Advanced Higher Maths 2011-2012 : Discussion and Help

Original post by Quintro

Algebra >>> Ok, but never understood those binomial coefficient proofs

What proofs are those?

Reply 902

Original post by ukdragon37

What proofs are those?

This monstrosity:

\displaystyle \binom{n}{r-1} + \binom{n}{r}= \binom{n+1}{r}

I believe there was a question on it in last years paper though, meaning we'll likely avoid it this year.

Blows my mind it does :tongue:

(edited 12 years ago)

Reply 903

Slumpy

Original post by Quintro

This monstrosity:

\displaystyle \binom{n}{r-1} + \binom{n}{r}= \binom{n+1}{r}

I believe there was a question on it in last years paper though, meaning we'll likely avoid it this year.

Blows my mind it does :tongue:

The proof's lovely though! Especially if you think of it set theoretically :p:

Reply 904

Original post by Quintro

This monstrosity:

\displaystyle \binom{n}{r-1} + \binom{n}{r}= \binom{n+1}{r}

I believe there was a question on it in last years paper though, meaning we'll likely avoid it this year.

Blows my mind it does :tongue:

This particular proof avoids much of the factorial work if you just expand both binomial coefficients and added them together:

From definition expand

\displaystyle\binom{n}{r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}

and

\displaystyle\binom{n}{r - 1} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}

Comparing the difference in factors of the numerators and denominators of the two lets us write the relation:

\displaystyle\binom{n}{r - 1} = \displaystyle\binom{n}{r}\dfrac{r}{{n - r + 1}}

The factor between the two binomial coefficients becomes obvious if you divide nC(r-1) by nCr.

Hence we write the sum in nCr only using our relation above and take out nCr as a common factor:

Unparseable latex formula:

\begin{array}{l}[br]\displaystyle\binom{n}{r - 1} + \displaystyle\binom{n}{r} = \displaystyle\binom{n}{r}\dfrac{r}{{n - r + 1}} + \displaystyle\binom{n}{r}\\\\[br] = \displaystyle\binom{n}{r}\left( {1 + \dfrac{r}{{n - r + 1}}} \right)\\\\[br] = \displaystyle\binom{n}{r}\dfrac{{n + 1}}{{n - r + 1}}\\\\[br] = \dfrac{{\left( {n + 1} \right)!}}{{r!\left[ {\left( {n + 1} \right) - r} \right]!}}\\\\[br] = \displaystyle\binom{{n + 1}}{r} \\\\[br]\end{array}

(edited 12 years ago)

Reply 905

Slumpy

Original post by ukdragon37

This particular proof avoids much of the factorial work if you just expand both binomial coefficients and added them together:

From definition expand

\displaystyle\binom{n}{r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}

and

\displaystyle\binom{n}{r - 1} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}

Comparing the difference in factors of the numerators and denominators of the two lets us write the relation:

\displaystyle\binom{n}{r - 1} = \displaystyle\binom{n}{r}\dfrac{r}{{n - r + 1}}

The factor between the two binomial coefficients becomes obvious if you divide nC(r-1) by nCr.

Hence we write the sum in nCr only using our relation above and take out nCr as a common factor:

Unparseable latex formula:

\begin{array}{l}[br]\displaystyle\binom{n}{r - 1} + \displaystyle\binom{n}{r} = \displaystyle\binom{n}{r}\dfrac{r}{{n - r + 1}} + \displaystyle\binom{n}{r}\\\\[br] = \displaystyle\binom{n}{r}\left( {1 + \dfrac{r}{{n - r + 1}}} \right)\\\\[br] = \displaystyle\binom{n}{r}\dfrac{{n + 1}}{{n - r + 1}}\\\\[br] = \dfrac{{\left( {n + 1} \right)!}}{{r!\left[ {\left( {n + 1} \right) - r} \right]!}}\\\\[br] = \displaystyle\binom{{n + 1}}{r} \\\\[br]\end{array}

Nah, mate.
Ways to pick r objects from n+1=number of ways to pick r objects from n+number of ways to pick r-1 objects from n(and the last is the n+1th object). Done! Much nicer :p:

Reply 906

Original post by Slumpy

Nah, mate.
Ways to pick r objects from n+1=number of ways to pick r objects from n+number of ways to pick r-1 objects from n(and the last is the n+1th object). Done! Much nicer :p:

Even nicer is to say (on the assumption that nCr is nth row and rth column of Pascal's triangle), this identity represents the way each number in Pascal's triangle is constructed :tongue:

Reply 907

Slumpy

Original post by ukdragon37

Even nicer is to say (on the assumption that nCr is nth row and rth column of Pascal's triangle), this identity represents the way each number in Pascal's triangle is constructed :tongue:

I am a big fan of that way, although I think you might struggle to sneak that assumption past people without any justification :p:

Edit-to my mind, the most obvious proof of that result comes back to what I said above also. But I may be missing something.

(edited 12 years ago)

Reply 908

laughylolly

Did 2002 past paper today. Realized I really need to revise second order differential equations, and maybe transformation matrices but other than that it was alright.

Reply 909

Original post by ukdragon37

...

It's still unbelievably difficult in comparison to everything else in the AH course. :frown:

I think I would be better focussing on the stuff that I actually have a chance of understanding, because that proof is way beyond my level.

I only found out today that logarithmic differentiation can be used to differentiate products and quotients of more than two terms easier. I always thought it could only be used when you had a variable to the power of a variable. :colondollar:

Reply 910

JordanR

12

Original post by Quintro

It's still unbelievably difficult in comparison to everything else in the AH course. :frown:

I think I would be better focussing on the stuff that I actually have a chance of understanding, because that proof is way beyond my level.

I only found out today that logarithmic differentiation can be used to differentiate products and quotients of more than two terms easier. I always thought it could only be used when you had a variable to the power of a variable. :colondollar:

Oh god, logarithmic differentiation is your best friend. I think it was last year's paper where there was a question that was gonna be crazy without it. I'd use it as often as possible, really.

(edited 12 years ago)

Reply 911

Original post by JordanR

Oh god, logarithmic differentiation is your best friend. I think it was last year's paper where there was a question that was gonna be crazy without it. I'd use it as often as possible, really.

I definitely will know to use it now. I was under the impression where it was something that you were restricted to doing under a very specific scenario.

And I think I know the question you mean.

Reply 912

JordanR

12

Original post by Quintro

I definitely will know to use it now. I was under the impression where it was something that you were restricted to doing under a very specific scenario.

And I think I know the question you mean.

Just anywhere you want to make life easier for yourself, I'm sure. Anything involving an exponential is probably a good place to use it.

Reply 913

TheUnbeliever

Original post by Quintro

It's still unbelievably difficult in comparison to everything else in the AH course. :frown:

I think I would be better focussing on the stuff that I actually have a chance of understanding, because that proof is way beyond my level.

Whilst ukd's proof requires less work, it's entirely possible to use the complete factorial expansion of the left hand side and bash away at fairly-straightforward but lengthy algebra until the identity falls out.

Reply 914

Original post by TheUnbeliever

Whilst ukd's proof requires less work, it's entirely possible to use the complete factorial expansion of the left hand side and bash away at fairly-straightforward but lengthy algebra until the identity falls out.

Yeah my friend sent me two insanely detailed methods of doing it, so I think I'm getting the hang of it.

Started proofs by induction today and it's actually surprisingly enjoyable. So far at least (we've only done the summation proofs).

Reply 915

laughylolly

Original post by JordanR

Oh god, logarithmic differentiation is your best friend. I think it was last year's paper where there was a question that was gonna be crazy without it. I'd use it as often as possible, really.

Logarithmic differentiation is amazingg. Makes things so easy! Still like the quotient rule and product rule though. I just use the logarithmic if it tells me to really.

Starting to storm through papers now, nearly done with 2003.

Oh by the way, anyone wanna have a look at this question for me? It's Q15 of the 2002 paper. Now I found it quite complicated and stuff but I think I over complicated it so would someone mind doing it or just writing out how they would tackle the question?

Find the general solution of the differential equation

\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+5y=4\: cosx

Hence determine the solution which satisfies y(0)=0 and y'(0)=1

Worth 10 marks in total

Reply 916

Original post by laughylolly

Logarithmic differentiation is amazingg. Makes things so easy! Still like the quotient rule and product rule though. I just use the logarithmic if it tells me to really.

Starting to storm through papers now, nearly done with 2003.

Oh by the way, anyone wanna have a look at this question for me? It's Q15 of the 2002 paper. Now I found it quite complicated and stuff but I think I over complicated it so would someone mind doing it or just writing out how they would tackle the question?

Find the general solution of the differential equation

\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+5y=4\: cosx

Hence determine the solution which satisfies y(0)=0 and y'(0)=1

Worth 10 marks in total

Set up Auxiliary equation to find the roots.

Doesn't factorise so use quadratic formula.

Roots are complex (

p \pm qi

) so use complimentary function

y=e^{px}(Acosqx + Bsinqx)

Substitute in values of p and q to form the complimentary function.

Set up particular integral as Rcosx+Ssinx

Differentiate PI twice to get y' and y''

Sub into original DE

Compare coefficients of sinx and cosx

Use simultaneous equations or otherwise to find values of R and S.

Solution is y=complimentary function + particular integral

Then for the final part substitute in x=0 y=0 and then y'=1 and x=0 to find values for A and B.

Plug those values into your solution to make it particular.

Hope I didn't screw up :redface:

(edited 12 years ago)

Reply 917

JordanR

12

Original post by laughylolly

Logarithmic differentiation is amazingg. Makes things so easy! Still like the quotient rule and product rule though. I just use the logarithmic if it tells me to really.

Starting to storm through papers now, nearly done with 2003.

Oh by the way, anyone wanna have a look at this question for me? It's Q15 of the 2002 paper. Now I found it quite complicated and stuff but I think I over complicated it so would someone mind doing it or just writing out how they would tackle the question?

Find the general solution of the differential equation

\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+5y=4\: cosx

Hence determine the solution which satisfies y(0)=0 and y'(0)=1

Worth 10 marks in total

r^2 + 2r + 5 = 0

Solutions are -1 + 2i and -1-2i

So y = e^-x (Asin2x + Bcos2x) + PI

Particular integral has the form y = C1cosx + C2sinx.

dy/dx = -C1sinx + C2cosx

d^2y/dx^2 = -C1cosx - C2sinx

Plug these into the original equation:

(-C1cosx - C2sinx) + 2(-C1sinx + C2cosx) + 5(C1cosx + C2sinx).

Solve to find C1 and C2 to get your full solution.

Spoiler

Then to find the values of the constants A and B by subbing in x = 0.

Sorry for not really being in depth, got stuff to do. :tongue:

Reply 918

laughylolly

Original post by Quintro

Set up Auxiliary equation to find the roots.

Doesn't factorise so use quadratic formula.

Roots are complex (

p \pm qi

) so use complimentary function

y=e^{px}(Acosqx + Bsinqx)

Substitute in values of p and q to form the complimentary function.

Set up particular integral as Rcosx+Ssinx

Differentiate PI twice to get y' and y''

Sub into original DE

Compare coefficients of sinx and cosx

Use simultaneous equations or otherwise to find values of R and S.

Solution is y=complimentary function + particular integral

Then for the final part substitute in x=0 y=0 and then y'=1 and x=0 to find values for A and B.

Plug those values into your solution to make it particular.

Hope I didn't screw up :redface:

Original post by JordanR

r^2 + 2r + 5 = 0

Solutions are -1 + 2i and -1-2i

So y = e^-x (Asin2x + Bcos2x) + PI

Particular integral has the form y = C1cosx + C2sinx.

dy/dx = -C1sinx + C2cosx

d^2y/dx^2 = -C1cosx - C2sinx

Plug these into the original equation:

(-C1cosx - C2sinx) + 2(-C1sinx + C2cosx) + 5(C1cosx + C2sinx).

Solve to find C1 and C2 to get your full solution.

Spoiler

Then to find the values of the constants A and B by subbing in x = 0.

Sorry for not really being in depth, got stuff to do. :tongue:

Yeah, never mind that's exactly what I did. Just seemed rather long and laborious. Well I guess it is worth 10 marks. Thanks guys though =)

Reply 919