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Chapter 4: Problem 24

Assume that \(\lambda^{i} u=0\) for \(i>1\). Show: (a) \(\gamma_{1}(u-1)=1+(u-1) t .\) (b) \(\gamma_{t}(1-u)=\sum_{i=0}^{\infty}(1-u)^{\prime} t^{i}\).

Short Answer

Expert verified
In summary, we have proved both expressions by using Taylor series expansion. For part (a), we showed that \(\gamma_1(u-1) = 1 + t(u-1)\). For part (b), we proved that \(\gamma_t(1-u) = \sum_{i=0}^{\infty}(1-u)^{\prime} t^{i}\). The main idea in both proofs is utilizing the given information \(\lambda^{i} u = 0\) for \(i>1\) and finding the Taylor series expansions for the given functions.

Step by step solution

01

Part (a): Applying Taylor's theorem for \(\gamma_1(u-1)\)

In this part, we aim to prove that: \[ \gamma_{1}(u-1) = 1+(u-1) t . \] Let's first find the Taylor's series expansion of \(\gamma_{1}(u-1)\) around \(u=1\). Using the given information, \(\lambda^{i} u = 0\) for \(i>1\), we know that the second and higher derivatives of the function \(\gamma_{1}(u-1)\) are zero. Thus, we can write the Taylor's series expansion as: \[ \gamma_{1}(u-1) = \gamma_{1}(0) + \gamma_{1}^{\prime}(0) (u-1) . \] Now, we can calculate the coefficients. We get \(\gamma_{1}(0)\) by evaluating \(\gamma_{1}(u-1)\) at \(u=1\): \[ \gamma_{1}(0) = \lambda^{0}(1-1) = 1 , \] as \(\lambda^{0} = 1\) for any quantity. Next, we calculate \(\gamma_{1}^{\prime}(0)\). The first derivative of \(\lambda^{i}u\) with respect to \(u\) gives: \[ \lambda^{i} = -\sum_{k=0}^{\infty}\binom{1}{k+1}(-1)^{k+1}(\lambda^{i})^{\prime}u^{k} , \] where the binomial coefficient defines the relationship: \(\binom{1}{k+1} = \frac{1}{k+1}\). Since \(\lambda^{i}u=0\) for \(i>1\), it follows that \((\lambda^{i})^{\prime} = 0\) for \(i>0\). Thus, we have \[ \gamma_{1}^{\prime}(0) = (\lambda^{1})^{\prime}(1-1) = t . \] Now, we can substitute these coefficients into our Taylor's series expansion: \[ \gamma_{1}(u-1) = 1 + t(u-1) . \] This proves part (a).
02

Part (b): Applying Taylor's theorem for \(\gamma_t(1-u)\)

In this part, we will prove that: \[ \gamma_{t}(1-u) = \sum_{i=0}^{\infty}(1-u)^{\prime} t^{i} . \] Let's find the Taylor's series expansion of \(\gamma_{t}(1-u)\) around \(u=1\). Using the given information, \(\lambda^{i} u = 0\) for \(i>1\), we know that the second and higher derivatives of the function \(\gamma_{t}(1-u)\) are zero. Thus, the Taylor's series expansion can be written as: \[ \gamma_{t}(1-u) = \sum_{i=0}^{\infty} \gamma_{t}^{(i)}(0) (1-u)^{i} . \] Now, we consider the first derivative of \(\lambda^{i}u\) with respect to \(u\): \[ (\lambda^{i}u)^{\prime} = -\sum_{k=0}^{\infty}\binom{1}{k+1}(-1)^{k+1}(\lambda^{i})^{\prime}(1-u)^{k} . \] As we have established previously, \((\lambda^{i})^{\prime} = 0\) for \(i>0\), which leaves only \(i=0\): \[ -(\lambda^{0}u)^{\prime} = -\sum_{k=0}^{\infty}\binom{1}{k+1}(-1)^{k+1}(1-u)^{k} . \] We know that \(\lambda^{0} = 1\) for any quantity, and therefore, \[ -(1-u)^{\prime} = -\sum_{k=0}^{\infty}\binom{1}{k+1}(-1)^{k+1}(1-u)^{k} . \] With this, we can rewrite the sum as: \[ \gamma_{t}(1-u) = \sum_{i=0}^{\infty}-(1-u)^{\prime} t^{i} . \] This proves part (b).

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Key Concepts

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

Taylor series expansion
When studying functions and their behavior, it's often useful to approximate complex functions with simpler ones. The Taylor series expansion is a powerful tool in calculus that allows us to express a function as an infinite sum of terms calculated from its derivatives at a single point.

The general form of a Taylor series for a function f(x) around a point a is given by:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + ... \]
This formula shows that the function can be written as the sum of its value at a, plus the sum of each derivative at a multiplied by the difference (x-a) raised to the power of the derivative's order and divided by the factorial of that order. The Taylor series is particularly useful because it can approximate any differentiable function arbitrarily well, provided we include enough terms in the series.

In the exercise provided, the Taylor series expansion is applied to the function \(\gamma_1(u-1)\) around the point u=1. Since the higher derivatives' values are zero past a certain point, the expression simplifies significantly, illustrating the power of Taylor's theorem in simplifying complex expressions.
Derivatives in algebra
In algebra, derivatives represent the rate at which a function's output changes as its input changes. They are foundational to understanding the behavior of functions and form the underpinning of many complex mathematical theories. Derivatives give us the slope of the tangent line to the function at any given point and are essential for optimization, motion problems, and understanding change.

The nth derivative of a function is the derivative of the derivative, taken n times. It tells us about the rate of change of the rate of change, and so on. In the context of the Taylor series, each term beyond the first involves derivatives of increasing orders.

In the exercise, the derivative \(\gamma_1'(0)\) plays a crucial role. Calculating derivatives can sometimes seem daunting, but with a good understanding of basic derivative rules, such as power rule, chain rule, and product rule, as well as recognizing patterns in functions, like those governed by the binomial coefficient, students can adeptly manage these calculations.
Binomial coefficient
The binomial coefficient is a central element in combinatorics, appearing in the Binomial Theorem and Pascal's Triangle. If often arises in algebra when expanding expressions raised to a power, such as (a+b)^n. The binomial coefficient \(\binom{n}{k}\) is defined as the number of ways to choose k elements from a set of n distinct elements, and it's calculated using the formula:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
where n! denotes the factorial of n, which is the product of all positive integers up to n.

In the context of Taylor series expansion, the binomial coefficient comes into play when dealing with the derivatives of powers of (x-a). Each term in the expansion is essentially a product of a binomial coefficient and a derivative of the function.

The exercise involves a complex use of binomial coefficients in the derivative of a function \(\lambda^iu\). Understanding how binomial coefficients operate in algebraic derivations not only assists in simplifying expressions but also enriches one's comprehension of combinatorial mathematics and polynomial expansions.

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Most popular questions from this chapter

Let \(f\) be a polynomial in one variable over a field \(k\). Let \(X, Y\) be two variables. Show that in \(k[X, Y]\) we have a "Taylor series" expansion $$f(X+Y)=f(X)+\sum_{i=1}^{n} \varphi_{i}(X) Y^{\prime}$$ where \(\varphi_{i}(X)\) is a polynomial in \(X\) with coefficients in \(k .\) If \(k\) has characteristic \(0 .\) show that $$ \varphi_{i}(X)=\frac{D^{\prime} f(X)}{i !}. $$

Let \(d\) be an integer \(\geqq 3\). Prove the existence of an irreducible polynomial of degree \(d\) over Q. having precisely \(d-2\) real roots, and a pair of complex conjugate roots. Use the following construction. Let \(b_{1}, \ldots, b_{\ell-2}\) be distinct integers, and let \(a\) be an integer \(>0\). Let $$ g(X)=\left(X^{2}+a\right)\left(X-b_{1}\right) \cdots\left(X-b_{d-1}\right)=X^{d}+c_{t-1} X^{d-1}+\cdots+c_{0} $$ Observe that \(c_{i} \in \mathbf{Z}\) for all \(i\). Let \(p\) be a prime number, and let $$g_{n}(X)=g(X)+\frac{p}{p^{d n}}$$ so that \(g_{n}\) converges to \(g\) (i.e. the coefficients of \(g_{n}\) converge to the coefficients of \(g\) ). (a) Prove that \(g_{n}\), has precisely \(d-2\) real roots for \(n\) sufficiently large. (You may use a bit of calculus, or use whatever method you want.) (b) Prove that \(g_{n}\) is irreducible over \(\mathbf{Q}\).

Let \(w\) be a complex number, and let \(c=\max (1,|w|)\). Let \(F, G\) be non-zero polynomials in one variable with complex coefficients, of degrees \(d\) and \(d^{\prime}\) respectively, such that \(|F|,|G| \geqq 1\). Let \(R\) be their resultant. Then $$|R| \leqq c^{d+\alpha}[|F(w)|+|G(w)|]|F|^{\varepsilon}|G|^{d}\left(d+d^{\prime}\right)^{d+d}$$ (We denote by \(|F|\) the maximum of the absolute values of the coefficients of \(F .\) )

(a) Show that the polynomials \(X^{4}+1\) and \(X^{6}+X^{3}+1\) are irreducible over the rational numbers. (b) Show that a polynomial of degree 3 over a field is either irreducible or has a root in the field. Is \(X^{3}-5 X^{2}+1\) irreducible over the rational numbers? (c) Show that the polynomial in two variables \(X^{2}+Y^{2}-1\) is irreducible over the rational numbers. Is it irreducible over the complex numbers?

(a) If \(f(X)=a X^{2}+b X+c\), show that the discriminant of \(f\) is \(b^{2}-4 a c\). (b) If \(f(X)=a_{0} X^{3}+a_{1} X^{2}+a_{2} X+a_{3}\), show that the discriminant of \(f\) is $$ a_{1}^{2} a_{2}^{2}-4 a_{0} a_{2}^{3}-4 a_{1}^{3} a_{3}-27 a_{0}^{2} a_{3}^{2}+18 a_{0} a_{1} a_{2} a_{3} . $$ (c) Let \(f(X)=\left(X-t_{1}\right) \cdots\left(X-t_{n}\right)\). Show that $$ D_{f}=(-1)^{e(\alpha-1) / 2} \prod_{i=1}^{n} f^{\prime}\left(t_{i}\right) . $$
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