/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Let $$f(x)=\sum_{k=0}^{\infty}... [FREE SOLUTION] | 91影视

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Let $$f(x)=\sum_{k=0}^{\infty} c_{k} x^{k} \text { and } g(x)=\sum_{k=0}^{\infty} d_{k} x^{k}$$a. Multiply the power series together as if they were polynomials, collecting all terms that are multiples of \(1, x,\) and \(x^{2} .\) Write the first three terms of the product \(f(x) g(x)\) b. Find a general expression for the coefficient of \(x^{n}\) in the product series, for \(n=0,1,2, \ldots\)

Short Answer

Expert verified
Based on the given step-by-step solution, provide a short answer. When we multiply two power series f(x) and g(x), the general expression for the coefficient of x鈦 in the product series f(x)g(x) is given by the formula: $$\sum_{k=0}^{n} c_{k} d_{n-k}$$ This expression allows us to find the coefficients of any x鈦 term in the product series.

Step by step solution

01

Multiply the power series

To multiply the power series f(x) and g(x), we need to perform a term-by-term multiplication. This means we will multiply every term of f(x) with every term of g(x) and sum the result: $$f(x)g(x) = \left(\sum_{k=0}^{\infty} c_{k} x^{k}\right) \left(\sum_{k=0}^{\infty} d_{k} x^{k}\right)$$
02

Collect the terms multiples of 1, x, and x虏

We have to find the coefficients of 1, x, and x虏 in the product series f(x)g(x). It is possible by examining all possible product pairs and to sum them up: Coefficients of 1 in f(x)g(x): \(c_0d_0\) Coefficients of x in f(x)g(x): \(c_1d_0+c_0d_1\) Coefficients of x虏 in f(x)g(x): \(c_2d_0+c_1d_1+c_0d_2\)
03

Write the first three terms of the product f(x)g(x)

Now, we can write down the first three terms of the product of the power series f(x)g(x): $$f(x)g(x) = (c_0d_0) + (c_1d_0+c_0d_1)x + (c_2d_0+c_1d_1+c_0d_2)x^2 + \cdots$$
04

Find the general expression for the coefficient of x鈦

To find the general expression for the coefficient of x鈦, we notice the pattern that is emerging in the first three terms. The coefficient of x鈦 can be obtained by summing the products of the coefficients of f(x) and g(x) at different indices, such that their sum of indices is equal to n: $$x^{n} \text{ coefficient in the product series: }\sum_{k=0}^{n} c_{k} d_{n-k}$$ So, the general coefficient expression for x鈦 in the product series f(x)g(x) is: $$\sum_{k=0}^{n} c_{k} d_{n-k}$$

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Key Concepts

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

Coefficient
Coefficients in power series play a crucial role, just as they do in polynomials. Each term of a power series has a coefficient, which is the constant part multiplied by the variable's power. In this context, when dealing with the power series \( f(x) = \sum_{k=0}^{\infty} c_{k} x^{k} \) and \( g(x) = \sum_{k=0}^{\infty} d_{k} x^{k} \), the coefficients are denoted by \( c_k \) and \( d_k \) respectively.
  • These coefficients specify how much each term contributes to the overall function.
  • The coefficient of the term \(x^n\) in a product series is derived by multiplying and summing certain coefficients from the individual series.
To simplify the concept, consider each term in the power series like a piece in a jigsaw puzzle. Together, these pieces (coefficients) help to form the entire picture (the polynomial expression). Understanding the pattern in forming these coefficients is vital when multiplying power series.
Polynomial Multiplication
Polynomial multiplication is a critical process where two polynomials are combined to form a new polynomial. This involves multiplying every term in one polynomial by every term in the other. For power series, this operation follows similar principles but involves an infinite number of terms.Here鈥檚 how it works in a general sense:
  • Each term from the first polynomial is multiplied by each term of the second polynomial.
  • After multiplying, we collect like terms to simplify the expression.
In the exercise involving series \( f(x) \) and \( g(x) \), the resultant series after multiplication is achieved by systematically multiplying and summing terms from these two series. This is similar to multiplying two finite polynomials, except you handle an infinite number of terms, focusing initially on the relevant lower powers like 1, \( x \), and \( x^2 \) as shown in the previous steps.
Term-by-term Multiplication
Term-by-term multiplication is a method where each term of one series is multiplied by every term of another series. This ensures that every possible combination of product terms is considered.
For power series multiplication \( f(x)g(x) \):
  • We take each term \( c_k x^k \) from \( f(x) \) and multiply it with terms \( d_j x^j \) from \( g(x) \).
  • The power of \( x \) in each resultant term is \( k+j \), and we accumulate these products for terms with the same power of \( x \).

This method allows us to derive expressions for coefficients at various powers as shown in Step 2 of the solution. By applying term-by-term multiplication, you map out all possible product contributions for terms like \( x^n \), and then sum them to form the coefficient of that term in the multiplied series. Keeping track of these calculations helps in finding the general expression for any power in the series.

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Most popular questions from this chapter

Taylor coefficients for \(x=a\) Follow the procedure in the text to show that the \(n\) th-order Taylor polynomial that matches \(f\) and its derivatives up to order \(n\) at \(a\) has coefficients $$c_{k}=\frac{f^{(k)}(a)}{k !}, \text { for } k=0,1,2, \ldots, n$$

Find power series representations centered at 0 for the following functions using known power series. $$f(x)=\frac{1}{1-x^{4}}$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The function \(f(x)=\sqrt{x}\) has a Taylor series centered at \(0 .\) b. The function \(f(x)=\) cse \(x\) has a Taylor series centered at \(\pi / 2\) c. If \(f\) has a Taylor series that converges only on \((-2,2),\) then \(f\left(x^{2}\right)\) has a Taylor series that also converges only on (-2,2) d. If \(p(x)\) is the Taylor series for \(f\) centered at \(0,\) then \(p(x-1)\) is the Taylor series for \(f\) centered at \(1 .\) e. The Taylor series for an even function centered at 0 has only even powers of \(x\).

Use the methods of this section to find the power series centered at 0 for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{-3 x}$$

Probability: tossing for a head The expected (average) number of tosses of a fair coin required to obtain the first head is \(\sum_{i=1}^{\infty} k\left(\frac{1}{2}\right)^{t} .\) Evaluate this series and determine the expected number of tosses. (Hint: Differentiate a geometric series.)

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