UoM Maths Thread

[Ballerina]

I am so happy I am not studying for your exams.

This is an excerpt from my statistics lecture: 'Remember, 0.05 is just the mathematical way of writing 5%.'

[Ballerina]

You are welcome for the free tutoring.

[hobbitle]

use partial fractions to integrate: (x+13)/(x^3+2x^2-5x-6)
i can't work out how to factories the bottom polynomial.

Probably not the most efficient way but chances are they've given you something easily factorisable by hand... So just choose a couple of simple values of x (x=1,-1,2,-2) and see when the solution = 0. In this case if you sub in x=-1 you get 0 so (x+1) is a factor. Use polynomial long division and you'll end up with a factorisable quadratic.

[lzxnl]

More specifically, if you have a polynomial with integer coefficients and the coefficient of the largest power of x is 1 (as it is here), any integer roots roots, or the x values that make the polynomial zero, must be factors of the constant term. So you only need to check +-1, +-2, +-3. This works in general. So if you had a -10 at the end, you wouldn't bother checking x=3, for instance.

[Inside Out]

ok thanks guys

[Captain Rascal]

You are welcome for the free tutoring.

I've still got my econometrics exam coming up next week, which i haven't started studying for so i might need some of your tutoring haha

On the brighter side, I've finished my last practice exam for eng mathsssssssssssssss. Can go in tomorrow with confidence (and come out crying when i realise that mechanics is in 6 days)

[Seacow]

RPN Calculator Master Race

awyiisss

EDIT: http://commerce.hpcalc.org/34s.php Damn this looks good! Why can't there be a scientific RPN with solar to help maintain battery life! Hate using CR2032 batteries.

[Deleted User]

How do I compute the sum of j*1.05^(-j-1) from j=1 to 15? The answer is 70.1597 btw. Thanks

[LOLs99]

Can someone help me with this question?
Find the inverse by reducing to r.r.ef?

[-8 -3 -1; 4 2 1; -1 -1 1]

I can't seem to get the right answer   

[kinslayer]

How do I compute the sum of j*1.05^(-j-1) from j=1 to 15? The answer is 70.1597 btw. Thanks

Easiest way is to just compute all 15 terms and add  them. Computers ftw

If you really need a closed formula, you can take the formula for finite geometric series and differentiate to get the 'j' out the front.

http://en.wikipedia.org/wiki/Geometric_progression

[Deleted User]

Easiest way is to just compute all 15 terms and add  them. Computers ftw

If you really need a closed formula, you can take the formula for finite geometric series and differentiate to get the 'j' out the front.

http://en.wikipedia.org/wiki/Geometric_progression

I differentiated the RHS of this equation:
http://upload.wikimedia.org/math/6/f/b/6fbd5a05358545c9be77e2b2afd1cd30.png
subbed in all the values, but I got 77 something, which isn't correct because the answer is 70 something.

[LOLs99]

[-8 -3 -1 | 1 0 0 ]
[ 4   2  1 | 0 1 0 ]
[-1 -1  1 | 0 0 1 ]

Reduce the left side to reduced row echelon form and the right side should be the inverse matrix

If you still need help, you know where to find me on Monday

Haha I just couldn't get the answer after all the row operations
I have been trying for 3 times and gave up.

[lzxnl]

Haha I just couldn't get the answer after all the row operations
I have been trying for 3 times and gave up.

So, what you can do is add twice the second row to the first row to clear the top 8. Then, add four times the last row to the second row to clear the 4. Then swap rows 1 and 3. And finally multiply your new row 1 by -1. You'll have a 1 in the top left corner and everything else in the column will be zero. So, move your attention now to the second column. Try that.

[notveryasian]

Haha I just couldn't get the answer after all the row operations
I have been trying for 3 times and gave up.

I recommend going through each row operation carefully when you start doing row reduction. After some practice you'll be able to apply the operations more quickly and develop good techniques to dealing with uglier matrices.

Also, keep trying. A lot of the time you won't get the answer on the first go since one mistake puts your entire answer off, but with perseverence and practice a lot of things in linear algebra become MUCH easier.

[LOLs99]

Thanks for the helping out