Official TSR Mathematical Society

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  1. ambience's Avatar
    • Respected Member
    • Posts: 163
    Re: Official TSR Mathematical Society
    \displaystyle\int_0^1 (-\mathrm{log}\ x)^n\ \mathrm{dx} =n!
    How can we prove this?
  2. Intriguing Alias's Avatar
    • TSR Idol
    • Location: Yorkshire
    Re: Official TSR Mathematical Society
    (Original post by ambience)
    \displaystyle\int_0^1 (-\mathrm{log}\ x)^n\ \mathrm{dx} =n!
    How can we prove this?
    Induction, probably and then using IBP? Not sure though - I'd need to actually try it



    EDIT: Yep that comes out pretty easily using induction and IBP
    Last edited by Intriguing Alias; 14-06-2012 at 00:41.
  3. Maths boy's Avatar
    • Junior Member
    • Posts: 32
    Re: Official TSR Mathematical Society
    Can someone help me with this maths problem or know anyone that can. thank you :-)
    In steady state heat transfer the temperature T satisfies the Laplace equation. In two dimensions with Cartesian coordinates x and y this equation is d2T/dx2 +d2T/dY2=0
    This is a partial differential equation, and general methods of solving it are beyond the scope of this module - wait for the second and final year! However there are ways of getting an approximate solution to the equation by realising that it is a mathematical statement of the law of conservation of energy - in this case heat energy. One such method is the finite difference method.

    One way to see how such methods work, is to consider a square area ABCD of a heat-conducting plate:
    Click image for larger version. 

Name:	M1 square.png 
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    Figure 1
    The values T0, T1, T2, T3, and T4 represent respectively the temperatures at the central point (or node) and the four surrounding equally distant nodes. T0 is at the centre of the sqaure, T1 goes through the middle of the points AD and conects to T0 on the sqaure. T3 goes through the middle ofthe points CB and connects to T0. T2 goes through the middle of the points DC and connects to T0 and T4 goes through the middle of the points AB and connects to T0. For the sake of simplicity suppose the length of side of the square is h, and that each of the temperature nodes are a distance h apart.

    The law of conservation of energy in this case says that the net heat flowing into ABCD must equal the net heat flowing out. The heat flowing across a boundary of ABCD is proportional to the temperature gradient and the length of that boundary. Thus conservation of energy says
    Heat in + Heat in = Heat out + Heat out
    across AC across CD across AB across BD

    Approximately this gives
    (T0-T1/h)h +(T0-T4/h)h = (T2-T0/h)h+(T3-T0/h)h
    or rearranging T0=1/4(T1+T2+T3+T4) .
    ie temperature at one node = average of temperatures at four surrounding nodes.
    Why is this approximate, and how can it be made more accurate? Write some words on this, and to see if you’ve really understood, do the following:
    Work out the equation when ABCD is a rectangle but not necessarily a square, with the lengths
    AB = CD = distance from T1 to T0 = distance from T0 to T3 = h
    and the lengths
    AC = BD = distance from T4 to T0 = distance from T0 to T2 = k

    A simple case
    A flat plate as shown in Figure 2 has the temperature on three edges held at T = 0; the left hand side and the top and bottom the fourth edge is held at a temperature of T = 100, which is on the right hand side. T1 and T2 are on the central nodes of the grid.
    Figure 2
    Click image for larger version. 

Name:	M1 rectangle for part b.jpg 
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ID:	156948

    An approximate value of the temperature at certain points can be obtained as follows:
    • Place a regular square grid over the plate as shown.
    • Name the temperatures at the two central nodes of the grid T1 and T2.
    • Then according to the finite difference model above, the values of T1 and T2, which are approximate values of the temperature at those points, can be found by noting that the temperature at each node is the average of the temperatures at the four surrounding nodes.

    This results in two simultaneous equations which can be solved for T1 and T2. The equations are:
    T1=(0+0+0+T2)/4and T2=(0+0+100+T1)/4

    Solve these equations in two ways. Verify your answers.
  4. SimonM's Avatar
    • TSR Idol
    • Posts: 9,192
    Re: Official TSR Mathematical Society
    (Original post by ambience)
    \displaystyle\int_0^1 (-\mathrm{log}\ x)^n\ \mathrm{dx} =n!
    How can we prove this?
    u = -log x

    Get \displaystyle \int_0^{\infty} u^n e^{-u} \, du = \Gamma(n+1)
  5. boromir9111's Avatar
    • TSR Legend
    • Location: Here and There
    • Posts: 10,801
    Re: Official TSR Mathematical Society
    (Original post by Maths boy)
    Can someone help me with this maths problem or know anyone that can. thank you :-)
    In steady state heat transfer the temperature T satisfies the Laplace equation. In two dimensions with Cartesian coordinates x and y this equation is d2T/dx2 +d2T/dY2=0
    This is a partial differential equation, and general methods of solving it are beyond the scope of this module - wait for the second and final year! However there are ways of getting an approximate solution to the equation by realising that it is a mathematical statement of the law of conservation of energy - in this case heat energy. One such method is the finite difference method.

    One way to see how such methods work, is to consider a square area ABCD of a heat-conducting plate:
    Click image for larger version. 

Name:	M1 square.png 
Views:	45 
Size:	2.5 KB 
ID:	156947
    Figure 1
    The values T0, T1, T2, T3, and T4 represent respectively the temperatures at the central point (or node) and the four surrounding equally distant nodes. T0 is at the centre of the sqaure, T1 goes through the middle of the points AD and conects to T0 on the sqaure. T3 goes through the middle ofthe points CB and connects to T0. T2 goes through the middle of the points DC and connects to T0 and T4 goes through the middle of the points AB and connects to T0. For the sake of simplicity suppose the length of side of the square is h, and that each of the temperature nodes are a distance h apart.

    The law of conservation of energy in this case says that the net heat flowing into ABCD must equal the net heat flowing out. The heat flowing across a boundary of ABCD is proportional to the temperature gradient and the length of that boundary. Thus conservation of energy says
    Heat in + Heat in = Heat out + Heat out
    across AC across CD across AB across BD

    Approximately this gives
    (T0-T1/h)h +(T0-T4/h)h = (T2-T0/h)h+(T3-T0/h)h
    or rearranging T0=1/4(T1+T2+T3+T4) .
    ie temperature at one node = average of temperatures at four surrounding nodes.
    Why is this approximate, and how can it be made more accurate? Write some words on this, and to see if you’ve really understood, do the following:
    Work out the equation when ABCD is a rectangle but not necessarily a square, with the lengths
    AB = CD = distance from T1 to T0 = distance from T0 to T3 = h
    and the lengths
    AC = BD = distance from T4 to T0 = distance from T0 to T2 = k

    A simple case
    A flat plate as shown in Figure 2 has the temperature on three edges held at T = 0; the left hand side and the top and bottom the fourth edge is held at a temperature of T = 100, which is on the right hand side. T1 and T2 are on the central nodes of the grid.
    Figure 2
    Click image for larger version. 

Name:	M1 rectangle for part b.jpg 
Views:	44 
Size:	16.3 KB 
ID:	156948

    An approximate value of the temperature at certain points can be obtained as follows:
    • Place a regular square grid over the plate as shown.
    • Name the temperatures at the two central nodes of the grid T1 and T2.
    • Then according to the finite difference model above, the values of T1 and T2, which are approximate values of the temperature at those points, can be found by noting that the temperature at each node is the average of the temperatures at the four surrounding nodes.

    This results in two simultaneous equations which can be solved for T1 and T2. The equations are:
    T1=(0+0+0+T2)/4and T2=(0+0+100+T1)/4

    Solve these equations in two ways. Verify your answers.
    Create a thread and ask there, not here.
  6. TheMagicMan's Avatar
    • Overlord in Training
    • Posts: 2,940
    Re: Official TSR Mathematical Society
    (Original post by ambience)
    \displaystyle\int_0^1 (-\mathrm{log}\ x)^n\ \mathrm{dx} =n!
    How can we prove this?
    n!=\gamma (n+1)

    =\displaystyle\int^{\infty}_0 e^{-t}t^n dt

    The substitution t=-lnx instantly renders the result you want.
  7. TickTackToe's Avatar
    • Adored and Respected Member
    • Posts: 501
    Re: Official TSR Mathematical Society
    Can someone explain to me how to calculate the ratios in this question, it shows the answer in the bottom table , not not actually how they arrived at the numbers, which is very frustrating.

    http://imageshack.us/f/688/screenshot20120614at175.png/
  8. Lord of the Flies's Avatar
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: Official TSR Mathematical Society
    (Original post by TickTackToe)
    Can someone explain to me how to calculate the ratios in this question, it shows the answer in the bottom table , not not actually how they arrived at the numbers, which is very frustrating.

    http://imageshack.us/f/688/screenshot20120614at175.png/
    A ratio is a relationship between two numbers: how much bigger/smaller one is compared to the other. So the ratio of average sales price to the average production cost is simply:

    \dfrac{\text{average sales price}}{\text{average production cost}}:1

    Example: for 2010 the fraction will give you 4.3 (to one decimal place), which means the average sales price was 4.3 times more than the average production cost.
  9. GreenLantern1's Avatar
    • Banned
    • Posts: 3,316
    • Warning points: 1000
    Re: Official TSR Mathematical Society
    (Original post by TheMagicMan)
    n!=\gamma (n+1)

    =\displaystyle\int^{\infty}_0 e^{-t}t^n dt

    The substitution t=-lnx instantly renders the result you want.
    Gut gemacht!
  10. nohomo's Avatar
    • Benevolent Member
    • Posts: 662
    Re: Official TSR Mathematical Society
    Does anyone have any new problems for people to try?
  11. MrBlueMo0n's Avatar
    • Adored and Respected Member
    • Location: London/York
    • Posts: 438
    Re: Official TSR Mathematical Society
    I've done my AS exams, and I'll be continuing A2 in lessons. I shall then be doing AS Further over the summer and up to Christmas, on my own. If I manage this, is it possible for me to carry on the full A2 Further? I.e. 3 more modules (independently) between Christmas and summer.
  12. boromir9111's Avatar
    • TSR Legend
    • Location: Here and There
    • Posts: 10,801
    Re: Official TSR Mathematical Society
    (Original post by MrBlueMo0n)
    I've done my AS exams, and I'll be continuing A2 in lessons. I shall then be doing AS Further over the summer and up to Christmas, on my own. If I manage this, is it possible for me to carry on the full A2 Further? I.e. 3 more modules (independently) between Christmas and summer.
    You should really be creating your own thread for this and not asking on here!
  13. boromir9111's Avatar
    • TSR Legend
    • Location: Here and There
    • Posts: 10,801
    Re: Official TSR Mathematical Society
    (Original post by nohomo)
    Does anyone have any new problems for people to try?
    Why not. It's a "free" world

     \displaystyle\int_0^1 \left( \dfrac{\ln x}{1-x} \right)^m \ \mathrm{d}x

    where, m>1
  14. mashmammad's Avatar
    • Respected Member
    • Location: LONDON
    • Posts: 212
    (Original post by nohomo)
    Does anyone have any new problems for people to try?
    I do. But I'm afraid It's a bit so easy.

    This was posted from The Student Room's Android App on my HTC Wildfire
  15. TheMagicMan's Avatar
    • Overlord in Training
    • Posts: 2,940
    (Original post by boromir9111)
    Why not. It's a "free" world

     \displaystyle\int_0^1 \left( \dfrac{\ln x}{1-x} \right)^m \ \mathrm{d}x

    where, m>1
    Something along the lines of a x=u+1 sub and power series?

    Ibp looks like a distinct possibility as well

    This was posted from The Student Room's iPhone/iPad App
    Last edited by TheMagicMan; 12-07-2012 at 23:36.
  16. MrBlueMo0n's Avatar
    • Adored and Respected Member
    • Location: London/York
    • Posts: 438
    Re: Official TSR Mathematical Society
    (Original post by boromir9111)
    You should really be creating your own thread for this and not asking on here!
    I'm not joking, I don't know how to create my own thread!

    Plus I thought it was fairly relevant, but that doesn't matter.
  17. pbsjohnz's Avatar
    • Benevolent Member
    • Posts: 651
    Re: Official TSR Mathematical Society
    Hi, I have a question. If I do engineering at undergraduate level and get a BEng degree, can I do a Masters in Maths (MSc) at a UK University? Any help would be hugely appreciated as I am not from the UK.
  18. nohomo's Avatar
    • Benevolent Member
    • Posts: 662
    Re: Official TSR Mathematical Society
    (Original post by boromir9111)
    Why not. It's a "free" world

     \displaystyle\int_0^1 \left( \dfrac{\ln x}{1-x} \right)^m \ \mathrm{d}x

    where, m>1
    Any hints?
  19. boromir9111's Avatar
    • TSR Legend
    • Location: Here and There
    • Posts: 10,801
    Re: Official TSR Mathematical Society
    (Original post by nohomo)
    Any hints?
    Consider that \dfrac{1}{(1-x)^m} = \displaystyle\sum_{n=0}^{\infty} \displaystyle\binom{m+n-1}{n} x^n

    And \displaystyle\int_0^1 (\log x)^m x^{n-1} \mathrm{d}x = \displaystyle\int_0^{\infty} x^m e^{-nx} \mathrm{d}x = \dfrac{1}{n^{m+1}} \displaystyle\int_0^{\infty} x^m e^{-x} \mathrm{d}x = \dfrac{\Gamma (m+1)}{n^{m+1}}

    where \Gamma (x) is the Gamma function [google it].

    I have essentially given the problem away, but oh well.
  20. Astronomical's Avatar
    • Overlord in Training
    • Location: England
    • Posts: 2,144
    Re: Official TSR Mathematical Society
    (Original post by boromir9111)
    Why not. It's a "free" world

     \displaystyle\int_0^1 \left( \dfrac{\ln x}{1-x} \right)^m \ \mathrm{d}x

    where, m>1
    That looks tough. :lolwut:
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