**Description**

Let $\vec{U}=\{u_0,u_1,\ldots,u_m\}$ denotes a non-decreasing sequence of real numbers, i.e, $u_i\leq u_{i+1} \quad i=0,1,2\ldots m-1$.

and the $i$-th B-spline basis function of $p$-degree, denoted by $N_{i,p}(u)$, is defined as below:

$$N_{i,0}(u)= \begin{cases} 1 & u_i\leq u<u_{i+1}\\ 0 & otherwise \end{cases} $$ $$N_{i,p}(u)=\frac{u-u_i}{u_{i+p}-u_i}N_{i,p-1}(u)+\frac{u_{i+p+1}-u}{u_{i+p+1}-u_{i+1}}N_{i+1,p-1}(u) $$

**Aim**

Now I want to solve the `maximal value point`

of the $N_{i,p}(u)$, the following steps is my initial trial.

For instance, now I have a knots vector $U=\{0, 0, 0, 0, 1/5, 1/4, 1/3, 1/2, 1, 1, 1, 1\}$. Namely, $m=11$ and I assume that the degree of the basis function is $3$.

- How to solve
**the maximal value point**of the $N_{3,3}(u)$?

**Step 1:** Calculate the mathematical expression of $N_{3,3}(u)$

To achieve the $N_{3,3}(u)$ expression, I draw the following schematic diagram

Then I ultilize the recursive formula to calculate the mathematical expression of $N_{3,3}(u)$ **by hand**.

$$ N_{3,3}(x)= \begin{cases} 60 x^3 & 0\leq x<\frac{1}{5} \\ -10 \left(119 x^3-75 x^2+15 x-1\right) & \frac{1}{5}\leq x<\frac{1}{4} \\ 10 \left(73 x^3-69 x^2+21 x-2\right) & \frac{1}{4}\leq x<\frac{1}{3} \\ -10 \left(8 x^3-12 x^2+6 x-1\right) & \frac{1}{3}\leq x\leq \frac{1}{2} \\ \end{cases} $$

**Step 2:** Differentiate the $N_{3,3}(x)$ expression with respect to the varible $x$
$$
\frac{d}{dx}N_{3,3}(x)=
\begin{cases}
180 x^2 & 0\leq x<\frac{1}{5} \\
-10 \left(357 x^2-150 x+15\right) & \frac{1}{5}\leq x<\frac{1}{4} \\
10 \left(219 x^2-138 x+21\right) & \frac{1}{4}\leq x<\frac{1}{3} \\
-10 \left(24 x^2-24 x+6\right) & \frac{1}{3}\leq x\leq \frac{1}{2} \\
\end{cases}
$$

**Step 3:** Set the $N'_{3,3}(x)$ to $0$. Namely,
$$
\begin{cases}
180 x^2=0 & 0\leq x<\frac{1}{5} \\
-10 \left(357 x^2-150 x+15\right)=0 & \frac{1}{5}\leq x<\frac{1}{4} \\
10 \left(219 x^2-138 x+21\right)=0 & \frac{1}{4}\leq x<\frac{1}{3} \\
-10 \left(24 x^2-24 x+6\right)=0 & \frac{1}{3}\leq x\leq \frac{1}{2} \\
\end{cases}
$$
$$
\Rightarrow \quad
\left(
\begin{array}{c}
x_1=0 \\
x_2=\frac{1}{119} \left(25-\sqrt{30}\right) \\
x_3=\frac{1}{73} \left(23-3 \sqrt{2}\right) \\
x_4=\frac{1}{2} \\
\end{array}
\right)
$$

**Step 4:** Compare to the value of $N_{3,3}(x_1),N_{3,3}(x_2),N_{3,3}(x_3)$ and $N_{3,3}(x_4)$

By the calculation, I discovered that the `maximal value point`

is

$$x_3=\frac{1}{73} \left(23-3 \sqrt{2}\right)$$

### QUESTION

- However, I think my initial trial is very
`complicated`

and very`time-consuming`

. I would like to know is there a`simple/elegant method`

to solve this question. Because in my actual work, I need to calculate`all the maximal value point`

of $N_{i,p}(u)$ where, $i \in [0, m-n-1]$. Thanks a lot sincerely!