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Published November 17th, 2006 in 6AHI am very sorry that I did not put an 6AH category in. Here’s one for you. Join us friends!
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Hello. I found your blog!
Great Steve-O. Please feel free to help the community where you can. I’m gald you’ve joined us.
Darn, i didnt make first response. oh well, look at what you get if you google stebbo, mr stebbing!http://www.*******ctionary.com/define.php?term=A+stebbo
Thanks for that Sam. Glad you’ve joined us.
i’ve found your blog at last
Hi Timothei. I’m glad you’ve joined the community. Please feel free to comment on posts as much as you like. You may feel that you want to help some of the pupils with problems on the higher posts. See how you feel.
Well hellooo there, Mr Stebb! This blog things a well gd idea i think, I shall keep my eye out for confused fourth and fifth years needing the help of highly skilled mathematicians such as us. Im off now cause im starting to sound a bit wierd, im not really used to this blogging doo-dah but ill get there!
Welcome Kat The Great. Your offer of help is gratefully received. I’m sure my higher class feel a little special with all the extra help on offer. Thanks again. mr S.
hello. I was wandering if anyone had done the question on the binomial theorem in the past prelim.
Find the term in x^3 in the expansion of (3x + 2/x)^5
The way i used didn’t work..any pointers as to where i should start?
Cheers. SJ
Well, SJ, the general term of that expansion will be 5Cr(3x)^(5-r)(2/x)^r
You get an expression for the overall power of x in terms of r, then find out what value of r makes the power of x become zero.
Then evaluate that therm
Your coeeficients will be 1, 5, 10, 10, 5, 1 form pascal’s triangle. expansion would be :
1a^5 b^0 + 5a^4b^1+10a^3b^2+10a^2b^3+5a^1b^4 + 1a^0b^5
where a = 3x and b =2/x
You may find that when you were cancelling terms that you made a wee boo boo. Go back and try it agian. If you want, email what you’ve done to the stebblog email address. I’ll have a look at it for you. You will of course have needed to take the Activstudio Student edition disc away with you on thursday. Did you? If not type it up in word and I’ll give it a look.
Of course Jonsies way is far simpler
Just emailed my workings so you can have a laugh.
Hi Steve-O. I haven’t received it yet. Did you manage it in the end?
here they are …
Coefficients of x^3 ;
5 x^(5-2r) . (2)^r 5-2r = 3
r r = 1
5 x^3 . 2 = 10x^3
1
the alignment haasnt worked 5-2r = 3 .. r=1 should be separate.
I got 5Cr {(3)^(5-r).2^r.x^(5-2r)} (I think!!!!) 5-2r =3 => r=1 thus evaluate
5C1 {(3)^4.2^1.X^3} and you shold get your answer. Can’t wait for LaTEX.
I hope that’s useful (and correct. I am working on a post it note. I’ll do it properly later if you want.)
Cool It actually works
Hello. Any help in starting this question would be good…
Given that 1 + i is a root of thereal polynomial
x^4 + 3x^2 - 6x + 10 = 0 prove that x^2 - 2x + 2 is a factor of the polynomial, and hence solve the equation completely.
Cheers. sj
We’ve not covered this yet Steve - O so if it’s in the prelim yer stuffed!
Don’t worry it’s not.
:) 
Ahh .. Good good! Thank you.
I hope i didn’t upset you there.
how are you meant to work out whether a graph has a horizontal asymptote?
Hi Timothei. Try having a look at this webapage. Hopefully it will remind you what to do.
http://library.thinkquest.org/10030/8grarfun.htm
carry out the long division in the function .. then work out what happens to F(x) as x=>infinity . I think :-s
thanks i remember now. thnks guys
Collaborationis the key innit.
Do you have to use partial fractions if f(x) denominator is a higher power than the numerator.
How would you go about question 2 in the complex number revision sheet:
Show that if z=cosQ+isinQ then z/(1-z^2) is purely imaginery
I thought if you expressed it in the form x+yi and showed that x was equal to zero would that work.
When i try it i just get alot of cosQ+isinQ and it makes my head hurt.
Hey gibbardo can you make sure the question about complex numbers is correct. I’m still working on it.
Are sequences and series arent in the prelim? sj
I’m not sure. All i can remember about is my own work. If you’ve done it a while back it will be. If it is still quite fresh it may be. That’s the best I can do for you. I’ve not got a copy of the prelim to check. What did Jonesy say?
Hmmm..I can’t remember. We’ve only been doing it the last week or two. I will assume it is. Cheers. sj
I’d go with knowing the stuff yoyu’ve done up to date. Mr S
steve-o i thought sequences and series weren’t in the prelim and that induction was the last thing in the prelim with jonesy
That’ll play .. can’t remember what we’ve done so far in sequences anyway. Fingers crossed I don’t get caught out
. SJ
Hi again gibbardo. Just to let you know I’ve solved your complex number problem. Stick with polar form. Get z^2 in polar form by doing what we did last thursday. You will get something a little ugly as a denominator. Make the denominator real by multiplying numerator and denominator by the complex conjugate of the denominator. Then you’ll have to work with compound angle formulae in the numerator to show that all the real parts cancel. Everything else is imaginary thus showing z / 1-z^2 is imaginary. I’ll give you a clue as well : cosQ = cos (-q) I’ll go through it with you first thing in the morning if you want.
See you all bright and early. It shall be fun. Sj
Will do Steve- O. Good luck
um just revising for the prelim and cant remember how to calculate the all the roots of a polynomial complex number equation
You can only do it if the coefficients are real. If you have a complex root the complex conjugate will also be a root. You can multiply these together and get a new quadratic real factor then use long division on the original polynomial to find the rest of the roots. Sometimes it just a case of using the quadratic formula to find the complex roots. If this doesn’t make sense post again and I’ll try and elaborate further for you. I’ll not be on until 8.30 ish tonight.
thanks yep that makes more sense
Cool
Hello. I was wondering if anyone could pointme in the right direction for part b of this q (appologies for the lack of latex);
a) Express 4 / (y^2 - 4) in partial fractions.
I got 1/(y-2) - 1/(y+2).
b) Show that the general solution of the differential equation 4(dy/dx + 1) = y^2 can be expressed in the form y= 2(1 + Ae^x) / (1 - Ae^x) , where A is a constant.
Any help wud be awsome. Cheers, sj.
Just doing the 2002 AH Maths Past Paper and came across a matrix question and was wondering if we were meant to be able to do it and if it might come up in the prelim.
Matrix A = 2 -1
-1 0
Prove by induction that A^n= n+1 n
-n 1-n
where n is any positive integer.
also just wondering whether you can plot Armand diagrams in Geograba with complex axis?
okay that came out bad i’ll write out the matrix again
A
2 -1
-1 0
Prove that A^n
n+1 n
-n 1-n
Steve-O i can only get it to y=2(1/Ae^-x)
Mr Stebbing, i was going to purchase a graphic calculator from amazon and was wondering whether i should get the TI-83 or the TI-84 calculator
Hi Tim. We have some ti84’s in school for sale now. I think they are £55 or thereabout. I would be getting the TI84. It’s they latest version but TI83 does the same job except you need a special cable to connect it to the computer. The TI84 uses usb. Hope this is helpful. Mr S. By the way can you get latex to work at the moment. I’m having trouble.
Nope latex hasn’t worked for me for a while
Thanks for that Tim. Do you know how long it’s not been working for? We’re trying to resolve the issue and any information may be helpful. Thanks Tim.
It didn’t work from about March. Hope that was useful. Tim
Hi. Just a quick question about complex numbers…How do i find the non real roots of Z^3 + 1 = 0 ? Can’t remember much of this stuff. Cheers.
slotted it. it’s all good.
I’m glad it’s “slotted”.
i’m stuck on question 12 on the 2005 paper part c)
(cos4θ/cos^2θ)=pcos^2θ + qsec^2θ+ r
stating the values of p,q and r
cancel my last comment, i worked it out, it was me being dumb thats all
Glad to hear it Timothei.
Having a ‘mare like!
Sounds painful SJ. Chill out. Go to bed and you’ll feel better in the morning.
Mr Stebbing, i find the 7th year studies syllabus very thin in content, can there be more complex numbers?
Just to give you an update on how uni’s going: all very good as am currently doing FOXY COMPLEX NUMBERS in my maths module! What could be better? Hope alls going well with you and that you are still finding time to do some MAGIC MATRICES!!!
Hey. It all seems a little quiet considering there are normally prelims around this time?! How you doing Stebbo? Missing me lots I should think. Hope all’s well. SJ
Hey Guy’s. Long time no see/hear. As you’ve probably worked out I haven’t been blogging that much recently. I’ve no doubt you think this is a little strange after how much fun we had last year. There is however a reason. I’m not actually at NBHS this year. I am working with the East Lothian Inclusion Service based at Meadowmill this year on secondment. I therefore don’t have any maths classes preparing for exams and haven’t been using Stebblog. I thought it best to keep Stebblog alive as some of the maths love contained within will still help people this year. After all the past paper exam questions haven’t changed. I’m glad that you are all well and still remember Stebblog. (This is the first time in a long time that I’ve been here as you can see). I hope you are all enjoying your new lives. I remember when I went to University. Meeting so many exciting new friends. Drinking coca cola at the students union with chums. I’m sure student life can’t have changed much since 1993! I’m sorry to hear your finding the course thin Timothei. Maybe you should do a second degree at the same time. I did ponder that for a while in first year. I chose not to in the end. Saviour of the Sharpener, I still have a set of batteries for you. I’m glad uni is going well. As for Steve-O, you’re gone but definitely not forgotten. I hope life treats you all well. Stebbo
I found the complex number song - and it is foxy!:
Mine eyes have seen the glory of the Argand diagram
They have seen the i’s and thetas of De Moivre’s mighty plan
Now I can find the complex roots with consummate elan
With the root of minus one
Complex numbers are so easy
Complex numbers are so easy
Complex numbers are so easy
With the root of minus one
In Cartesian co-ordinates the complex plane is fine
But the grandeur of the polar form this beauty doth outshine
You be raising i+40 to the power of 99
With the root of minus one
You’ll realise your understanding was just second rate
When you see the power and magic of the complex conjugate
Drawing vectors corresponding to the roots of minus eight
With the root of minus one
Woohoo Timothei. Is there music to go with it? In fact why don’t you sing it for us and post it on google video or something. Hope life is good at uni.
Hi Stebbo, hope your new job is going well as well. I’ve obviously forgotten to mention that i’m doing a gap year at Procter & Gamble until the end of august down in Newcastle. So i’ve got more reasons to do complex numbers!