Fp2 integration

That's a nice analytic approach. I was looking into getting the inequality bounds by looking at the Laurent series for the function and deducing then from there but I don't have quite enough experience to know whether that's viable or not.

Argument makes perfect sense up to the last two lines, granted I don't have much experience with Taylor expansions so that's no great suprise, it looks 'nicer' than the way I used though

I don't think he used Taylor expansions anywhere.

https://en.wikipedia.org/wiki/Taylor_series Do some reading guys.

Instructions not clear enough, ended up with 10 different tabs open


Posted from TSR Mobile

Argument makes perfect sense up to the last two lines, granted I don't have much experience with Taylor expansions so that's no great suprise, it looks 'nicer' than the way I used though


Posted from TSR Mobile

BTW: Look at Q3 here.

I don't use a Taylor expansion at all. The argument is as follows:

1) For the upper line, i.e. the tangent, $f'(x)$ is more negative over the whole range, so $f(x)$ must always lie below that line as it decreases more quickly. So that part of the inequality is done.

2) We need to show that $f(x)$ always lies above the other line, the chord. I split this into two parts.

First, I show that, since $f'(x)$ eventually exceeds the gradient of the chord, then there is some point $x_0$ where $f'(x)$ must be equal to that gradient - so in the interval between 0 and $x_0$, the average gradient of $f(x)$ is less than than of the chord, so it decreases more slowly in this region, so is always above the chord in $(0,x_0)$

Then we have to show that it is above the chord even in the region where $f(x)$ is decreasing faster than then chord. I do this by contradiction: essentially I say that if at some point $f(x)$ was below the chord in this region, then it would stay below it thereafter, since $f(x)$ is decreasing faster than the chord - the chord could never "catch it up" again.

But we know that the chord and $f(x)$ meet at $\pi/4$ i.e. we know that $f(x)$ is *not* below the chord, at the end of the region in question. Hence, there can be no point in the region where it was below the chord i.e. the chord is always below $f(x)$. QED.