Monday, 31 December 2012
AQA GCSE Additional Maths June 09 Question 3
In the following calculations each letter represents a different digit.
\(A \times A = BC\)
\(BC \times BC = DEC\)
Which digit does each letter represent?
=========================================
\(BC\) is a two digit square so is one of \(16,\ 25,\ 36,\ 49,\ 64,\ 81\), but \(BC \times BC\) is a three digut number so \( BC\lt \sqrt{1000} \), hence \(B\lt 4 \).
Which leaves us with the possibilities BC is one of \(16, \ 25, \ 36\), In which case the corresponding \(A\)s would be \(4, \ 5,\ 6\) but the latter two are not possible since that would make \(A=C\) which is not permitted.
Hence the solution is: \(A=4,\ B=1,\ C=6,\ D=2,\ E=5\).
Friday, 30 November 2012
David's Fourier Transform Problem from Y!Answers
Here's my problem;
Find the Fourier transform \(P(\omega)\) of the function;
\[ p(t)=\left\lbrace \begin{array}{ll} e^{-9t} & \text{for } t \ge 0 \\ e^{9t} & \text{for } t \lt 0 \end{array} \right.\]
Hence (use one of the shift theorems) find the inverse Fourier transform of; \( \frac{6}{(\omega+2)^2 + 9^2} \)
many thanks!
----------------------------------------------------------------------
Well we start by guessing which definition of the Fourier transform you are working with. I will assume you want: \[P(\omega)=\int_{-\infty}^{\infty}p(t)e^{-i \omega t}\;dt\]
With our function this becomes:
\[ \begin{array}{ll} P(\omega) & =\int_{0}^{\infty}e^{-9t}e^{-i \omega t}\;dt + \int_{-\infty}^{0} e^{9t}e^{-i \omega t}\; dt \\ &=\left. \frac{e^{-(9+i \omega ) t}}{-9 - i \omega } \right|_0^{\infty}+\left. \frac{e^{-(-9+i \omega )t}}{+9-i \omega } \right|^0_{-\infty} \\ &= \frac{1}{9+i\omega}+\frac{1}{9-i\omega} \\ &=\frac{18}{9^2+\omega^2} \end{array} \]
To complete this the shift theorem you need is:
\[\mathfrak{F}\left[ e^{-i a t}p(t) \right]=P(\omega+a)\]
CB
Find the Fourier transform \(P(\omega)\) of the function;
\[ p(t)=\left\lbrace \begin{array}{ll} e^{-9t} & \text{for } t \ge 0 \\ e^{9t} & \text{for } t \lt 0 \end{array} \right.\]
Hence (use one of the shift theorems) find the inverse Fourier transform of; \( \frac{6}{(\omega+2)^2 + 9^2} \)
many thanks!
----------------------------------------------------------------------
Well we start by guessing which definition of the Fourier transform you are working with. I will assume you want: \[P(\omega)=\int_{-\infty}^{\infty}p(t)e^{-i \omega t}\;dt\]
With our function this becomes:
\[ \begin{array}{ll} P(\omega) & =\int_{0}^{\infty}e^{-9t}e^{-i \omega t}\;dt + \int_{-\infty}^{0} e^{9t}e^{-i \omega t}\; dt \\ &=\left. \frac{e^{-(9+i \omega ) t}}{-9 - i \omega } \right|_0^{\infty}+\left. \frac{e^{-(-9+i \omega )t}}{+9-i \omega } \right|^0_{-\infty} \\ &= \frac{1}{9+i\omega}+\frac{1}{9-i\omega} \\ &=\frac{18}{9^2+\omega^2} \end{array} \]
To complete this the shift theorem you need is:
\[\mathfrak{F}\left[ e^{-i a t}p(t) \right]=P(\omega+a)\]
CB
Thursday, 15 November 2012
Muffin's GCSE Additional Maths Moments Question (from Y!Answers)
Question:
Two painters of mass 50kg and 70 kg stand on a uniform horizontal plank ABCDEFG of mass 40kg and length 8m.
B, C, D and E are respectively 1m, 2m, 5m and 6m from A. The 50kg painter stands at B, the 70kg painter at D and the plank's supported at C and E.
Find the reaction at each support I got the answers to be 1053.5 and 514.5 but my answer book tells me otherwise (although it has been wrong before) can someone please check this and explain where I've gone wrong please?
Solution:
Since the reaction forces must sum to the total load (the two painters and the weight of the plank acting at the centre of mass of the plank) which is \(160g = 1569.6 \, {\rm{N}}\) (taking \(g\) to be \(9.81\, {\rm{m/s^2}}\) ) you answer is not impossible so far (using \(g=9.8\, {\rm{m/s^2}}\) we have exact agreement).
You need to set up a pair of simultaneous equations under the assumption that the system is in equilibrium, the first is that the reaction forces sum to the load forces:
\(R_C+R_E = 160 g \)
The second is obtained by taking moments about some convenient point, as the system is in equilibrium these must sum to zero. A convenient point (as it eliminates one of the unknowns) is either C or E.
Taking moments about E:
\(4 R_C -2 (40 g) -5 (50 g) - 1 (70 g) =0\)
(the load forces give anti-clockwise moments and the reaction a clockwise moment, I am taking clockwise as positive here)
So:
\(R_C = 100 g\, {\rm{N}}\) ...
Two painters of mass 50kg and 70 kg stand on a uniform horizontal plank ABCDEFG of mass 40kg and length 8m.
B, C, D and E are respectively 1m, 2m, 5m and 6m from A. The 50kg painter stands at B, the 70kg painter at D and the plank's supported at C and E.
Find the reaction at each support I got the answers to be 1053.5 and 514.5 but my answer book tells me otherwise (although it has been wrong before) can someone please check this and explain where I've gone wrong please?
Solution:
Since the reaction forces must sum to the total load (the two painters and the weight of the plank acting at the centre of mass of the plank) which is \(160g = 1569.6 \, {\rm{N}}\) (taking \(g\) to be \(9.81\, {\rm{m/s^2}}\) ) you answer is not impossible so far (using \(g=9.8\, {\rm{m/s^2}}\) we have exact agreement).
You need to set up a pair of simultaneous equations under the assumption that the system is in equilibrium, the first is that the reaction forces sum to the load forces:
\(R_C+R_E = 160 g \)
The second is obtained by taking moments about some convenient point, as the system is in equilibrium these must sum to zero. A convenient point (as it eliminates one of the unknowns) is either C or E.
Taking moments about E:
\(4 R_C -2 (40 g) -5 (50 g) - 1 (70 g) =0\)
(the load forces give anti-clockwise moments and the reaction a clockwise moment, I am taking clockwise as positive here)
So:
\(R_C = 100 g\, {\rm{N}}\) ...
Sunday, 28 October 2012
Linda's Stationary points question from Y!A
Question
2)(a) The deflection \(y\) of a structural member is given by the formula:
\[y=x^3-2x^2 –x\]
Where x is the distance from one end.
(i) Find the value of \(x\) for which a turning point occur
(ii) Determine the value of \(y\) at the turning point
(b)(i) Find the stationary points of the surface
\[z=x^3 +xy+y^2\]
(ii)Determine their nature
Answers:
2.a The stationary points are given by solutions to \(y'(x)=0\), but from the nature of the question we may assume we are only interested in positive solutions.
\[y'(x)=3x^2-4x-1\]Descartes' rule of signs tells us that this quadratic has exactly one positive root (and in consequence also one negative root, but we are not interested in that). We can find this using the quadratic formula:
\[x=\frac{4\pm\sqrt{16+12}}{6}=\frac{2\pm\sqrt{7}}{3}\] and we are interested only in the positive root:\[x=\frac{2+\sqrt{7}}{3}\]
So for part (i) we have the stationary point of interest occurs when \(\displaystyle x=(2+\sqrt{7})/3\) which gives for part (ii) the deflection at the stationary point is:
\[ \begin{aligned} y&=\frac{{\left( \sqrt{7}+2\right) }^{3}}{27}-\frac{2\,{\left( \sqrt{7}+2\right) }^{2}}{9}-\frac{\sqrt{7}+2}{3}\\ &=-\frac{2\times {7}^{\frac{3}{2}}+34}{27}\end{aligned} \]
2.b.i The stationary points are given by solutions to: \[ \frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=0\], which give the pair of equations:\[\begin{aligned} \frac{\partial z}{\partial x}&=3x^2+y=0 \\ \frac{\partial z}{\partial y}&=x+2y=0\end{aligned} \]
From the second of these we find that \(y=-x/2\) which we substitute into the first to get \(3x^2-x/2=0\) which has solutions \(x=0, 1/6\) with corresponding \(y=0,-1/12\), so the stationary points are \( (0,0) \) and \( (\frac{1}{6},-\, \frac{1}{12}) \)
2.b.ii The classification of the stationary points can start with \( (0,0)\), where since along the line \(y=0\) function behaves like \(z=x^3\) which has a point of inflection at \(x=0\), we conclude that \( (0,0)\) is a saddle point.
For the other stationary point we use the classification rule:
\[D=\left[ \frac{\partial^2z}{\partial x^2}\frac{\partial^2z}{\partial y^2}-\left(\frac{\partial^2z}{\partial x \partial y} \right) \right]_{ (\frac{1}{6},-\, \frac{1}{12})}\].
If \(D \lt 0\) then we have a saddle point.
If \(D \gt 0\) and \(\partial^2z/\partial x^2 \gt 0 \) (or \(\partial^2z/\partial y^2 \gt 0 \) ) then we have a local minimum
If \(D \gt 0\) and \(\partial^2z/\partial x^2 \lt 0 \) (or \(\partial^2z/\partial y^2 \lt 0 \) ) then we have a local maximum
Otherwise the test is inconclusive.
I will leave the application of this test to the point \( (\frac{1}{6},-\, \frac{1}{12}) \) as an exercises for the reader.
(note I treated the stationary point \((0,0)\) differently because (a) it was easier, (b) the above test would have been inconclusive.)
Answers:
2.a The stationary points are given by solutions to \(y'(x)=0\), but from the nature of the question we may assume we are only interested in positive solutions.
\[y'(x)=3x^2-4x-1\]Descartes' rule of signs tells us that this quadratic has exactly one positive root (and in consequence also one negative root, but we are not interested in that). We can find this using the quadratic formula:
\[x=\frac{4\pm\sqrt{16+12}}{6}=\frac{2\pm\sqrt{7}}{3}\] and we are interested only in the positive root:\[x=\frac{2+\sqrt{7}}{3}\]
So for part (i) we have the stationary point of interest occurs when \(\displaystyle x=(2+\sqrt{7})/3\) which gives for part (ii) the deflection at the stationary point is:
\[ \begin{aligned} y&=\frac{{\left( \sqrt{7}+2\right) }^{3}}{27}-\frac{2\,{\left( \sqrt{7}+2\right) }^{2}}{9}-\frac{\sqrt{7}+2}{3}\\ &=-\frac{2\times {7}^{\frac{3}{2}}+34}{27}\end{aligned} \]
2.b.i The stationary points are given by solutions to: \[ \frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=0\], which give the pair of equations:\[\begin{aligned} \frac{\partial z}{\partial x}&=3x^2+y=0 \\ \frac{\partial z}{\partial y}&=x+2y=0\end{aligned} \]
From the second of these we find that \(y=-x/2\) which we substitute into the first to get \(3x^2-x/2=0\) which has solutions \(x=0, 1/6\) with corresponding \(y=0,-1/12\), so the stationary points are \( (0,0) \) and \( (\frac{1}{6},-\, \frac{1}{12}) \)
2.b.ii The classification of the stationary points can start with \( (0,0)\), where since along the line \(y=0\) function behaves like \(z=x^3\) which has a point of inflection at \(x=0\), we conclude that \( (0,0)\) is a saddle point.
For the other stationary point we use the classification rule:
\[D=\left[ \frac{\partial^2z}{\partial x^2}\frac{\partial^2z}{\partial y^2}-\left(\frac{\partial^2z}{\partial x \partial y} \right) \right]_{ (\frac{1}{6},-\, \frac{1}{12})}\].
If \(D \lt 0\) then we have a saddle point.
If \(D \gt 0\) and \(\partial^2z/\partial x^2 \gt 0 \) (or \(\partial^2z/\partial y^2 \gt 0 \) ) then we have a local minimum
If \(D \gt 0\) and \(\partial^2z/\partial x^2 \lt 0 \) (or \(\partial^2z/\partial y^2 \lt 0 \) ) then we have a local maximum
Otherwise the test is inconclusive.
I will leave the application of this test to the point \( (\frac{1}{6},-\, \frac{1}{12}) \) as an exercises for the reader.
(note I treated the stationary point \((0,0)\) differently because (a) it was easier, (b) the above test would have been inconclusive.)
Wednesday, 17 October 2012
Louis's Question from YahooAnswers:Fp1 Polynomial and roots question! Help!?
Question:
1.Find the range of values of \(a\) for which \[(2-3a)x^2+(4-a)x+2=0\]has real roots.
2. If the roots of the equation \(4x^3+7x^2-5x-1=0\) are \(\alpha\) , \(\beta\) and \( \gamma\),find the equation whose roots are:
(a) \( \alpha+1,\beta+1\) and \(\gamma+1\)
(b) \(\alpha^2 \beta^2\) and \( \gamma^2 \)
Answer:
Part 1.
For a quadratic \(ax^2+bx+c\) the discriminant is \(b^2-4ac\) this is the term that appears under the square root sign in the quadratic formula: \[x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}\]for the solution of \(ax^2+bx+c=0\). The quadratic equation has real roots precisely when the discriminant is non-negative.
The discriminant of \((2-3a)x^2+(4-a)x+2\) is \((4-a)^2-4(2-3a)2=a^2+16a\). Because the discriminant is positive for large \(|a|\) the discriminant is negative between the roots of \(a^2+16a\) which is on the interval \( (-16,0)\). Hence the discriminant is non-negative when \(a \lt -16\) or \(a \gt 0\) and so \((2-3a)x^2+(4-a)x+2=0\) has real roots when \(a \lt -16\) or \(a \gt0\).
Part 2.
You may know Vieta's relations for the roots and coefficients of polynomials or cubics in particular or we can derive them:
For a cubic \(ax^3+bx^2+cx+d\) with roots \(\alpha,\beta\) and \(\beta\) we have:\[\begin{aligned}a(x-\alpha)(x-\beta)(x-\gamma)&=ax^3-a(\alpha+\beta+\gamma)x^2+a(\alpha\beta+\alpha\gamma+\beta\gamma)x-a(\alpha\beta\gamma)\\ &=ax^3+bx^2+cx+d \end{aligned}\] means that \(b=-a(\alpha+\beta+\gamma)\), \(c=a(\alpha\beta+\alpha\gamma+\beta\gamma)\) and \(d=-a(\alpha\beta\gamma)\).
So for the cubic \(4x^3+7x^2-5x-1\) we have:
\(\alpha+\beta+\gamma=-7/4\)
\(\alpha\beta+\alpha\gamma+\beta\gamma=-5/4\)
\(\alpha\beta\gamma=1/4\)
(a) Let the equation with roots \( \alpha+1,\beta+1\) and \(\gamma+1\) be:\[x^3+ux^2+vx+w=0\] (we can of course multiply this by any constant we want if we want the left hand side to not be monic). Now from the relations between coefficients and roots we know that \[\begin{aligned}w&=-( \alpha+1)(\beta+1)(\gamma+1)\\ &=-(\alpha \beta \gamma + \alpha\beta + \alpha \gamma + \beta \gamma +\alpha+\beta+\gamma +1) \\ &= - \frac{1}{4} + \frac{5}{4} + \frac{7}{4} -1=\frac{7}{4} \end{aligned}\]
\[\begin{aligned}v&=( \alpha+1)(\beta+1)+(\alpha+1)(\gamma+1)+(\beta+1)(\gamma+1) \\
&=\alpha\beta+\alpha+\beta+1+\alpha\gamma+\alpha+\gamma+1+\beta\gamma+\beta+\gamma+1\\
&=(\alpha\beta+\alpha\gamma+\beta\gamma)+2(\alpha+\beta+\gamma)+3\\
&=\dots
\end{aligned}\]
(There is at least one other method of doing part (a), but I use Viete's relations here because they will be needed for part (b) and I don't want to provide two different approaches for the two parts.
...to be continued? Probably not, there is enough here to indicate how to complete the solution.
1.Find the range of values of \(a\) for which \[(2-3a)x^2+(4-a)x+2=0\]has real roots.
2. If the roots of the equation \(4x^3+7x^2-5x-1=0\) are \(\alpha\) , \(\beta\) and \( \gamma\),find the equation whose roots are:
(a) \( \alpha+1,\beta+1\) and \(\gamma+1\)
(b) \(\alpha^2 \beta^2\) and \( \gamma^2 \)
Answer:
Part 1.
For a quadratic \(ax^2+bx+c\) the discriminant is \(b^2-4ac\) this is the term that appears under the square root sign in the quadratic formula: \[x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}\]for the solution of \(ax^2+bx+c=0\). The quadratic equation has real roots precisely when the discriminant is non-negative.
The discriminant of \((2-3a)x^2+(4-a)x+2\) is \((4-a)^2-4(2-3a)2=a^2+16a\). Because the discriminant is positive for large \(|a|\) the discriminant is negative between the roots of \(a^2+16a\) which is on the interval \( (-16,0)\). Hence the discriminant is non-negative when \(a \lt -16\) or \(a \gt 0\) and so \((2-3a)x^2+(4-a)x+2=0\) has real roots when \(a \lt -16\) or \(a \gt0\).
Part 2.
You may know Vieta's relations for the roots and coefficients of polynomials or cubics in particular or we can derive them:
For a cubic \(ax^3+bx^2+cx+d\) with roots \(\alpha,\beta\) and \(\beta\) we have:\[\begin{aligned}a(x-\alpha)(x-\beta)(x-\gamma)&=ax^3-a(\alpha+\beta+\gamma)x^2+a(\alpha\beta+\alpha\gamma+\beta\gamma)x-a(\alpha\beta\gamma)\\ &=ax^3+bx^2+cx+d \end{aligned}\] means that \(b=-a(\alpha+\beta+\gamma)\), \(c=a(\alpha\beta+\alpha\gamma+\beta\gamma)\) and \(d=-a(\alpha\beta\gamma)\).
So for the cubic \(4x^3+7x^2-5x-1\) we have:
\(\alpha+\beta+\gamma=-7/4\)
\(\alpha\beta+\alpha\gamma+\beta\gamma=-5/4\)
\(\alpha\beta\gamma=1/4\)
(a) Let the equation with roots \( \alpha+1,\beta+1\) and \(\gamma+1\) be:\[x^3+ux^2+vx+w=0\] (we can of course multiply this by any constant we want if we want the left hand side to not be monic). Now from the relations between coefficients and roots we know that \[\begin{aligned}w&=-( \alpha+1)(\beta+1)(\gamma+1)\\ &=-(\alpha \beta \gamma + \alpha\beta + \alpha \gamma + \beta \gamma +\alpha+\beta+\gamma +1) \\ &= - \frac{1}{4} + \frac{5}{4} + \frac{7}{4} -1=\frac{7}{4} \end{aligned}\]
\[\begin{aligned}v&=( \alpha+1)(\beta+1)+(\alpha+1)(\gamma+1)+(\beta+1)(\gamma+1) \\
&=\alpha\beta+\alpha+\beta+1+\alpha\gamma+\alpha+\gamma+1+\beta\gamma+\beta+\gamma+1\\
&=(\alpha\beta+\alpha\gamma+\beta\gamma)+2(\alpha+\beta+\gamma)+3\\
&=\dots
\end{aligned}\]
(There is at least one other method of doing part (a), but I use Viete's relations here because they will be needed for part (b) and I don't want to provide two different approaches for the two parts.
...to be continued? Probably not, there is enough here to indicate how to complete the solution.
Sunday, 14 October 2012
Ken's Question from Yahoo Answers: Probability Question For Verification?
Question:
"A point \({\rm{P}}(x,y)\) is chosen at random in a unit disc, centred at \((0,0)\).
The probability required is that the point chosen is such that both \(| x -y| \lt 1\) and \(|x+y| \lt 1\) .
Is the answer \(2/\pi\) or \(1-2/\pi\)?
Thank you."
Answer:
I take disc to be a disc of radius 1 centred at the origin
The region defined by the inequalities \(|x-y| \lt 1\) and \(|x+y| \lt 1\) is an inscribed square to the circle, which has side \(\sqrt{2}\) and hence area \(2\). The area of the circle is \(\pi\), so the probability that a point sampled uniformly on the unit disc satisfies the inequalities is the ratio of these two area: \(2/\pi\).
To convince yourself that the required region is the interior of the square rather than the exterior consider the point \((0,0)\), does it satisfy the inequalities. It it does then you want the interior of the square rather than the exterior.
Below is a scatter plot showing random points uniformly sampled on the unit disc and in black those satisfying the inequalities:
.
"A point \({\rm{P}}(x,y)\) is chosen at random in a unit disc, centred at \((0,0)\).
The probability required is that the point chosen is such that both \(| x -y| \lt 1\) and \(|x+y| \lt 1\) .
Is the answer \(2/\pi\) or \(1-2/\pi\)?
Thank you."
Answer:
I take disc to be a disc of radius 1 centred at the origin
The region defined by the inequalities \(|x-y| \lt 1\) and \(|x+y| \lt 1\) is an inscribed square to the circle, which has side \(\sqrt{2}\) and hence area \(2\). The area of the circle is \(\pi\), so the probability that a point sampled uniformly on the unit disc satisfies the inequalities is the ratio of these two area: \(2/\pi\).
To convince yourself that the required region is the interior of the square rather than the exterior consider the point \((0,0)\), does it satisfy the inequalities. It it does then you want the interior of the square rather than the exterior.
Below is a scatter plot showing random points uniformly sampled on the unit disc and in black those satisfying the inequalities:
.
Saturday, 13 October 2012
CrazyCat's MATHS QUESTION HELP 3? from Yahoo Answers
CrazyCat's Question:
Find the discrimnant of \(kx^2 - 4x + k\) in terms of \(k\), hence find possible values of \(k\) given that \(kx^2 -4x + k = 0\) has equal roots.
Answer:
For a quadratic \(ax^2+bx+c\) the discriminant is \(b^2-4ac\) this is the term that appears under the square root sign in the quadratic formula:
\[x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}\]for the solution of \(ax^2+bx+c=0\). The quadratic equation has equal roots precisely when the discriminant is zero.
Now for the problem at hand \(a=k\), \(b=-4\) and \(c=k\) so the discriminant is \(D=b^2-4ac=16-4k^2\), and when \(D=0\) we have \(16-4k^2=0\) which we may solve for \(k\) to find: \(k=\pm2\).
http://www.localtutor.co.uk/Tuition/Havant_Maths_Tuition.html
.
Find the discrimnant of \(kx^2 - 4x + k\) in terms of \(k\), hence find possible values of \(k\) given that \(kx^2 -4x + k = 0\) has equal roots.
Answer:
For a quadratic \(ax^2+bx+c\) the discriminant is \(b^2-4ac\) this is the term that appears under the square root sign in the quadratic formula:
\[x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}\]for the solution of \(ax^2+bx+c=0\). The quadratic equation has equal roots precisely when the discriminant is zero.
Now for the problem at hand \(a=k\), \(b=-4\) and \(c=k\) so the discriminant is \(D=b^2-4ac=16-4k^2\), and when \(D=0\) we have \(16-4k^2=0\) which we may solve for \(k\) to find: \(k=\pm2\).
http://www.localtutor.co.uk/Tuition/Havant_Maths_Tuition.html
.
Friday, 5 October 2012
Q3 from STEP 1998 Paper1
Which of the following are true and which false? Justify your answers
(i) \(a^{\ln(b)}=b^{\ln(a)}\), for all \(a,b>0\).
(ii) \(\cos(\sin(\theta))=\sin(\cos(\theta))\), for all real \(\theta\).
(iii) There exists a polynomial \(P\) such that \(|P(\theta)-\cos(\theta)| \lt 10^{ -6 } \) for all real \(\theta\)
(iv) \(x^4+3+x^{-4} \ge 5\) for all \(x\gt 0\).
Solution:
(i) True, since we know for \(a,b \gt 0\) that \( \ln(b)\ln(a)=\ln(a)\ln(b) \) and by the laws of logarithms this may be rewritten: \(\ln(a^{\ln(b)})=\ln(b^{\ln(a)})\) and as the equality of logarithms implies the equality of the arguments we can conclude that \(a^{\ln(b)}=b^{\ln(a)}\) for all \(a,b \gt 0\).
(ii) False, putting \(\theta=0\) the left hand side is \(1\) and the right hand side is \(\sin(1)\ne 1\).
(iii) False, polynomials are unbounded on the reals while \(|\cos(\theta)| \le 1\).
(iv) True, The function \(f(x)=x^4+3+x^{-4}\) goes to infinity as \(x\) goes to \(0\) from above, and as \(x\) goes to infinity. It also in continuous and differentiable and has one stationary point on \((0,\infty)\). This stationary point therefore must be a global minimum and it occurs at \(x=1\), and \(f(1)=5\).
(i) \(a^{\ln(b)}=b^{\ln(a)}\), for all \(a,b>0\).
(ii) \(\cos(\sin(\theta))=\sin(\cos(\theta))\), for all real \(\theta\).
(iii) There exists a polynomial \(P\) such that \(|P(\theta)-\cos(\theta)| \lt 10^{ -6 } \) for all real \(\theta\)
(iv) \(x^4+3+x^{-4} \ge 5\) for all \(x\gt 0\).
Solution:
(i) True, since we know for \(a,b \gt 0\) that \( \ln(b)\ln(a)=\ln(a)\ln(b) \) and by the laws of logarithms this may be rewritten: \(\ln(a^{\ln(b)})=\ln(b^{\ln(a)})\) and as the equality of logarithms implies the equality of the arguments we can conclude that \(a^{\ln(b)}=b^{\ln(a)}\) for all \(a,b \gt 0\).
(ii) False, putting \(\theta=0\) the left hand side is \(1\) and the right hand side is \(\sin(1)\ne 1\).
(iii) False, polynomials are unbounded on the reals while \(|\cos(\theta)| \le 1\).
(iv) True, The function \(f(x)=x^4+3+x^{-4}\) goes to infinity as \(x\) goes to \(0\) from above, and as \(x\) goes to infinity. It also in continuous and differentiable and has one stationary point on \((0,\infty)\). This stationary point therefore must be a global minimum and it occurs at \(x=1\), and \(f(1)=5\).
Saturday, 29 September 2012
Q2 from the Further Maths FPM1 AS 2006 AQA Exams
2. A curve satisfies the differential equation \[\frac{dy}{dx}=\log_{10} (x)\] Starting at the point \( (2, 3)\) on the curve, use a step length or \(0.2\) to estimate the value of \(y\) at \(x=2.4\). Give your answer to three decimal places.
Solution:
Doing this sort of thing professionally leaves one with a large number of available methods, so it is necessary to consult the A-level specification to determine what method you are expected to use. From the 2012 specification it is clear that you are expected to use what is commonly known as the Euler method. This may be summarised as:
\[ \begin{aligned} x_{i+1}&=x_i+ \Delta x\\ y_{i+1}&=y_i+\Delta x \times \left[ y'(x_i)\right]=y_i+\Delta x \times \log_{10}(x_i) \end{aligned} \] where \(\Delta x=0.2 \) starting from initial condition \(x_0=2,\ \ y_0=3\).
\[ \begin{array}{ccccc}i&x_i&y_i&\log_{10}(x_i)&y_{i+1}=y_i+\Delta x \log_{10}(x_i)\\ 0&2&3&0.30103&3.06021\\1&2.2&3.06021&0.34242&3.12869\\2&2.4&3.12869 \end{array} \]
We stop at this point as we have reached \(x=2.4\) and have the corresponding estimate \(y=3.129\ (3DP)\).
The arithmetic has been worked to 5 decimal places as I have assumed this will be sufficient to give the final answer to 3 places. (I have done an error analysis and know that four DP are sufficient, but looking at the mark scheme indicates that an error analysis is not expected and you are expected to assume that 5 DP for the calculations will be adequate to give 3 DP in the final solution).
Discussion:
The Euler method used here is probably the crudest method for numerical integration of Ordinary Differential Equations (ODEs) with initial values. The essense of the method is that we step forwards from \(x_i\) to \(x_{i+1}\) using the trauncated Taylor series: \[y(x_i+\Delta x)=y(x_i)+\Delta x \times y'(x_i)\] We can improve this stepping procedure in many ways but two simple ones are to increase the number of terms in the Taylor approximation, which in the case of our problem above would give (for a single step with \(\Delta x=0.4\) ):
\[y(2.4)=y(2)+0.4 \times \log_{10}(2)+0.4^2\frac{1}{4\ln(10)} \approx 3.138 \ \ (3DP)\]
Another method of improving the approximation is instead of using \(y'(x_i)\) we approximate the average value of the derivative over the step. The simplest way of doing this is to use \((y'(x_i)+y'(x_{i+1})/2\) as our estimate of the derivative over the interval. Then again using \(\Delta x=0.4\) we get:\[y(2.4)=y(2)+0.4\frac{\log_{10}(2)+\log_{10}(2.4)}{2}\approx 3.136 \ \ (3DP)\]
Note we use larger steps in the improved methods above since at each stage we are doing longer computations, so to keep the computational effort approximately constant we increase the step length.
If we are interested in the actual value of the solution at \(x=2.4\) to three decimals we can integrate the ODE with either much more powerful methods and/or smaller step sizes, doing this we find the solution to 3DP is \(3.137\).
We may also note that the exact solution to the initial value problem is:\[y(x)=3+\frac{x(\ln(x)-1)}{\ln(10)}-\frac{2(\ln(2)-1)}{\ln(10)}=3+x\log_{10}(x/e)-2\log_{10}(2/e),\ \ \ x>0\] which can be verified by checking the initial condition and differentiating to show that it gives the right derivative.
The numerical integration of first order ordinary differential equations is of greater interest than we might expect since a differential equation of the form:\[\frac{d^n}{dx^n}y(x)=f(x,y,y',...,y^{(n-1)})\] can always be rewritten as a first order system of ordinary differential equations. These can then be integrated by methods developed for simple first order equations. I will illustrate this with a second order equation:\[y''=f(x,y,y')\]
We introduce \(u(x)=y(x)\) and \(v(x)=y'(x)\), then our second order ODE may be written as the first order system:\[\begin{aligned}v' &=f(x,u,v)\\u' &= v \end{aligned}\] This in turn may be integrated by a numerical method intended for first order ODEs. Using Eulers method we get: \[\begin{aligned}x_{i+1}&=x_i+\Delta x \\ v_{i+1}&=v_i+\Delta x\times v'(x_i) = v_i+\Delta x\times f(x_i,u_i,v_i) \\ u_{i+1}&=u_i+\Delta x\times u'(x_i) = u_i+\Delta x\times v_i \end{aligned}\]
Solution:
Doing this sort of thing professionally leaves one with a large number of available methods, so it is necessary to consult the A-level specification to determine what method you are expected to use. From the 2012 specification it is clear that you are expected to use what is commonly known as the Euler method. This may be summarised as:
\[ \begin{aligned} x_{i+1}&=x_i+ \Delta x\\ y_{i+1}&=y_i+\Delta x \times \left[ y'(x_i)\right]=y_i+\Delta x \times \log_{10}(x_i) \end{aligned} \] where \(\Delta x=0.2 \) starting from initial condition \(x_0=2,\ \ y_0=3\).
\[ \begin{array}{ccccc}i&x_i&y_i&\log_{10}(x_i)&y_{i+1}=y_i+\Delta x \log_{10}(x_i)\\ 0&2&3&0.30103&3.06021\\1&2.2&3.06021&0.34242&3.12869\\2&2.4&3.12869 \end{array} \]
We stop at this point as we have reached \(x=2.4\) and have the corresponding estimate \(y=3.129\ (3DP)\).
The arithmetic has been worked to 5 decimal places as I have assumed this will be sufficient to give the final answer to 3 places. (I have done an error analysis and know that four DP are sufficient, but looking at the mark scheme indicates that an error analysis is not expected and you are expected to assume that 5 DP for the calculations will be adequate to give 3 DP in the final solution).
Discussion:
The Euler method used here is probably the crudest method for numerical integration of Ordinary Differential Equations (ODEs) with initial values. The essense of the method is that we step forwards from \(x_i\) to \(x_{i+1}\) using the trauncated Taylor series: \[y(x_i+\Delta x)=y(x_i)+\Delta x \times y'(x_i)\] We can improve this stepping procedure in many ways but two simple ones are to increase the number of terms in the Taylor approximation, which in the case of our problem above would give (for a single step with \(\Delta x=0.4\) ):
\[y(2.4)=y(2)+0.4 \times \log_{10}(2)+0.4^2\frac{1}{4\ln(10)} \approx 3.138 \ \ (3DP)\]
Another method of improving the approximation is instead of using \(y'(x_i)\) we approximate the average value of the derivative over the step. The simplest way of doing this is to use \((y'(x_i)+y'(x_{i+1})/2\) as our estimate of the derivative over the interval. Then again using \(\Delta x=0.4\) we get:\[y(2.4)=y(2)+0.4\frac{\log_{10}(2)+\log_{10}(2.4)}{2}\approx 3.136 \ \ (3DP)\]
Note we use larger steps in the improved methods above since at each stage we are doing longer computations, so to keep the computational effort approximately constant we increase the step length.
If we are interested in the actual value of the solution at \(x=2.4\) to three decimals we can integrate the ODE with either much more powerful methods and/or smaller step sizes, doing this we find the solution to 3DP is \(3.137\).
We may also note that the exact solution to the initial value problem is:\[y(x)=3+\frac{x(\ln(x)-1)}{\ln(10)}-\frac{2(\ln(2)-1)}{\ln(10)}=3+x\log_{10}(x/e)-2\log_{10}(2/e),\ \ \ x>0\] which can be verified by checking the initial condition and differentiating to show that it gives the right derivative.
The numerical integration of first order ordinary differential equations is of greater interest than we might expect since a differential equation of the form:\[\frac{d^n}{dx^n}y(x)=f(x,y,y',...,y^{(n-1)})\] can always be rewritten as a first order system of ordinary differential equations. These can then be integrated by methods developed for simple first order equations. I will illustrate this with a second order equation:\[y''=f(x,y,y')\]
We introduce \(u(x)=y(x)\) and \(v(x)=y'(x)\), then our second order ODE may be written as the first order system:\[\begin{aligned}v' &=f(x,u,v)\\u' &= v \end{aligned}\] This in turn may be integrated by a numerical method intended for first order ODEs. Using Eulers method we get: \[\begin{aligned}x_{i+1}&=x_i+\Delta x \\ v_{i+1}&=v_i+\Delta x\times v'(x_i) = v_i+\Delta x\times f(x_i,u_i,v_i) \\ u_{i+1}&=u_i+\Delta x\times u'(x_i) = u_i+\Delta x\times v_i \end{aligned}\]
Sunday, 22 July 2012
Question 1 AQA AS Maths Pure Core 1, May 2011
1. The line \(AB\) has equation \(7x+3y=13\).
(a) Find the gradient of \(AB\). (2 marks)
(b) The point \(C\) has coordinates \((-1,3)\).
(c) The line \(AB\) intersects the line with equation \(3x+2y=12\) at the point \(B\). Find the coordinates of \(B\) (3 marks)
Answer:
(a) We rewrite the equation of the line in the form \(y=mx+c\), then \(m\) is the gradient:
\[3y=-7x+13 \\ y=-\;\frac{7}{3}x+\frac{13}{3}\],
so the gradient is \(-\frac{7}{3}\)
(b)(i) The line through \(C: (-1,3)\) has gradient \(-\frac{7}{3}\), and so is of the form: \( y=-\;\frac{7}{3}x+c\), and as it passes through \(C\) we have:\[3=-\;\frac{7}{3}\times (-1)+c\], so \(c=3-\frac{7}{3}=\frac{2}{3}\) and the required equation of the lone through \(C\) parallel to \(AB\) is:\[y=-\;\frac{7}{3}x+\frac{2}{3}\]. Or multipling through by \(3\):\[3y=-7x+2\]
(b)(ii) As \(M: (1\frac{1}{2},-1) \) is the mid-point of \(AC\) we have the coordinates of \(M\) are the average of the coordinates of \(A\) and \(C\), so if \(A:(a_1,a_2)\) we have:\[\begin{aligned} 3/2&=(a_1+(-1))/2 \\ (-1)&=(a_2+3)/2 \end{aligned}\]. Hence \(a_1=4\) and \(a_2=-5\), so \(A\) is the point \((4,-5)\)
(c) As the line \(AB\) intersects the line \(3x+2y=12\) at \(B\) we may substitute \(y\) from the equation for \(AB\) into this to get: \[3x+2\left(-\frac{7}{3}x+\frac{13}{3}\right)=12\]or \[\left( 3-\frac{14}{3}\right)x+\frac{26}{3}=12\]simplifying some more: \[\left( -\;\frac{5}{3}\right)x=\frac{10}{3}\], so \(x=-2\), and substituting back into either equation gives \(y=9\) so \(B\) is the point \((-2,9)\).
(a) Find the gradient of \(AB\). (2 marks)
(b) The point \(C\) has coordinates \((-1,3)\).
(i) Find an equation of the line which passes through the point \(C\) and which is parallel to \(AB\). (2 marks)
(ii) The point \( (1\frac{1}{2},-1)\) is the mid point of \(AC\). Find the coordinates of the point \(A\) (2 marks)
(c) The line \(AB\) intersects the line with equation \(3x+2y=12\) at the point \(B\). Find the coordinates of \(B\) (3 marks)
Answer:
(a) We rewrite the equation of the line in the form \(y=mx+c\), then \(m\) is the gradient:
\[3y=-7x+13 \\ y=-\;\frac{7}{3}x+\frac{13}{3}\],
so the gradient is \(-\frac{7}{3}\)
(b)(i) The line through \(C: (-1,3)\) has gradient \(-\frac{7}{3}\), and so is of the form: \( y=-\;\frac{7}{3}x+c\), and as it passes through \(C\) we have:\[3=-\;\frac{7}{3}\times (-1)+c\], so \(c=3-\frac{7}{3}=\frac{2}{3}\) and the required equation of the lone through \(C\) parallel to \(AB\) is:\[y=-\;\frac{7}{3}x+\frac{2}{3}\]. Or multipling through by \(3\):\[3y=-7x+2\]
(b)(ii) As \(M: (1\frac{1}{2},-1) \) is the mid-point of \(AC\) we have the coordinates of \(M\) are the average of the coordinates of \(A\) and \(C\), so if \(A:(a_1,a_2)\) we have:\[\begin{aligned} 3/2&=(a_1+(-1))/2 \\ (-1)&=(a_2+3)/2 \end{aligned}\]. Hence \(a_1=4\) and \(a_2=-5\), so \(A\) is the point \((4,-5)\)
(c) As the line \(AB\) intersects the line \(3x+2y=12\) at \(B\) we may substitute \(y\) from the equation for \(AB\) into this to get: \[3x+2\left(-\frac{7}{3}x+\frac{13}{3}\right)=12\]or \[\left( 3-\frac{14}{3}\right)x+\frac{26}{3}=12\]simplifying some more: \[\left( -\;\frac{5}{3}\right)x=\frac{10}{3}\], so \(x=-2\), and substituting back into either equation gives \(y=9\) so \(B\) is the point \((-2,9)\).
Saturday, 16 June 2012
Draft answer to a MathHelpBoards Question
[quote]
so i am stuck at one improper integral question. By the way its from further pure 3 module a-level.
first the question asks to substitute y=1/x to transform the the integral lnx^2 / x^3 dx into
integral 2y ln(y) dy. Then it asks to evaluate the integral 2y ln(y) dy with limit (1 , 0) by showing the limiting process.
i did it up to there and i got -0.5 but at last it then asks to find the value of the integral lnx^2 / x^3 dx with limit (infinity , 1). Somehow the answer in the markscheme is 0.5.
can someone please explain how the two integral are connected and how the sign changed?
thanks.[/quote]
When you make the change of variable you have:
\[ \int_a^b \frac{\ln(x^2)}{x^3} dx = \int_{1/a}^{1/b} 2y \ln(y) dy \]
So:
\[ \int_A^1 \frac{\ln(x^2)}{x^3} dx = \int_{1/A}^{1} 2y \ln(y) dy \]
and so the limits of both sides if they as \(A\) goes to infinity are equal and hence:
\[ \int_{\infty}^1 \frac{\ln(x^2)}{x^3} dx = \int_{0}^{1} 2y \ln(y) dy = \int_1^0 2y \ln(y) dy =-(-0.5)=0.5\]
CB
so i am stuck at one improper integral question. By the way its from further pure 3 module a-level.
first the question asks to substitute y=1/x to transform the the integral lnx^2 / x^3 dx into
integral 2y ln(y) dy. Then it asks to evaluate the integral 2y ln(y) dy with limit (1 , 0) by showing the limiting process.
i did it up to there and i got -0.5 but at last it then asks to find the value of the integral lnx^2 / x^3 dx with limit (infinity , 1). Somehow the answer in the markscheme is 0.5.
can someone please explain how the two integral are connected and how the sign changed?
thanks.[/quote]
When you make the change of variable you have:
\[ \int_a^b \frac{\ln(x^2)}{x^3} dx = \int_{1/a}^{1/b} 2y \ln(y) dy \]
So:
\[ \int_A^1 \frac{\ln(x^2)}{x^3} dx = \int_{1/A}^{1} 2y \ln(y) dy \]
and so the limits of both sides if they as \(A\) goes to infinity are equal and hence:
\[ \int_{\infty}^1 \frac{\ln(x^2)}{x^3} dx = \int_{0}^{1} 2y \ln(y) dy = \int_1^0 2y \ln(y) dy =-(-0.5)=0.5\]
CB
Saturday, 21 April 2012
Friday, 20 April 2012
Part 2 of Pharaoh's Taylor series and modified Euler question from Yahoo Answers
consider van der pol's equationy" - 0.2(1-y^2)y' + y = 0 y(0)=0.1 y'(0)=0.1
2)Write down the above problem as a system of first order differential equations.Calculate the numerical solution at x = 0.2 using the Modified Euler method. Take thestep-length h = 0.1 and work to 6 decimal places accuracy. Compare with your solutionin part (i) and comment on your answers.
There is a standard method of converting higher order ODEs into first order systems, in this case it is to introduce the state vector:
\(Y(t)=\left[ \begin{array}{c} y(t) \\ y'(t) \end{array} \right] \)
Then the ODE becomes:
\( Y'(t)= F(t,Y)=\left[ \begin{array}{c} y'(t) \\ y''(t) \end{array} \right]= \left[ \begin{array}{c} y'(t) \\ 0.2 \left(1-(y(t))^2 \right) y'(t)-y(t) \end{array} \right] \)
Now you can use the standard form of the modified Euler method on this vector first order ODE.
.
Pharaoh's Taylor series and modified Euler question from Yahoo Answers
consider van der pol's equationy" - 0.2(1-y^2)y' + y = 0 y(0)=0.1 y'(0)=0.1
1)You are asked to find the approximate solution for this problem using the Taylor seriesmethod. Your expansion should include the first three non-zero terms and you shouldwork to six decimal places accuracy. First find the approximate solutions for both y (0.1)and y’(0.1) using the first three non-zero terms of Taylor series expansion for eachfunction and then use this information to calculate the approximate solution at x = 0.2.
The Taylor series expansion about \(t=0\) is of the form:
\(y(t)=y(0)+y'(0)t+\frac{y''(0)t^2}{2}+.. \)
We are given \(y(0)\) and \(y'(0)\) in the initial condition, and so from the equation we have:
\(y''(0) = 0.2(1-(y(0))^2)-y(0)=0.2(1-0.1^2)-0.1=-0.0802\)
So the Taylor series about \(t=0\) is:
\(y(t)=0.1+0.1t-0.0401t^2+... \)
and using the first three terms of this we have \(y(0.1)\approx 0.109599\),
Also:
\(y'(t)=0.1-0.0802t+...\)
and so \(y'(0.1) \approx 0.09198\)
Now the Taylor expansion about \(t=0.1\) is:
\(y(t)=y(0.1)+(t-0.1)y'(0.1)+\frac{(t-0.1)^2y''(0.1)}{2}+...\)
where \(y''(0.1)=0.2\left(1-(y(0.1))^2\right)y'(0.1)-y(0.1)=-0.08178796\).
So:
\(y(0.2)\approx 0.109599+0.1\times 0.0198-0.1^2\times 0.8178796 = 0.108789 \) to 6 six DP
.
Sunday, 15 April 2012
Another POTW proposal
Prove that \(|\sin(nx)|\le n|\sin(x)|\) for all \(x \in \mathbb{R}\) and \(n \in \mathbb{N}_+\).
==========================
Solution: This is just an excesise in induction.
The base case \(n=1\) holds obviously and trivialy.
Now suppose that for some \(k \in \mathbb{N}_+\):
\[|\sin(kx)| \le k |\sin(x)|\].
Now consider:
\[|\sin((k+1)x)|=| \sin(kx)\cos(x)+\cos(kx)\sin(x)| \].
Then by the triangle inequality we get:
\[|\sin((k+1)x)|\le |\sin(kx)\cos(x)|+|\cos(kx)\sin(x)| \\ \phantom{[ \sin((k+1)x)xxx}\le |\sin(kx)| + |\sin(x)|=(k+1)|\sin(x)|\].
QED
==========================
Solution: This is just an excesise in induction.
The base case \(n=1\) holds obviously and trivialy.
Now suppose that for some \(k \in \mathbb{N}_+\):
\[|\sin(kx)| \le k |\sin(x)|\].
Now consider:
\[|\sin((k+1)x)|=| \sin(kx)\cos(x)+\cos(kx)\sin(x)| \].
Then by the triangle inequality we get:
\[|\sin((k+1)x)|\le |\sin(kx)\cos(x)|+|\cos(kx)\sin(x)| \\ \phantom{[ \sin((k+1)x)xxx}\le |\sin(kx)| + |\sin(x)|=(k+1)|\sin(x)|\].
QED
Monday, 9 April 2012
David's question from Yahoo Answers
This is the prototype for the posting of this question/solution to MathHelpBoards: http://www.mathhelpboards.com/showthread.php?824-David-s-question-from-Yahoo-Answers
2. Let X be a random variable that follows a Uniform(0; 1) distribution.
(a) Show that E(X) = 1/2 and Var(X) = 1/12.
(b) Using Chebyshev's inequality fi nd an upper bound on the prob-
ability that X is more than k standard deviations away from its
expected value.
(c) Compute the exact probability that X is more than k standard
deviations from its expected value.
(d) Compare the bound to the exact probability.
Thanks
============================================
(a) By definition of the expectation: \[E(X)= \int_{-\infty}^{\infty} x p(x)\; dx\]
but \(X \sim U(0,1)\) so \(p(x)=1\) for \(x\) in \([0,1]\) and zero otherwise this becomes:
\[E(X) = \int_0^1 x dx\]
Hence \[E(X)= \biggl[ \frac{x^2}{2} \biggr]_0^1= \frac{1^2}{2}-\frac{0^2}{2}=1/2 \]
Similarly for the variance:
\[V(X)=E( (X - \overline{X})^2 ) = \int_0^1 (x-1/2)^2\; dx = \frac{1}{12}\]
(b) I don't know what is being asked for here, Chebyshev's inequality is just this:
\[P( |X-\overline{X}| \ge k \sigma) \le \frac{1}{k^2}\]
(c) There are a number of ways of doing this, the easiest involves a diagram, but that is not convienient to use here, so we take the definition of the required probability:
\[P( |X-\overline{X}| \ge k \sigma)= P( X-\overline{X} \le - k \sigma) + P( X-\overline{X} \ge k \sigma)\]
Which may be written as a sum of integrals:
\[P( X-\overline{X} \le - k \sigma)=P(X \le \overline{X}-k \sigma)=\int_0^{\frac{1}{2}-\frac{k}{\sqrt{12}}} dx\] when \(k < \sqrt{12}/2\) and zero otherwise, and:
\[P( X-\overline{X} \ge - k \sigma)=P(X \ge \overline{X}+k \sigma)=\int_{\frac{1}{2}+\frac{k}{\sqrt{12}}}^1 dx\] when \(k < \sqrt{12}/2\) and zero otherwise.
CB
2. Let X be a random variable that follows a Uniform(0; 1) distribution.
(a) Show that E(X) = 1/2 and Var(X) = 1/12.
(b) Using Chebyshev's inequality fi nd an upper bound on the prob-
ability that X is more than k standard deviations away from its
expected value.
(c) Compute the exact probability that X is more than k standard
deviations from its expected value.
(d) Compare the bound to the exact probability.
Thanks
============================================
(a) By definition of the expectation: \[E(X)= \int_{-\infty}^{\infty} x p(x)\; dx\]
but \(X \sim U(0,1)\) so \(p(x)=1\) for \(x\) in \([0,1]\) and zero otherwise this becomes:
\[E(X) = \int_0^1 x dx\]
Hence \[E(X)= \biggl[ \frac{x^2}{2} \biggr]_0^1= \frac{1^2}{2}-\frac{0^2}{2}=1/2 \]
Similarly for the variance:
\[V(X)=E( (X - \overline{X})^2 ) = \int_0^1 (x-1/2)^2\; dx = \frac{1}{12}\]
(b) I don't know what is being asked for here, Chebyshev's inequality is just this:
\[P( |X-\overline{X}| \ge k \sigma) \le \frac{1}{k^2}\]
(c) There are a number of ways of doing this, the easiest involves a diagram, but that is not convienient to use here, so we take the definition of the required probability:
\[P( |X-\overline{X}| \ge k \sigma)= P( X-\overline{X} \le - k \sigma) + P( X-\overline{X} \ge k \sigma)\]
Which may be written as a sum of integrals:
\[P( X-\overline{X} \le - k \sigma)=P(X \le \overline{X}-k \sigma)=\int_0^{\frac{1}{2}-\frac{k}{\sqrt{12}}} dx\] when \(k < \sqrt{12}/2\) and zero otherwise, and:
\[P( X-\overline{X} \ge - k \sigma)=P(X \ge \overline{X}+k \sigma)=\int_{\frac{1}{2}+\frac{k}{\sqrt{12}}}^1 dx\] when \(k < \sqrt{12}/2\) and zero otherwise.
CB
Tuesday, 27 March 2012
Draft PotW
\(\mathbb{R}^3\) is coloured so that each point is either red, blue or green.
Prove that for the set of points of at least one of the colours it is the case that for all \(r \in \mathbb{R}^3\) there are points in the set separated by distance \(r\).
============================================================
Suppose otherwise. Then for each colour set \(R, B, G\) there is an \(r_r, r_b, r_g\ge0\) such that there are no points of that colour separated by that distance. Also suppose that without loss of generality that \(r_r\ge r_b \ge r_g\) and so \(r_r>0\).
Choose any point \(x_r\) in \(R\) and consider the sphere centred on \(x_r\) of radius \(r_r\). This sphere (which to be clear is the 2D surface of the ball of radius \(r_r\) centred at \(x_r\) ) has no points coloured red, in fact it must have points coloured blue and others coloured green. Choose a point \(x_b\) on the sphere coloured blue, and consider the intersection of the first sphere with a sphere of radius \(r_b\) centred at \(x_b\). This circle cannot be coloured red or blue so must be green.
Now choose any point on this green circle \(x_g\) and consider the sphere of radius \(r_g\) centred there, this meets the green circle at two points, which therefore cannot be green - a contradiction.
Prove that for the set of points of at least one of the colours it is the case that for all \(r \in \mathbb{R}^3\) there are points in the set separated by distance \(r\).
============================================================
Suppose otherwise. Then for each colour set \(R, B, G\) there is an \(r_r, r_b, r_g\ge0\) such that there are no points of that colour separated by that distance. Also suppose that without loss of generality that \(r_r\ge r_b \ge r_g\) and so \(r_r>0\).
Choose any point \(x_r\) in \(R\) and consider the sphere centred on \(x_r\) of radius \(r_r\). This sphere (which to be clear is the 2D surface of the ball of radius \(r_r\) centred at \(x_r\) ) has no points coloured red, in fact it must have points coloured blue and others coloured green. Choose a point \(x_b\) on the sphere coloured blue, and consider the intersection of the first sphere with a sphere of radius \(r_b\) centred at \(x_b\). This circle cannot be coloured red or blue so must be green.
Now choose any point on this green circle \(x_g\) and consider the sphere of radius \(r_g\) centred there, this meets the green circle at two points, which therefore cannot be green - a contradiction.
Draft:CaptainBlacks PoW #2 for MathHelpBoards
Posted on MathHelpBoards
This problem I think is a bit tedious and can be hard work. It comes from the Purdue Maths Dept PoW, only slightly modified.
Given that \(\cos(36^\circ)=\frac{1}{4}(1+\sqrt{5}) \) find \( \tan^2(18^\circ)\, \tan^2(54^\circ) \)
=============================================================
Solution 1 (this is mine and I must admit I used Maxima to handle the algebra):
First observe that from the quadrant corresponding to te angles here:
\[\tan(A/2)=\sqrt{\frac{1-\cos(A)}{1+\cos(A)}}\]
and:
\[ \tan(3A/2)=\frac{3\tan(A/2)-\tan^3(A/2)}{1-3\tan^2(A/2)} \]
So if we put \( A=36^\circ \) and allowing Maxima to do the algebra we get:
\[ \tan^2(18^\circ)\, \tan^2(54^\circ)=\frac{1}{5} \]
--------------------------------------------------------------------
Solution 2 (this is the solution give on the originating site):
\[ \cos(72^\circ)=2\cos^2(36^\circ)-1=\frac{1}{4}(\sqrt{5}-1) \]
and:
\(\displaystyle \phantom{xxxx} \tan^2(18^\circ)\, \tan^2(54^\circ) =\frac{\sin^2(54^\circ)\sin^2(18^\circ)}{\cos^2(54^\circ)\cos^2(18^\circ)} =\left[ \frac{(1/2)(\cos(36^\circ)-\cos(72^\circ))}{(1/2)(\cos(36^\circ)+\cos(72^\circ))} \right]^2\)
\(\displaystyle \phantom{xxxx} \phantom{\tan^2(18^\circ)\, \tan^2(54^\circ)}=\left[ \frac{\frac{\sqrt{5}+1}{4} -\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{5}+1}{4}+\frac{\sqrt{5}-1}{4}} \right] =\frac{1}{5}\)
This problem I think is a bit tedious and can be hard work. It comes from the Purdue Maths Dept PoW, only slightly modified.
Given that \(\cos(36^\circ)=\frac{1}{4}(1+\sqrt{5}) \) find \( \tan^2(18^\circ)\, \tan^2(54^\circ) \)
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Solution 1 (this is mine and I must admit I used Maxima to handle the algebra):
First observe that from the quadrant corresponding to te angles here:
\[\tan(A/2)=\sqrt{\frac{1-\cos(A)}{1+\cos(A)}}\]
and:
\[ \tan(3A/2)=\frac{3\tan(A/2)-\tan^3(A/2)}{1-3\tan^2(A/2)} \]
So if we put \( A=36^\circ \) and allowing Maxima to do the algebra we get:
\[ \tan^2(18^\circ)\, \tan^2(54^\circ)=\frac{1}{5} \]
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Solution 2 (this is the solution give on the originating site):
\[ \cos(72^\circ)=2\cos^2(36^\circ)-1=\frac{1}{4}(\sqrt{5}-1) \]
and:
\(\displaystyle \phantom{xxxx} \tan^2(18^\circ)\, \tan^2(54^\circ) =\frac{\sin^2(54^\circ)\sin^2(18^\circ)}{\cos^2(54^\circ)\cos^2(18^\circ)} =\left[ \frac{(1/2)(\cos(36^\circ)-\cos(72^\circ))}{(1/2)(\cos(36^\circ)+\cos(72^\circ))} \right]^2\)
\(\displaystyle \phantom{xxxx} \phantom{\tan^2(18^\circ)\, \tan^2(54^\circ)}=\left[ \frac{\frac{\sqrt{5}+1}{4} -\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{5}+1}{4}+\frac{\sqrt{5}-1}{4}} \right] =\frac{1}{5}\)
Monday, 19 March 2012
Draft: CaptainBlack's Problem of the Week #1 for MHB
This is the draft for the problem posted on MathHelpBoards
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Let \(x_1, ..., x_n\) be real numbers such that:
\(\sum_{i=1}^n x_i=0\) and \(\sum_{i=1}^n x_i^2=1\)
Prove that for some \( k,l \) both in \( \{1, .. , n\} \) that \(x_k x_l\le -1/n\)
CB
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Solution
Let \(a=\max(x_1, ... , x_n) \) and \( b=\min(x_1, ... , x_n) \), and put:
\[ f(x)=(x-a)(x-b)=x^2-(a+b)\; x+ab \]
Then by construction \( f(x_i)\le 0,\;i=1, ... , n \) . Now consider:
\[ \sum_{i=1}^n f(x_i)=\sum_{i=1}^n x^2 -(a+b) \sum_{i=1}^n x_i +n\;ab = 1+n\;ab \le 0\]
hence:
\[ab \le -\frac{1}{n} \]
QED
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Let \(x_1, ..., x_n\) be real numbers such that:
\(\sum_{i=1}^n x_i=0\) and \(\sum_{i=1}^n x_i^2=1\)
Prove that for some \( k,l \) both in \( \{1, .. , n\} \) that \(x_k x_l\le -1/n\)
CB
===========================================================
Solution
Let \(a=\max(x_1, ... , x_n) \) and \( b=\min(x_1, ... , x_n) \), and put:
\[ f(x)=(x-a)(x-b)=x^2-(a+b)\; x+ab \]
Then by construction \( f(x_i)\le 0,\;i=1, ... , n \) . Now consider:
\[ \sum_{i=1}^n f(x_i)=\sum_{i=1}^n x^2 -(a+b) \sum_{i=1}^n x_i +n\;ab = 1+n\;ab \le 0\]
hence:
\[ab \le -\frac{1}{n} \]
QED
Sunday, 18 March 2012
Math Help Boards Progress
After MathHelpBoards (www.mathhelpboards.com ) was established the former site came back online under new ownership. The new owner moved it to a new server and reorganised it to his liking. Discussions over a potential merger between the new owner and leading members MHB did not lead anywhere so both sites continue in existence. The new owner has in the past made zero contribution to the work of the site, and has little idea how it used to be run.
The former site has the advantage of a higher position in any Google search for maths help, but retains fewer of the former regulars. MHBs has the advantage of a large number of experienced helpers. As a result the former site has a larger number of new posts per day, but also a large number of unanswered or slowly answered posts. MHB has a smaller number of new posts per day but few go unanswered for very long and is better moderated.
If I were in the business of making predictions I would expect both sites to remain in business for some time with the same relative traffic levels, but the former site will probably fail in the longer term when the new owner loses interest.
The former site has the advantage of a higher position in any Google search for maths help, but retains fewer of the former regulars. MHBs has the advantage of a large number of experienced helpers. As a result the former site has a larger number of new posts per day, but also a large number of unanswered or slowly answered posts. MHB has a smaller number of new posts per day but few go unanswered for very long and is better moderated.
If I were in the business of making predictions I would expect both sites to remain in business for some time with the same relative traffic levels, but the former site will probably fail in the longer term when the new owner loses interest.
Thursday, 26 January 2012
Replacement for Math Help Forum now online
Math Help Boards a replacement for the late Math Help Forum is in the process of commissioning at:
http://www.mathhelpboards.com/forum.php
http://www.mathhelpboards.com/forum.php
Monday, 23 January 2012
© Nur Al Hayat's question for Yahoo Answers
7(a) (i) Solve the differential equation: \[ \frac{dx}{dt}=\sqrt{x}\sin(t/2) \] to find \(x \) in terms of \(t\).
Well we may observe that this DE is of variables separable type so we may write: \[\int \frac{1}{\sqrt{x}} \; dx = \int \sin(t/2)\; dt \]
We may now integrate to get: \[ \sqrt{x}= -\cos(t/2)+C \] and squaring we get \[x=(C-\cos(t/2))^2\]
Now the remaining parts of this question can be solved by applying the relevant initial conditions etc.
Well we may observe that this DE is of variables separable type so we may write: \[\int \frac{1}{\sqrt{x}} \; dx = \int \sin(t/2)\; dt \]
We may now integrate to get: \[ \sqrt{x}= -\cos(t/2)+C \] and squaring we get \[x=(C-\cos(t/2))^2\]
Now the remaining parts of this question can be solved by applying the relevant initial conditions etc.
Friday, 20 January 2012
Sample Question (pulled off Yahoo Answers): Linear Algebra Question: Finding the Dimension?
Sample Question (pulled off Yahoo Answers): Linear Algebra Question: Finding the Dimension? (Original question posted as comment on the Questions Page)
If S is a subspace in R^4 and S = {(a,b,c,d) in R^4 , a+c = b , a+c+d=0} what is the dimension of S? I have the answers and it says that S is generated by the vectors (1,0,-1,0) and (1,1,0,-1) and that the dimension of S is 2. How do we find those 2 vectors that generate S? Please explain!!
I trust that it is obvious that \( S \) is of dimension 2? This is because we can assign values to \(a\) and \(c\) essentially arbitarily and construct a unique element of \(S\).
In fact we can construct the mapping:
\[ s=(a,b,c,d)=(u,v)A \]
where: \( A= \left[ \begin{array}{cccc}1&1&0&-1\\0&1&1&-1 \end{array} \right] \), which gives us a 1-1 linear mapping from \( \mathbb{R}^2 \) to \(S\).
Hence the image of any basis of \( \mathbb{R}^2 \) is a basis of \(S\) so in particular is:
\( \{ (1,0)A, (0,1)A \}=\{ (1,1,0,-1),(0,1,1,-1) \} \).
One of the elements in the basis found above is in the basis you quote and the other is not. This is because the basis is not unique, any pair of linearly independent vectors in \( \mathbb{R}^2 \) will produce a basis for \(S\). The other element of your quoted basis is \( (1,-1)A=(1,0,-1,0) \).
(If you are more used to working with column vectors just take the transpose of everything above)
If S is a subspace in R^4 and S = {(a,b,c,d) in R^4 , a+c = b , a+c+d=0} what is the dimension of S? I have the answers and it says that S is generated by the vectors (1,0,-1,0) and (1,1,0,-1) and that the dimension of S is 2. How do we find those 2 vectors that generate S? Please explain!!
I trust that it is obvious that \( S \) is of dimension 2? This is because we can assign values to \(a\) and \(c\) essentially arbitarily and construct a unique element of \(S\).
In fact we can construct the mapping:
\[ s=(a,b,c,d)=(u,v)A \]
where: \( A= \left[ \begin{array}{cccc}1&1&0&-1\\0&1&1&-1 \end{array} \right] \), which gives us a 1-1 linear mapping from \( \mathbb{R}^2 \) to \(S\).
Hence the image of any basis of \( \mathbb{R}^2 \) is a basis of \(S\) so in particular is:
\( \{ (1,0)A, (0,1)A \}=\{ (1,1,0,-1),(0,1,1,-1) \} \).
One of the elements in the basis found above is in the basis you quote and the other is not. This is because the basis is not unique, any pair of linearly independent vectors in \( \mathbb{R}^2 \) will produce a basis for \(S\). The other element of your quoted basis is \( (1,-1)A=(1,0,-1,0) \).
(If you are more used to working with column vectors just take the transpose of everything above)
Thursday, 19 January 2012
First Post
This is the opening post for this blog. The purpose of which is to provide free help for maths homework. It is not here to help you cheat on assessed work, if something will constitute part of your course assessment don't post it here.
Post your question in a comment on the top post on the Questions Needing Answers Page
This blog is not here to help with significant work related problems, for that you need to employ a paid for consultant (with appropriate insurance cover?). A question that can be dealt with in a few minutes or if it interests me sufficiently may be answered..
LaTeX test:
[Tex] \int_0^{\xi} \frac{1}{\sin(x)}\; dx[/Tex]
well that's not working! For some reason it needs the page reloading to display properly on this machine/browser! Will need to check this on another machine/browser. Seems to work under Chrome :) but not under Opera :( or IE8 :((
\[ \int_0^{\xi} \frac{1}{\sin(x)}\; dx \]
\( \int_0^{\xi} \frac{1}{\sin(x)}\; dx \)
OK seems to work with the MathJax default delimiters!
Post your question in a comment on the top post on the Questions Needing Answers Page
This blog is not here to help with significant work related problems, for that you need to employ a paid for consultant (with appropriate insurance cover?). A question that can be dealt with in a few minutes or if it interests me sufficiently may be answered..
LaTeX test:
[Tex] \int_0^{\xi} \frac{1}{\sin(x)}\; dx[/Tex]
well that's not working! For some reason it needs the page reloading to display properly on this machine/browser! Will need to check this on another machine/browser. Seems to work under Chrome :) but not under Opera :( or IE8 :((
\[ \int_0^{\xi} \frac{1}{\sin(x)}\; dx \]
\( \int_0^{\xi} \frac{1}{\sin(x)}\; dx \)
OK seems to work with the MathJax default delimiters!