Archive

Posts Tagged ‘proof’

Compound tan – tan(A+B)

September 24, 2009 Leave a comment

We can use expressions for cos(A+B) and sin(A+B) to help us find tan(A+B).

Using the identity tanx = sinx / cosx we can write

tan(A+B) = \frac{sin(A+B)}{cos(A+B)}

we can now substitue in
sin(A+B) = sinAcosB + sinBcosA
and
cos(A+B) = cosAcosB – sinAsinB
to get

tan(A+B) = \frac{sinAcosB + sinBcosA}{cosAcosB - sinAsinB}

We can now divide both the top and bottom by cosAcosB to get

tan(A+B) = \cfrac{\cfrac{sinAcosB + sinBcosA}{cosAcosB}}{\cfrac{cosAcosB - sinAsinB}{cosAcosB}}
or
tan(A+B) = \cfrac{\cfrac{sinAcosB}{cosAcosB} + \cfrac{sinBcosA}{cosAcosB}}{\cfrac{cosAcosB}{cosAcosB} - \cfrac{sinAsinB}{cosAcosB}}

We can now simplify this by cancelling any cosA and cosB to get

tan(A+B) = \cfrac{\cfrac{sinA}{cosA} + \cfrac{sinB}{cosB}}{1 - \cfrac{sinA sinB}{cosA cosB}}

finally by substituting the identity tan(x) = \frac{sinx}{cosx} we find our result

tan(A+B) = \cfrac{ tanA + tanB}{1 - tanAtanB}

And it can be shown that this result can be extended to

tan(A \pm B) = \cfrac{tanA \pm tanB}{1 \mp tanAtanB}

Why the proof 2=1 is wrong

September 17, 2009 Leave a comment

There are a number of apparent proof that 2=1. An example of one such proof is below

Suppose a=b then
a2 = ab
2a2=a2+ab
2a2-2ab = a2+ab-2ab
2a2-2ab=a2-ab
2(a2-ab) = 1(a2-ab)
2 = 1

However this is clearly not true which mean there must be a step in the above algebra which isnt valid.

The invalid step

The invalid step is  in fact the very last step in the proof, cancelling both sides by a2-ab. This is because a2-ab = 0 since we initially assumed a=b hence

ab = aa = a<sup>2</sup>

This then means that the second last line states

2 x 0 = 1 x 0

which is obviously true but you cant “cancel” both sides by 0 since any number divided by zero is meaningless and doesnt have a values. The trick involved in this proof is hiding the divide by zero which most people would recognise as invalid with a more complicated looking expression.

What this means

This example demonstrates the dangers of allowing divide by zeros to occur in your calculations even if its in the form of algebra. It also demonstrates the problems of considering infinty (x/0) as a number that can be used in calculations.

Categories: algebra, maths Tags: , , , ,

Fibonacci Sequence and PHI

September 12, 2009 Leave a comment

The Fibonacci sequence is a famous sequence of numbers starting with 1,1 where each term (from the third onwards) is the sum of the previous two terms. The first few numbers in the sequence are as follows:

1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,
46368,75025,121393,196418,317811,514229,832040,1346269….

which quickly become very large.

The Fibonacci numbers, fr can be defined by
f1 = f2 = 1
and
fk = fk-1 + fk-2 for k≥3

PHI the golden ratio

PHI or Φ is said to be the golden ratio since so many things in nature seem to naturally arrange themselves in this ratio. It is approximately equal to 1.618033988749895.

Phi is also the positive solution to the equation

x2 = 1 + x

which has the irrational solution

phi = (1+√5)/2

Interestingly an irrational number is one which cant be written as the ratio of two integers so the golden ratio is not in fact a ratio meaning ratios in nature can only become very close to it but cant actually equal it.

PHI and Division of Fibonacci numbers

It is interesting to calculate the ratio of consecutive numbers in the Fibonacci sequence. These ratios quickly approach the golden ratio know as PHI or Φ.

The start of list of value obtained from these divisions are:
1
2
1.5
1.6666666666666667
1.6
1.625
1.6153846153846154
1.619047619047619
1.6176470588235294
1.6181818181818182
1.6179775280898876
1.6180555555555556
1.6180257510729614
1.6180371352785146

Calculation of PHI from limit of Fibonacci divisions

If we suppose that the ratio of consecutive terms in the Fibonacci sequence do approach a limit we can use this to find the value of phi.

IF we denote the nth Fibonacci number by fn and the nth ratio as rnthen

rn = fn+1 / fn

But using the definition of a Fibonacci number:

fn+1 = fn + fn-1

then the ratio is

rn = (fn + fn-1 )
fn

Which can be simplified to be

rn = 1 + fn-1/fn

but fn-1/ f = 1/rn-1

so

rn = 1 + 1/rn-1

Now supposing that the ratios tend to a limit p as n tends to infinity then p is the solution of the equation

p = 1 + 1/p

which can be re-arranged to give

p2 – p – 1 = 0

and can be solved using the quadratic equation to give

p = (1 + √5)/2
and
p = (1 – √5)/2

The first of these solutions happens to be the golden ratio PHI or Φ

If anyone has a proof that these ratios do infact approach a limit please include it in the comments of this post.

Categories: maths Tags: , , , ,

Differentiate Logs with Proof

In order to differentiate logs we must use the chain rule. The simplest type of log to differentiate is a natural log this can be done as shown below.

Differentiate Natural Logs
A natural log is a log to the base e.
d/dx (ln x) = 1/x

However if we want to differentiate ln(f(x)) we must use the chain rule to get

d/dx (ln(f(x)) = f'(x)/f(x)

Proof of Derivative of Natural Logs
Consider

y=ln(x)

then from the definition of a log we get

ey = x                   –(1)

Differentiate each side with respect to x (you need to use implicit differentiation for the left to get ey dy/dx) to get

ey dy/dx = 1

but from (1) we know that ey = x which we can substitute to get

x dy/dx =1

giving the derivative

dy/dx = 1/x

Compound Angles: Cos(A+B) = CosACosB – SinASinB

July 20, 2009 2 comments

Compound angles are angles made by adding two other angles together. When using trigonometry unfortunately you cant just “times out” the trig function but have to use an identity. This post will consider how we get the identity for cos(A+B):


Compound angle of A+B showing how they relate

Compound angle of A+B showing how they relate

From the definition of cos we find

cos(A+B) = OT/OR

but
OT = OP – PT
and PT = SQ so
OT = OP – SQ

so

cos(A+B) = ( OP – SQ ) / OR
so
cos(A+B) = OP/OR – SQ/OR

if we now times the both the top and bottom of the first term by OQ and do the same for the second term but with RQ we can get

compcos1

but OP/OQ = cosB,
OQ/OR = cosA,
SQ/RQ = sinB,
RQ/OR = sinA

so we get, when these are substituted in and re arranged

cos(A+B) = cosAcosB-sinAsinB

Proof by Mathematical Induction

Proof by induction involves proving that if a statement is true in one case then it must also be true for the next case. Then by showing that it is true for a base case, eg 0, you can conclude that it is true for all positive integral cases, since if it is true for case 0 it must be true for case 1 and if it is true for case 1 it must also be true for case 2 etc

To do this it is normal to let Pn be the statement you are trying to prove in a general case n. Let us use the example Pn: 9n – 1 is divisible by 8 so we may right 9n-1 = 8a where a is an integer

Then we must prove that if Pk is true Pk+1 is also true. This will require using the expression Pk otherwise it isn’t induction since you are not showing that the truth of each statement can be found from the previous one.

so consider 9(k+1) – 1
This is equal to 9x9k -1=9x9k – 9+ 8=9(9k-1)+8
but from Pk we know 9k-1 = 8a so
9k+1-1 = 9(8a)+8=8(9a+1)
which is of the form 8b so is divisible by 8
Hence if Pk “9k-1 is divisible by 8″ is true then Pk+1 is also true

Finally we must prove that P1 is true and we then know that Pn is true for all k >=1 where k is an integer.

P1 is the statement 91 – 1 is divisible by eight
91-1=9-1=8 so is true

therefore Pn is true for all positive integral value of n

Area of a Triangle

Here is a general formula to calculate the area, A, of a triangle with width w and height h as shown in the diagram.

A = ½ wh

triangle of width w and height h

triangle of width w and height h

Proof

In order to work out the area of a triangle we can draw it within a rectangle that touches all three corners and has a base of equal width as shown below.

triangle inside rectangle

triangle inside rectangle

From this diagram we can now consider the are of the triangle on the left and the one on the right seperatley and then add them together to get the area of the whole triangle. We can see that each of these triangles cuts the rectangle they are in in half so they half the area of that rectangle,

ie) the left hand triangle has the area ha/2 and the right hand one has the area hb/2 so the total are of the triangle is

A = ha/2 + hb/2

Now we can take out the h/2 as a common factor to get

A =(a+b) h/2

but a+b = w since that was how the triangle was constructed hence we get the formula for the area

A = wh/2