Mega A Level Maths Thread MK II

Original post by joostan

Stole this from the old thread courtesy of Felix :lol:

Spoiler

Don't know what e and ln is :s I know that I have to use:

\frac{a}{1-r}

Original post by Felix Felicis

It doesn't take that long! And it's peculiar :tongue:

But fine, fine... :wink:

How about this...

If

\displaystyle\prod_{r=1}^{n} f(r)

denotes

f(1) \cdot f(2) \cdot f(3) \cdots f(n)

then evaluate

\displaystyle\prod_{r=1}^{n} \left( \dfrac{r+1}{r} \right)

Ahh...I see...

Spoiler

Reply 2181

Original post by joostan

I mean exactly what I said. . . And whole number has no real meaning in this context. :ahee:

You troll you :tongue:

Reply 2182

No one ever wants to do my questions. :emo:

Reply 2183

Original post by Felix Felicis

No one ever wants to do my questions. :emo:

Wasn't that part of a STEP question? I remember the look of it :tongue:

Reply 2184

Original post by MathsNerd1

Wasn't that part of a STEP question? I remember the look of it :tongue:

Haha yeah but come on, admittedly it's pretty straightforward xD

Reply 2185

Original post by tigerz

Don't know what e and ln is :s I know that I have to use:

\frac{a}{1-r}

e is defined such that:

\dfrac{d}{dx}(e^x) = e^x

In series form it's.

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + . . . + \frac{x^n}{n!}+ . . .

or more formally.

e^x = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n!}

Alternatively this is also:

e = \lim{n \to \infty} (1+\frac{1}{n})^n

\ln(x) = \log_e(x)

Original post by tigerz

Ahh...I see...

Spoiler

There's a link in one of my previous posts if you want to have a look. :tongue:

(edited 10 years ago)

Reply 2186

Original post by Felix Felicis

No one ever wants to do my questions. :emo:

I would...If i could :ashamed:

Reply 2187

Original post by Felix Felicis

Haha yeah but come on, admittedly it's pretty straightforward xD

Yeah it's rather simple once you think about it but the sight of it would probably scare some people off, that was a nice question in general :tongue:

Reply 2188

Original post by Felix Felicis

No one ever wants to do my questions. :emo:

They are doing your question, reubenkinara did your other one.
And I'd already done the product one :tongue:

Original post by Felix Felicis

You troll you :tongue:

Hehe -

Spoiler

Reply 2189

I just broke maths. Come at me bros.

Spoiler

(edited 10 years ago)

Reply 2190

Original post by joostan

e is defined such that:

\dfrac{d}{dx}(e^x) = e^x

In series form it's.

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + . . . + \frac{x^n}{n!}

or more formally.

Unparseable latex formula:

e^x = \displaystyle\sum_{n=0}^\ifnty \dfrac{x^n}{n!}

Alternatively this is also:

Unparseable latex formula:

e = \lim{n \to \ifnty} (1+\frac{1}{n}

\ln(x) = \log_e(x)

There's a link in one of my previous posts if you want to have a look. :tongue:

Ahh, my target is to solve one your questions, and one from felix felices, after the 10th of June though LOOOL
For now I don't think I can solve that :'(

Reply 2191

Original post by tigerz

I would...If i could :ashamed:

You can. xD It's honestly not as daunting as it seems and is entirely possible with C2 knowledge.

Reply 2192

Original post by Felix Felicis

You can. xD It's honestly not as daunting as it seems and is entirely possible with C2 knowledge.

Ahh, I do understand it but I think it'll take me a while, maybe i'll try this one tomorrow :biggrin:

Reply 2193

Original post by tigerz

Ahh, my target is to solve one your questions, and one from felix felices, after the 10th of June though LOOOL
For now I don't think I can solve that :'(

lol, I tidied up the latex in my post - hopefully it's clearer now.
Felix's one's not too bad :tongue:

Reply 2194

Original post by tigerz

Ahh, I do understand it but I think it'll take me a while, maybe i'll try this one tomorrow :biggrin:

If you work through his question it should only take about 5 minutes, once you 'see' it. :smile:

Reply 2195

Original post by Felix Felicis

I just broke maths. Come at me bros.

Spoiler

This implies that:

\infty - \infty = -1

Which doesn't really mean much :lol:

just like:

\dfrac{\infty}{\infty} = ?

Reply 2196

Original post by MathsNerd1

If you work through his question it should only take about 5 minutes, once you 'see' it. :smile:

Okays

Thank youu

Original post by joostan

lol, I tidied up the latex in my post - hopefully it's clearer now.
Felix's one's not too bad :tongue:

Ahh thanks, for now i'll just do the simple ones :wink:

I don't even know how many conversations i'm having anymore :s

Reply 2197

Original post by joostan

This implies that:

\infty - \infty = -1

Which doesn't really mean much :lol:

just like:

\dfrac{\infty}{\infty} = ?

Yeah, I know. xD

Reply 2198

Kvothe the Arcane

Original post by joostan

Personally? No, I just dive straight into it :tongue:

Hint: - If you want one.

Spoiler

I give up. What I have is a stupid mess. I'll post in spoilers

Spoiler

(edited 10 years ago)

Reply 2199