Exponential Function

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For Euler’s number holds \[e\approx 2.718281828459046.\] We already saw as an example application of the quotient criterion that the above defined series converges for all \(x\in\mathbb{K}\). Now we present an estimate for \(\exp(x)\) if the series is replaced by a finite sum.

Proof:

1. Already shown as an example application of the cauchy product of series.

2. Since we have \(\overline{x_1+x_2}=\overline{x_1}+\overline{x_2}\) and \(\overline{x_1x_2}=\overline{x_1} \cdot \overline{x_2}\) for all \(x_1,x_2\in\mathbb{C}\), we have \[\exp(\bar{x})=\sum_{k=0}^\infty\frac{\bar{x}^k}{k!}=\overline{\sum_{k=0}^\infty\frac{{x}^k}{k!}}=\overline{\exp(x)}.\]

3. By definition of \(\exp\), we have \(\exp(0)=1\). From (i), we get \(\exp(x)\cdot\exp(-x)=\exp(x-x)=\exp(0)=1\). Then the statement follows.

4. From \(x\geq0\), we get \[\exp(x)=\sum_{k=0}^\infty\frac{x^k}{k!}=1+\sum_{k=1}^\infty\frac{x^k}{k!}\geq1.\] It is immediately clear that equality only holds for \(x=0\).

5. If \(x<0\), we get from (iii) that \(\exp(-x)>1\). Then, \(\exp(x)=\frac1{\exp(-x)}<1\).

6. First we show that \(\exp\) is continuous at \(0\).

7. Let \(x_1,x_2\in\mathbb{R}\) with \(x_1> x_2\). Then we have \(x_1-x_2>0\) and thus, by (iv), \(\exp(x_1-x_2)>1\). Since \(\exp(x_2)>0\), we have \[\exp(x_1)=\exp(x_2)\exp(x_1-x_2)>\exp(x_2).\]

8. The fact \(\lim_{x\to\infty}\exp(x)=\infty\) follows, since we have for \(x>0\) that \[\exp(x)=\sum_{k=0}^\infty\frac{x^k}{k!}=1+x+\sum_{k=2}^\infty\frac{x^k}{k!}>1+x.\] The statement (iii) then directly implies \(\lim_{x\to-\infty}\exp(x)=0\).

9. The injectivity of \(\exp:\mathbb{R}\to(0,\infty)\) follows from (vii). It remains to show surjectivity. Let \(y\in(0,\infty)\).
   Since \(\lim_{x\to-\infty}\exp(x)=0\), there exists some \(x_0\in\mathbb{R}\) such that \(\exp(x_0)<y\).
   Since \(\lim_{x\to\infty}\exp(x)=\infty\), there exists some \(x_1\in\mathbb{R}\) such that \(\exp(x_1)>y\).
   Now, by the Mean Value Theorem, (remember that \(\exp\) is continuous), there exists some \(x\in\mathbb{R}\) with \(\exp(x)=y\).

10. Let \(x=x_1+ix_2\) with \(x_1=\Re(x)\), \(x_2=\Im(x)\). Then \[| \exp(x) | = |\exp(x_1+ix_2)|=|\exp(x_1)\exp(ix_2)|=|\exp(x_1)||\exp(ix_2)|.\] If we now show that \(|\exp(ix_2)|=1\), the result is proven: Making use of (ii), we obtain \[|\exp(ix_2)|^2=\exp(ix_2)\cdot\overline{\exp(ix_2)}=\exp(ix_2-ix_2)=\exp(0)=1.\]\(\Box\)