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46

Yes. They are the same thing. When exponents get really really complicated, mathematicians tend to start using $\exp(\mathrm{stuff})$ instead of $e^{\mathrm{stuff}}$. For example: $e^{x^5+2^x-7}$ is kind of hard to read. So instead one might write: $\exp(x^5+2^x-7)$. Note: For those who use Maple or other computer algebra systems, e^x is not usually the ...


29

Yes. The purpose for the notation $\exp$ is twofold: It allows one to talk about the exponentiation function itself, without specifying a particular input. For example, one can write that $\exp$ is a homomorphism from the additive group on $\mathbb{R}$ to the multiplicative group on $\mathbb{R}$. One may also say that $\exp$ and $\log$ are inverses. It ...


17

As other answers say, in your homework (and, indeed, in most places in mathematics) there is no difference. I have seen a beginning textbook first defining a certain function $\exp(x)$, then proving certain properties of it, and finally using those properties to motivate calling it $e^x$.


15

You may just write $$\begin{align} \int_{-\sqrt3}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx &=\int_{-\sqrt3}^{0}{e^x\over(e^x+1)(x^2+1)}dx+\int_{0}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx\\\\ &=\int_0^{\sqrt3}{e^{-x}\over(e^{-x}+1)(x^2+1)}dx+\int_{0}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx\\\\ ...


10

No such $F$ and $G$ exist. The fact that $F(0) + G(y) = e^{y}$ for all real $y$ would imply $G(y) = e^{y} - F(0)$. Similarly, $F(x) = e^{x} - G(0)$ for all real $x$. No matter how $F(0)$ and $G(0)$ are chosen, you're sunk.


9

The confusion might be that here $\log$ denotes the logarithm to base $e$ the Euler constant. So it is the natural logarithm, often also denoted $\ln$. Then that $n = \log (e^n)$ is true is just the definition of the logarithm.


8

Here is another solution. Applying $\dfrac{\partial^2}{\partial x \partial y}$ to the both sides of $F(x)+G(y)=e^{x+y}$, we have $0=e^{x+y}$. This never holds. Therefore the functions you want do not exist.


8

First, I assume that $b>0$. Since $y>0$, and since $\ln$ is an increasing and one-to-one function, when $y$ is maximized, $\ln y$ is also maximized and vice versa. Therefore take $\ln$ and maximize, i.e. \begin{align} f(x)&=\ln e^{-\frac{(x-a)^2}{b}}\\ &=-\frac{(x-a)^2}{b} \end{align} now since we assumed that $b>0$ we have that ...


7

I agree with these two answers, but I want to add one thing: Well defines. $e$ is some (positive) number, so (without knowing the function $\exp$), you can compute $e^n$ for $n \in \mathbb{N}$ – just multiply $e$ $n$ times with itself. You can also compute $e^{-n} = \frac{1}{e^n}$ and even $e^\frac{p}{q} = \sqrt[q] e^p$ (for $n, q \in \mathbb{N}, p \in ...


7

A commonly used set of algorithms is known as CORDIC: COordinate Rotation DIgital Computer. Basically, it uses a bunch of bitshifts, adds, subtracts and look up tables. Its good for cases where you don't have hardware multipliers and what not. You can also truncate series as well. An interesting thing is that you can often do some math and identify which ...


5

Hint. You may transform the equation $$ \frac{x-1}{e^x-1} = y \tag1 $$ with a little algebra into $$ -(x+y-1)e^{-(x+y-1)}=-ye^{1-y},\tag2 $$ set $X:=-(x+y-1)$ obtaining $$ Xe^X=-ye^{1-y} \tag3 $$ then use the Lambert function W to get $$ x=1-y-W\left(-ye^{1-y}\right). \tag4 $$


5

There is another way to solve. Let $$ A=\int_{-\sqrt3}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx, B=\int_{-\sqrt3}^{\sqrt3}{1\over(e^x+1)(x^2+1)}dx. $$ Clearly $$ A+B=\int_{-\sqrt3}^{\sqrt3}\frac{1}{x^2+1}=2\frac{\pi}{3}.$$ Changing variable $x\to -x$, it is easy to get $A=B$. Thus $$ A=\frac{\pi}{3}. $$


5

You can apply strictly monotonic increasing functions to both sides of an inequality. This is the case for $\log_2$, so just apply $\log_2$ to both sides. (A strictly monotonic decreasing function reverses the inequality direction.) Wikipedia has an article on monotonicity if you want to learn more about it.


5

Consider the function $$ g(s) = \exp(-sA)\frac{\partial}{\partial x}\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0} $$ We have, taking derivatives \begin{align*} g'(s) &= -A\exp(-sA)\frac{\partial}{\partial x}\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0}+ \exp(-sA)\frac{\partial}{\partial x}(A+xB)\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0}\\ &= ...


5

First, this result is false if $\Bbb C$ is replaced by $\Bbb R$ (even after making reasonable additional hypotheses to avoid the most obvious counterexamples) as Travis remarks, so one should use something specific for the complex numbers. Given a linear operator $\phi$ that we wish to realise as in the image of $\exp$, one may decompose the vector space ...


5

Let $|z| \le 1$ and $N \in \Bbb N$. Then $$ \left|\sum_{n = 1}^N \frac{z^n}{n!}\right| \le \sum_{n = 1}^N \frac{|z|^n}{n!} \le \sum_{n = 1}^N \frac{|z|^n}{2^{n-1}} \le |z| \sum_{n = 1}^\infty \left(\frac{1}{2}\right)^{n-1} = 2|z|.$$ Letting $N\to \infty$ results in $$|e^z - 1| \le 2|z|.$$


5

We can prove it by induction. For $n\geq 7$, let $S(n)$ denote the statement $$ S(n) : 3^n < n!. $$ Base case ($n=7$): $S(7)$ says that $3^7 = 2187<5040=7!$, and this is true. Induction step: Fix some $k\geq 7$, and assume that $S(k)$ is true where $$ S(k) : 3^k < k! $$ To be shown is that $S(k+1)$ is true where $$ S(k+1) : 3^{k+1} < (k+1)! $$ ...


4

$\frac {e^z - 1}{z}= 1 + z/2 + z^2/3! + \cdots .$ For $|z|\le 1,$ the absolute value of this is $\le 1 + 1/2 + 1/3! + \cdots = e - 1 <2.$


4

If that were the case, you would have for all $y$, $$F(1) - F(0)=F(1)+G(y) -(F(0)+G(y)) = e^{1+y}-e^y=e^y(e-1),$$ so $e^{y}(e-1)$ would be constant. Motivation: Initially I thought of taking a partial derivative, but with more generality, it suffices to take a difference $f(0+1,y) -f(0,y)$.


4

Although I agree with the answers already provided that in this situation (and indeed in most other ones in mathematics) there is no difference between the two notations, I would like to add the following for completeness: In manifold theory (most particularly Lie Group theory or Riemannian geometry), the exponential map $\exp$ is a map from a tangent space ...


4

Note that $1/a_n=e^n-e^{-n}=2\sinh n$. Since the function $\sinh$ is increasing, so is the sequence $1/a_n$, and therefore, $a_n$ is decreasing. Another way: $$\frac{a_{n+1}}{a_n}=\frac{e^{n+1}(e^{2n}-1)}{e^n(e^{2n+2}-1)}\leq e\frac{e^{2n}-1}{e^{2n+2}-e^2}=\frac1e<1$$


4

Look at the ratio:$$\frac{a_{n+1}}{a_n} = \frac{e^{n+1}}{e^{2n+2}-1}\frac{e^{2n}-1}{e^n}=\frac{e^{2n+1}-e}{e^{2n+2}-1}.$$Since $e^{2n+1}<e^{2n+2}$ and $e>1,$ the numerator is less than the denominator. So this ratio is less than $1.$


4

It applies the exponential function $\exp(y) = \sum \frac{1}{n!}y^n$ to the differential operator $a\frac{d}{dx}$ to give $$\exp\left(a\frac{d}{dx}\right) = \sum \frac{1}{n!}a^n \frac{d^n}{dx^n}.$$ This, when applied to $f(x)$, gives your middle expression.


4

I would be surprised if they actually used Taylor series. For example, the basic 80x87 "exponential" instruction is F2XM1 which computes $2^x-1$ for $0 \le x \le 0.5$. I don't think the implementation is documented, but if I were programming this, I might use a minimax polynomial or rational approximation: the following polynomial of degree $9$ ...


4

Let $$ D=\mathrm{diag}(2k_1\pi \mathrm{i},\ldots,2k_n\pi \mathrm{i}) \in\mathbb C^{n\times n}, $$ where $k_1,\ldots, k_n\in\mathbb Z$, and $U\in\mathbb C^{n\times n}$ invertible. Then, if $X=U^{-1}DU$, $$ \mathrm{e}^{X}=I. $$ If you are looking for a matrix with real elements, try $$ X=\left(\begin{array}{cc} 0 & 2\pi \\ -2\pi & 0\end{array}\right). ...


4

$$3^{(x^x)} = 1000 \to x^x \ln 3 = 3\ln 10 \to \ln(x^x) + \ln(\ln3)=\ln 3 + \ln(\ln 10) \to\\x\ln x = \ln 3 + \ln(\ln 10)-\ln(\ln 3) $$ that is how far i can go without resorting to some numerical approximation. my ti-83 gives me an approximate solution $x = 2.25772568.$


3

For $0 <|z| < 1$, $$\sum_{n = 2}^\infty \frac{|z|^n}{n!} \le \sum_{n = 2}^\infty \frac{|z|}{n!} = (e - 2)|z| < \frac{3}{4}|z|.$$


3

The tangent to $y=e^{bx}$ at $x=a$ is $$y-e^{ab}=be^{ab}(x-a)\ ,$$ which simplifies to $$y=be^{ab}x+(1-ab)e^{ab}\ .$$ For this to be the required line we need $$be^{ab}=10\ ,\quad (1-ab)e^{ab}=0\ ,$$ so $ab=1$ and hence $$b=\frac{10}{e^{ab}}=\frac{10}{e}\ .$$


3

No, it is not possible using elementary functions. After substituting $y= i \sqrt{x}$ you will get the integral of a Gaussian.


3

For any real $n\geq 0$, we have $$ n=\log_e\left(e^n\right) $$ $$ n=n\log_e\left(e\right) $$ $$ n=n\cdot 1$$ $$ n=n$$



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