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1

$$\int_0^N\frac{x}{1+x}dx=\int_0^N\left(1-\frac{1}{1+x}\right) dx=N-\ln(1+N)$$ Hence, $$H_N-N+\int_0^N\frac{x}{1+x}dx=H_N-N+N-\ln(1+N)=H_N-\ln(N)-\ln\left(1+\frac{1}{N}\right)$$ $H_N-\ln(N)\rightarrow \gamma$ and $\ln\left(1+\frac{1}{N}\right)\rightarrow 0$.


3

$$H_N-N+\int_0^N\left(1-\frac1{1+x}\right)dx=H_N-N+N-\log(N+1)=$$ $$=H_N-\log( N+1)\xrightarrow[N\to\infty]{}\gamma.$$


1

The error of both explicit and implicit Euler are $O(h)$. So $$f(x-h) = f(x) - h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) + \cdots$$ and $$f(x+h) = f(x) + h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots$$ So the backward Euler is $$f(x) - f(x-h) = h f'(x) - \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) - \cdots$$ $$f'(x) = ...


1

If for some $a<0<b$ we have $f\colon(a,b)\to\Bbb R$ with $f'(x)=f(x)$ for all $x\in(a,b)$ and $f(0)=1$, note that for any fixed $c\in(a,b)$ with $f(c)\ne 0$, we can define $$g_c(x)=\frac{f(x+c)}{f(c)} $$ which gives us a function $g_c\colon (a-c,b-c)\to \Bbb R$ with $$g_c'(x)=\frac{f'(x+c)}{f(c)}=\frac{f(x+c)}{f(c)}=g_c(x)$$ for all $x\in(a-c,b-c)$. In ...


0

Reading your long post, I come to the conclusion that you have some working definition of $a^{b}$ for $a, b \in \mathbb{R}$ and $a > 0$ (that's why you are talking about limit $e = \lim_{h \to 0}(1 + h)^{1/h}$). Although a working definition of $a^{b}$ without using any knowledge of $e^{x}$ is hard, I will assume that you have already managed to solve ...


0

1) you correctly got as far as $e^{ax}((C_1+C_2)\cos bx+i(C_1-C_2)\sin bx)$. But $C_1,C_2$ are arbitrary. So take $C_1=\frac{1}{2}B_1-\frac{1}{2}iB_2,C_2=\frac{1}{2}B_1+\frac{1}{2}iB_2$. Then $C_1+C_2=B_1,i(C_1-C_2)=B_2$, so your solution becomes $e^{ax}(B_1\cos bx+B_2\sin bx)$, which has the required form.


0

This identity can be derived using the Taylor Series expansions for $e^{ix}$, $\sin(x)$, and $\cos(x)$: $$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$$ $$i\sin(x)=i(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots)$$ $$e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots=1+ix-\frac{x^2}{2!}+\frac{ix^3}{3!}+\frac{x^4}{4!}-\cdots$$ ...


0

$e^y$ is a function whose range (output) are positive real numbers when the real numbers is its domain. However when we say $e^{ix}$ we are using the function $e^y$ on the complex number $ix$, thus we suddenly have a bit of a different function, acting on complex numbers, and thus we do not necessarily need to have a real number as output.


0

It is quite easy to construct such a function from a function, $f(x)$, whose limit at infinity is a finite rational number, $q$. If $r$ is an irrational number, then the limit of $g(x) = f(x) + (r - q)$ is clearly $r$. Similarly, $h(x) = \frac {f(x)r} q$ also has a limit of $r$ as $x$ goes to infinity.


3

Here's one you know: $\arctan x,$ whose limit at $\infty$ is $\pi/2.$


3

If $f(x) = 1/x^2 + \pi$, then $\lim_{x \to \infty} f(x) = \pi$ (and same with $\lim_{x \to -\infty} f(x) = \pi$), which is irrational.



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