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Advanced Higher Maths 2012-2013 : Discussion and Help Thread

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Reply 80
I was wondering if anyone could give me a bit of help with some homework questions I have.

First one is Gaussian Elimination. I have worked it out to be redundant however the answers at the back say that it is solveable. I was wondering if someone could point out where I went wrong or if the answers are.

Q)3x + 2y +5z = 0
2x + y - 2z = 5
7x + 4y + z = 10

3 2 5 | 0
2 1 -2 | 5
7 4 1 | 10

R2 =3R2 -2R1
R3 = 3R3 -7R1

3 2 5 | 0
0 -1 -16 | 15
0 -2 -32 | 30

R2 = 2R2

3 2 5 | 0
0 -2 -32 | 30
0 -2 -32 | 30

R3 = R3 - R2


3 2 5 | 0
0 -2 -32 | 30
0 0 0 | 0

Row of 0's = Redundant therefore infinite solutions , given in terms of K.

So I have given my answers as z =k etc but the back says the answer is x=1, y= 1, z=-1.

So where is the mistake? Thanks
(edited 11 years ago)
Original post by NicNacCola
I was wondering if anyone could give me a bit of help with some homework questions I have.

First one is Gaussian Elimination. I have worked it out to be redundant however the answers at the back say that it is solveable. I was wondering if someone could point out where I went wrong or if the answers are.

Q)3x + 2y +5z = 0
2x + y - 2z = 5
7x + 4y + z = 10

3 2 5 | 0
2 1 -2 | 5
7 4 1 | 10

R2 =3R2 -2R1
R3 = 3R3 -7R1

3 2 5 | 0
0 -1 -16 | 15
0 -2 -32 | 30

R2 = 2R2

3 2 5 | 0
0 -2 -32 | 30
0 -2 -32 | 30

R3 = R3 - R2


3 2 5 | 0
0 -2 -32 | 30
0 0 0 | 0

Row of 0's = Redundant therefore infinite solutions , given in terms of K.

So I have given my answers as z =k etc but the back says the answer is x=1, y= 1, z=-1.

So where is the mistake? Thanks


I see no mistakes in your working.


This was posted from The Student Room's iPhone/iPad App
Original post by NicNacCola
I was wondering if anyone could give me a bit of help with some homework questions I have.

First one is Gaussian Elimination. I have worked it out to be redundant however the answers at the back say that it is solveable. I was wondering if someone could point out where I went wrong or if the answers are.

Q)3x + 2y +5z = 0
2x + y - 2z = 5
7x + 4y + z = 10

3 2 5 | 0
2 1 -2 | 5
7 4 1 | 10

R2 =3R2 -2R1
R3 = 3R3 -7R1

3 2 5 | 0
0 -1 -16 | 15
0 -2 -32 | 30

R2 = 2R2

3 2 5 | 0
0 -2 -32 | 30
0 -2 -32 | 30

R3 = R3 - R2


3 2 5 | 0
0 -2 -32 | 30
0 0 0 | 0

Row of 0's = Redundant therefore infinite solutions , given in terms of K.

So I have given my answers as z =k etc but the back says the answer is x=1, y= 1, z=-1.

So where is the mistake? Thanks


There is no mistake. I think the book just set k to a particular value (probably z = k = 1) and had that as the canonical solution.

Note that redundant and solvable are not mutually exclusive. Redundant just means it's solvable with infinite solutions. Redundant and uniquely solvable are exclusive, and so are inconsistent and solvable.
Reply 83
Original post by adam_johnson


I see no mistakes in your working.



Original post by ukdragon37
There is no mistake. I think the book just set k to a particular value (probably z = k = 1) and had that as the canonical solution.

Note that redundant and solvable are not mutually exclusive. Redundant just means it's solvable with infinite solutions. Redundant and uniquely solvable are exclusive, and so are inconsistent and solvable.


Thanks
Reply 84
hi can some one tell me if these answers are correct:
Q:differentiate x/(1-2x)
A:2x-2

Q:differentiate x^2/(x^2+1)
A:1/(x^4+1)

Q:differentiate x(x-1)^(1/2)
A: (3/2)x-1
Original post by hazzm2
hi can some one tell me if these answers are correct:
Q:differentiate x/(1-2x)
A:2x-2

Q:differentiate x^2/(x^2+1)
A:1/(x^4+1)

Q:differentiate x(x-1)^(1/2)
A: (3/2)x-1


Question 1, nope, you should be using the quotient rule here!

Question 2, quotient rule again, wrong answer

Question 3, product rule here, wrong answer

Quotient rule:
if f(x)=g(x)h(x)f(x)=g(x)h(x)g(x)h(x)h2(x)if \ f(x) = \frac{g(x)}{h(x)} \Rightarrow f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h^2(x)}

Product rule:
if f(x)=g(x)h(x)f(x)=g(x)h(x)+g(x)h(x) if \ f(x) = g(x)h(x) \Rightarrow f'(x) = g'(x)h(x) + g(x)h'(x)

Solutions

Question 1:

Spoiler



Question 2:

Spoiler



Question 3:

Spoiler

(edited 11 years ago)
Reply 86
Original post by Hype en Ecosse
Question 1, nope, you should be using the quotient rule here!

Question 2, quotient rule again, wrong answer

Question 3, product rule here, wrong answer

Quotient rule:
if f(x)=g(x)h(x)f(x)=g(x)h(x)g(x)h(x)h2(x)if \ f(x) = \frac{g(x)}{h(x)} \Rightarrow f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h^2(x)}

Product rule:
if f(x)=g(x)h(x)f(x)=g(x)h(x)+g(x)h(x) if \ f(x) = g(x)h(x) \Rightarrow f'(x) = g'(x)h(x) + g(x)h'(x)

Solutions

Question 1:

Spoiler



Question 2:

Spoiler



Question 3:

Spoiler



thanks for your help i was using the quotiont and product rules but have just gotten really rusty over the summer. just need more practice
Reply 87
I really need to practice before Tuesday :frown: It seems like we are a bit behind


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Reply 88
I was wondering if someone could help me with this question:

n + n = 28
1. 2

I don't understand what I have to do to solve the problem
Original post by NicNacCola
I was wondering if someone could help me with this question:

n + n = 28
1. 2

I don't understand what I have to do to solve the problem


It means that you must find n such that the equation

(n1)+(n2)=28\displaystyle\binom{n}{1} + \displaystyle\binom{n}{2} = 28

is true.

Perhaps you should try using the fact that

(n1)+(n2)=(n+12)\displaystyle\binom{n}{1} + \displaystyle\binom{n}{2} = \displaystyle\binom{n+1}{2}

And then equating the RHS with 28. Afterwards try expanding the binomial coefficient into its factorial definition.

Solution

(edited 11 years ago)
Reply 90
Original post by ukdragon37
It means that you must find n such that the equation

(n1)+(n2)=28\displaystyle\binom{n}{1} + \displaystyle\binom{n}{2} = 28

is true.

Perhaps you should try using the fact that

(n1)+(n2)=(n+12)\displaystyle\binom{n}{1} + \displaystyle\binom{n}{2} = \displaystyle\binom{n+1}{2}

And then equating the RHS with 28. Afterwards try expanding the binomial coefficient into its factorial definition.

Solution



Thank you! This makes so much more sense then my book.
Original post by ukdragon37
It means that you must find n such that the equation

(n1)+(n2)=28\displaystyle\binom{n}{1} + \displaystyle\binom{n}{2} = 28

is true.

Perhaps you should try using the fact that

(n1)+(n2)=(n+12)\displaystyle\binom{n}{1} + \displaystyle\binom{n}{2} = \displaystyle\binom{n+1}{2}

And then equating the RHS with 28. Afterwards try expanding the binomial coefficient into its factorial definition.

Solution



I've seen you do this question (and the proof of the identity you used) so many times in the last year I can do it as soon as I see it now. I didn't even encounter it when I did AH. Got it in one past paper. :tongue:
(edited 11 years ago)
Original post by Hype en Ecosse
I've seen you do this question so many times in the last year I can do it as soon as I see it now. I didn't even encounter it when I did AH. Got it in one past paper. :tongue:


I don't think I actually ever did a question like this for real when I sat AH Maths :tongue:

God that was a long time ago... :frown: Time flies like an arrow, fruit flies like a banana.
Original post by ukdragon37
I don't think I actually ever did a question like this for real when I sat AH Maths :tongue:

God that was a long time ago... :frown: Time flies like an arrow, fruit flies like a banana.


Last year at Cambridge...I wish you the best of luck. Still planning on PhDing? In category theory?
Original post by Hype en Ecosse
Last year at Cambridge...I wish you the best of luck. Still planning on PhDing? In category theory?


Thanks :smile: Yes and yes, although now I'm slightly regretting declining Yale and Columbia. :tongue:
Original post by ukdragon37
Thanks :smile: Yes and yes, although now I'm slightly regretting declining Yale and Columbia. :tongue:


25026438.jpg
Original post by Hype en Ecosse
25026438.jpg


Correction: Ivy League + no tuition fees + $1500 stipend per month + opportunity to teach undergrads of said Ivy League for more stipend :tongue:
Original post by ukdragon37
Correction: Ivy League + no tuition fees + $1500 stipend per month + opportunity to teach undergrads of said Ivy League for more stipend :tongue:


:eek3:
Original post by Ecosse_14
:eek3:


Pfft, I'd be living in NYC for the next 5 years of my life. Too expensive! :tongue:

Although maybe I should have given it more thought. :colonhash:
Original post by ukdragon37
Correction: Ivy League + no tuition fees + $1500 stipend per month + opportunity to teach undergrads of said Ivy League for more stipend :tongue:


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