what's the formula for the sequence
7, 11, 7, 11, 7, 11
What type of sequence is this?
Thanks
Answer
$$a_n=9+2\cdot(-1)^n\ \ (n=1,2,\cdots).$$
what's the formula for the sequence
7, 11, 7, 11, 7, 11
What type of sequence is this?
Thanks
Answer
$$a_n=9+2\cdot(-1)^n\ \ (n=1,2,\cdots).$$
Is there a way to show the sum of any different square root of prime numbers is irrational? For example, $$\sqrt2+\sqrt3+\sqrt5 +\sqrt7+\sqrt{11}+\sqrt{13}+\sqrt{17}+\sqrt{19}$$ should be a irrational number.
One approach I used is to let the sum be a solution of an even polynomial $f(x)$with integer coefficients and prove by induction that by adding another $\sqrt{p_{k+1}}$. The new polynomial can be written as $$f(x+\sqrt{p_{k+1}})f(x-\sqrt{p_{k+1}})$$
where $$f(x+-\sqrt{p_{k+1}})=P(x)+- Q(x)\sqrt{p_{k+1}},$$
where $P(x)$ is an even plynomial and $Q(x)$
is an odd polynomial.
The new polynomial can be written as $$P^{2}(x)- Q^{2}(x)p_{k+1}.$$
Assume it has a rational solution $a$, we must have$$P(a)=Q(a)=0.$$
My calculation stopped here since I can't find any contradiction result from this. Can anyone continue this proof, or has other better way to solve this? Thanks!
I have read many articles on this confusion but i am still confused...
My simple question is -
What is $0^0$?
What is the present agreement to this?
I feel that it should be 1 as anything to the power zero is one....
I am currently a school student so i would like a more of a school based answer..
So incase it comes in my exam i should know what to write:)
Answer
$0^0$ is most often undefined. The reason is that it is not possible to define it in a good enough way. Notice the following examples:
$0^x$
Whenever $x \neq 0$ then this expression should equal to 0. However
$x^0$
should be $1$ whenever $x\neq 0$. Thus, if we define $0^0$ to either $0$ or $1$ then we get problems with these functions not being continous (without jumps if you plot them) where they are defined, which is why we keep $0^0$ undefined in most cases.
I am unable to calculate the expression of the sum of the series $1^{3/2} + 2^{3/2} + \cdots + n^{3/2}$. Could you please help me finding the answer.
Suppose that $f:(a,b] \rightarrow \mathbb{R}$ is continuous and that the limit as $\lim\limits_{x \rightarrow a}f(x)$ exists. Show that $f$ is uniformly continuous.
I am really struggling with this one. HELP
I solved this equation something like this, as shown in the photo:

Is it correct?
If I put $x=2$ I get weird results!
Answer
As a lot of the comments and the other answers point out, this only works for $-1
$$
-1 - \frac 1x - \frac{1}{x^2} - \frac{1}{x^3} - \cdots
$$
which is a sequence that converges as long as $|x|> 1$, and it gives $\frac{x}{1-x}$ (or really, $\frac{1}{1/x-1}$, which ammounts to the same thing) if you do the same trick as you've done in your question. This is what is called "the series expansion of $\frac x{1-x}$ around $\infty$" (since the series converges as long as $x$ is large enough), while $x + x^2 + x^3 + \cdots$ is the series expansion around $0$ (since it converges as long as $x$ is close enough to $0$ [there is a bit more to it than that, but that's details]). Together, they give you geometric series that evaluate to $\frac{x}{x-1}$ on the whole number line except at $-1$ and $1$.
The problem is to find the summary of this statement:
$$\sin(x) + \sin(3x) + \sin(5x) + \dotsb + \sin(2n - 1)x = $$
I've tried to rewrite all sinuses as complex numbers but it was in vain. I suppose there is much more complicated method to do this. I think it may be solved somehow using complex numbers or progressoins.
How it's solved using complex numbers or even without them if possible?
Thanks a lot.
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...