Thread Starter


Joined Apr 11, 2009
I'm stuck with an extremely difficult numerical analysis problem and all help will be much appreciated.

Heres the question:

In certain signal detection problems (eg, radar or sonar) the probability of false alarm (FA) (ie of saying that a certain signal is present in the data when it actually is not) is given by:

pFA = ∫ _______1_______x^p/2-1 e-x/2 dx
.........η ...Γ(p/2) 2^p/2

Where η is called the detection threshold. If p is an even number, it can be shown that reduces to the finite series:

pFA = e^(-1/2 η) ... Σ ...... _1_..(_) ^k
...........................k=0 .......k! .... 2

. are used to align the values correctly

The detection threshold η is a very important design parameter in signal detectors. Often it is desired to specify an acceptable value for pFA (where 0 < pFA <1), and then it is necessary to solve nonlinear equation for η. Let p = 6. Use the bisection method to find η for:

a) pFA = 0.001
b) pFA = 0.01
c) pFA = 0.1

Instructions on how problem was solved will be appreciated as well so I can undertand what was done


Joined Nov 9, 2007
The bisection algorithm is about the simplest possible numerical method for solving equations of the type f(\(\eta\)) = 0.

What you have to do is rewrite your expression for pFA in this form and then find a range of \(\eta\) which includes a sign change

ie find a value of \(\eta\) for which f(\(\eta\)) is negative and then one for which it is positive.
The correct value lies somewhere between these two, so you keep halving the range, choosing the half containing the sign change each time.

Since p=6, (p/2) - 1 is only 2 and your summed series is very short.

Thread Starter


Joined Apr 11, 2009
Thanks for replying.

I already understand the concept of bisection method, what is unclear to me is the actual steps to solve this problem.
Where exactly do I start????
And although the summed series lies between 0-2 how do i arrive at actually finding the summed series???
And how do i link those answers to using the bisection method??

Awaiting more replies...


Joined Nov 9, 2007
What you have to do is rewrite your expression for pFA in this form
That is for pFA = 0.001, 0.01 and 0.1

0.001 = exp(-1/2η) { η/2 + η^2/2.2 }

rearrange this expression

0.001 *exp(+1/2η) = { η/2 + η^2/2*2 }

(please check my arithmetic - it's late here)


Joined Nov 9, 2007
What did you not understand?

I have taken your formula for pFA, and substituted in one set of the values you gave (pFA=0.001 and p=6) You will need to do this three times, once for each value of pFA.

I did get the summation wrong as the series should start with a 1

That is

0.001 = e^(-1/2 η) ... Σ ...... _1_..(_η_) ^k = 0.001 = exp(-1/2η) { 1 + η/2 + η^2/2.2 }

...........................k=0 .......k! .... 2

Now e^(-1/2 η) means divided by e^(+1/2 η) so multiply both sides of the equation by e^(+1/2 η) to get my second line (with the series corrected for the missing first term)

This gives you an equation in η which you can plug into your numerical method of choice.

Can you continue from here?


Joined Nov 9, 2007
e^(-1/2 η)
To answer you questions about starting values. I am not sure if you mean

exponential a half η or

exponential 1 over 2η

Either way start with η = 1. Here the exponential will be smaller than the series.

If you meant my first statement the larger η the larger the expnential so it will cross the series somewhere between 1 and 10

If you meant my second statement the smaller ( <1) η the larger the exponential so search between 0 and 1.