Recent content by KungPeng Zhou

Ok, I have solved it. I need to defferential f(tx, ty, tz) with respect to t

For the first equation: ##f(tx, ty, tz)=f(u, v, w) ##, ##u=tx, v=ty, w=tz##,##k=f(u, v, w) #### t^{n}f_{x}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}## As the same calculation ##xf_{x}+yf_{y}+zf_{z}=[\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}...

Ok, there is good way to proof this sequence is bounded. From##a_{n}=\sqrt{2a_{n-1}}## ##a_{n}<2## as ## a_{n-1}<2## However we know##a_{1}=\sqrt{2}<2## So we kown##a_{2}<2... a_{n}<2##

First, we can know ##a_{n}=\sqrt{2a_{n-1}}## When##n\rightarrow \infty## ##a_{n}=\sqrt{2a_{n-1}}## And we can get the answer is 2. Is this solution right? And is any other way to solve the question?

Yes, it's tighter. It seems like the definition of limits.

It's very good. We can evaluate it easier with the way.

Sorry,maybe there is an error ##\int_{0}^{1}x^{4n}dx=\frac{1}{1+4n}## So, we know $$0<\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx<\frac{1}{1+4n}$$

Maybe I have found a better way to solve the question. ##a_n=\int_{0}^{2-\sqrt{3}}\frac{1-x^{4n}}{1+x^{2}}dx=\int_{0}^{2-\sqrt{3}}\frac{1}{1+x^{2}}dx-\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx=arctan(2-\sqrt{3})...

In my opinion , if it can be shown that this is a monotonically bounded sequence, one can confirm that there is a limit. First，we know $$ \frac{1-x^{4n}}{1+x^{2}}dx=(1-x^{2}) (1+x^{2}) ^{n-1}=(1-x^{4}) ^{n-1}(1+x^{2}).$$ According to the integral median theorem，we can get $$a_n=(2- \sqrt{3} )...

Yes，you are right.

Ithe seems that we still can't solve it with this way...

Sorry，but I can't understand you.Could you please tell how to show my math code properly?Now I just can use these math code.

Yes.Thank you

\frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{sint}\sqrt{1+u^{4}}du)dt=\frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du then we let m=sinx，so x=arcsinx，then we get \frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du=\frac{dm}{dx}\frac{d}{dm}\int_{0}^{m}(\sqrt{1+u^{4}})du=\sqrt{1+m^{4}}\frac{dm}{dx}，then we...

When I encountereD this kind of question before.For example \int x\sqrt{2+x^{2}}dx We make the Substitution t=x^{2}+2，because its differential is dt=2xdx，so we get \int x\sqrt{2+x^{2}}=1/2\int\sqrt{t}dt，then we can get the answer easily But the question，it seems that I can't use the way to...