The book I'm studying gives the answer as $\frac{1-e^{j\frac{x}{N}N}}{1- e^{j\frac{x}{N}}} = 0$ but doesn't explain how.
1806
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:  $$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$  There is a slight difference in case of $\sin$, which is: $$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2  a + (n-1)\cdot d}{2}\biggr)$$   How do we prove the above two identities?
114117
A friend left this teaser for me. He asked me to first compute:  $$ \lim_{n \to \infty} \frac{\binom{2n}{n}}{2^{2n}}$$  Using Stirling's approximation (and another method), I got the answer as $0$. Then he gave the following "proof" for me to check:  Let $X_i$ be iid Bernoulli$(1/2)$. Then we have  $$ P\{\text{n out of 2n $X_i$ are 1} \} = P\{ \sum_{i=1}^{2n}X_i = n\} = \frac{\binom{2n}{n}}{2^{2n}}$$ $$ = P\left\{\frac{\sum_{i=1}^{2n}X_i}{2n} =\frac{1}{2}\right \}$$ $$ = P\left\{\frac{\sum_{i=1}^{2n}(X_i - \frac{1}{2})}{2n} = 0\right \}$$  Denote $E_n := \frac{\sum_{i=1}^{2n}(X_i - \frac{1}{2})}{2n}$  He argues according to Weak law of large numbers, $\forall \epsilon &gt; 0$ $$ \lim_{n \to \infty} P(|E_n| &lt; \epsilon) = 1 \Rightarrow \lim_{n \to \infty} P(|E_n| =0) = 1 \Rightarrow \lim_{n \to \infty} P(E_n =0) = 1$$  Hence the limit must be $1 \neq 0$.   I, for some strange reason, was unable to refute his proof although I suspect that the bug is in the last step somewhere. I would appreciate any help/hints on this.
67934
$$\frac{\sum_{i=0}^nx_iy_i-\bar y\sum_{i=0}^nx_i}{\sum_{i=0}^nx_i^2-\bar x\sum_{i=0}^nx_i}= \frac{\sum_{i=0}^n(x_i-\bar x)(y_i-\bar y)}{\sum_{i=0}^n(x_i-\bar x)^2}$$  Maybe it refers some calculation rules about summation. I cannot solve that. Thanks!
119803
I'm trying to prove that $\frac{\mathrm{d} }{\mathrm{d} x}\ln x = \frac{1}{x}$.  Here's what I've got so far: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &amp;= \lim_{h\to0} \frac{\ln(x + h) - \ln(x)}{h} \\ &amp;= \lim_{h\to0} \frac{\ln(\frac{x + h}{x})}{h} \\ &amp;= \lim_{h\to0} \frac{\ln(1 + \frac{h}{x})}{h} \\ \end{align} $$ To simplify the logarithm: $$ \lim_{h\to0}\left (1 + \frac{h}{x}\right )^{\frac{1}{h}} = e^{\frac{1}{x}} $$ ^This is the line I have trouble with. I can see that it is true by putting numbers in, but I can't prove it. I know that $e^{\frac{1}{x}} = \lim_{h\to0}\left (1 + h \right )^{\frac{1}{xh}}$, but I can't work out how to get from the above line to that. $$ \lim_{h\to0}\left ( \left (1 + \frac{h}{x}\right )^{\frac{1}{h}}\right )^{h} = e^{\frac{h}{x}} $$ Going back to the derivative: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &amp;= \lim_{h\to0} \frac{\ln(e^{\frac{h}{x}})}{h} \\ &amp;= \lim_{h\to0} \frac{\frac{h}{x}\ln(e)}{h} \\ &amp;= \lim_{h\to0} \frac{h}{x} \div h\\ &amp;= \frac{1}{x} \\ \end{align} $$  This proof seems fine, apart from the middle step to get $e^{\frac{1}{x}}$. How could I prove that part?
33255
Let's say I want to use Maclaurin series to get the series expansion $S(x)$ for $f(x) = e^x$ where $S(x) = c_0x^0 + c_1x^1 + c_2x^2 + c_3x^3 + ...$  $f(0) = S(0) = c_0 = e^0 = 1$ so that's fine.  $f'(0) = S'(0) = c_1 = e^0 = 1$ again, fine.  $f''(0) = S''(0) = 2c_2 = e^0 = 1$ so $c_2 = 1/2$, fine.  $f'''(0) = S'''(0) = 6c_3 = e^0 = 1$ so $c_3 = 1/6$, fine.  $f''''(0) = S''''(0) = 24c_4 = e^0 = 1$ so $c_4 = 1/24$, fine.  And so on, so we conclude that $c_k = \frac{1}{k!}$ so then:  $$S(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$  But now what I don't understand is what allows us to go "Furthermore, $S(x) = f(x)$ for all $x$"! In fact what makes it valid for us to plug any other number other than $x=0$ into this? I see this equation used as a straight-up equivalent to $e^x$ even though we used $x=0$ and nothing else.  So I thought I would try it with a neighborhood of $1$ instead to see what happens:  $S(x) = c_0(x-1)^0 + c_1(x-1)^1 + c_2(x-1)^2 + c_3(x-1)^3 + ...$  $f(1) = S(1) = c_0 = e^1 = e$ so that's fine.  $f'(1) = S'(1) = c_1 = e^1 = e$ again, fine.  $f''(1) = S''(1) = 2c_2 = e^1 = e$ so $c_2 = e/2$, fine.  $f'''(1) = S'''(1) = 6c_3 = e^1 = e$ so $c_3 = e/6$, fine.  $f''''(1) = S''''(1) = 24c_4 = e^1 = e$ so $c_4 = e/24$, fine.  Looks pretty similar:  $$S(x) = \sum_{k=0}^{\infty} \frac{e(x-1)^k}{k!}$$  and, in fact, for a neighborhood of $x=a$:  $$S(x) = \sum_{k=0}^{\infty} \frac{e^a(x-a)^k}{k!}$$  Now we have $e$ inside the function expansion for $e^x$ which seems a little circular?  I guess my question is this: We created the function $S$ to have the same-valued $n$th derivatives as $f(x)$ but only at $x=0$ or $x=1$, or $x=a$, etc, but the series representations look different depending on which neighborhood I pick, and in some cases the expansions seem to include the very number we're trying to describe.  How do we know which is "right" or that it is even "right" to use for all $x$? I know the usual response to this is that it's equivalent when $f(x)$ is analytic but that doesn't help me at all because it says the function is analytic when the Taylor series around $x_0$ converges to the function in a neighborhood around $x_0$ for all $x_0$ in the function's domain. But how do I know this?  Yet again feels circular... just working through these two examples makes me wonder, how do I know these series converge to the function itself? How do I know they're equivalent representations? Am I supposed to avoid the self-referencing?  I'm looking for some context behind how to make sense of these two Taylor series and how I am supposed to know that $e$ is analytic or that I can use $S(x)$ for any $x$ I want even if I only computed it for the neighborhood around $x=0$.
12639
$\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$  The general term seems to be $$T_r= \frac{r+2}{r(r+1)(r+3)}.$$  I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.  Do I have to use something else? Or am I missing some easy manipulation?
74621
As my question says,   $$\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...$$  As far as $\frac 1{1.2}+\frac 1{2.3}+\frac 1{3.4}+...$ is concerned, we can write it as $\frac 1{n-1} - \frac 1{n}$ What can we do here?
122170
Suppose $f(0) =0 $ and  $0&lt;f''(x)&lt;\infty (\forall$  $x&gt;0)$, then $\frac{f(x)}{x}$ strictly increases as $x$ increases.    I have shown that $f'(x)-\frac{f(x)}{x}  = \frac{1}{2}xf''(c)$, for some $c\in (0,x)$. How do I proceed from here?
73841
So taking the derivative of the latter half we get  $$\left(\frac{2}{x^{3}}\right)e^{\frac{-1}{x^{2}}}$$ by chain rule. Im not sure how to find the derivative at x=0, and show that it equals 0 as it only produces more '1/0' expressions. Additionally I assume to show it is not analytic at $x_0=0$, we use the taylor series for $e^{x}$ which is $$1+x+\frac{x^2}{2}+\cdots+\frac{x^{n}}{n!}$$ Where we substitute $\frac{-1}{x^{2}}$ for x to get  $$e^{\frac{-1}{x^{2}}}=1-\frac{1}{x^{2}}+\frac{1}{x^{4}2!}-\frac{1}{x^{6}3!}\cdots$$ So when we plug in x=0 this is invalid and thus is not analytic.  Is my latter attempt sufficient or do I need to add more? And how do I reach the conclusion that the function is differentiable at x=0.
51238
If $x + \frac{1}{x} = 1$, then what is  $$ x^{2000} + \frac{1}{x^{2000}} = ?$$
49156
In an , I stated $$\sum\limits_{j=0}^M \frac{M \choose j}{N+M \choose j} = \frac{N+M+1}{N+1}.$$  It is clearly true when $N=0$ since you add up $M+1$ copies of $1$, and when $M=0$ since you add up one copy of $1$.  And, for example, with $M=4$ and $N=9$ you get $\frac{1}{1}+\frac{4}{13}+\frac{6}{78}+\frac{4}{286}+\frac{1}{715} = \frac{14}{10}$ as expected.  But how might you approach a general proof?
72097
$$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$   by directly substituting $x=0$ i got $\infty$ by using L-H's rule, i got $-1/8$   the given options are  $a)\frac{24e}{11}$    $b)\frac{11e}{24}$    $c)\frac{e}{11}$    $d)\frac{e}{24}$    may be the question would be  $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
29542
Let $J_{\alpha}(x)$be the Bessel function. Show the equality:  $$\frac{d}{dx} \left( xJ_{\alpha}(x)J_{\alpha +1}(x)\right) = x \left( J_{\alpha}^2(x) -J_{\alpha +1}^2(x)\right)$$  How to start with it?
78897
$\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$  I let, $x = \frac{1}{n}$, then as   $\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty$  L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty$  Again, if we apply L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}$. This is also going to $\infty$ as $x \to 0$. But, I know the answer is $\frac{-e}{2}$, and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule.
11387
On , I was bored and asked for  and the result was (indeterminate). Another two that give the same result are  and .  From what I know, given $x$ being any number, excluding $0$, $\frac{x}{x} = 1$ is true.  So just what, exactly, is $\infty$?
119760
Consider the functiong $g(x) =\frac{e^x−1}{x}$.  Find a general formula for $g^{(n)}(x)$and prove that this formula is correct.  If you want it as a  finite sum,  Based on guess and check, I think this one would work:  $$\frac{d^n}{dx^n}\frac{e^x−1}{x}=\frac{e^xn!(-1)^n+n!(-1)^{n+1}}{x^{n+1}}+ \frac{e^x}{x^{n+1}}\sum_{j=1}^{n}\frac{x^jn!(-1)^{n-j}}{j!}$$
36720
I know when evaluated it gives one but I can't figure out how to prove it.  Can anyone help? I believe it requires L'Hopitals rule and taking a natural log but i cannot figure out the exact math.