CATEGORY / Learning

02Dec

2016

Eugene / Dale Carnegie Sales, Learning / 0 comment

Compelling Sales Presentation

Sales presentation flow

Research
Rapport (Agenda – my side, your side, timeline, next action, confirmation)
Interest
Solution
Commitment
Follow through

Research

Name and title
How long have they been with your company
Common ground, contacts or referral
Communication style (formal or informal)
Challenges that they are facing
How could they benefit from me

Listening principles

Maintain contact with the person talking either visually or with focused listening.
Practice pure listening. Remove all distractions and minimise internal and external filters. Refrain from multi-tasking.
Be sensitive to what is not being said. Listen for voice tone and observe body language for incongruent messages.
Practice patience. Do not interrupt, finish the speaker’s sentence, or change the subject.
Listen empathetically and listen to understand. Try to see things from his or her perspective.
Act a if there will be a quiz at the end.
Clarify any uncertainties after he or she has spoken. Make sure you understood what was said by rephrasing what you heard.
Don’t jump to conclusions or make assumptions. Keep an open and accepting attitude.
Turn off your mind and “be with” the speaker.

Power questions

Require the customer to think about things they have never considered before
Usually open-ended
Thought-provoking
Questions provide us with essential information about the customer’s current or desired situation and their buying motives

Examples:

What would it mean to you, if you were able to …
How do you think that … will affect your business?
What do you look for …
What has been your experience …
How do you determine …
What does your competitor do about …
What are you doing to ensure …
How do your customers react to …
What is one thing you would improve about …

Response generators:

In what way?
Give me an example …
How so?
Tell me more …
That word has many definitions, what way do you mean it?
Oh?

Rephrase

So what I hear you saying is …
Let me make sure I understand …

The three E’s of compelling sales presentations

Earned the right through study and experience
Excited with positive feelings about your subject
Eager to share to project the value to your listeners

Presenting solutions & closing deal

Compelling question
Picture this
Evaluative question

Deliver the message

Voice
Words
Body language

01Dec

2016

Eugene / Learning, Stanford Machine Learning / 0 comment

Control Statements (Octave)

1. for i=1:10,
v(i) = 2^i;
end;
v
$v=\begin{pmatrix} 2\\ 4\\ 8\\ 16\\ 32\\ 64\\ 128\\ 256\\ 512\\ 1024 \end{pmatrix}$

2. i=1;
while i<=5,
v(i) = 100;
i = i+1;
end;
v
$v=\begin{pmatrix} 100\\ 100\\ 100\\ 100\\ 100\\ 64\\ 128\\ 256\\ 512\\ 1024 \end{pmatrix}$

3. i=1;
while true,
v(i) = 999;
i = i+1;
if i==7;;
break;
end;
end;
v
$v=\begin{pmatrix} 999\\ 999\\ 999\\ 999\\ 999\\ 999\\ 128\\ 256\\ 512\\ 1024 \end{pmatrix}$

4. v(1) = 2;
if v(1) == 1,
disp('The value is one');
elseif v(1) == 2,
disp('The value is two');
else
disp('The value is not one or two');
end;
The value is two

26Nov

2016

Eugene / Learning, Stanford Machine Learning / 0 comment

Plotting Data (Octave)

1. t=[0:0.01:0.98]
$t=\begin{pmatrix} 0 & 0.01 & 0.02 & ... & 0.98 \end{pmatrix}$

2. y1 = sin(2*pi*4*t)
plot(t,y1)
sine

3. y2 = cos(2*pi*4*t)

plot(t,y2)
cosine

4. plot(t,y1);
hold on;
plot(t,y2,'r');
xlabel('time')
ylabel('value')
legend('sin','cos')
title('my plot')
myplot

26Nov

2016

Eugene / Learning, Stanford Machine Learning / 0 comment

1. A = [1 2; 3 4; 5 6]
$A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}$
B = [11 12; 13 14; 15 16]
$B=\begin{pmatrix} 11 & 12 \\ 13 & 14 \\ 15 & 16 \end{pmatrix}$
C = [1 1; 2 2]
$C=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$

2. A*C
$\begin{pmatrix} 5 & 5\\ 11 & 11 \\ 17 & 17 \end{pmatrix}$

3. A .* B take each element of A to multiply by each element of B
$\begin{pmatrix} 11 & 24\\ 39 & 56 \\ 75 & 96 \end{pmatrix}$

4. A .^ 2 square each element of A
$\begin{pmatrix} 1 & 4\\ 9 & 16 \\ 25 & 36 \end{pmatrix}$

5. v = [1; 2; 3]
$v=\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}$

6. 1 ./ v element-wise reciprocal of v
$v=\begin{pmatrix} 1.00000\\ 0.50000\\ 0.33333 \end{pmatrix}$

7. log(v) element-wise logarithm of v
exp(v) element-wise exponential of v
abs(v) element-wise absolute value of v
-v element-wise negative value of v
v+1 element-wise addition of 1 to v

8. A = [1 2; 3 4; 5 6]
$A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}$
A' transpose of A
$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$

9. w = [1 15 2 0.5]
$w=\begin{pmatrix} 1 & 15 & 2 & 0.5 \end{pmatrix}$

10. max (w) maximum value of w
val = 15

11. [val, ind] = max(w) maximum value of w and index where it is located
val = 15
ind = 2

12. w < 3 element-wise comparison of whether w is less than 3
$\begin{pmatrix} 1 & 0 & 1 & 1 \end{pmatrix}$

13. find(w < 3) find which elements that variable w is less than 3
$\begin{pmatrix} 1 & 3 & 4 \end{pmatrix}$

14. sum(w) sum of w
ans = 18.5

15. prod(w) product of w
ans = 15

16. floor(w) rounds down elements of w
$\begin{pmatrix} 1 & 15 & 2 & 0 \end{pmatrix}$

17. ceil(w) rounds down elements of w
$\begin{pmatrix} 1 & 15 & 2 & 1 \end{pmatrix}$

18. A = magic(3) magic square of 3 by 3
$\begin{pmatrix} 8 & 1 & 6\\ 3 & 5 & 7\\ 4 & 9 & 2 \end{pmatrix}$

19. [r,c] = find(A >= 7) find rows and columns of A greater than or equal to 7
$r=\begin{pmatrix} 1\\ 3\\ 2 \end{pmatrix}$
$c=\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}$

20. A(2,3)
$\begin{pmatrix} 7 \end{pmatrix}$

21. max(A,[],1) column-wise maximum of A
$\begin{pmatrix} 8 & 9 & 7 \end{pmatrix}$

22. max(A,[],2) row-wise maximum of A
$\begin{pmatrix} 8 \\ 7 \\ 9 \end{pmatrix}$

23. max(max(A))
$\begin{pmatrix} 9 \end{pmatrix}$

24. pinv(A) inverse of A
$\begin{pmatrix} 0.147 & -0.144 & 0.064 \\ -0.061 & 0.022 & 0.106 \\ -0.019 & 0.189 & -0.103 \end{pmatrix}$

16Nov

2016

Eugene / Learning, Stanford Machine Learning / 0 comment

Moving Data (Octave)

Matrices
1. A = [1 2; 3 4; 5 6]
$A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}$

2. size(A) size of matrix
$\begin{matrix} 3 & 2 \end{matrix}$

3. size(A,1) number of rows
ans = 3

4. size(A,2) number of columns
ans = 2

5. A(3,2) $A_{32}$
ans = 6

6. A(2,:) every element along row 2
$\begin{pmatrix} 3 & 4 \end{pmatrix}$

7. A(:,1) every element along column 1
$\begin{pmatrix} 1\\ 3\\ 5 \end{pmatrix}$

8. A([1 3],:) every element along rows 1 and 3
$\begin{pmatrix} 1 & 2\\ 5 & 6 \end{pmatrix}$

9. A(:,2) = [10; 11; 12] replace column 2 with new elements
$\begin{pmatrix} 1 & 10 \\ 3 & 11 \\ 5 & 12 \end{pmatrix}$

10. A = [A, [100; 101; 102]] append new column vector to the right
$\begin{pmatrix} 1 & 10 & 100 \\ 3 & 11 & 101\\ 5 & 12 & 102 \end{pmatrix}$

11. A(:) put all elements of A into a single vector
$\begin{pmatrix} 1\\ 3\\ 5\\ 10\\ 11\\ 12\\ 100\\ 101\\ 102 \end{pmatrix}$

12. A = [1 2; 3 4; 5 6]
B = [11 12; 13 14; 15 16]
C = [A B] concatenating A and B
$C=\begin{pmatrix} 1 & 2 & 11 & 12 \\ 3 & 4 & 13 & 14\\ 5 & 6 & 15 & 16 \end{pmatrix}$

13. C = [A; B] putting A on top of B
$C=\begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \\ 11 & 12 \\ 13 & 14 \\ 15 & 16 \end{pmatrix}$

14. v = [1 2 3 4]
$v=\begin{pmatrix} 1 & 2 & 3 & 4 \end{pmatrix}$

15. length(v) length of vector v
ans = 4

Loading files
1. path: pwd shows where Octave location is

2. change directory: cd '/Users/eugene/desktop'

3. list files: ls

4. load files: load featuresfile.dat

5. list particular file: featuresfile

6. check saved variables: who

7. check saved variables (detailed view): whos

8. clear particular variable: clear featuresfile

9. clear all: clear

10. restrict particular variable: v = featuresfile(1:10) only first 10 elements from featuresfile

11. save variable into file: save testfile.mat v variable v is saved into testfile.mat

12. save variable into file: save testfile.txt v -ascii variable v is saved into text file

16Nov

2016

Eugene / Learning, Stanford Machine Learning / 0 comment

Basic Operations (Octave)

Logical operations
1. 1 == 2 1 equals to 2
ans = 0 false

2. 1 ~= 2 1 is not equal to 2
ans = 0 true

3. 1 && 2 AND
ans = 0

4. 1 || 2 OR
ans = 1

5. xor(1,0)
ans = 1

Change default octave prompt: PS1('>> ');

Assign variables
1. a = 3 printing out a = 3

2. a = 3; suppress print out

3. b = 'hi' for strings

4. c = (3>=1) true

Printing
1. disp(a) to show a

2. a = pi
disp(sprintf('2 decimals: %0.2f', a)) 2 decimal places
2 decimals: 3.14

2. disp(sprintf('2 decimals: %0.6f', a)) 6 decimal places
2 decimals: 3.141593

3. format long
a
a = 3.14159265358979

4. format short
a
a = 3.1416

Matrices
1. A = [1 2; 3 4; 5 6] 3 by 2 matrix
$A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}$

2. v = [1 2 3] 1 by 3 matrix (row vector)
$v=\begin{pmatrix} 1 & 2 & 3 \end{pmatrix}$

3. v = [1; 2; 3] 3 by 1 matrix (column vector)
$v=\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}$

4. v=1:0.1:2 1 by 11 matrix (row vector)
$v=\begin{pmatrix} 1 & 1.1 & 1.2 & ... & 1.9 & 2 \end{pmatrix}$

5. v=1:6 1 by 6 matrix (row vector)
$v=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \end{pmatrix}$

6. ones(2,3) 2 by 3 matrix of ones
$\begin{pmatrix} 1 & 1 & 1\\ 1 & 1 & 1 \end{pmatrix}$

7. 2*ones(2,3)
$\begin{pmatrix} 2 & 2 & 2\\ 2 & 2 & 2 \end{pmatrix}$

8. zeros(2,3) 2 by 3 matrix of zeroes
$\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$

9. rand(2,3) 2 by 3 matrix of random numbers between 0 and 1

10. randn(2,3) 2 by 3 matrix of random numbers drawn from a Gaussian distribution with mean 0 and variance

11. eye(4) 4 by 4 identity matrix
$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$

Plot histogram: w = -6 + sqrt(10)*(randn(1,10000))
hist(w) histogram
hist(w,50) histogram with 50 bins

Help: help rand help function

15Nov

2016

Eugene / Learning, Stanford Machine Learning / 0 comment

Normal Equation

$\theta = (X^T X)^{-1}X^T y$

Advantages and disadvantages:
1. No need to choose $\alpha$
2. Don’t need to iterate
3. Need to compute $(X^TX)^{-1}$
4. Slow if $n$ is very large

If $X^TX$ is non-invertible:
1. Redundant features (linearly dependent)
2. Too many features (e.g. $m\le n$ )
a. delete some features
b. use regularisation

14Nov

2016

Eugene / Learning, Stanford Machine Learning / 0 comment

Application Of Gradient Descent

Feature scaling: get every feature into approximately a $-1 \le x_i \le 1$ range

Mean normalisation: replace $x_i$ with $x_i-\mu_i$ to make features have approximately zero mean (do not apply to $x_0=1$ )

$x_i := \dfrac{x_i - \mu_i}{s_i}$
$\mu_i$ : average value of $x_i$ in the training set
$s_1$ : standard deviation

Points to note:
1. If gradient descent is working correctly, $J(\theta)$ should decrease after each iteration.
2. If $\alpha$ is too small, we will have slow convergence.
3. If $\alpha$ is too large, $J(\theta)$ may not converge.

Advantages and disadvantages:
1. Need to choose $\alpha$
2. Needs many iterations
3. Works well even when $n$ is large

14Nov

2016

Eugene / Learning, Stanford Machine Learning / 0 comment

Gradient Descent For Multiple Variables

Cost function: $J(\theta) = \dfrac {1}{2m} \displaystyle \sum_{i=1}^m \left (h_\theta (x^{(i)}) - y^{(i)} \right)^2$
$J(\theta) = \dfrac {1}{2m} \displaystyle \sum_{i=1}^m \left (\theta^Tx^{(i)} - y^{(i)} \right)^2$
$J(\theta) = \dfrac {1}{2m} \displaystyle \sum_{i=1}^m \left ( \left( \sum_{j=0}^n \theta_j x_j^{(i)} \right) - y^{(i)} \right)^2$

Gradient descent:

$ <img src="https://teach.sg/wp-content/ql-cache/quicklatex.com-180d38ebaaff2b28806bdd8feee8e5ee_l3.png" height="50" width="632" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \text{repeat until convergence:} \; \lbrace \newline \; & \theta_j := \theta_j - \alpha \frac{1}{m} \sum\limits_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x_j^{(i)} \; & \text{for j := 0..n} \newline \rbrace \end{align*}" title="Rendered by QuickLaTeX.com"/>$

which breaks down into

$ <img src="https://teach.sg/wp-content/ql-cache/quicklatex.com-ade19e57866347ec87a2a0df45d0c145_l3.png" height="50" width="1189" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \text{repeat until convergence:} \; \lbrace \newline \; & \theta_0 := \theta_0 - \alpha \frac{1}{m} \sum\limits_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x_0^{(i)}\newline \; & \theta_1 := \theta_1 - \alpha \frac{1}{m} \sum\limits_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x_1^{(i)} \newline \; & \theta_2 := \theta_2 - \alpha \frac{1}{m} \sum\limits_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x_2^{(i)} \newline & \cdots \newline \rbrace \end{align*}" title="Rendered by QuickLaTeX.com"/>$

14Nov

2016

Eugene / Learning, Stanford Machine Learning / 0 comment

Linear Regression With Multiple Variables

Notation:
$n$ : number of features
$x^{(i)}$ : input (features) of $i^{th}$ training example
$x^{(i)}_j$ : value of feature $j$ in $i^{th}$ training example

$h_\theta (x) = \theta_0 + \theta_1 x_1 + \theta_2 x_2 + \theta_3 x_3 + \cdots + \theta_n x_n$

$ <img src="https://teach.sg/wp-content/ql-cache/quicklatex.com-06f129ca3a97efd2c74e813c392da568_l3.png" height="32" width="384" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} h_\theta(x) = \begin{bmatrix} \theta_0 \hspace{2em} \theta_1 \hspace{2em} ... \hspace{2em} \theta_n \end{bmatrix} \begin{bmatrix} x_0 \newline x_1 \newline \vdots \newline x_n \end{bmatrix} = \theta^T x \end{align*}" title="Rendered by QuickLaTeX.com"/>$

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