MATH2701 Abstract Algebra/Fundamental Analysis (1 Viewer)

leehuan

Well-Known Member
Don't need to show me how to do the entire question if it's way too long. Suggestions are plenty

$\bg_white f(x)=\int_x^{x+2\pi}\frac{\sin t}{t}dt$

$\bg_white \text{Prove that }f(x)=O(x^{-2})\text{ as }x\to \infty$

My starting point was just saying f(x) < integrand being 1/t instead of sint/t, but working backwards it just gave me $\bg_white 1+\frac{2\pi}x \le \exp\left(\frac{M}{x^2}\right)$

He-Mann

Vexed?
This may help, $\bg_white \ln(x) \leq x - 1.$

InteGrand

Well-Known Member
Don't need to show me how to do the entire question if it's way too long. Suggestions are plenty

$\bg_white f(x)=\int_x^{x+2\pi}\frac{\sin t}{t}dt$

$\bg_white \text{Prove that }f(x)=O(x^{-2})\text{ as }x\to \infty$

My starting point was just saying f(x) < integrand being 1/t instead of sint/t, but working backwards it just gave me $\bg_white 1+\frac{2\pi}x \le \exp\left(\frac{M}{x^2}\right)$
You can try using integration by parts.

-insert title here-
Don't need to show me how to do the entire question if it's way too long. Suggestions are plenty

$\bg_white f(x)=\int_x^{x+2\pi}\frac{\sin t}{t}dt$

$\bg_white \text{Prove that }f(x)=O(x^{-2})\text{ as }x\to \infty$

My starting point was just saying f(x) < integrand being 1/t instead of sint/t, but working backwards it just gave me $\bg_white 1+\frac{2\pi}x \le \exp\left(\frac{M}{x^2}\right)$
I'm getting O(x⁻¹)

Differentiating under the integral sign yields:

$\bg_white f'(x) = \sin{x} \left(\frac{1}{x+2\pi} - \frac{1}{x}\right) = \sin{x} \left(-\frac{2\pi}{x^2} + \frac{4\pi^2}{x^3} + \mathcal{O}\left(x^{-4} \right) \right)$

Ignore sin x and integrate both sides to obtain an asymptotic...

seanieg89

Well-Known Member
I'm getting O(x⁻¹)

Differentiating under the integral sign yields:

$\bg_white f'(x) = \sin{x} \left(\frac{1}{x+2\pi} - \frac{1}{x}\right) = \sin{x} \left(-\frac{2\pi}{x^2} + \frac{4\pi^2}{x^3} + \mathcal{O}\left(x^{-4} \right) \right)$

Ignore sin x and integrate both sides to obtain an asymptotic...
They are both true, your statement is just weaker. IBP usually gains you stuff when you are analysing oscillatory expressions...if you just look at the size of things via absolute values, you ignore a lot of the "cancellation" that comes from the oscillation. IBP picks this up.

Edit: Note that you can recover this improvement from your final expression by integrating the leading order term and NOT ignoring the sin. (The remaining terms have size small enough to not matter.) To deal with this leading oscillatory term you still need to use IBP or something similar, and the differentiation under the integral sign has not saved any time.

Last edited:

leehuan

Well-Known Member
This was in the final exam and I never figured it out.

$\bg_white \text{Let }f:\mathbb{R}\to \mathbb{R}\text{ be a convex function. Prove that for any }x,y,z\in \mathbb{R},$

$\bg_white \frac{ f(x)+f(y)+f(z) }3 + f \left( \frac{x+y+z}3 \right) \ge \frac{2}{3} \left[ f \left( \frac{x+y}{2} \right) + f \left( \frac{x+z}{2} \right) + f \left( \frac{y+z}{2 }\right) \right]$

with the hint x + y + z = (x+y)/2 + (x+z)/2 + (y+z)/2

$\bg_white \text{Let }f:\mathbb{R}\to \mathbb{R}\text{ be a convex function. Prove that for any }x,y,z\in \mathbb{R},$
$\bg_white \frac{ f(x)+f(y)+f(z) }3 + f \left( \frac{x+y+z}3 \right) \ge \frac{2}{3} \left[ f \left( \frac{x+y}{2} \right) + f \left( \frac{x+z}{2} \right) + f \left( \frac{y+z}{2 }\right) \right]$