Advanced Higher Maths 2013-2014 :: Discussion and Help Thread

from what I understand, you have:

\displaystyle \frac{3e^{3x}(e^{3x}+1)-e^{3x}3e^{3x}}{e^{3x}+1}

is that correct?

you can take a common term out of the denominator:

\displaystyle \frac{3e^{3x}(e^{3x}+1-e^{3x})}{e^{3x}+1}

=....

(edited 10 years ago)

Reply 62

Proud_Student

thanks and one last question

iv been simplifying this partial fraction but im getting weird answers

x^2 + 2 divided by x^2 - 1

I factorized the bottom to (x-1)(x+1) and well from there i was going fine but I got strange answers for one of my values, like 2/3

im not sure where i went wrong ><

Reply 63

I am Ace

4

Original post by Proud_Student

thanks and one last question

iv been simplifying this partial fraction but im getting weird answers

x^2 + 2 divided by x^2 - 1

I factorized the bottom to (x-1)(x+1) and well from there i was going fine but I got strange answers for one of my values, like 2/3

im not sure where i went wrong ><

You haven't went wrong, there's nothing wrong with fractions and negative values.

Reply 64

the easiest way to look at this one is to stop and notice - hey! - x^2+2 is "abit like" x^2-1, so you can put the numerator as

x^2+2= (x^2-1)+3

which means you get:

\frac{x^2-1}{x^2-1}+\frac{3}{x^2-1}

.

The first thing you have to do when you see this type of function is use synthetic or long division, because the degree of the numerator and denominator are equal. This brings it into a form which is easier to deal with.

what you`ve got above is then:

1+\frac{3}{x^2-1}

. and you need only split the 2nd part using the denominator (x-1)(x+1)

(edited 10 years ago)

Reply 65

Proud_Student

Epic thanks :biggrin:

My teacher says its cuz im very shy and passive about maths, not to mention not very confident, that reassured me..i got my work down...how well it was done...ill be lucky to get half of them right :rolf:

I guess its my fault for being away at Open days, not to mention those pesky PPAs :biggrin:

Reply 66

this is a very good website, and everyone is always willing to help, so my advice is (after you`ve tried first of course) don`t hesitate to get help here.

like a lot of things, confidence comes with practice. Try hard, believe in yourself, and don`t get annoyed or downhearted with maths.

(I do exercises in books, but now, I hardly ever look up the answers (unless I`m utterly stumped!), as I believe in my own ability.)

believe in yourself
:smile:

(hippy rant over)

Reply 67

ukdragon37

Original post by Hasufel

Maths in action - advanced higher maths book 3 question:

would you be prepared to attempt this question?:

Show that all of the following iterations have the same fixed points?:

\displaystyle x_{n+1}=2-n_{n}^{2}, x_{n+1}=(x_{n})^{2}+2x_{n}-2, x_{n+1}= \frac{2-x_{n}}{x_{n}}

the answer is easy enough. It`s the next part that annoys me slightly:

With a starting value of 0.5, one of the sequences converges to one of the fixed points. Investigate this.

it`s the 3rd one, which converges to -2 - but just the 3rd one alone requires 25 iterations!

I would like to know what you think about this question? Is it a fair question to get anyone to attempt given that it is so long? Is this book just full of far too complex things you would never be given?

What do you think? (p.s. i can "see" the answer before i attempt it, using cobweb graphing plots)

Original post by Simarilli

Stick it in your calculator using ANS and spam the equals key. Takes about 4 seconds.

It's a perfectly reasonable question, you are just not using the right method. Even the calculator "method" is not a proof that it converges. The course expects you to know the following method (quoted from the HSN course summary page 7):

To calculate a root using an iterative scheme you must decide whether the scheme converges or diverges.

Convergence is decided by differentiating and substituting in the approximate root.

If the answer is between –1 and 1 [exclusive], the scheme will converge. The closer the
value is to zero the quicker the scheme will converge.

From the first part of the question we know that the fixed point of the schemes in fact corresponds to computing a root of the equation

x^2+x-2 = 0

and it is either

-2

or

1

. Differentiating each of the schemes respectively gives:

1.

$\dfrac{dx_{n+1}}{dx_n} = -2x_n$

2.

$\dfrac{dx_{n+1}}{dx_n} = 2x_n + 2$

3.

$\dfrac{dx_{n+1}}{dx_n} = \dfrac{-2}{x_n^2}$

Hence plugging

-2

and

1

into the the above gives only the third result is between

-1

and

1

when

-2

is tried. Therefore the third method will converge for the fixed point / root

-2

.

(Further, note that if the result is outside the range then the scheme would diverge for the root tested, whereas if it is exactly

-1

or

1

then the test fails and the scheme may or may not converge, although if it does converge then it is likely to be very slow.)

(edited 10 years ago)

Reply 68

-Neuro-

12

Watching this thread to help as much as I can. :woo:

AH maths is very interesting guys, I enjoyed much more than Higher. Don't let some of the insane mathmos on here put you off by the way, this is TSR after all!

I think I'm the only girl who has posted on this thread so far. :ninja:

Reply 69

I am Ace

4

Original post by -Neuro-

Watching this thread to help as much as I can. :woo:

AH maths is very interesting guys, I enjoyed much more than Higher. Don't let some of the insane mathmos on here put you off by the way, this is TSR after all!

I think I'm the only girl who has posted on this thread so far. :ninja:

Na you're not the first girl, UKD's posted above

:colone:

Reply 70

Original post by ukdragon37

It's a perfectly reasonable question, you are just not using the right method. Even the calculator "method" is not a proof that it converges. The course expects you to know the following method (quoted from the HSN course summary page 7):

From the first part of the question we know that the fixed point of the schemes in fact corresponds to computing a root of the equation

x^2+x-2 = 0

and it is either

-2

or

1

. Differentiating each of the schemes respectively gives:

1.

$\dfrac{dx_{n+1}}{dx_n} = -2x_n$

2.

$\dfrac{dx_{n+1}}{dx_n} = 2x_n + 2$

3.

$\dfrac{dx_{n+1}}{dx_n} = \dfrac{-2}{x_n^2}$

Hence plugging

-2

and

1

into the the above gives only the third result is between

-1

and

1

when

-2

is tried. Therefore the third method will converge for the fixed point / root

-2

.

(Further, note that if the result is outside the range then the scheme would diverge for the root tested, whereas if it is exactly

-1

or

1

then the test fails and the scheme may or may not converge, although if it does converge then it is likely to be very slow.)

Thanks so much! - my book does not give ANY sort of explanation as you have given - authors missed explaining themselves there!

Thanks again!

(just realised, this is in another book I have {but it`s on non-linear dynamics as the "fixed point theorem" i.e: | F ' (-2)|<1 for only the 3rd one}

(edited 10 years ago)

Reply 71

CSM1996

I have unit 1 maths NAB on Wednesday and extension test on Monday :frown:

Reply 72

ukdragon37

Original post by I am Ace

Na you're not the first girl, UKD's posted above

:colone:

Though as much as you'd like me to be one, I am not a girl. :tongue:

Reply 73

ukdragon37

Original post by Hasufel

Thanks so much! - my book does not give ANY sort of explanation as you have given - authors missed explaining themselves there!

Thanks again!

(just realised, this is in another book I have {but it`s on non-linear dynamics as the "fixed point theorem" i.e: | F ' (-2)|<1 for only the 3rd one}

You're welcome :smile:

Iterative schemes is one of the weakly-taught topics in AH I've found so this confusion is quite common.

Reply 74

username970964