Recent content by YouAreAwesome

Ok, I think there are errors in the textbook. I just tried a simple substitution of ## a=1, b=2, c=3, d=4 ## and the conclusion is false. ##4abcd = 96## while the ##LHS= 74 ## approximately. So the RHS should have ## \sqrt{abcd}## and I've been lead down a rabbit hole. Apologies. I wonder if I...

Great thanks, I was satisfied it was a typo, or that the writer recognized the product was equal to 1 so thought it didn't matter. But then in a later proof he writes: $$ \frac{\frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{d} + \frac{d^3}{a}}{4} \geq \sqrt{\frac{a^3}{b} \times \frac{b^3}{c}...

Hi, Can someone explain why we can use the square root rather than the 4th root in this application of the AM-GM? (a/b + b/c + c/d + d/a) / 4 ≥ (a/b × b/c × c/d × d/a)^(1/2) Thanks

Let ##u=4-x \implies dx=-du## \begin{align} \int_1^3 \frac{\cos^2\frac{\pi x}{8}}{x(4-x)}dx & = \int_3^1 \frac{\cos^2\frac{\pi (4-u)}{8}}{(4-u)u} -du \nonumber \\[8pt] & = -\int_3^1 \frac{\cos^2\left(\frac{4\pi}{8} -\frac{\pi u}{8}\right)}{u(4-u)}du \nonumber \\[8pt] & = \int_1^3...

Fair enough. The way I see it, integration can be defined as the (signed) area under a curve. The way we evaluate the total area is by summing the areas of the infinite rectangles between ##x=a## and ##x=b## where each rectangle has height ##f(x)## and width $$\frac{b-a}{n}$$ and ##n \rightarrow...

@SammyS @FactChecker Again, correct me if I've missed something. But when we use ##u=4-x## the bounds from ##x=1## to ##x=3## become ##u=4-1=3## to ##u=4-3=1## leaving us with ##\int_3^1##. We then take the negative of the integral and swap them back to ##\int_1^3##. Perhaps you say this...

This is the sentence you are referring to: Ok perhaps if I keep explaining my understanding you might find a mistake along the way that you can correct. The question I'm answering is: Part ii. We have two equal integrals expressed differently. The given equation has numerator ##cos^2(\frac...

@renormalize @FactChecker Ok I think it just clicked!!! Thank you. (First I'll double check it did click, but I'm pretty confident it did). Because we are integrating over a defined domain (the bounds), when we integrate, the variable will be completely removed from the expression --...

Wow, so even though ##u=4-x##, we can still use ##u=x## because the area will be same no matter what variable we use for the integral that is currently in terms of ##u##? So is this a true statement then? $$\int_a^b sin^2xdx+\int_a^b cos^2udu= \int_a^b1dx$$

This is the question I am answering: Are you saying we can just replace ##u## with ##x## even though ##u=4-x##? This is just not clicking for me. 😒

Thanks, you are saying that ##\frac{\sin^{2}\left(\frac{\pi}{8}u\right)}{u\left(4-u\right)}=\frac{\sin^{2}\left(\frac{\pi}{8}x\right)}{x\left(4-x\right)}## But don't we have to take into account that ##u=4-x##?

Sorry, I did mention it in my original post, it's at the end where I wrote ##u=4-x##. I understand what you mean by 'dummy variable', and that's the reason this is true: $$\int_{a}^{b}f(u)du=\int_{a}^{b}f(x)dx$$ But we are dealing with this $$\int_{a}^{b}f(u)du=\int_{a}^{b}g(x)dx$$ So I...

Nope lol. When I say: I am meaning I recognize these as equivalent integrals, and therefore they are equal in value. But it seems to me you are saying this implies ##u=x##? But have we not already established that ##u=4-x##?

Looking at only the numerators and substituting ##u=4-x## then $$\int_1^3 cos^2(\frac{\pi}{8}x)+sin^2(\frac{\pi}{2}-\frac{\pi}{8}x) dx=2\int_1^3 cos^2(\frac{\pi}{8}x) dx$$ Sorry I'm not seeing why the integrand is equal to 1.

Hmmm I understand ##\intop_{a}^{b}f\left(u\right)du=\intop_{a}^{b}f\left(x\right)dx## because each is evaluating the same expression. But how does this relate to numerators of integrands being able to be added together when one is ##\cos^2(x)## and the other is ##\sin^2(4-x)##. How do these add...