Pedantic On Mathematical Software (Middle School Algebra)

Pedantic On Mathematical Software (Middle School Algebra)

17:37:00  

This time Maple get it wrong while Wolfram get it flawlessly. Consider the following middle-school level problem:

Solve the following equation on $x$

\[\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\]

Solution:

The equation is equivalent to $\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}+3
\right)^2}=1$ so that we have

\[\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\]

We divide into three cases:

i) When $\sqrt{x-1}< 2$ then we have $\sqrt{x-1}< 3$ so that  the above equation becomes
\[2-\sqrt{x-1}+3-\sqrt{x-1}=1\]
that is $\sqrt{x-1}=2$, contradicting $\sqrt{x-1}< 2$.

ii) When $\sqrt{x-1}> 3$ then we have $\sqrt{x-1}> 2$ so that the equation becomes
\[\sqrt{x-1}-2+\sqrt{x-1}-3=1\]
that is $\sqrt{x-1}=3$, contradicting $\sqrt{x-1}> 3$.

ii) When $2\leq \sqrt{x-1}\leq 3$ then we have
\[\sqrt{x-1}-2+3-\sqrt{x-1}=1\]
which is equivalent to $1=1$, and so all $x$ on this case are the solutions, those are $x\in [5,10]$.

However, if you input the command

\[solve(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}-1,x)\]

into Maple 16, it gives you an output

\[x=x\]

Other math  software like Wolfram Mathematica gives us the correct answer $5\leq x \leq 10$.

Using the method of squaring you'll get $1=1$ as well.
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Pedantic on Mathematical Software [Trap Limit]

Pedantic on Mathematical Software [Trap Limit]

08:11:00  

Wolfram Alpha result for

\[\lim_{x\rightarrow 0}\frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}\]

is "No result found in terms of standard mathematical functions" while Maple clearly says  the limit is  "Undefined".  Notice that Wolfram Alpha will say "The Limit Does Not Exist" for $\lim_{x\rightarrow 0} \sin \left(\frac{1}{x}\right)$.

Indeed Maple is right, the limit does not exist. It is important to notice that the following solution is wrong

\[\lim_{x\rightarrow 0} x\sin \frac{1}{x}=0 \qquad \text{and} \qquad \lim_{x\rightarrow 0} \frac{\sin{x}}{x}=1\] Then \[\lim_{x\rightarrow 0}\frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}=\lim_{x\sin \frac{1}{x} \rightarrow 0} \frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}=\lim_{y\rightarrow 0} \frac{\sin y}{y}=1\]

First, we cannot use the composition law of limit here, the law says that:

[Substitution Rule] Suppose that as $x\rightarrow a$, $g(x)$ converges properly to $A$ and that \[\lim_{y\rightarrow A} f(y)=L.\]
Then
\[\lim_{x\rightarrow a} f(g(x))=L\]

Though, $\lim_{x\rightarrow 0} x \sin\frac{1}{x}=0$, the convergent is not proper, thus the above rule can't be applied.

To see why the limit does not exist, We look back at the definition of the limit, we say that such a limit exists if we can find a real number $L$, such that:

\[\forall \epsilon > 0 \exists \delta > 0 , \forall x, \, \, \,  0< |x-c| < \delta \Rightarrow \left|f(x)-L\right| < \epsilon \]

The problem is:  for every $L \in \mathbb{R}$,  given an $\epsilon > 0$,  no matter how small we choose a neighbourhood of $0$ (by choosing a sufficiently small $\delta$), the function $x\sin \frac{1}{x}$ can be equal to zero for infinitely many values of $x$ on that neighbourhood, thus the function
\[\frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}\]
is undefined (infinitely many times) on that neighbourhood.

In other words, it is impossible for us to choose a neighbourhood around $0$ such that the function
\[\frac{\sin \left(x\sin\left(\frac{1}{x}\right)\right)}{x\sin \left(\frac{1}{x}\right)}\]
is defined on all points on that neighbourhood, hence no such $\delta$ can be chosen on the definition of limit, thus the limit does not exist.

More formally, for any $\delta > 0$, by Archimedean Principle there exists a positive integer $N$ such that $N>\frac{1}{2\pi \delta}$, thus we have $x_k=\frac{1}{2k\pi} \in (0,\delta)$ for every $k\geq N$, and on these points we have $x_k \sin \frac{1}{x_k} = 0$.

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Soal Persamaan Fungsi

Soal Persamaan Fungsi

20:07:00   Olimpiade Matematika,

Sebenarnya post ini cuma mau nge-test LaTex di Dynamic View nya blogger. Tapi ya sekalian kasi soal-Soal Fungsi berikut saya anggap susah.

Source nya nggak disebutkan, ada yang saya tahu ada yang saya lupa  (itung-itung buat yang mau latihan bisa ngilangin temptation untuk nge-browse solusinya di Mathlinks :D )
 ------------------------------------------------------
1. Tentukan semua fungsi $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ yang memenuhi
\[f(x)f(y)=f(y)f(xf(y))+\frac{1}{xy}\]

2. Tentukan semua fungsi $f: \mathbb{R} \rightarrow \mathbb{R}$ yang memenuhi : untuk setiap $x,y\in \mathbb{R}$
\[f(y+f(x))=f(x)f(y)+f(f(x))+f(y)-xy\]
untuk setiap $x,y \in \mathbb{R}$.

3. Tentukan semua fungsi $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ yang memenuhi: untuk setiap $x,y\in \mathbb{R}^+$
\[f(x)f(y)=2f(x+yf(x))\]

4. Misalkan $f(x) \in \mathbb{Z}[x]$ dengan $\text{deg}(f(x))=4$, buktikan terdapat $M \in \mathbb{N}$ sedemikian sehingga \[f(n!) \equiv \pmod {2^n-1}\] untuk setiap $n>M$

5. Tentukan semua fungsi $f: \mathbb{N} \rightarrow \mathbb{N}$ yang memenuhi: untuk setiap $m,n \in \mathbb{N}$
\[f(m^2+f(n))=f(m)^2+n\]

6. Tentukan semua fungsi $f: \mathbb{R}\rightarrow \mathbb{R}$ yang memenuhi: untuk setiap $x,y\in \mathbb{R}$
\[f(xf(x+y))=f(yf(x))+x^2\]

7. Tentukan semua fungsi $f:\mathbb{R} \rightarrow \mathbb{R}$ yang memenuhi: untuk setiap $x,y\in \mathbb{R}$
\[f(x-f(y))=f(f(y))+xf(y)+f(x)-1\]

8. Tentukan semua  fungsi $f: \mathbb{N} \rightarrow \mathbb{N}$ yang memenuhi untuk setiap $m,n \in \mathbb{N}$
\[f(f(m)^2+2f(n)^2)=m^2+2n^2\]

9. Tentukan semua fungsi $f:\mathbb{R} \rightarrow \mathbb{R}$ yang memenuhi: untuk setiap $x\in \mathbb{R}$ berlaku
\[f(x^2)=f(x)^2 \qquad \text{dan} \qquad f(x+1)=f(x)+1\]

10.Misalkan $f:\mathbb{R} \rightarrow \mathbb{R}$ fungsi monoton tak turun yang memenuhi
\[f(x+f(y))=f(f(x))+f(y)\]
untuk setiap $x,y \in \mathbb{R}$

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