BEC'S methods questions

[Collin Li]

This was done using the 2 standard deviations from the mean 95% estimate. Don't use this, the answer was nearer to 0.023.

Find out how to use your calculator soon, I can tell you tried to use the 1 standard deviation from the mean 68.5% estimate. That is extremely inaccurate, especially when you're looking for 1.5 standard deviations from the mean, LOL

[bec]

Are you sure you're not looking at the wrong part of the answers?

um.... ok yes. i was.
i checked so many times though!

for the record, you were right of course (it's 0.0228 for the first one and 6.7% for the second)
A+.

ok you won't be hearing from me for a while! i'm going away tomorrow and not looking at my maths book until i'm sitting in a classroom. thanks so much for all your help, it's so much easier to learn if you can ask questions as you go instead of letting them accumulate.

hopefully next year i can actually help other people instead of just being at the receiving end!

[Ahmad]

You definitely won't need to buy another one bec, you can get the application transferred. I found it a bit weird that the ti-89/ti-92+ didn't have those probability functions.

You could calculate it 'by hand' using the integral , but it's most accurate, easier and faster if you use a program, which would have specialised routines to compute this.

[/0]

But... isn't it impossible to evaluate over a finite interval?

[enwiabe]

Nope! My UMEP teacher showed me how using double integrals.

EDIT: Sorry! He didn't show me how to evaluate it, he showed me how to prove that the area under a normal distribution curve = 1.

[Ahmad]

The gaussian integral can be evaluated over -infinity to infinity by first converting it to a double integral and transforming it into polar coordinates, then squeezing it between the incircle and circumcircle of a square. However, it is "impossible" to evaluate it over a finite interval exactly. That doesn't mean it can't be evaluated numerically, this can be done, for example, by converting it into a power series which can be evaluated until the required degree of accuracy is reached. Or you could use simpson's rule, or some other numeric integration technique which your calculator would use.

[bec]

i'm back...
I've just got a couple of easy questions; if anyone could help that'd be great

If x² = a(x + 2)² + b(x + 2) + c for all values of x, find the values of a, b and c.
I got the correct answer but I think my method involved assumptions...I did this to find a:
ax² + 4ax + 4a +bx + 2b + c = x²

On RHS, coefficient of x²=1, therefore a=1  <-----can I be sure of this assumption?

[Neobeo]

Since you are given that the equation holds for all values of x, it becomes an identity.
You already simplified it to , and since it is an identity over x, you can actually equate all coefficients of x with the corresponding one on the other side.



Then comparing coefficients gives:


And finally solving for a, b and c as required.

Alternatively, you can note the RHS to be a polynomial over (x+2),so let or equivalently, .
This simplifies the original equation to

Remember that the equation holds for all x, therefore it should also hold for all X.


Comparing the coefficients as before will give a, b and c as required.

[Collin Li]

On RHS, coefficient of x²=1, therefore a=1  <-----can I be sure of this assumption?

Yes. It is because of the linear independence of (basically means you can't express any of these elements as another one of these elements even when multiplied by any constant). The consequence of it is that you can equate the coefficients in the way you and Neobeo have.

[bec]

great thanks for that

how about this one

Find the exact solution to the following equation:
x⁴ + 5x² − 36 = 0

I completed the square, then factorised using difference of two squares and somehow came up with:
OR
So I got some hideously complicated value for x but in the text solns it's x=-2,2
how do you do it?

[/0]

But the first solution gives complex numbers, so if you're sticking to reals, only look at the second:

[bec]

whoa i was wayyy off
how did you factorise that? just with inspection or is there another method?

[aiisha]

to factorise it you can do the thing that's like the criss-cross method but not. So you times the first and the last term so you get -36x^4. then you think of 2 terms that multiply to give -36^4 and add to give the middle term (5x^2). so the two numbers are 9x^2 and -4x^2. then you substitute these terms in place of the 5x^2. So x^4+9x^2-4x^2-36.then you group two and two and you get (x^2-9)(x^2-4)

[aiisha]

sorry i mean x^2+9 not -9

[Collin Li]

I think you meant , not .


I would just use the quadratic formula, letting





Since , therefore , so the only solution is



You don't need to factorise if you're only asked to find the solutions. Completing the square is a pain in the ass - using the quadratic formula is preferred.