In the Previous Three Problems, the Solution of the Original $u ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) t }$

question 13

Multiple Choice

In the previous three problems, the solution of the original problem is

• A) \(u ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) t }\) , where \(c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x\)
• B) \(u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) \cos ( n \pi t /L )\) , where \(c _ { 0 } = \int _ { 0 } ^ { L } f ( x ) d x / L , c _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / L\)
• C) \(u ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sin ( n \pi t / L )\) , where \(c _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / L\)
• D) \(u ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - nxt / L }\) , where \(c _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / L\)
• E) \(u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) e ^ { - n x t / L }\) , where \(c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x\)

Master the concept and computation of the slope of a regression line.

Gain the ability to predict outcomes using regression analysis.

Understand the principles of least squares regression.

Learn to create and interpret side-by-side bar charts.

---

Definitions:

Crime Rates

The number of crimes committed per unit of population within a given time frame.

Economic Growth

An increase in a country's output of goods and services over time, often measured as the rate of change in real GDP.

Specialized Equipment

Machinery or tools designed for a specific task or industry, often crafted to perform particular functions that general equipment cannot.

Layout

The arrangement of elements or components within a given space, often relating to design, manufacturing, or architectural planning.