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Mega A Level Maths Thread MK II

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Following on from DJ's q

Spoiler

Kvothe the Arcane

Original post by MAyman12

Why is that? :redface:

:redface:

This is a lame explanation but the algorithm required to compute

^nC_r

requires that you get the factorial of n which can only occur if it's a positive integer.

Original post by Felix Felicis

Following on from DJ's q

Spoiler

Spoiler

Original post by reubenkinara

This is a lame explanation but the algorithm required to compute

^nC_r

requires that you get the factorial of n which can only occur if it's a positive integer.

Thanks

:biggrin:

Kvothe the Arcane

Original post by Felix Felicis

Following on from DJ's q

Spoiler

I think you should have left the 2 as

\dfrac{6}{3}

. It's makes the pattern easier to see :biggrin:

:biggrin:

Original post by reubenkinara

I think you should have left the 2 as

\dfrac{6}{3}

. It's makes the pattern easier to see :biggrin:

:biggrin:

Hmmm, I didn't want to give it away that easily. :L

Original post by DJMayes

Spoiler

Sehr gut

:wink:

Kvothe the Arcane

Original post by Felix Felicis

Hmmm, I didn't want to give it away that easily. :L

Sehr gut :wink:

:wink:

My last Q for you before I retire :smile:

:smile:

Original post by reubenkinara

My last Q for you before I retire :smile:

:smile:

I = \displaystyle\int \dfrac{dx}{(x-1) \sqrt{x^{2} - 1}}

Let

x = 1 + u^{-1} \Rightarrow I = -\displaystyle\int \dfrac{du}{u^{-1} \sqrt{2u^{-1} + u^{-2}}} \dfrac{1}{u^{2}}

= - \displaystyle\int (2u + 1)^{-\frac{1}{2}} du = - \sqrt{2u + 1} + \mathcal{C}

= - \sqrt{\dfrac{x+1}{x-1}} + \mathcal{C}

as required

\displaystyle\int_{\sec \alpha}^{\sec \beta} \dfrac{1}{(x-1) \sqrt{x^{2} - 1}} dx = \left[ \sqrt{ \dfrac{x+1}{x-1}} \right]_{\sec \beta}^{\sec \alpha}

= \sqrt{ \dfrac{\sec \alpha + 1}{\sec \alpha - 1}} - \sqrt{ \dfrac{\sec \beta + 1}{\sec \beta - 1}}

= \sqrt{ \dfrac{ (\cos \alpha + 1)^{2}}{\sin^{2} \alpha}} - \sqrt{ \dfrac{ (\cos \beta + 1)^{2}}{\sin^{2} \beta}}

= \cot \dfrac{\alpha}{2} - \cot \dfrac{\beta}{2}

(edited 10 years ago)

Kvothe the Arcane

Original post by Felix Felicis

Unparseable latex formula:

I = \displaystyle\int \dfraC{1}{(x-1) \sqrt{x^{2} - 1}} dx

Let

x = 1 + u^{-1} \Rightarrow I = -\displaystyle\int \dfrac{du}{u^{-1} \sqrt{2u^{-1} + u^{-2}}} \dfrac{1}{u^{2}}

= - \displaystyle\int (2u + 1)^{-1/2} du = - \sqrt{2u + 1} + \mathcal{C} = - \sqrt{\dfrac{x+1}{x-1}} + \mathcal{C}

as required

\displaystyle\int_{\sec \alpha}^{\sec \beta} \dfrac{1}{(x-1) \sqrt{x^{2} - 1}} dx = \left[ \sqrt{ \dfrac{x+1}{x-1}} \right]_{\sec \beta}^{\sec \alpha}

= \sqrt{ \dfrac{\sec \alpha + 1}{\sec \alpha - 1}} - \sqrt{ \dfrac{\sec \beta + 1}{\sec \beta - 1}}

= \sqrt{ \dfrac{ (\cos \alpha + 1)^{2}}{\sin^{2} \alpha}} - \sqrt{ \dfrac{ (\cos \beta + 1)^{2}}{\sin^{2} \beta}}

= \cot \dfrac{\alpha}{2} - \cot \dfrac{\beta}{2}

correct. Sorry :frown:

:frown:

Just realised that it must've been a pain to type out in latex! Thanks

Original post by reubenkinara

correct. Sorry :frown:

:frown:

Just realised that it must've been a pain to type out in latex! Thanks

I've had worse :L

Lord of the Flies

Very accessible question, as long as you know what a polynomial is. Let's see who will crack it the quickest! Show that if two polynomials are equal over a given interval, then they are the same polynomial.

Original post by Lord of the Flies

Very accessible question, as long as you know what a polynomial is. Let's see who will crack it the quickest! Show that if two polynomials are equal over a given interval, then they are the same polynomial.

Spoiler

Original post by Lord of the Flies

Very accessible question, as long as you know what a polynomial is. Let's see who will crack it the quickest! Show that if two polynomials are equal over a given interval, then they are the same polynomial.

Spoiler

(edited 10 years ago)

Messed up my maths A-Level so gonna have to retake next year :frown:

:frown:

Posted from TSR Mobile

Original post by DJMayes

Spoiler

Ah, that is a nice proof.

Suppose that x,y and z are natural numbers satisfying xy=z²+1. Prove that there exist integers a,b,c,d such that x=a²+b²,y=c²+d² and z=ac+bd.

Hint

Spoiler

:colone:

:colone:

:colone:

(edited 10 years ago)

Original post by MAyman12

Suppose that x,y and z are natural numbers satisfying xy=z²+1. Prove that there exist integers a,b,c,d such that x=a²+b²,y=c²+d² and z=ac+bd.

Hint

Spoiler

:colone:

:colone:

:colone:

Brain dead after 3 hours of bio so I'll do this the lazy way - Proof by example. :tongue:

:tongue:

Spoiler

(edited 10 years ago)

Original post by MAyman12

Suppose that x,y and z are natural numbers satisfying xy=z²+1. Prove that there exist integers a,b,c,d such that x=a²+b²,y=c²+d² and z=ac+bd.

Hint

Spoiler

:colone:

:colone:

:colone:

What does the i mean? :s-smilie:

:s-smilie:

Original post by tigerz

What does the i mean? :s-smilie:

:s-smilie:

i^2=-1

:smile:

Original post by tigerz

What does the i mean? :s-smilie:

:s-smilie:

It's an imaginary number

it is denoted as

i=\sqrt{-1}

If you do FP1 onwards you'll come across them all the time

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