New answers tagged

4

Another way. $\exp(-z^2)$ is an entire function. Its power series at the origin converges for all $z$. Take the anti-derivative of that series term-by-term. It still converges for all $z$. (Use the formula for radius of convergence.) So the series of antiderivatives is an antiderivative for $\exp(-z^2)$. That argument works for all entire functions. ...


3

Define, for some $z_0\in \mathbb{C}$ $$g(z)=\int_{z_0}^z e^{-t^2}dt$$ Now for any $2$ homotopic (within the domain where $e^{-z^2}$ is analytic) curves the value of this integral will be the same. Since $e^{-z^2}$ is analytic and $\mathbb{C}$ is simply connected, we find that the value of $g(z)$ is independent of the chosen contours. Hence this is a well ...


2

The composition of functions with an antiderivative has an antiderivative. $\exp$ has an antiderivative because it is its own derivative (and hence its own antiderivative) everywhere. $z \mapsto -z$ has an antiderivative $z \mapsto \frac{-z^2}{2}$. $z \mapsto z^2$ has antiderivative $z \mapsto \frac{z^3}{3}$.


1

Hint. Assume $x>0$. Then you get $$ f'(x)=\frac{-2x\times (1+2 \ln x)}{x^{2 x^2}} . $$ Can you take it from here?


0

This is not mainly a site for programming specific help, so don't be surprised if people sometimes vote down questions that are too specific about computer language. You can use the fsolve command and an anonymous function f = @(x)[60 14 63 34 21 12 11]' - 4*exp(0.034.*x); x_solution = fsolve(f,rand(1,7)');


2

If $M$ is complete, then $D_1$ is open, which follows essentially from the fact that $p \mapsto r_p$ is continuous. For instance, consider the map $f : TM \to TM$ given by $$ f : v \mapsto \frac{v}{r_p}, $$ where $p$ is the footpoint of $v$; $f$ is continuous, and your $D_1$ is the inverse image under $f$ of the open set $$ \lbrace v \in TM : ||v|| < ...


1

Set $s^{2\alpha+1}=w$. Then $$ I = \int_0^t s^{2\alpha - 1} \exp\left(\frac{i \sqrt{2} \left(t^{2 \alpha + 1} - s^{2 \alpha + 1}\right)}{ 2 \alpha + 1}\right)\mbox{d}{s}=(1+2\alpha)^{-1}\exp\left(\frac{i\sqrt{2}t^{2\alpha+1}}{2\alpha+1}\right)\int_0^{t^{2\alpha+1}}\frac{dw}{w^{\frac{1}{2\alpha+1}}} \exp\left(\frac{-i\sqrt{2}w}{2\alpha+1}\right) $$ $$ ...


0

Let $y=e^\lambda$, then we have $$py + (1-p)y^{-2} = 1. $$ Multiplying by $y^2$ and simplifying yields $$py^3 -y^2+1-p=0. $$ Factoring, we have $$(1-y)(1-p+(1-p)y + py^2)=0. $$ The solution $y=1$ has $\lambda =0$, so taking the other root, we have $$py^2 + (1-p)y + 1-p = 0. $$ Completing the square yields $$p\left(y -\frac12(1-p)\right)^2 - ...


1

HINT: Enforce the substitution $s\to s^{-1/b}$ to yield the integral $$\frac1b \int_{t^{-1/b}}^\infty e^{-\frac12 a^2s^s}\,ds$$


0

Only homogeneous equations with logarithms are invariant to change of base. Examples: $4 \log 2 \log(xy) = 2 \log 4 \log(x) + \log 16 \log (y)$ is base independent but $\log 1000000 = 6$ is true only for $\log_{10}$.


1

You're just misunderstanding what the article (in Wikipedia, here) says. The quote that explains why the natural log is uniquely called "natural" is this one: The natural logarithm can be defined for any positive real number $a$ as the area under the curve y = $1/x$ from $1$ to $a$ (the area being taken as negative when $a<1$). The simplicity of this ...


1

Repeating the quotation from Wikipedia: However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and are usually defined in terms of the latter. There is nothing here that says this property of the natural logarithm is unique. Indeed, if \begin{align} \log_a(x) &= A\log_e(x) &&\text{and}& ...


0

The intent of the proof is to relate the derivative at $x$ to the derivative at $0$, which is possible thanks to the property $a^{x+y}=a^xa^y$. It states $$f'(x)=(a^x)'=\lim_{h\to0}\frac{a^{x+h}-a^x}h=\lim_{h\to0}a^x\frac{a^{h}-1}h=a^x\lim_{h\to0}\frac{a^{h}-1}h=a^xL_a.$$ And obviously, setting $x=0$, you have $f'(0)=(a^x)'|_{x=0}=a^0L_a=L_a$. This is a ...


1

The limit is the same as the one that defines $f'(0)$ using the definition of derivative applied to this function $f$. The harder part of the proof, not written in the question, is to show that this limit exists and to determine its value (a logarithm) as a function of the base $a$. The calculation posted in the question is the easier algebraic step of ...


1

By definition, $$f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h}.$$ For $a > 0$ and $f(t) = a^t$, we get $$f'(t) = \lim_{h \to 0} \frac{a^{t+h} - a^t}{h}.$$ Now let $t = 0$: $$f'(0) = \lim_{h \to 0} \frac{a^{0+h} - a^0}{h} = \lim_{h \to 0} \frac{a^h - 1}{h}.$$


3

There is a mistake in your algebra. From your third equation $$\int\frac{\log_a(e)}{x}\,dx=\log_a(x)+C\tag1$$ you can deduce that for any $c$ $$\int\frac cx\,dx={c\over\log_a(e)}\log_a(x)+C.\tag2$$ (This is equivalent to the Wolfram result). Comparing (2) to your second equation, it doesn't follow that $c=\log_a(e)$.


0

the signs depend on the solutions of $k^x=x^k$, which cannot be solved by elementary means. Rewrite $$x=\left(\sqrt[k]k\right)^x=e^{x\ln(k)/k}$$ and set $u=-x\ln(k)/k$. Then $$e^{-u}=-\frac k{\ln(k)}u,\\ ue^u=-\frac{\ln(k)}k,\\ u=W\left(-\frac{\ln(k)} k\right),\\ x=-\frac k{\ln(k)}W\left(-\frac{\ln(k)} k\right).$$ where $W$ denotes the Lambert function. ...


4

You haven't really proven anything. You have used something that needs proving, to "prove" that the derivative of $\mathrm e^x$ is itself. It's really just a circular argument. I would suggest that you prove it from first principles: $$\mathrm f'(x) = \lim_{h \to 0} \left(\frac{\mathrm f(x+h)-\mathrm f(x)}{h}\right)$$ In your case, $\mathrm f(x) = \mathrm ...


1

For the record, I agree with @EthanBolker's answer above. With regards to the question of whether a proof of the fact exists without assuming the derivative of $e^x$ is $e^x$, the answer is yes. https://en.wikipedia.org/wiki/Logarithmic_derivative https://proofwiki.org/wiki/Derivative_of_Natural_Logarithm_Function The equality follows from the chain rule, ...


4

The answer to your question depends on what you expect your readers to know in advance. Since you're writing about derivatives, they should know calculus. Then the fact that $e^x$ is its own derivative is likely to be a better known prerequisite than the statement about logarithmic derivatives you want to use to prove it.


0

Let : $$ \alpha_{\text{min}} = \min \lbrace \alpha_{i}, \; 1 \leq i \leq n \rbrace. $$ Then, for all $x > 0$ and for all $i \in \lbrace 1,\ldots,n \rbrace$, $e^{-\alpha_{i}x} \leq e^{-\alpha_{\text{min}}x}$. Therefore : $$ \forall x > 0, \; f(x) = \Big( \sum_{j=1}^{N} e^{-\alpha_{j}x} \Big)^{1/2} \leq \sqrt{N}e^{-\alpha_{\text{min}}x/2}. $$


0

Here is a possible reduction(?) of complexity, that someone with more experience may be able to turn into a solution. Consider the integral $$ I = \int_\gamma \; g^{-(a+1)}\;\exp\left\{-\left(\frac{b}{g} + \frac{1}{2} \sum_{i=1}^{n} \frac{t_i^2}{g+\lambda_i^{\psi}} \right) \right\} \log g \;\mathrm{d}g $$ Choose the branch cut of $\log g$ to run along the ...


2

The base is the one that satisfies $$\begin{align*} \frac1x &= \frac{d}{dx}\log_a x\\ &=\lim_{h\to0}\frac{\log_a(x+h)-\log_a x}{h}\\ &= \lim_{h\to0}\log_a\left(1+\frac hx\right)^{1/h}\\ &= \lim_{h\to0}\log_a\left(1+\frac hx\right)^{x/hx}\\ &= \frac1x\log_a\left[\lim_{h\to0}\left(1+\frac hx\right)^{x/h}\right]\\ \end{align*}$$ It ...


2

It is the natural log! Think about it this way. Let $y=\ln(x)$. It follows that $e^y=x$. From there, we can differentiate implicity; $y'*e^y=1$. Now, $e^y=x$, so $x*y'=1$, and $y'=1/x$.


1

Because it is exponential decay, I would prefer to use $A(t)=A(0)e^{-kt}$. Then $\frac{300}{375}=e^{-72k}$, so $k=\frac{1}{72}\ln(375/300)$. Now for the half-life $\tau$, we have $\frac{1}{2}A(0)=A(0)e^{-k\tau}$, so $k\tau=\ln 2$. It follows that the half-life, in hours, is $72\frac{\ln 2}{\ln(375/300)}$.


0

Use that $$\exp (z) = \exp (\Re z)(\cos (\Im z) +i\sin (\Im z)).$$


0

Many trigonometric problems may be easily solved if you replaced the trig with exponentials. For example, $\sum_{n=0}^x\cos(n)$ can be changed into the "real" part of $\sum_{n=0}^xe^{ni}$, which is then found to be a very simple geometric sum. We can do the exact same thing with sine. Also, many calculus related trig problems can be easily solved as ...


0

That is true if and only if $x\in\mathbb{Z}$, else, I recommend you read on roots of unity.


0

Assume $\tau>0$ and $x,\xi \in \mathbb{R}$. Then using $\left|e^{-ia}\right|=1 $ for any $a \in \mathbb{R}$, we get $$ 0<\left|e^{-x(1/\tau - i\xi)}\right|=\left|e^{-x/\tau}\right|\times \underbrace{\left|e^{ ix\xi}\right|}_{\color{red}{=1}}\leq e^{-x/\tau} $$ giving $$ 0\leq\lim_{x \to \infty}\left|e^{-x(1/\tau - i\xi)}\right|\leq \lim_{x \to ...


0

Since you are calling for intuition, you may know that a fundamental property of the natural exponential is that it is equal to its derivative. Suppose that $$ f(x) = \sum_{n=0}^\infty a_n{x^n}\,,$$ is equal to its derivative. Then formally, $$ f'(x) = \sum_{n=1}^\infty n a_n{x^{n-1}}\,,$$ thus $$ f'(x) = \sum_{n=0}^\infty (n+1) a_{n+1}{x^{n}}\,,$$ thus, ...


0

If $k=\alpha+i\beta$ with $\beta>1$ then $$\left|e^{(ik+1)x}\right|=\left|e^{i\alpha x}\>e^{(1-\beta) x}\right|= \bigl|e^{(1-\beta) x}\bigr|\to0\qquad(x\to\infty)\ .$$ It follows that the improper integral $$\int_a^\infty e^{(ik+1)x}\>dx:=\lim_{b\to\infty}\int_a^b e^{(ik+1)x}\>dx\tag{1}$$ is convergent, and there is no "multiplication with ...


0

Consider using $P(t)=P_0\cdot 3^{t/3}$ where $t$ is in days. Then use $P(7)=10,000,000=10^7$ to find $P_0$.


0

Remember that $e^{x}$ is the inverse function of $\ln(x)$, so they "undo" each other: $e^{\ln(x)}=x$ and $\ln(e^x)=x$. Therefore you can rearrange your equation by doing this: \begin{equation} e^x = e^{\ln20} \\ \therefore e^x = 20 \end{equation} which is the "exponential form" of the equation.


0

Let's calculate it: $g(t s) = f(e^{t s},e^{t s}) = e^{\frac{t s}{2}} + 2 e^{\frac{t s}{2}} + \frac{3e^{t s}}{2e^{\frac{t s}{2}}} = \frac{9}{2}e^{\frac{t s}{2}} \not= t^n (\frac{9}{2}e^\frac{s}{2})$ for any $n$. Hence $g$ does not satisfy the definition: https://en.wikipedia.org/wiki/Homogeneous_function.


1

The number $e^{\text{something}}$ can always be written as $3^{\text{something else}}$, where “something” and “something else” only differ by a constant factor ($\ln(3)$, to be precise). So it doesn't really matter what base you use for your exponentials. It's just that $e^x$ is much more convenient when computing derivatives, for example. And almost all ...


3

This is what's known as a transcendental equation, which has no elementary closed form. It can, however be solved in terms of the Lambert W function. $\begin{align*}z + e^{-z} &= a\\ze^{z} + 1 &= ae^z\\e^z(z-a) &= -1\\e^{z-a}(z-a)&= -e^{-a}\\ z-a &= W(-e^{-a})\\z &= W(-e^{-a}) + a\end{align*}$


0

As noted in the comments, you can write $$z=W(-e^{-a})+a$$ but this fails to shed any real light on the problem. There aren't really any algebraic manipulations that you can do to work out anything useful. You can use numerical approximation techniques if you want a decimal answer though.


0

There is no elementary solution to that. Wolfram Alpha produces $$z = W(-e^{-a})+a $$ which involves the Lambert W function, but this probably doesn't make you much wiser in practice -- you would just solve it numerically.


1

You can see $e$ as the unique real number such that $$ \ln (e)=1 $$ giving $$ e=2.71828182845904523536028747135266249775724709370\ldots. $$ One may prove that $$ \ln(e^x)=x, \quad x \in \mathbb{R}, $$$$ e^{\ln (x)}=x, \quad x \in (0,\infty), $$$$ \frac{d}{dx}e^x =e^x \quad x \in \mathbb{R}, $$ $$ e^x =\sum_{n=0}^\infty\frac{x^n}{n!}\quad x \in \mathbb{R}. $$ ...


0

I find that using the Laplace transform is the easiest way for me to solve this. (That, I now remember, is what I used in the past.) \begin{align} \frac{dx}{dt} &= \frac{-x}{\tau}\\ -x(0) + sx_L &= \frac{-x_L}{\tau}\\ x_L(s+1/\tau)&=x(0)\\ x_L &= \frac{x(0)}{s + 1/\tau}\\ x&=x(0)e^{-t/\tau}. \end{align} Note the presence of the initial ...


1

$$\frac{\mathrm{d}x }{\mathrm{d} t} = \frac{-x}{\tau }$$ Seperate Variables $$\frac{\mathrm{d}x }x=\frac{\mathrm{d}t }{-\tau}$$ Integrate both sides $$\int \frac{\mathrm{d}x }x=\int \frac{\mathrm{d}t }{-\tau}$$ Antiderivative: $$\ln \left | x \right |= \frac{-t}{\tau} + c_{1}$$ Solve for Variable $$\left | x \right |=e^{(-t/\tau)+c_{1}}$$ $$\left | x \right ...


1

You want to integrate via separation of variables. So, $\dfrac{dx}{dt}=-\dfrac{x}{\tau}$ $\dfrac{dx}{x}=-\dfrac{dt}{\tau}\quad\Rightarrow\quad \int\dfrac{dx}{x}=-\int\dfrac{dt}{\tau} \quad \Rightarrow \quad \ln(x)=-\dfrac{t}{\tau}$ Thus, $e^{-\frac{t}{\tau}}=x$


6

I don't know if this is what you are looking for, but with $$ T_1(f) = f''', \quad T_2(f) = f'', \quad T_3(f) = f' $$ your given equations are satisfied.


1

$x\rightarrow e^{-A(x)}$ is bounded iff $g:x\rightarrow tr(e^{-A(x)}e^{-A^T(x)})$ has an upper bound iff the greatest singular value of $e^{-A(x)}$ has an upper bound; you cannot say anything more except if, for example, $A(x)$ is normal; then $g(x)=tr(e^{-(A+A^T)(x)})=\sum_i e^{-\sigma_i(x)}$ where $spectrum(A+A^T)=(\sigma_i)$. Thus $g(x)$ has an upper ...


0

Suppose that $A(x)$ is bounded, then you have $\|e^{-A(x)}\| \leq e^{\|-A(x)\|}$, therefore is bounded.


2

I let you finish...


2

If $n$ is a non-negative integer you can put $\mu = \lambda + b $ so that the integral becomes $$ e^b \int_b^{\infty} \frac{(\mu-b)^n e^{-\mu}}{\mu^2} d\mu = e^b \sum_{k=0}^n \binom{n}{k}(-b)^{n-k} \int_b^{\infty} \mu^{k-2} e^{-\mu} d\mu$$ Then, except for when $k=0$ or $1$, the integrals in the sum can be solved using integration by parts whereas $$ ...


3

Even for $n=0$, the integral requires special functions, see Wolfram Alpha, so there probably is no simple way of computing it. For $n=0$, the result is $$-e^b \text{Ei}(-b-\lambda )-\frac{e^{-\lambda }}{b+\lambda }$$


0

Look at the series expansion: $$e^{cx} = 1 + \sum_{n=1}^\infty \frac{(cx)^{n}}{n!} $$ Subtract one and dividing by x leaves: $$\sum_{n=1}^\infty \frac{(cx)^{n-1}}{n!} = c + \sum_{n=2}^\infty \frac{(cx)^{n-1}}{n!}$$ All terms for n>=2 contain some nonzero power of x and approach zero as x approaches zero.



Top 50 recent answers are included