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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/
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:

average sales priceaverage production cost:1\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.
Original post by TheMagicMan
n!=γ(n+1)n!=\gamma (n+1)

=∫0∞e−ttndt=\displaystyle\int^{\infty}_0 e^{-t}t^n dt

The substitution t=−lnxt=-lnx instantly renders the result you want.


Gut gemacht!
Reply 3743
Avatar for nohomo
nohomo
5
Does anyone have any new problems for people to try?
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.
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!
Original post by nohomo
Does anyone have any new problems for people to try?


Why not. It's a "free" world

∫01(ln⁡x1−x)m dx \displaystyle\int_0^1 \left( \dfrac{\ln x}{1-x} \right)^m \ \mathrm{d}x

where, m>1
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
Original post by boromir9111
Why not. It's a "free" world

∫01(ln⁡x1−x)m dx \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
(edited 11 years ago)
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.
Reply 3750
Avatar for thiccc
thiccc
17
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.
Reply 3751
Avatar for nohomo
nohomo
5
Original post by boromir9111
Why not. It's a "free" world

∫01(ln⁡x1−x)m dx \displaystyle\int_0^1 \left( \dfrac{\ln x}{1-x} \right)^m \ \mathrm{d}x

where, m>1


Any hints?
Original post by nohomo
Any hints?


Consider that 1(1−x)m=∑n=0∞(m+n−1n)xn\dfrac{1}{(1-x)^m} = \displaystyle\sum_{n=0}^{\infty} \displaystyle\binom{m+n-1}{n} x^n

And∫01(log⁡x)mxn−1dx=∫0∞xme−nxdx=1nm+1∫0∞xme−xdx=Γ(m+1)nm+1 \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}}

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

I have essentially given the problem away, but oh well.
Original post by boromir9111
Why not. It's a "free" world

∫01(ln⁡x1−x)m dx \displaystyle\int_0^1 \left( \dfrac{\ln x}{1-x} \right)^m \ \mathrm{d}x

where, m>1


That looks tough. :lolwut:
Original post by boromir9111
Consider that 1(1−x)m=∑n=0∞(m+n−1n)xn\dfrac{1}{(1-x)^m} = \displaystyle\sum_{n=0}^{\infty} \displaystyle\binom{m+n-1}{n} x^n

And∫01(log⁡x)mxn−1dx=∫0∞xme−nxdx=1nm+1∫0∞xme−xdx=Γ(m+1)nm+1 \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}}

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

I have essentially given the problem away, but oh well.


Now it looks scary. :lolwut:

I don't even get it; the integrand isn't defined at x=0 (what is ln⁡0\ln 0?), so how is the integral valid?
(edited 11 years ago)
hello everyone, I hope your all in good health and doing well...

Please could I have help with getting full solutions for the following 2 problems, which I am stuck on thanks.

Simplify following expressions
1) root 5 minus 1 divided by root 5 minus 1 divided by root 125.

2) root 8 times root 10 times root 12 And then divide all that by root 2 times root 20 times root 24

Appreciate the help and will definitely rep.
Reply 3756
Avatar for nohomo
nohomo
5
Original post by boromir9111
Consider that 1(1−x)m=∑n=0∞(m+n−1n)xn\dfrac{1}{(1-x)^m} = \displaystyle\sum_{n=0}^{\infty} \displaystyle\binom{m+n-1}{n} x^n

And∫01(log⁡x)mxn−1dx=∫0∞xme−nxdx=1nm+1∫0∞xme−xdx=Γ(m+1)nm+1 \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}}

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

I have essentially given the problem away, but oh well.


Nice problem.

Here's another:

For which real numbers aa does the sequence defined by the initial condition u0=au_0 = a and the recursion un+1=2un−n2u_{n+1} = 2u_n-n^2 have un>0u_n > 0 for all n≥0n \ge 0?
Original post by boromir9111
Consider that 1(1−x)m=∑n=0∞(m+n−1n)xn\dfrac{1}{(1-x)^m} = \displaystyle\sum_{n=0}^{\infty} \displaystyle\binom{m+n-1}{n} x^n

And∫01(log⁡x)mxn−1dx=∫0∞xme−nxdx=1nm+1∫0∞xme−xdx=Γ(m+1)nm+1 \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}}

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

I have essentially given the problem away, but oh well.


Very nice :clap2:
hello everyone, I hope your all in good health and doing well...

Please could I have help with getting full solutions for the following 2 problems, which I am stuck on thanks.

Simplify following expressions
1) root 5 minus 1 divided by root 5 minus 1 divided by root 125.

2) root 8 times root 10 times root 12 And then divide all that by root 2 times root 20 times root 24

Appreciate the help and will definitely rep.
Original post by nohomo

For which real numbers aa does the sequence defined by the initial condition u0=au_0 = a and the recursion un+1=2un−n2u_{n+1} = 2u_n-n^2 have un>0u_n > 0 for all n≥0n \ge 0?


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