Author Topic: 4U Maths Question Thread (Read 665304 times) Tweet Share

cutiepie30 · « **Reply #975 on:** March 13, 2017, 08:38:40 pm »

Quote from: RuiAce on March 13, 2017, 06:56:15 am

But then again, if you get it in terms of only cosh it's still in terms of both cosh and sinh. This is because you're basically adding $0\sinh x$ anyhow.

Thanks RuiAce

basically so mahans answer should be fine right?

is there a way you could get cosh(3x) in terms of cosh(x) and sinh(x), with both sinh(x) and cosh(x) in the final answer?

RuiAce · « **Reply #976 on:** March 13, 2017, 08:40:22 pm »

Quote from: cutiepie30 on March 13, 2017, 08:38:40 pm

Thanks RuiAce

basically so mahans answer should be fine right?

is there a way you could get cosh(3x) in terms of cosh(x) and sinh(x), with both sinh(x) and cosh(x) in the final answer?

Should be, yeah.
$\text{You could always use the hyperbolic Pythagorean identity}\\ \cosh^2x-\sinh^2x=1\\ \text{(Note the subtle change in sign with the trigonometric ones)}$

cutiepie30 · « **Reply #977 on:** March 13, 2017, 09:11:53 pm »

Quote from: RuiAce on March 13, 2017, 08:40:22 pm

Should be, yeah.
$\text{You could always use the hyperbolic Pythagorean identity}\\ \cosh^2x-\sinh^2x=1\\ \text{(Note the subtle change in sign with the trigonometric ones)}$

Thanks RuiAce

chelseam · « **Reply #978 on:** March 14, 2017, 08:57:25 pm »

Hi! Can someone please help me with part 3 of this question

I've read the board of studies solutions but I still don't get what I'm supposed to do!

RuiAce · « **Reply #979 on:** March 14, 2017, 10:19:47 pm »

Quote from: chelseam on March 14, 2017, 08:57:25 pm

Hi! Can someone please help me with part 3 of this question I've read the board of studies solutions but I still don't get what I'm supposed to do!

$\text{Suppose that }z\text{ is real}\\ \text{Obviously the square of a real number can never be negative}\\ \text{Only the square of an imaginary number can be negative.}$
$\text{Since }z\text{ and }k\text{ are both real numbers}\\ \text{The above observation gives}\\ \begin{align*}z^2(z-k)^2 &\ge 0\\ (kz-1)^2&\ge 0\end{align*}$
$\text{Adding these up, we see that}\\ z^2(z-k)^2 + (kz-1)^2 \ge 0$
$\text{But we are only interested in }z^2(z-k)^2+(kz-1)^2 =0\\ \text{if two numbers are (strictly) positive, they must add to a positive number}\\ \text{Only a positive and negative number can add to 0}\\ \text{but clearly }z^2(z-k)^2\text{ and }(kz-1)^2\text{ can't be negative}\\ \text{We established that at the start}$
$\text{Hence, the only possibility for }z^2(z-k)^2+(kz-1)^2 = 0\\ \text{is if both the terms included are 0. i.e.}\\\begin{align*}z^2(z-k)^2 &=0\\ (kz-1)^2&=0\end{align*}$
_____________________________
$\text{By considering }z^2(z-k)^2=0\\ \text{Quite clearly, }P(0) = 1 \text{ from the original equation so }z\neq 0\\ \text{This solves to give }z=k$
$\text{Doing simultaneous equations with }(kz-1)^2=0\text{ gives}\\ k^2=1 \implies k = \pm 1$
$\text{Hence, }P(z)\text{ has a real 0}\\ \text{only when }k=-1\text{ or }k=1\\ \text{Substituting either of these into the expression in ii}\\ \text{gives the desired result.}$

chelseam · « **Reply #980 on:** March 14, 2017, 10:56:10 pm »

Thank you so much!!

Quote from: RuiAce on March 14, 2017, 10:19:47 pm

$\text{Suppose that }z\text{ is real}\\ \text{Obviously the square of a real number can never be negative}\\ \text{Only the square of an imaginary number can be negative.}$
$\text{Since }z\text{ and }k\text{ are both real numbers}\\ \text{The above observation gives}\\ \begin{align*}z^2(z-k)^2 &\ge 0\\ (kz-1)^2&\ge 0\end{align*}$
$\text{Adding these up, we see that}\\ z^2(z-k)^2 + (kz-1)^2 \ge 0$
$\text{But we are only interested in }z^2(z-k)^2+(kz-1)^2 =0\\ \text{if two numbers are (strictly) positive, they must add to a positive number}\\ \text{Only a positive and negative number can add to 0}\\ \text{but clearly }z^2(z-k)^2\text{ and }(kz-1)^2\text{ can't be negative}\\ \text{We established that at the start}$
$\text{Hence, the only possibility for }z^2(z-k)^2+(kz-1)^2 = 0\\ \text{is if both the terms included are 0. i.e.}\\\begin{align*}z^2(z-k)^2 &=0\\ (kz-1)^2&=0\end{align*}$
_____________________________
$\text{By considering }z^2(z-k)^2=0\\ \text{Quite clearly, }P(0) = 1 \text{ from the original equation so }z\neq 0\\ \text{This solves to give }z=k$
$\text{Doing simultaneous equations with }(kz-1)^2=0\text{ gives}\\ k^2=1 \implies k = \pm 1$
$\text{Hence, }P(z)\text{ has a real 0}\\ \text{only when }k=-1\text{ or }k=1\\ \text{Substituting either of these into the expression in ii}\\ \text{gives the desired result.}$

Rathin · « **Reply #981 on:** March 15, 2017, 06:13:53 pm »

For the question: Factorise 4x^4+1 asa product of real polynomials.
I got (2x^2+1)(2x^2-1)
But the answer is (2x^2+2x+1)(2x^2-2x+1), why is it so?

jamonwindeyer · « **Reply #982 on:** March 15, 2017, 06:21:45 pm »

Quote from: Rathin on March 15, 2017, 06:13:53 pm

For the question: Factorise 4x^4+1 asa product of real polynomials.
I got (2x^2+1)(2x^2-1)
But the answer is (2x^2+2x+1)(2x^2-2x+1), why is it so?

Hey Rathin! It's been a while

$(2x^2+1)(2x^2-1)=4x^2-1$

I think you've used the difference of two squares shortcut when you shouldn't have, or otherwise misread the question?

I'll leave Rui/Jake to tag in with the actual method if you need it (I'm not sure of the 4U approach to this)

Edit: Nevermind, you'd just complete the square (sort of, not the way you'd normally do) - Pretty sure you'd just do this

$4x^4+1=4x^4+4x^2+1-4x^2=(2x^2+1)^2-4x^2\\=(2x^2-2x+1)(2x^2+2x+1)\text{ by the difference of two squares}$

I hope that helps

Rathin · « **Reply #983 on:** March 15, 2017, 06:29:09 pm »

Quote from: jamonwindeyer on March 15, 2017, 06:21:45 pm

Hey Rathin! It's been a while

$(2x^2+1)(2x^2-1)=4x^2-1$

I think you've used the difference of two squares shortcut when you shouldn't have, or otherwise misread the question?

I'll leave Rui/Jake to tag in with the actual method if you need it (I'm not sure of the 4U approach to this)

Edit: Nevermind, you'd just complete the square (sort of, not the way you'd normally do) - Pretty sure you'd just do this

$4x^4+1=4x^4+4x^2+1-4x^2=(2x^2+1)^2-4x^2\\=(2x^2-2x+1)(2x^2+2x+1)\text{ by the difference of two squares}$

I hope that helps

That is probably the stupidest thing I have ever done lol
Thanks for the help

chelseam · « **Reply #984 on:** March 15, 2017, 08:53:24 pm »

Can someone please help me with the last two parts of this question? Thank you

RuiAce · « **Reply #985 on:** March 15, 2017, 09:08:09 pm »

Quote from: chelseam on March 15, 2017, 08:53:24 pm

Can someone please help me with the last two parts of this question? Thank you

Part iv) needs the result of b), which you need to provide.
$\text{In part v), clearly we set }n=5\\ \text{so we can use the result of iv)}$
$\text{We know that }x=1\text{ is a double zero of }P(x)\\ \text{and we know that the third zero satisfies }-1 < \alpha \le -\frac12\\\text{So these three zeroes obviously satisfy the criteria to verify.} \\\text{It falls to check that the last two zeroes satisfy this as well.}$
$\text{Note that }P(x)\text{ only has those real zeroes from part iii)}\\ \text{Hence, the last two zeroes are complex}\\ \text{Because the coefficients on }P(x)\text{ are real, they are complex conjugates.}$
$\text{Let }\beta\text{ and }\overline{\beta}\text{ be the other zeroes}\\ \text{From the product of roots:}\\ 1\times 1\times \alpha\times \beta \times \overline{\beta} = -\frac{1}{4}$
$\text{Note that }z\times \overline{z}=|z|^2\\ \text{So we have}\\ \alpha |\beta|^2=-\frac{1}{4}\implies |\beta|^2=-\frac{1}{4\alpha}$
$\text{Rearranging the result of part iv) gives}\\ \frac12 \le -\alpha < 1\\ \therefore 1 < -\frac{1}{\alpha} \le 2\\ \text{so by dividing by 4, substituting in the above result:}\\ \frac14 < |\beta|^2 \le \frac12\\ \text{which clearly will result in }|\beta| < 1$

chelseam · « **Reply #986 on:** March 15, 2017, 10:09:18 pm »

Thank you!

This is part b)

Quote from: RuiAce on March 15, 2017, 09:08:09 pm

Part iv) needs the result of b), which you need to provide.

RuiAce · « **Reply #987 on:** March 16, 2017, 07:27:27 am »

Quote from: chelseam on March 15, 2017, 10:09:18 pm

Thank you! This is part b)

Hint: in 3U, when you were doing Newton's method, you learnt that a root is between -1 and -1/2 if f(-1) and f(-1/2) change signs. The result of part b) is supposed to help make your life easier when you have to deal with f(-1/2).

There are many other solutions better than the BoS ones which you should consider reviewing. However if those don't make sense, post up your interpretation of whichever solutions you looked at and we will fill in the remainder of the puzzle.

bluecookie · « **Reply #988 on:** March 16, 2017, 12:20:02 pm »

a) Show that the roots of y^4+y^3+y^3+y+1=0 are y=cos(2kpi/5)+1sin(2kpi/5), k=1,2,3,4.
b) Hence deduce that cos36 deg = 1/2 + cos 72 deg.
^ I need help with part b.

bluecookie · « **Reply #989 on:** March 16, 2017, 01:14:17 pm »

Suppose that z^7=1, z is not equal to 1.
i) deduce that z^3+z^2+z+1+1/z+1/z^2+1/z^3=0.
ii) by letting x=z+1/z, reduce the equation in i to a cubic equation in x
iii) Hence deduce that cospi/7cos2pi/7cos3pi/7=1/8

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