Give a precise meaning to evaluate the following:
$$\large{\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dotsb}}}}}$$
Since I think it has a recursive structure (does it?), I reduce the equation to
$$
p=\sqrt{1+p}
$$
$$
p^2=1+p
$$
$$
p^2-p-1=0
$$
$$
p=\frac{1\pm\sqrt{5}}{2}
$$
Did I do this right?
My university lecturer told me that:
If $$F=m\dfrac{dv}{dt}$$ it's incorrect to write $$F\,dt=m\,dv\tag{1}$$ but it is okay to write $$\int F\,dt=\int m\,dv$$ for Newtons' second law.
But never explained why $(1)$ is mathematically incorrect.
My high school teacher told me that:
Derivatives with respect to one independent variable can be treated as fractions.
So this implies that $(1)$ is valid.
This is clearly a contradiction as my high school teacher and university lecturer cannot both be correct. Or can they?
Another example of this misuse of derivatives uses the specific heat capacity $c$ which is defined to be $$c=\frac{1}{m}\frac{\delta Q}{dT}\tag{2}$$
Now in the same vain another lecturer wrote that $$\delta Q=mc\,dT$$ by rearranging $(2)$.
Another contraction to the first lecturer. I this really allowed or if it's invalid then which mathematical 'rule' has been violated here?
In my question here I have used formulae that belong to Physics but these were just simple examples to illustrate the point. My question is much more general and applies to any differential equation in mathematics involving the treatment of derivatives with respect to one independent variable as fractions.
Specifically; Why is it 'strictly' incorrect to rearrange them without taking the integral of both sides?
Answer
It is possible that your lecturer is telling you that, on its own, the expression $dt$ is meaningless, whereas $\int...dt$ does mean something quite specific, i.e. an operator or instruction to integrate with respect to $t$.
In contrast, $\delta t$ does mean something specific, i.e. a small increment in the value of $t$.
However, most people are fairly casual about this sort of thing.
Here is a theorem on expectation of a random variable in terms of its distribution function
Theroem: Let $X$ be a (continuous or discrete) non-negative random variable with distribution function $F$. Then, $E(|X|) < \infty$ if and only if $\displaystyle \int_0^\infty 1-F(x)dx <\infty$, and in that case,
$$E(X) = \displaystyle \int_0^\infty1-F(x)dx$$.
Then, a corollary of the Theorem is given as:
Corollary: For any random variable $X$, $E(|X|) <\infty $ if and only if the integrals $\displaystyle \int_0^\infty 1-F(x)dx$ and $\displaystyle \int_{-\infty}^0 F(x)dx $ both converge, and in that case
$$E(X) = \displaystyle \int_0^\infty 1-F(x)dx - \displaystyle \int_{-\infty}^0 F(x)dx$$
I understand the Theorem, but I do not see how the Corollary follows from the Theorem. I understand the first claim of the Corollary, but I do not see why
$$E(X) = \displaystyle \int_0^\infty 1-F(x)dx - \displaystyle \int_{-\infty}^0 F(x)dx \tag{1}$$
holds in that case.
I have that:
$$E(|X|) = \displaystyle \int_0^\infty P\{|X| > x\}dx \\
= \displaystyle \int_0^\infty P\{X > x\} + \displaystyle \int_0^\infty P\{X < -x\}dx \\
= \displaystyle \int_0^\infty P\{X > x\} - \displaystyle \int_0^\infty P\{X < x\}dx \tag{2}$$,
but then I could not conclude (1) since the integrand in the second integral of the last line in (2) is $P\{X < x\}$, which is equal to $F(x)$ if X is a continuous random variable, but not equal to $F(x)$ if X is discrete variable.
What am I missing?
Here $n_i$ are integral numbers, and not all of them are zero.
It is natural to conjecture that similar statement holds for even more prime numbers. Namely,
$$ n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} + n_5 \sqrt{11} +n_6 \sqrt{13} $$ is never zero too.
I am asking because this is used in some numerical algorithm in physics
I'm only doing this at GSCE and I'm really only asking here because of an interesting email conversation between my Grandfather and I regarding the fact that 0.(9) equals 1, so I'd appreciate it if you could make any explanation as simple as possible.
Basically, I have proven to my Grandfather that 0.(9) must equal 1, using the following method:
Let x = 0.(9)
So, 10x will equal 9.(9); 10x - x is 9x which is the same as 9.(9) - 0.(9) = 9, and therefore 9 / 9 is 1!
However, he has questioned the fact that 0.(9) * 9 equals 9, as he rightly stated that it equals 8.(9). I do remember learning in my maths lesson a rule regards to upper and lower bounds that meant that 8.(9) was actually the same as 9, or something along those lines, but I can not remember the correct statement to inform my Grandfather - so any suggestions would be appreciated.
Thanks in advance
How would you go about proving the recursion
$$T(n) = T\left(\frac n4\right) + T\left(\frac{3n}4\right) + n$$is $\mathcal O(n\log n)$ using induction?
Thanks!
How to find the following limit without using l'Hospital rule
$$\lim_{x\rightarrow0}\frac{\tan x-\sin x}{x^3}$$
Using l'Hospital I got $1\over2$. Thanks for your help.
Answer
Hints:
$$\frac{\tan x-\sin x}{x^3}=\frac{\sin x-\sin x\cos x}{x^3\cos x}=\frac1{\cos x}\frac{\sin x}x\frac{1-\cos x}{x^2}$$
Now, use arithmetic of limits and also
$$\frac{1-\cos x}{x^2}=\frac{\sin^2x}{x^2(1+\cos x)}=\left(\frac{\sin x}x\right)^2\frac1{1+\cos x}$$
