \fbox{1}, we get: $$\begin{aligned} f(x)&=f(0)+xf^'(0)+\frac{1}{2!}x^2f''(0)+\cdots+\frac{1}{k!}x^kf^{(k)}(0)\\ &+\frac{1}{k!(k+1)!}x^{k+1}f^{(k+1)}(0)-\frac{1}{(k+1)!}\int_0^x f^{(k+2)}(t)(x-t)^{k+1} dt+\cdots \end{aligned}$$ Which matches the identity in \fbox{2} for n=k+1. Therefore, by induction, the given identity holds true for all integer values of n, and we have obtained the Taylor series formula for \(f\) at \(x=0\).

Key concepts

Integration by Parts

Integration by parts is a vital technique used to integrate products of functions. When you have an integral of two functions multiplied together, and a direct approach doesn't work, integration by parts can be particularly useful. It's based on the product rule for differentiation and is essential in finding solutions to some integrals.

The formula for integration by parts is given by: \[ \int u \, dv = uv - \int v \, du \] Here, the functions \(u\) and \(dv\) are chosen strategically to simplify the integral. The formula resembles the product rule used in differentiation and allows us to transform the given integral into a potentially simpler one.

For instance, when solving part (a) of the exercise, we use integration by parts on \( \int_{0}^{x} f^{\prime}(t) dt \) by letting \( u = f^{\prime}(t) \) and \( dv = dt \). This approach transforms the integral into a form that's more manageable and easier to evaluate. The outcome helps us understand how a Taylor series begins to develop around a specific point.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of calculus, bridging the concepts of differentiation and integration. In essence, it states two central ideas about antiderivatives and definite integrals:

• The first part of the theorem links the derivative of a function to its integral. Specifically, if \( F \) is an antiderivative of \( f \) on an interval, then the integral of \( f \) over that interval can be solved as \( F(b) - F(a) \).
• The second part states that if \( f \) is continuous over an interval \([a, b]\), the function \( F(x) = \int_{a}^{x} f(t) \, dt \) is an antiderivative of \( f \). This means differentiating \( F \) gives back the original function \( f \).

This theorem is essential in calculus as it provides a way to evaluate definite integrals without having to compute limits of Riemann sums. Moreover, it's a fundamental rationale behind many integral manipulations, such as those involved in deriving the Taylor series for functions, as demonstrated in the exercise.

Induction in Calculus

In calculus, mathematical induction is a proof technique often employed to prove statements or formulas for all natural numbers. It consists of two main steps:

• **Base Case**: Verify that the statement holds for the initial value, usually \( n=0 \) or \( n=1 \).
• **Inductive Step**: Assume the statement is true for \( n=k \), and then prove it for \( n=k+1 \). This step involves showing that the truth of the statement for one number implies its truth for the next number.

In the context of calculus and, specifically, Taylor series, induction can be used to prove that a pattern exists for the expansion of a function. By employing induction, you confirm that repeated application of integration by parts results in a deductive pattern forming the Taylor series.

So, in the given problem, induction helps us see how starting from an assumption based on the cases we've already shown leads to a general rule for all \( n \). This rule helps express functions as series, assisting in their approximation and serving as a powerful tool in calculus.