Quiz: Big Data Matrix
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Name:
Roll Number:
Prepared by: Brajesh Jha
Instructions: Enter your details, answer all questions, and click Submit Quiz.
Name:
Roll Number:
Prepared by: Brajesh Jha
Let f(x,y) be defined in a neighborhood of the point (a,b), except possibly at the point itself. We say the limit of f(x,y) as (x,y) → (a,b) is L, and write:
lim(x,y)→(a,b) f(x,y) = L
if for every ε>0, there exists δ>0 such that:
0 < √((x-a)² + (y-b)²) < δ ⇒ |f(x,y)-L| < ε
Function: f(x,y) = x + y
Find: lim(x,y)→(1,2) f(x,y)
Solution: f(1,2) = 1 + 2 = 3
Answer: 3
Function: f(x,y) = xy / (x² + y²)
Find: lim(x,y)→(0,0) f(x,y)
Solution:
Answer: Limit does not exist (different values along different paths)
Function: f(x,y) = x² + y²
Find: lim(x,y)→(1,1) f(x,y)
Solution: Direct substitution: f(1,1) = 1² + 1² = 2
Answer: 2
Function: f(x,y) = (x² - y²) / (x² + y²)
Find: lim(x,y)→(0,0) f(x,y)
Solution:
Answer: Limit does not exist
Function: f(x,y) = 3x²y / (x² + y²)
Find: lim(x,y)→(0,0) f(x,y)
Solution:
Answer: 0
| Function | Point | Limit | Exists? |
|---|---|---|---|
| x + y | (1,2) | 3 | ✔ |
| xy / (x² + y²) | (0,0) | – | ✘ |
| x² + y² | (1,1) | 2 | ✔ |
| (x² - y²) / (x² + y²) | (0,0) | – | ✘ |
| 3x²y / (x² + y²) | (0,0) | 0 | ✔ |
Function of Two variable: Let u be a symbol which has a definite value for every pair of values of x and y, then u is called a function of two independent variable x and y and is written as
u=f(x,y)
A function of two variables is written as:
z = f(x, y)
z
|
| •
| •
| •
| •
|•____________ y
/
/
x
Explanation: This shows a surface in 3D space where z depends on x and y.
z
|
| /
| /
| /
|/________ y
/
/
x
Application: Cost, temperature variation.
z
|
__|__
/ | \
/ | \
/_______|_______\ y
|
x
Application: Heat distribution, potential energy.
z
|
___/ \___
/ \
----/-------------\---- y
\ /
\___ ___/
x
Application: Profit–loss analysis.
y
|
○ ○ ○
○ ○
○ ○ ○
○ ○
○ ○
|
x
Application: Topographic maps, weather maps.
Exam Note:
The graph of a function of two variables is a surface in three-dimensional space.
A function of three variables is a function that depends on three independent variables.
Mathematical Form:
w = f(x, y, z)
A function of three variables cannot be drawn directly because it requires four dimensions. Hence, it is represented using level surfaces or cross-sections.
For: x² + y² + z² = c (Sphere)
z
|
___|___
.-' | '-.
.' | '.
| | |
'. | .'
'-._____|_____.-'
|
x
/
y
Explanation: Each surface represents points where the function has the same value.
Fix one variable (e.g., z = k) and draw the 3D surface in x–y plane.
z = k
|
____|____
/ | \
| | |
\____|____/
|
x
/
y
Exam Note:
A function of three variables is represented graphically using level surfaces or cross-sections.
A function of two variables depends on two independent variables x and y and produces a single dependent variable z:
z = f(x, y)
A level curve is given by f(x,y) = c, e.g., x² + y² = 1.
A function of three variables depends on x, y, z and produces a single output w:
w = f(x, y, z)
Calculus of Several Variables
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1
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2
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3
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Continuity of function of two variables
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4
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Partial derivatives
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5
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Partial derivatives
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6
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Partial derivatives
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7
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Total derivatives
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8
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Total derivatives
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9
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Maxima and minima
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. 10
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Lagrange multipliers method.
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A Big Data Matrix is a very large data structure arranged in rows and columns, where:
A Numerical Data Matrix is a matrix in which all elements are numerical values.
$$ X = \begin{bmatrix} x_{11} & x_{12} & \cdots & x_{1n} \\ x_{21} & x_{22} & \cdots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} & x_{m2} & \cdots & x_{mn} \end{bmatrix} $$
$$ M = \begin{bmatrix} 78 & 85 & 90 \\ 88 & 76 & 84 \\ 92 & 89 & 95 \end{bmatrix} $$
Average Marks of Student i:
$$ \mu_i = \frac{1}{n}\sum_{j=1}^{n} M_{ij} $$
$$ W = \begin{bmatrix} 30.5 & 78 & 12 \\ 29.8 & 82 & 5 \\ 31.2 & 75 & 0 \end{bmatrix} $$
Mean Temperature:
$$ \bar{T} = \frac{1}{m}\sum_{i=1}^{m} W_{i1} $$
$$ S = \begin{bmatrix} 1200 & 1350 & 1100 \\ 900 & 1000 & 980 \\ 1500 & 1600 & 1700 \end{bmatrix} $$
Total Sales:
$$ \text{Total} = \sum_{i=1}^{m}\sum_{j=1}^{n} S_{ij} $$
$$ R = \begin{bmatrix} 22.4 & 101.3 \\ 22.6 & 101.1 \\ 22.5 & 101.2 \end{bmatrix} $$
Variance of Temperature:
$$ \sigma^2 = \frac{1}{m}\sum_{i=1}^{m}(R_{i1}-\mu)^2 $$
$$ I = \begin{bmatrix} 0 & 128 & 255 \\ 64 & 192 & 128 \\ 255 & 128 & 0 \end{bmatrix} $$
Normalization:
$$ I' = \frac{I}{255} $$
Types and worked examples (converted from the uploaded notes).
When the denominator is expressible as a product of distinct linear factors.
Evaluate \( \displaystyle \int \frac{x-1}{(x+1)(x-2)} \, dx \).
Write
\( \dfrac{x-1}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2} \)
Then
\( x-1 = A(x-2) + B(x+1) \)
Putting \(x=2\): \(1 = 3B \Rightarrow B=\tfrac{1}{3}\).
Putting \(x=-1\): \(-2 = -3A \Rightarrow A=\tfrac{2}{3}\).
Thus
\( \dfrac{x-1}{(x+1)(x-2)} = \dfrac{2/3}{x+1} + \dfrac{1/3}{x-2} \)
Integrate:
\( \int \frac{x-1}{(x+1)(x-2)} dx = \frac{2}{3}\ln|x+1| + \frac{1}{3}\ln|x-2| + C \)
Evaluate \( \displaystyle \int \frac{2x-1}{(x-1)(x+2)(x-3)} \, dx \).
Assume
\( \dfrac{2x-1}{(x-1)(x+2)(x-3)} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3} \)
Multiply through and compare or use substitution at roots.
Putting \(x=-2\): \(-5 = ( -3)(-5)B \Rightarrow B=-\tfrac{1}{3}\).
Putting \(x=1\): \(1 = (3)(-2)A \Rightarrow A=-\tfrac{1}{6}\).
Putting \(x=3\): \(5 = 2\cdot5\;C \Rightarrow C=\tfrac{1}{2}\).
So
\( \dfrac{2x-1}{(x-1)(x+2)(x-3)}
= -\tfrac{1}{6}\dfrac{1}{x-1} -\tfrac{1}{3}\dfrac{1}{x+2} + \tfrac{1}{2}\dfrac{1}{x-3}
\)
Integrate:
\( \int \dfrac{2x-1}{(x-1)(x+2)(x-3)} dx
= -\tfrac{1}{6}\ln|x-1| -\tfrac{1}{3}\ln|x+2| + \tfrac{1}{2}\ln|x-3| + C
\)
When denominator contains repeating linear factors.
Evaluate \( \displaystyle \int \frac{3x+1}{(x-2)^2(x+2)} \, dx \).
Assume decomposition of the form
\( \dfrac{3x+1}{(x-2)^2(x+2)} = \dfrac{A}{x-2} + \dfrac{B}{(x-2)^2} + \dfrac{C}{x+2} \)
Solving (substitute roots & compare coefficients) gives \(B=\tfrac{4}{7}\), \(C=-\tfrac{5}{16}\), and \(A=\tfrac{5}{16}\).
Thus integrate to get logarithms and simple rational terms. (Detailed algebra steps in original notes.)
This example in the notes follows the same repeated-factor method — substitute roots and compare coefficients to find constants, then integrate term-by-term to obtain log terms and rational terms.
When denominator has irreducible quadratic factors, use linear numerators for those parts.
Evaluate \( \displaystyle \int \frac{8}{(x+2)(x^2+4)} \, dx \).
Decompose as
\( \dfrac{8}{(x+2)(x^2+4)} = \dfrac{A}{x+2} + \dfrac{Bx+C}{x^2+4} \)
Solving constants yields \(A=1\), \(B=-1\), \(C=2\).
So
\( \dfrac{8}{(x+2)(x^2+4)} = \dfrac{1}{x+2} + \dfrac{-x+2}{x^2+4} \)
\end{pre>
Integration gives
\( \int \dfrac{8}{(x+2)(x^2+4)} dx
= \ln|x+2| - \tfrac{1}{2}\ln(x^2+4) + \arctan\left(\tfrac{x}{2}\right) + C
\)
Notes converted from the uploaded file: partial fraction.pdf. :contentReference[oaicite:1]{index=1}
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A concise reference listing common matrix types with a short definition and two examples for each. Perfect for students, teachers, and blog readers.
Examples
[ 3 5 7 ] [ -2 4 9 1 ]
Examples
[ 4 -1 2 ] [ 7 0 ]
Examples
[ 1 2 3 4 ] [ 5 6 7 1 0 2 3 4 8 ]
Examples
[ 1 2 3 4 5 6 ] [ 7 8 9 10 11 12 ]
Examples
[ 0 0 0 0 ] [ 0 0 0 ]
Examples
I2 = [ 1 0
0 1 ]
I3 = [ 1 0 0
0 1 0
0 0 1 ]
Examples
[ 3 0 0 0 5 0 0 0 7 ] [ 1 0 0 -4 ]
Examples
[ 5 0 0 5 ] [ -3 0 0 0 -3 0 0 0 -3 ]
Upper triangular examples
[ 1 2 3 0 5 6 0 0 7 ] [ 4 -2 0 9 ]
Lower triangular examples
[ 2 0 0 3 4 0 5 6 7 ] [ 1 0 -3 8 ]
Examples
[ 2 3 3 5 ] [ 1 4 7 4 2 6 7 6 3 ]
Examples
[ 0 -2 2 0 ] [ 0 3 -1 -3 0 5 1 -5 0 ]
Examples
[ 1 2 2 4 ] [ 3 6 1 2 ]
Examples
[ 1 2 3 4 ] [ 2 5 1 3 ]
Examples
[ 1 0 0 1 ] [ 0 1 -1 0 ]
Examples
[ 0 0 5 0 0 0 2 0 0 ] [ 0 3 0 0 0 0 ]
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