MTH603 Quiz 3 Solution and Discussion
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Given that dydt=y−ty+tdydt=y−ty+t with the initial condition y=1,t=0y=1,t=0 find the 3rd term in Taylor series when t=0.3 and y//= 0.2.
@zaasmi said in MTH603 Quiz 3 Solution and Discussion:
Given that dydt=y−ty+tdydt=y−ty+t with the initial condition y=1,t=0y=1,t=0 find the 3rd term in Taylor series when t=0.3 and y//= 0.2.
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Given that dydt=t+y√dydt=t+y with the initial condition y0=1att0=0y0=1att0=0 find the 2nd term in Taylor series when t=1, y/ =0.2, and h=0.1.
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Given that dydt=t+y√dydt=t+y with the initial condition y0=1att0=0y0=1att0=0 find the 2nd term in Taylor series when t=1, y/ =0.2, and h=0.1.
@zaasmi said in MTH603 Quiz 3 Solution and Discussion:
Given that dydt=t+y√dydt=t+y with the initial condition y0=1att0=0y0=1att0=0 find the 2nd term in Taylor series when t=1, y/ =0.2, and h=0.1.
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Euler’s Method numerically computes the approximate ________ of a function.
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@zaasmi said in MTH603 Quiz 3 Solution and Discussion:
Euler’s Method numerically computes the approximate ________ of a function.
Euler’s method is a numerical tool for approximating values for solutions of differential equations.
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the area of a trapeziod is obtained by adding the area of a … and a triangle.
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@cyberian said in MTH603 Quiz 3 Solution and Discussion:
the area of a trapeziod is obtained by adding the area of a … and a triangle.
The area of a trapezoid can be obtained by adding the area of a rectangle and a triangle.
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In double integration, the process involves integrating a function of two variables over a two-dimensional region. The procedure typically follows these steps:
Keep One Variable Fixed: Select one of the variables (say
𝑥
x) to be integrated first while keeping the other variable (
𝑦
y) fixed. This creates an inner integral.Integrate with Respect to the Fixed Variable: Perform the integration with respect to the selected variable (
𝑥
x), treating the other variable (
𝑦
y) as a constant. This is known as the inner integral.Integrate the Result with Respect to the Remaining Variable: After integrating with respect to
𝑥
x, integrate the resulting expression with respect to the remaining variable (
𝑦
y). This is known as the outer integral. -
which of the following reason lead towards the numerical integration methods?
Analytical evaluation of integral is very complicated
All above choices are true
Integrand is given in tabular form
Analytical evaluation of integral is impossible -
which of the following reason lead towards the numerical integration methods?
Analytical evaluation of integral is very complicated
All above choices are true
Integrand is given in tabular form
Analytical evaluation of integral is impossible@cyberian said in MTH603 Quiz 3 Solution and Discussion:
which of the follwing reason lead towards the numerical integration methods?
Numerical integration methods are often employed for a variety of reasons, including:
Complex or Non-Analytic Functions:
Some functions are too complex to integrate analytically. They may involve complicated expressions, special functions, or be defined only by tabulated data.
Lack of Closed-Form Solutions:Many integrals do not have closed-form solutions, meaning they cannot be expressed in terms of elementary functions. In such cases, numerical methods provide approximate solutions.
Data-Driven Integrals:In practical applications, the function to be integrated may be known only at discrete points (e.g., experimental data). Numerical integration methods are necessary to approximate the integral from such tabular data.
High-Dimensional Integrals:In higher dimensions, integrals become increasingly difficult to solve analytically. Numerical methods are often the only feasible approach for evaluating multi-dimensional integrals.
Efficiency:Even when an analytical solution exists, it may be cumbersome or computationally expensive to evaluate. Numerical methods can provide a more efficient way to approximate the integral, especially for repeated calculations.
Adaptive Techniques:Numerical methods can adapt to the behavior of the integrand, allocating more computation effort where the integrand has higher variability. This adaptability can lead to more accurate results than fixed analytical approaches.
Real-Time Applications:In real-time systems or simulations, quick approximations of integrals may be required. Numerical methods can provide sufficiently accurate results in a timely manner.
To summarize, the main reasons leading towards the use of numerical integration methods include:Complexity and non-analytic nature of functions
Absence of closed-form solutions
Data-driven integrals requiring numerical approximation
High-dimensional integrals that are infeasible to solve analytically
Efficiency in computational resources
Adaptive techniques that provide accurate results
Real-time application needs
These reasons collectively justify the need for and advantages of numerical integration methods in various scientific, engineering, and mathematical applications. -
in newton-cotes formula for finding the definite of a tabular function, which of the following taken as an approximate function then find the desire integral?
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in newton-cotes formula for finding the definite of a tabular function, which of the following taken as an approximate function then find the desire integral?
@cyberian said in MTH603 Quiz 3 Solution and Discussion:
in newton-cotes formula for finding the definite of a tabular function, which of the following taken as an approximate function then find the desire integral?
In the Newton-Cotes formulas for numerical integration, a polynomial function is typically used as the approximate function to estimate the integral of a given tabular function. The Newton-Cotes formulas include several specific methods, such as the Trapezoidal Rule, Simpson’s Rule, and higher-order polynomial approximations.
Here are some of the commonly used Newton-Cotes formulas:
Trapezoidal Rule:
Approximates the function as a first-degree polynomial (a straight line) between each pair of points.Simpson’s Rule:
Approximates the function as a second-degree polynomial (a parabola) over pairs of intervals.Simpson’s 3/8 Rule:
Uses a cubic polynomial approximation over three subintervals.Higher-Order Newton-Cotes Formulas:
Use higher-degree polynomials for approximation over more points.To find the desired integral using one of these formulas, you would follow these general steps:
Select the appropriate Newton-Cotes formula:
Choose the formula based on the degree of accuracy you need and the number of data points available.Divide the interval into subintervals:
Split the interval
[
𝑎
,
𝑏
]
[a,b] into
𝑛
n equal subintervals, where
𝑛
n depends on the chosen formula.Apply the formula:
Use the selected Newton-Cotes formula to compute the integral. Here are the basic forms for the Trapezoidal Rule and Simpson’s -
in Simpson’s 1/3 rule, the global error is of …?
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@cyberian said in MTH603 Quiz 3 Solution and Discussion:
in Simpson’s 1/3 rule, the global error is of …?
In Simpson’s 1/3 rule for numerical integration, the global error (also known as the total error or the error over the entire interval) is of order
𝑂(ℎ4) -
givren that dy/dx= y-t/y+1 with the intial condition y=1, t=0 using euler’s method y at h=0.01; the value of y(0.01) is?
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givren that dy/dx= y-t/y+1 with the intial condition y=1, t=0 using euler’s method y at h=0.01; the value of y(0.01) is?
@cyberian said in MTH603 Quiz 3 Solution and Discussion:
givren that dy/dx= y-t/y+1 with the intial condition y=1, t=0 using euler’s method y at h=0.01; the value of y(0.01) is?
To solve the differential equation
𝑑
𝑦
𝑑
𝑡𝑦
−
𝑡
𝑦
+
1
dt
dy
y+1
y−t
with the initial condition
𝑦
(
0
)1
y(0)=1 using Euler’s method with a step size
ℎ0.01
h=0.01, we can follow these steps:Identify the differential equation and initial condition:
𝑑
𝑦
𝑑
𝑡𝑦
−
𝑡
𝑦
+
1
dt
dy
y+1
y−t
𝑦
(
0
)1
y(0)=1
Euler’s method formula:𝑦
𝑛
+
1𝑦
𝑛
+
ℎ
⋅
𝑓
(
𝑡
𝑛
,
𝑦
𝑛
)
y
n+1
=y
n
+h⋅f(t
n
,y
n
)
where
𝑓
(
𝑡
,
𝑦
)𝑦
−
𝑡
𝑦
+
1
f(t,y)=
y+1
y−t
.Initial values:
𝑡
00
,
𝑦
01
t
0
=0,y
0
=1
Calculate
𝑦
1
y
1
using Euler’s method with
ℎ0.01
h=0.01:𝑡
1𝑡
0
+
ℎ0
+
0.010.01
t
1
=t
0
+h=0+0.01=0.01𝑦
1𝑦
0
+
ℎ
⋅
𝑓
(
𝑡
0
,
𝑦
0
)1
+
0.01
⋅
1
−
0
1
+
11
+
0.01
⋅
1
21
+
0.01
⋅
0.51
+
0.0051.005
y
1
=y
0
+h⋅f(t
0
,y
0
)=1+0.01⋅
1+1
1−0
=1+0.01⋅
2
1
=1+0.01⋅0.5=1+0.005=1.005
Therefore, the value of
𝑦
(
0.01
)
y(0.01) using Euler’s method with a step size of
ℎ0.01
h=0.01 is approximately
1.005
1.005.