16
votes
log of summation expression
Sometimes we have asymptotics with most significant term $M \to \infty$ we do this
$$
\log(M+A) = \log(M\cdot(1+S))
$$
so that $S=A/M$ is "small" in the sense $S=o(1)$, and then
$$
\log(M\cdot(1+S)) = ...
- 107k
15
votes
Accepted
Show that $ \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=0}^{n-1}e^{ik^2}=0$
Gauss sums
Your sum is strongly related to the Gauss sum. The usual trick is to compute the modulus. This works particularly smoothly over $\mathbf{Z}/p\mathbf{Z}$ as with usual Gauss sums, but ...
- 9,525
10
votes
log of summation expression
There's this:
$$\log\left(\sum_{i=0}^n x_i\right) = \log(x_0) + \log\left(1+\sum_{i=1}^n\left(\frac{x_i}{x_0}\right)\right)$$
$x_0$ must be the biggest in the series. It didn't help me with what I'm ...
- 111
9
votes
log of summation expression
You may note that this is equivalent to trying to solve $$\log(a+b+c)$$You realize you can't do much about $a,b,$ or $c$.
The only way this can be simplified, is if you can factor something out and ...
9
votes
Accepted
Prove the following series $\sum\limits_{s=0}^\infty \frac{1}{(sn)!}$
It's $\sum\limits_{r=0}^{n-1} e^{i\frac{2\pi k}{n}r}=\frac{e^{i\frac{2\pi k}{n}n}-1}{e^{i\frac{2\pi k}{n}}-1}$ with $=0$ for $k\neq l\cdot n$ and $=n$ for $k=l\cdot n$, $l\in\mathbb{Z}$.
It follows $\...
- 11.4k
9
votes
Accepted
what are the zeroes of $f(x) = 2^x + 3^x - 5^x?$
Notice that $f(x)=0 \iff f(x)5^{-x}=0$. Therefore, it suffices to study $g(x)=f(x)5^{-x}$
$$g(x)= \left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x-1 $$
The function $g$ is decreasing as a sum of ...
- 15.2k
8
votes
Accepted
exponential equation: $6^x+8^x+15^x=9^x+10^x+12^x$
Let $a=3^x$ and $b=2^x$ and $c=5^x$.
Then we have that
$$ab+b^3+ac=a^2+bc+ab^2$$
$$ab+b^3+ac-a^2-bc-ab^2=0$$
$$a(b-a)+b^2(b-a)-c(b-a)=0$$
$$(b-a)(a+b^2-c)=0$$
Now $a=b \implies x=0$ and $a+b^2-c=0 \...
- 24.4k
8
votes
Accepted
How do I find the real and imaginary part of $z+ e^z $
HINT:
You are on the right track. Note that $e^{iy}=\cos(y)+i\sin(y)$.
- 176k
7
votes
Number of zeros in polynomial-exponential sums
The original questioner asked if there was a version of Descartes' rule of signs for sums of exponential functions. The answer is yes.
Theorem. For $n\ge 0$, let $p_0>p_1>\cdots > p_n > ...
- 119
7
votes
Accepted
Limit $\lim_{x\to \infty}\left(\frac{1^x+2^x+3^x+...n^x}{n}\right)^{\dfrac{a}{x}}$
Let $f(x)$ denote the argument of the limit.
The following inequalities trivially hold:
$$
\left(\frac{n^x}{n}\right)^{a/x} \leq f(x) \leq
\left(\frac{n \cdot n^x}{n}\right)^{a/x}.
$$
Hence
$$
(n^a)^{...
- 14.3k
7
votes
Accepted
Evaluating $\sum_{n=0}^{\infty}ne^{1-n}$ using calculus
Thanks to motivation from @Gerry Myerson. I'm attempting to answer my own question. Let the integral in question be denoted by $\mathrm I$
.$$\begin{aligned}\mathrm I &=\int_{0}^{\infty}xe^{1-x}\...
- 6,449
7
votes
Accepted
Number of zeros in difference of exponential sums: $\sum\limits_{i=1}^n a_i^x - \sum\limits_{i=1}^n b_i^x$
Here is an argument (revised, with nontrivial input from the OP) that the number of roots cannot exceed $n$. It has a calculus formula piece, a variation diminishing piece, and a combinatorics piece....
- 23.6k
7
votes
Accepted
A nested double sum(to do with e?)
Another way of looking at this expression:
Let your sum be $s=\sum_{i=0}^\infty\frac{1}{i!}\sum_{j=0}^i\frac{1}{j!}$. Then (summations are interchangeable because of absolute convergence):
\begin{...
- 5,131
6
votes
If $x = 2\log_39 + \log_{27}5,$ then $3^x = ??$
$$\begin{align}x= 2\log_{3}9+\log_{27}5\\
=2\log_{3}9+{1\over3}\log_{3}5 \end{align}$$
$$\begin{align}3^x=3^{[2\log_{3}9+1/3\log_{3}5]}&
&\\=3^{2\log_{3}9}3^{1/3\log_{3}5}
&\\=9^{2\log_{...
- 1,379
6
votes
Accepted
What is the formula for finding the summation of an exponential function?
You can recognize your sum as a geometric sum which has the basic formula:
$$
\sum_{n=0}^N r^n = \frac{r^{N+1} - 1}{r-1}
$$
To apply this to your sum
$$
\sum_{n=1}^{50} e^{-0.123(n)}
$$
recognize that ...
- 1,045
6
votes
All real number for which $n$ in $5^n+7^n+11^n=6^n+8^n+9^n$
Consider the function $f(x)=x^n$ for positive $x$. Its second derivative is $n(n-1)x^{n-2}$ and therefore for $n > 1$ or $n<0$ $f$ is strictly convex while for $0 < n < 1$ f is strictly ...
- 2,077
6
votes
Accepted
Finding value of $\int_{0}^{1} \int_{y}^{1} e^{x^2} dx dy$
Note that, since the region of integration is the triangle whose vertices ae $(0,0)$, $(1,0)$, and $(1,1)$, your integral is equal to$$\int_0^1\int_0^xe^{x^2}\,\mathrm dy\,\mathrm dx=\int_0^1xe^{x^2}\,...
- 421k
6
votes
Accepted
Why is $\frac{|1 - e^{2 \pi i \alpha N}|}{|1 - e^{2 \pi i \alpha}|} \leq \frac{\sin(\pi \alpha N)}{\sin(\pi \alpha)} \leq \frac{1}{||\alpha||}$?
We have that by $\sin x=\frac{e^{ix}-e^{-{ix}}}{2i}$
$$ \frac{|1 - e^{2 \pi i \alpha N}|}{|1 - e^{2 \pi i \alpha}|} = \frac{|e^{ \pi i \alpha N}|}{|e^{ \pi i \alpha }|}\frac{|\sin( \pi \alpha N)|}{|\...
- 151k
6
votes
Accepted
Standard way to evaluate $\sum_{k=0}^\infty \frac{x^k}{(k+1/2)!}$?
$$S(x)=\sum_{k=0}^\infty \frac{x^k}{(k+1/2)!}=\sum_{k=0}^\infty \frac{(2x)^k (2k)!!}{(2k+1)!}=\sum_{k=0}^\infty \frac{(4x)^k k!}{(2k+1)!}=\sum_{k=0}^\infty \frac{(4x)^k }{k!}\frac{(k!)^2}{(2k+1)!}$$
...
- 12.8k
5
votes
exponential equation: $6^x+8^x+15^x=9^x+10^x+12^x$
Hint: your equation can be factorized as $$\left(2^x-3^x\right) \left(2^{2 x}+3^x-5^x\right)=0$$
5
votes
How do you differentiate $\ln(\exp(z))$
$$\begin{array}{rcl}
\displaystyle \frac{\mathrm d}{\mathrm dz} \ln(\exp(z))
&=& \displaystyle \frac{\mathrm d}{\mathrm d\exp(z)} \ln(\exp(z)) \cdot \frac{\mathrm d\exp(z)}{\mathrm dz} \\
&...
- 24.9k
5
votes
Accepted
What is the summation of the series $1 + \frac{1+2} {2!} + \frac{1+2+2^2} {3!} + \frac{1+2+2^2+2^3} {4!}$ + . . . . till infinity?
$$\sum_{n=1}^\infty\frac1{n!}\sum_{k=0}^{n-1}2^k=\sum_{n=1}^\infty\frac1{n!}(2^n-1)=\sum_{n=1}^\infty\frac{2^n}{n!}-\sum_{n=1}^\infty\frac{1^n}{n!}=(e^2-1)-(e-1)=e^2-e$$
- 64.3k
5
votes
Accepted
Summation over multiple arguments
Proceeding from right to left:
$$\begin{align*}\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =\sum_{s_1=\pm1} \sum_{s_2=\pm1}e^{-{s_1s_2}}\left[ e^{{s_2}}+e^{-{s_2}} ...
- 5,604
5
votes
Accepted
Sum of series with exponential denominator
Let $$f(q) =\sum_{j=1}^{\infty}\frac{j^3q^j}{1-q^j},\,|q|<1$$ and $$g(q) =\sum_{j=1}^{\infty} \frac{j^3q^j}{1+q^j},\,|q|<1$$ then we have $$f(q^2)=\frac{f(q)-g(q)}{2}$$ or $$g(q) =f(q) - 2f(q^2)$...
- 85.1k
4
votes
How to find the value of this expression?
Hint You may try with the Sophie Germain identity. As $(10^4+324)=(10^4+4\times3^4)$
- 7,111
4
votes
Accepted
Prove $\sum_{i=0}^n {{n \choose i} \times 2^i} = 3^n $
In the binomial theorem, $$(x+y)^n={n \choose 0}x^{n-0}y^0+{n \choose 1}x^{n-1}y^1+{n \choose 2}x^{n-2}y^2+\cdots+{n \choose n}x^{n-n}y^n=\sum_{i=1}^{n} {n \choose i} x^{n-i}y^i$$
Put $x=1$ and $y=2$....
- 24.4k
4
votes
Accepted
Generate exponential weights (sum of all = 1)
Let $x_i = a^i$, then setting $w_i = \dfrac{x_i}{\sum_{i=1}^{500} x_i}$, we obtain what we want. We have
$$\sum_{i=1}^{500} a^i = a \left(\dfrac{1-a^{500}}{1-a}\right)$$
Hence, we have
$$w_i = \left(\...
- 20k
4
votes
Accepted
Closed form of a series
The typical way to "skip terms" in a power series is to use roots of unity. Here, let $\zeta=e^{i\pi/3}$ be a primitive $6$th root of unity, so that
$$
\frac16 \sum_{j=0}^5 (\zeta^\ell)^j = \begin{...
- 73.6k
4
votes
Prove the following series $\sum\limits_{s=0}^\infty \frac{1}{(sn)!}$
$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\...
- 88.1k
4
votes
Solving limits with ln without L’Hôpital’s rule
Hint:
$$
\ln(1+e^{ax})=\ln(e^{ax}(e^{-ax}+1))=ax+\ln(e^{-ax}+1)
$$
If $a>0$, then $\lim_{x\to\infty}e^{-ax}=0$.
However, you have to distinguishing between the cases $a>0$, $a=0$, $a<0$ and ...
- 236k
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