Posts Tagged ‘maths’

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Categories: maths Tags: , ,

Integration by Parts

July 19, 2009 1 comment

Integration by parts is a method that allows us to integrate the product of two function such as


where 2x is one function and e3x is another
To do this we use the formula

∫(u d/dx) dx = uv – ∫(v du/dx) dx

where u and v are both functions of x, 2x and e3x in the above example.

Proof of Integration by Parts Formula

This can be shown by considering the product rule for differentiation as shown below
the product rule states that
d(uv)/dx = u dv/dx + v du/dx

If we now integrate both side we get

uv = ∫( u dv/dx) dx + ∫(v du/dx) dx
since integration is the opposite of differentiation

now we can simply rearrange this to get

∫(u d/dx) dx = uv – ∫(v du/dx) dx

Simple Example (without limits)

To demonstrate this formula we shall integrate the example above


To do this we first need to pick which function (2x or e3x) to allocate to u and which to dv/dx. Picking the correct function is crucial to integration by parts since the method works by allowing us the differentiate the part of the function which is difficult to integrate. In order for this to work it is useful to use a function that will become simpler in some way. For example differentiating e3x gives 3e3x so making the product no easier to integrate, however integrating 2x gives 2 which does make it easier to integrate since we can take the 2 outside as a factor.

So we will let u=2x and dv/dx=e3x
we then differentiate 2x to get 2
and integrate e3x to e3x/3
(it can often be useful to put these in a table with columns u,v and the function on the first row and derivative on the second row)

These can then be put into the formula to find the integral
∫2xe3xdx = 2xe3x/3 – ∫e3x2/3 dx
∫2xe3xdx = 2xe3x/3 – 2e3x/9 + c

Example with Limits

usually, however integration is required to be carried out between limits. To perform integration by parts between limits is quite simple. You takes the value of uv as usual between limits and then set the limits of  ∫v du/dx dx to the limits of the original integral. To demonstrate this we will integrate the above example between 0 and 1


10 2x e3x dx = [2xe3x/3]10 – ∫10 e3x2/3 dx
=2e3 – 0 – 2e3/9 + 2/9 = (16e3 + 2) / 9

Confindence Intervals

Confidence intervals are a range of values within which you can say an unknown value is expected to lie with a specified degree of certainty or probability. For example, from a sample of 10 journeys you might say that the you are 95% certain that the average time it takes me to get to school (from all journeys I have made to school not just the sample of 10) lies within the range 10-12.5 minutes.

When taking a random sample it is much better to use a confidence interval for your results rather than just giving the mean, because it gives an idea of how reliable your mean is. This is because you can calculate an average value from a set of completely random results but this doesn’t mean that you can have any certainty that the next result will be similar to the mean (since we have stated that the results are completely random they are no more likely to be close to the mean than any other value).

Calculating a confidence interval (for a normal distribution)

When calculating a confidence interval you must first decide on the percentage certainty that you are going to use for the interval (a common value to use is a 95% confidence interval)

You then use the reverse tables for the normal distribution to work the value of the standardised normally distributed variable to use.
Note: You must use the value half way between the certainty level and 100%, ie if you want a 95% confidence interval use 97.5% since you only want 2.5% on either side of the distribution.

You can now calculate the interval. To do this you need the standard deviation and mean of the sample. However to correct the standard deviation for the entire sample of possible tests divide by the square root of the number of items in your sample

ie) if you sample of n items is Y and the entire sample of possible results is X
sd(X) = sd(Y)/√n

Now the confidence interval is

X – z sd(X), X + z sd(X)

Where z is the value obtained from the inverse tables.

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


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

Proof by Contradiction

The aim of proof by contradiction is to prove a statement is true by assuming it is false and showing that this leads to a contradiction so that the statement must be true. This is often easier than proving the statement directly.

For example consider the proof √2 is irrational that follows

Assume root 2 is rational, ie that it can be written as r/s where s≠0 and r and s are both integers. We can choose r and s such that they have no common factors, since any common factors can be cancelled out.

then 2=r2/s2
so r2 = 2s2 —(1)
hence 2 is a factor of r2 and since 2 is prime 2 must also be a factor of r so we can write r = 2k where k is an integer.

From (1) we can now write
so s = 2k2
so 2 is also a factor of s
But we assumed r and s had no common factors, Contradiction therefore root 2 must be irrational.

So to conclude, here we have taken the statement root 2 is irrational, assumed it to be false, shown this leads to a contradiction and therefore concluded that root 2 must be irrational.

Categories: maths Tags: , , ,

Proof there are infinitley many prime numbers


Prime numbers are natural numbers (1,2,3 etc) whose only factors are one and themselves. The first few primes are:


note that 1 isn’t a prime.

Primes are important because every number can be written uniquely as a product of primes. This idea can also be used to prove that there infinitely many primes. The proof which follows is an example of proof by contradiction.


Assume that there are a finite number of primes.

Then all these primes can be written in a list

P0, P1, P2 … Pn

Then we could multiply all these numbers together and add one to get a new number Q

Q = P0P1P2…Pn + 1

This means that Q doesn’t have any of the primes as a factor as it will leave a remainder of one when divided by any of the primes, because that was how it was constructed. But every number can be written as a product of primes. Therefore Q is either a new prime or there is a prime which isn’t on the list which is a factor of Q. So the list doesn’t contain all the primes because it doesn’t contain this new prime. This is a contradiction because we assumed that the list contained all the primes, therefore there cant be a finite number of prime so there must be an infinite number of them, hence they cannot be written in a list.

By David Woodford

Categories: maths Tags: , , , ,

Auxiliary Angle Method for Solving Trigonometry Equations

April 19, 2009 2 comments

This is a method of solving equations in the form asinx+bcosx = c where a and b are constants and c is another expression.

It involves rewriting letting asinx + bcosx = rsin(x+y) (or you could use cos(x+y)) where y is acute and then finding values for r and y, then with only one trig function to deal with the equation can be solved more easily.

For example

consider 2sinx + 3cosx = 3

Let 2sinx + 3cosx = rsin(x+y)

Now expand the sin(x+y) to get

2sinx + 3cosx = rsinx cosy + rcosx siny

Since y is constant and therefore cosy and sin y are constant we can compare the coefficients to get

2 = rcosy —–(1)
3 = rsiny ——(2)

We can solve these to find values for r and y.
To find y consider (2)/(1) to get

3/2 = tany
since sin/cos = tan and the r’s cancel
so y = 56.3 °

To find r consider (1)2+(2)2 to get
22+32 = r2
since sin2+cos2 = 1
so r =√13

So we can write

2sinx + 3 cosx = √13 cos(x+56.3) = 3

so x = cos<sup>-1</sup>(3/√13) -56.3

so x = cos-1(3/√13) -56.3

since cos-1(3/√13) = 33.7 for solutions between 0° and 90°

x = ±33.7 -56.3 + 180n where n is an integer

In General

asinx + bcosx = √(a2+b2) sin(x+tan-1(b/a))

If you have any questions, comments or corrections please leave them as a comment below

By David Woodford