algebra precalculus - Find the sum to n terms of the series $frac{1}{1.2.3}+frac{3}{2.3.4}+frac{5}{3.4.5}+frac{7}{4.5.6}+cdots$..

Question :

Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$

What I have done :

nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2)$ respectively.

Therefore the nth term of given series is :

$\frac{2r-1}{r(r+1)(r+2)} =\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}$ .....(1)

By using partial fraction :

and solving for A,B and C we get A = 1/2, B = -1, C =1/2

Putting the values of A,B and C in (1) we get :

$\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)}$

But by putting $r =1,2,3, \cdots$ I am not getting the answer. Please guide how to solve this problem . Thanks.

trigonometry - Application Of De Moivre & Euler's Formulae

Can anyone show me how I can prove that $\sin x \cos (3x) = \frac{1}{4} \sin (7x) - \frac{1}{4}\sin (5x) + \frac{1}{2}\sin x$?
I tried using Euler's formulae
$$\sin x= \frac{e^{ix}-e^{-ix}}{2i}$$
and
$$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$
but the simplification didn't help at all.
PS: Simplify starting from the left.

Elementary proof that the derivative of a real function is continuous somewhere

One can use the Baire category theorem to show that if $f:\mathbb{R} \to \mathbb{R}$ is differentiable, then $f'$ is continuous at some $c \in \mathbb{R}$. Is there an elementary proof of this fact? By "elementary" I mean at the level of intro real analysis.

Edit: In spite of the decent response this question has gotten, after more than two and a half months there are still no answers. It's perhaps possible that there's some "deep" reason we should not expect an elementary proof of this. I will therefore also accept a well reasoned discussion as to why such a proof is unlikely.

Answer

One can actually prove with elementary tools something stronger, namely the following

Theorem: If $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable in a (non-degenerate) interval $[a,b]$ then $f'$ is continuous at some point $c\in(a,b)$ (a corollary being that if $f$ is differentiable throughout $\mathbb{R}$, then $f’$ if continuous on a dense subset of $\mathbb{R}$).

Proof: For any $[u,t]\subseteq[a,b]$, define $osc(u,t)=\sup_{[w,z]\subseteq[u,t]}\left|\frac{f(t)-f(u)}{t-u} - \frac{f(z)-f(w)}{z-w}\right|$ (with $osc(u,t)=+\infty$ if the right-hand side is unbounded; this can easily be made more formal with some more verbiage). Informally, $osc(u,t)$ tells us how much the slope of $f$ can "oscillate" in the interval $[u,t]$; the essence of the proof is showing that $osc(u,t)$ must converge to $0$ as $u$ and $t$ converge to some point $c$, and that $c$ is then a point at which $f'$ is continuous.

Consider a generic sequence of “concentric and convergent” intervals $[u_i,t_i]$, i.e. one that satisfies $u_i\leq u_{i+1}

Let us now show that there is a sequence of concentric and convergent intervals $[u_i,t_i]\subseteq (a,b)$ for which $osc(u_i,t_i)\rightarrow 0$. Suppose it were not the case. Then, starting with an arbitrary $[w_0, z_0]\subseteq(a,b)$ there would be an $\epsilon>0$ such that given $[w_i,z_i]$ we could always find a non-degenerate $[w_{i+1},z_{i+1}]\subseteq [w_i,z_i]$ such that $\left|\frac{f(z_i)-f(w_i)}{z_i-w_i} - \frac{f(z_{i+1})-f(w_{i+1})}{z_{i+1}-w_{i+1})}\right|>\epsilon$ (note that $\epsilon$ is independent of $i$). Furthermore, such $[w_i,z_i]$ could always be chosen arbitrarily small, since if $g:\mathbb{R}\rightarrow\mathbb{R}$ is continuous in a non-degenerate interval $[\alpha,\beta]$, then for any $\delta>0$ there exists a non-degenerate interval $[\alpha',\beta']\subseteq(\alpha,\beta)$ with $|\beta'-\alpha'|<\delta$ and $\frac{g(\beta)-g(\alpha)}{\beta-\alpha}=\frac{g(\beta')-g(\alpha')}{\beta'-\alpha'}$ (the proof is essentially identical to that of Rolle's theorem, but stopping before taking the differentiation limit). Thus $\frac{f(z_i)-f(w_i)}{z_i-w_i}$ would not converge to a (finite) limit, contradicting the differentiability of $f$ in $\lim w_i =
\lim z_i$.

Then, consider a sequence of concentric and convergent intervals $[u_i,t_i]\subseteq(a,b)$, with $u_i,t_i\rightarrow c$, for which $osc(u_i,t_i)\rightarrow 0$. It is immediate to see that $osc(u_i,t_i)=max\left(\left(\left(\sup_{[w,z]\subseteq[u_i,t_i]} \frac{f(z)-f(w)}{z-w}\right) - \frac{f(t_i)-f(u_i)}{t_i-u_i}\right), \left(\frac{f(t_i)-f(u_i)}{t_i-u_i} - \left(\inf_{[w,z]\subseteq[u_i,t_i]} \frac{f(z)-f(w)}{z-w}\right)\right)\right)$, so if $osc(u_i,t_i)\rightarrow 0$ then $\sup_{[w,z]\subseteq[u_i,t_i]}\frac{f(z)-f(w)}{z-w}, \inf_{[w,z]\subseteq[u_i,t_i]
} \frac{f(z)-f(w)}{z-w}\rightarrow \lim \frac{f(t_i)-f(u_i)}{t_i-u_i}
= f’(c)$. And since in any interval $[u_i,t_i]$ we have that $\sup_{[w,z]\subseteq[u_i,t_i]}\frac{f(z)-f(w)}{z-w} \geq f' \geq \inf_{[w,z]\subseteq[u_i,t_i]

} \frac{f(z)-f(w)}{z-w}$, then $f’(x)\rightarrow f’(c)$ as $x\rightarrow c$, i.e. $f’$ is continuous at $c$.

real analysis - $0^0$ -- indeterminate, or $1$?

One of my teachers argued today that 0^0 = 1. However, WolframAlpha, intuition(?) and various other sources say otherwise... 0^0 doesn't really "mean" anything..

can anyone clear this up with some rigorous explanation?

Answer

Short answer: It depends on your convention and how you define exponents.

Long answer: There are a number of ways of defining exponents. Usually these definitions coincide, but this is not so for $0^0$: some definitions yield $0^0=1$ and some don't apply when both numbers are zero (leaving $0^0$ undefined).

For example, given nonnegative whole numbers $m$ and $n$, we can define $m^n$ to be the number of functions $A \to B$, where $A$ is a set of size $n$ and $B$ is a set of size $m$. This definition gives $0^0=1$ because the only set of size $0$ is the empty set $\varnothing$, and the only function $\varnothing \to \varnothing$ is the empty function.

However, an analyst might not want $0^0$ to be defined. Why? Becuase look at the limits of the following functions:
$$\lim_{x \to 0^+} 0^x = 0, \qquad \lim_{x \to 0} x^0 = 1, \qquad \lim_{x \to 0^+} (e^{-1/t^2})^{-t} = \infty$$
All three limits look like $0^0$. So when this is desired, you might want to leave $0^0$ undefined, so that it's a lack of definition rather than a discontinuity.

Typically this is resolved by:

• If you're in a discrete setting, e.g. considering sets, graphs, integers, and so on, then you should take $0^0=1$.



• If you're in a continuous setting, e.g. considering functions on the real line or complex plane, then you should take $0^0$ to be undefined.

Sometimes these situations overlap. For example, usually when you define functions by infinite series
$$f(x) = \sum_{n=0}^{\infty} a_nx^n$$
problems occur when you want to know the value of $f(0)$. It is normal in these cases to take $0^0=1$, so that $f(0)=a_0$; the reason being that we're considering what happens as $x \to 0$, and this corresponds with $\lim_{x \to 0} x^0 = 1$.

trigonometry - Generalization of Euler's Formula

Euler's formula states that, for any real number x:

$$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$

Can it be generalized in that way?

$$ae^{ix}+be^{-ix}=c\cos(x+d)$$

where $a,b\in \mathbb{C}$ and $c,d\in \mathbb{R}$.
Of course if $a=b=1$ and $c=2$, $d=0$ this is the common Euler's fomula, but it is true that for every $a,b$ I can rewrite a sum of complex exponentials as a single cosine? If it is, what is the relationship between these constants?

Answer

Suppose your formula is true for any $x \in \mathbb R$. You can write it as

$$ ae^{ix}+be^{-ix}- \frac{c}{2}(e^{id} e^{ix}+ e^{-id}e^{-ix}) = 0,$$

that is

$$ \left(a - \frac{c}{2}e^{id}\right) e^{ix}+ \left( b - \frac{c}{2}e^{-id}\right) e^{-ix} =0.$$

Now we can use the fact that $e^{ix}$ and $e^{-ix}$ are linearly independent to get

$$a = \frac{c}{2}e^{id}, \quad b = \frac{c}{2}e^{-id}.$$

This implies

$$ c = 2 (ab)^{1/2}, \quad \cos d = \frac{a+b}{2 (ab)^{1/2}}.$$

Since we require $c$ and $d$ to be real, then $ab$ has to be a positive real number and $a+b$ has to be real. This is possible only if $a$ and $b$ are complex conjugates: $b = \bar a$.

divisibility - Gcd number theory proof: $(a^n-1,a^m-1)= a^{(m,n)}-1$

Prove that if $a>1$ then $(a^n-1,a^m-1)= a^{(m,n)}-1$

where $(a,b) = \gcd(a,b)$

I've seen one proof using the Euclidean algorithm, but I didn't fully understand it because it wasn't very well written.
I was thinking something along the lines of have $d= a^{(m,n)} - 1$ and then showing
$d|a^m-1$ and $d|a^n-1$ and then if $c|a^m-1$ and $c|a^n-1$, then $c\le d$.

I don't really know how to show this though...

I can't seem to be able to get $d* \mathbb{K} = a^m-1$.

Any help would be beautiful!

real analysis - What is $lim_{ntoinfty} root{2n+1} of {-1} ?$

First of all, I'm sorry if this question has been already asked and answered, as far as I searched, I couldn't find such a question on this site.
So, I've been thinking about the limit of the sequence $\left(\root{2n+1}\of{-1}\right)_{n\geq 0}$. Since the order of the root is odd for every $n$, this sequence, is obviously a constant sequence with the general term $a_n = -1$. So, from this follows that $$ \lim_{n\to\infty} \root{2n+1}\of{-1} = -1 $$.
We can even do an $\epsilon-N$ proof to show this (and it's realy easy actually): $$ \forall \epsilon > 0 \hspace{0.5cm} \exists N \geq 0 \hspace{0.3cm} \text{s.t.} \hspace{0.3cm} \left|\root{2n+1}\of{-1}+1\right|<\epsilon \hspace{0.5cm} \forall n \geq N \\ \left|\root{2n+1}\of{-1} + 1\right| = \left|-1 + 1\right| = 0 < \epsilon \hspace{0.5cm} \forall n \geq 0 \\ N = 0 \ _\blacksquare $$.

However, if we use tehniques ussualy used for solving limits, we end up with a different result:
$$ \begin{align*}
\lim_{n\to\infty} \root{2n+1}\of{-1} &= \lim_{n\to\infty} (-1)^{1\over 2n+1} \\
&= \left(\lim_{n\to\infty}-1\right)^{\lim_{n\to\infty}{1\over 2n+1}} \\ &= (-1)^0 \\ &= 1 \end{align*} $$.

What is wrong here? Why do the two methods give different results?

Edit: To make everything clear, I'm assuming the real root as defined by $\root n \of {} : \mathbb{R} \to \mathbb{R}$ for odd $n$ and treating this as a real-analysis problem. Also, it's pretty explicit from my question that I'm working with a sequence and not with a function. The limit only goes through natural values of $n$

Edit 2: I've figured it out. Thank you all for your answers esspecially to @Jack who pointed out the theorem I've been using $\lim_{n\to\infty}(a_n^{b_n}) = (\lim_{n\to\infty} a_n)^{(\lim_{n\to\infty} b_n)}$ is not true in general. I've consulted my textbook again and saw that I've missed the part where they said $a_n > 0, \forall n \in \mathbb{N}$. Of course, we can think of this problem also from the viewpoint of functions and the fact that the function $(-1)^x$ is not continuous is another gap in using something like the above theorem. Thank you all again for being so kind and giving me so many answers.

Answer

Your expression $\sqrt[2n+1]{-1}$ (for any nonnegative integers $n$) is defined to be, as you stated in the post, the unique real number $y$ such that $y^{n+1}=-1$. Since by your definition, $\sqrt[2n+1]{-1}=-1$, there is no doubt that
$$
\lim_{n\to\infty}\sqrt[2n+1]{-1}=\lim_{n\to\infty}(-1)=-1.
$$

There is no problem for the limit itself.

What goes wrong here is in your second "method":

if we use techniques usually used for solving limits, we end up with a different result:
$$ \begin{align*}
\lim_{n\to\infty} \root{2n+1}\of{-1}
&= \lim_{n\to\infty} (-1)^{1\over 2n+1} \\
&= \left(\lim_{n\to\infty}-1\right)^{\lim_{n\to\infty}{1\over 2n+1}} \\
&= (-1)^0 \\ &= 1 \end{align*} $$.

The following step is problematic:
$$
\lim_{n\to\infty} (-1)^{1\over 2n+1}
= \left(\lim_{n\to\infty}-1\right)^{\lim_{n\to\infty}{1\over 2n+1}}
$$

What you use here is
$$
\lim_{n\to\infty}{a_n}^{b_n}=(\lim_{n\to\infty}a_n)^{(\lim_{n\to\infty}b_n)} \tag{1}
$$

where $a_n=-1$ is the constant sequence and $b_n=\frac{1}{2n+1}$. But (1) is NOT true in general.

---

[Added]
In real analysis, one rarely writes expression like $a^b$ for $a\leq 0$ and arbitrary real number $b$, unless one specifically defines such expression for some particular $a$ and $b$. For instance, you define $(-1)^{1/n}$ for only $n$ being an odd positive integer and let $(-1)^{1/n}$ be the unique number $y$ such that $y^{n}=-1$. In such situation, $(-1)^{1/n}$ is nothing but the real number $-1$.

One definition for the expression $a^b$ with $a>0$ and $b\in\mathbb{R}$ is $e^{b\ln a}$. And one has the following statement

Suppose $\{a_n\}$ is a positive sequence of real numbers such that $\lim_{n\to \infty}a_n=a$. Assume in addition that $\{b_n\}$ is a real sequence with $\lim_{n\to\infty}b_n=b$. Then
$$
\lim_{n\to \infty}a_n^{b_n}=\lim_{n\to \infty} e^{b_n\ln a_n}=\lim_{n\to\infty}e^{b\ln a}=a^b.
$$

If one does want to consider the expression $a^b$ for negative real number $a$, then one would

• either stick to the definition for the some specific $a$ one has,





• or unavoidably talk about the complex logarithm. See also this Wikipedia article.