Showing piece-wise function continuous

member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this,
1680662250322.png
,
The solution is,
1680662269391.png

However, should they not write ##f(x) = \cos x## on ##[\frac{pi}{4}, \infty)##

Many thanks!
 
Last edited by a moderator:
Physics news on Phys.org
Do you mean right after the "Similarly,"? It wouldn't hurt, but I think that it is easy enough to follow the logic without saying that. Initially, you should be in the habit of stating everything. After a while, that becomes tedious and both you and the reader will be happy if you skip obvious things. You must be careful though, what you skip.
 
  • Like
Likes member 731016
FactChecker said:
Do you mean right after the "Similarly,"? It wouldn't hurt, but I think that it is easy enough to follow the logic without saying that. Initially, you should be in the habit of stating everything. After a while, that becomes tedious and both you and the reader will be happy if you skip obvious things. You must be careful though, what you skip.
Thank you for you reply @FactChecker!

No sorry I meant right after the "Since f(x) = Sinx on ..."

Many thanks!
 
ChiralSuperfields said:
Thank you for you reply @FactChecker!

No sorry I meant right after the "Since f(x) = Sinx on ..."

Many thanks!
Oh. The reason for not including the point ##\pi/4## is that the rest of the sentence is only about continuity on ##(-\infty, \pi/4)\cup(\pi/4, \infty)##. So there was no need to include ##\pi/4##. It wouldn't have hurt to include it.
 
  • Like
Likes member 731016
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top