A Level Maths Lecture - Matt Parker

Matt Parker

Mr Parker is the Public Engagement in Mathematics Fellow at Queen Mary, University of London, and a sell-out comic. In his talk; he made maths really, really fun. 

First off he showed off his ability to be able to nearly instantly calculate the cube root of any number (as long as the original number was 2 digit). However I think that I have found how he did this; by using the 'Vedic Maths Trick', which relates the final digit of the number to the cube root. Still very impressive.

He continued by bringing about the topic of Barcodes. He asked a random person in the audience to read out the barcode, and then he calculated (using his very brain heavy trick) the last digit of the code. Then he expanded and revealed that he was actually using the mathematical pattern that barcode readers use, to calculate the final digit. He showed that error checking and error correction has actually made most checkouts and self checkouts possible. He then went on to explain how this is much like text messages. When a symbol such as the letter 'M' is typed, it is converted to binary. The binary equivalent of 'M' is 01001101, where the starting '010' is used to define the letter as upper-case. Text messages, when received by mobile phones, have a mathematical pattern that will allow the receiving mobile to generate lost data if not all of the message was received. He said that text messages actually resemble a simple suduko puzzle where with little parts of missing information, it is able to calculate the remains of the text. However this can only work up to a certain amount of lost data, as there is a point in which the problem becomes unsolvable.

Finally, he showed the audience how all pictures are actually just a 'spreadsheet' because he physically showed us all, a picture of him, made from an excel document. Crazy. 

After the talk I went and met him briefly,  talking about him on the Youtube Channel 'NumberPhile' and then he signed my calculator "To 'Mitch', Matt Parker", in binary. Awesome :)

Manipulation of Logarithms

With Logarithms, there aren't that many ways that they can manipulated. Here are the 6 different rules that will allow you to manipulate most equations:

Where 'b' is the base
Where 'a' is the part in the brackets

1)  The log can be manipulated  to remove the log if it is given a value: here such as 'c'.

Logb(a)=z       Then b^z=a

2)  The Coefficient of the Log can be moved to the power:

Y Logb(a) = Logb(a)^Y

3)  The Log of (ac) can be written as two different Logs; where they are split by an addition sign.

Logb(as) = Logb(a) + Logb(s)

4) The Division of the Log can be changed to a minus sign, keeping the base of the log the same.

Logb(a/s) = Logb(a) - Logb(s)

5) Any Log with the base the same as the part in the brackets will = 1. This is because anything to the power of 1 will equal itself. 

Logb(b) = 1

6) When changing the base of the Log: Rewrite the same thing but change the base to whatever you want; then divide this by Log with the new base, with in brackets the old base.

Where c is the new base

Logb(a) = Logc(a)/Logc(b)

Trigonometric Identities

Trigonometric Identities can be used to simplify an equation before solving it. They can be used for any values of x.

IDENTITIES:

1)      Tan x = Sin x / Cos x

2)      Sin^2 x + Cos^2 x = 1

Using these two identities; equations will be able to be manipulated to allow you so simplify / solve them.

For Quadratic Equations always remember:

1) Try and make the equation only (sin^2 x / sin x) or (cos^2 x / cos x). Eliminate all other terms using the identities above.

2) An equation can only be solved if    sin x = k    or    cos x = k     if     1 ≥  k  ≥ -1.

Solutions to     tan x = k     exist for ALL VALUES of k.

Example 1:

"Show that    Tan x  +  (1/Tan x)  can be written in the form (1/Cos x Sin x)".

Tan x  +  (1/Tan x)

This is the original starting point given in the question.

(Sin x / Cos x) + (Cos x / Sin x)

 Tan x = (Sin x / Cos x)      and    (1/ Tan x) =  (Cos x / Sin x)

(Sin^2 x + Cos^2 x/Cos x Sin x)

Cross multiply and make the two fractions into one.

(1/Cos x Sin x)

Sin^2 x + Cos^2 x = 1    Therefore the top part of the bracket becomes one and the denominator stays the same.

Example 2:

"Show that    5cos x = 1 + 2Sin^2 x    can be written in the form 2cos^2 x + 5Cos x -3 = 0".

5cos x = 1 + 2Sin^2 x

This is the original starting point given in the question.

5cos x = 1 + 2(1-Cos^2 x)

Sin^2 x + Cos^2 x = 1         REARRANGED =        Sin^2 x = 1 - Cos^2 x 

Sin^2 x   has been substituted for     1 - Cos^2 x.

5cos x = 1 + 2 - 2 Cos^2 x

The brackets have been expanded from 2(1-Cos^2 x) to 2 - 2 Cos^2 x

2 Cos^2 x + 5Cos x -3 = 0

The terms have been grouped and formatted as positioned in the question.