MHB Gaussian Quadrature Formula for Integrating Polynomials of Degree 6

[mathmari]

Hey!

I want to calculate the integral $$\int_0^1\frac{1}{x+3}\, dx$$ with the Gaussian quadrature formula that integrates exactly all polynomials of degree $6$.

The gaussian quadrature integrates exactly polynomials $\Phi (x)$ with maximum degree $2n-1$. In this case we consider $n=4$.

The formula is \begin{equation*}\int_{0}^1\frac{1}{x+3}\, dx\approx \sum_{i=1}^nf(x_i)\cdot w_i=f(x_1)\cdot w_1+f(x_2)\cdot w_2+f(x_3)\cdot w_3+f(x_4)\cdot w_4\end{equation*}

For the calculations of $w_i$ we need the following:
\begin{align*}&P_n(x)=(a_n+x)P_{n-1}(x)+c_nP_{n-2}(x) \\ &P_0=1, \ P_{-1}=0\end{align*} with \begin{equation*}a_n=-\frac{\langle x\cdot P_{n-1}, P_{n-1}\rangle_w}{\langle P_{n-1}, P_{n-1}\rangle_w}, \ \ \ , c_n=-\frac{\langle P_{n-1}, P_{n-1}\rangle_w}{\langle P_{n-2}, P_{n-2}\rangle_w}\end{equation*} right? What is in this case the weight function $w(x)$ ?

(Wondering)

 

[I like Serena]

Hey mathmari!

Can't we do a substitution on the integral so that it's on [-1,1]?
Then the standard weights apply, and we have the points and weights as given in Gaussian quadrature.
If we want to, afterwards we can do a reverse substitution to find the points on [0,1]. (Thinking)

 

[mathmari]

Can't we do a substitution on the integral so that it's on [-1,1]?
Then the standard weights apply, and we have the points and weights as given in Gaussian quadrature.
If we want to, afterwards we can do a reverse substitution to find the points on [0,1]. (Thinking)

Ah ok, so we get \begin{equation*}\int_0^1\frac{1}{t+3}\, dt=\int_{-1}^1\frac{1}{\frac{1}{2}+\frac{x}{2}+3}\, \frac{1}{2}dx=\int_{-1}^1\frac{1}{x+7}\, dx\end{equation*}

Since we want that the Gaussian quadrature formula integrates exactly all polynomials of degree 6, do we consider $n=4$ ? (Wondering)

 

[I like Serena]

Ah ok, so we get \begin{equation*}\int_0^1\frac{1}{t+3}\, dt=\int_{-1}^1\frac{1}{\frac{1}{2}+\frac{x}{2}+3}\, \frac{1}{2}dx=\int_{-1}^1\frac{1}{x+7}\, dx\end{equation*}

Since we want that the Gaussian quadrature formula integrates exactly all polynomials of degree 6, do we consider $n=4$ ?

Yes.
And with $n=4$ we get exact integration of polynomials up to degree $2n-1=7$, so that includes indeed degree $6$. (Nod)

 

[mathmari]

Yes.
And with $n=4$ we get exact integration of polynomials up to degree $2n-1=7$, so that includes indeed degree $6$. (Nod)

Ah ok! (Nerd)

If the values of $x_i$ and $w_i$ are not given and we have to calcualte them by ourself, how can we do that? In Wiki there is a formula for $w_i$, do we use this one? But what about $x_i$ ? (Wondering)

 

[I like Serena]

Ah ok!

If the values of $x_i$ and $w_i$ are not given and we have to calcualte them by ourself, how can we do that? In Wiki there is a formula for $w_i$, do we use this one? But what about $x_i$ ?

Looks like it yes.

Wiki also mentions:

the associated orthogonal polynomials are Legendre polynomials, denoted by P_n(x). With the n-th polynomial normalized to give P_n(1) = 1, the i-th Gauss node, x_i, is the i-th root of P_n​

(Thinking)

 

[I like Serena]

I want to calculate the integral $$\int_0^1\frac{1}{x+3}\, dx$$ with the Gaussian quadrature formula that integrates exactly all polynomials of degree $6$.

For the calculations of $w_i$ we need the following:
\begin{align*}&P_n(x)=(a_n+x)P_{n-1}(x)+c_nP_{n-2}(x) \\ &P_0=1, \ P_{-1}=0\end{align*} with \begin{equation*}a_n=-\frac{\langle x\cdot P_{n-1}, P_{n-1}\rangle_w}{\langle P_{n-1}, P_{n-1}\rangle_w}, \ \ \ , c_n=-\frac{\langle P_{n-1}, P_{n-1}\rangle_w}{\langle P_{n-2}, P_{n-2}\rangle_w}\end{equation*}

If the values of $x_i$ and $w_i$ are not given and we have to calcualte them by ourself, how can we do that? In Wiki there is a formula for $w_i$, do we use this one? But what about $x_i$ ?

Wiki also mentions:

The integration problem can be expressed in a slightly more general way by introducing a positive weight function ω into the integrand, and allowing an interval other than [−1, 1]. That is, the problem is to calculate
$$\int _{a}^{b}\omega (x)\,f(x)\,dx$$​

I believe we can freely pick $\omega(x)$. Whatever makes the calculations the easiest. (Wondering)

In our case we have $a=0$, $b=1$, and $\omega (x)\,f(x)=\frac{1}{x+3}$.
So I think we should either pick $\omega (x)=1$ or $\omega (x)=\frac{1}{x+3}$.

Suppose we pick $\omega (x)=1$.
Then:
\begin{equation*}a_1=-\frac{\langle x\cdot P_0, P_0\rangle_\omega}{\langle P_0, P_0\rangle_\omega}
= -\frac{\langle x\cdot 1, 1\rangle_\omega}{\langle 1, 1\rangle_\omega}
= -\frac{\int_0^1 \omega(x)x\,dx}{\int_0^1 \omega(x)\,dx}
= -\frac{\int_0^1 x\,dx}{\int_0^1 dx}
= -\frac {\frac 12}1 \\
P_1(x)=(a_1+x)P_0(x)+c_1P_{-1}(x)
= \left(-\frac 12+x\right)1+c_1\cdot 0 = x-\frac 12
\end{equation*}
(Thinking)

If we would pick $n=1$, then $x_i$ would be the $i$-th root of $P_1(x)$. That is, we would have $x_1=\frac 12$.
For $n=4$ we need to do a bit more work. (Sweating)

 

[mathmari]

Wiki also mentions:

The integration problem can be expressed in a slightly more general way by introducing a positive weight function ω into the integrand, and allowing an interval other than [−1, 1]. That is, the problem is to calculate
$$\int _{a}^{b}\omega (x)\,f(x)\,dx$$​

I believe we can freely pick $\omega(x)$. Whatever makes the calculations the easiest. (Wondering)

In our case we have $a=0$, $b=1$, and $\omega (x)\,f(x)=\frac{1}{x+3}$.
So I think we should either pick $\omega (x)=1$ or $\omega (x)=\frac{1}{x+3}$.

Ahh ok! So, do we not have to shift an arbitrary interval into $[-1,1]$ ? I thought that these formulas would hold only at $[-1,1]$. (Wondering)

 

[I like Serena]

Ahh ok! So, do we not have to shift an arbitrary interval into $[-1,1]$ ? I thought that these formulas would hold only at $[-1,1]$.

No, the interval [-1,1] is merely a convenient choice, and tables are readily available for the resulting $P_n(x)$, $x_i$, and $w_i$.
Also, it's not so easy to find the roots of $P_n(x)$ for $n=4$, which is a 4-th degree polynomial after all. (Thinking)

 

[mathmari]

No, the interval [-1,1] is merely a convenient choice, and tables are readily available for the resulting $P_n(x)$, $x_i$, and $w_i$.

So, do you suggest to let the interval of the integral at $[0,1]$ or to shift it to $[-1,1]$ ? (Wondering)

Also, it's not so easy to find the roots of $P_n(x)$ for $n=4$, which is a 4-th degree polynomial after all. (Thinking)

But in this case we have to calculate the roots of the $4$th degree polynomial, or is there also an other way? (Wondering)

 

[I like Serena]

So, do you suggest to let the interval of the integral at $[0,1]$ or to shift it to $[-1,1]$ ?

It's easiest to approximate the integral with a Gaussian quadrature by shifting to $[-1,1]$ and using the tabulated data.
You had already almost done that. It makes sense to finish it, so that we can use it for comparisons.

However, I think the purpose of the problem is to learn about weight functions, weights, and how Gaussian quadrature really works. For that I think we should try both choices for $\omega(x)$ and see what we get.
Maybe one of them works out really well. (Wait)

But in this case we have to calculate the roots of the $4$th degree polynomial, or is there also an other way?

I'm not sure yet. Maybe we will find a 4-th degree polynomial that is not so hard.
And anyway, we can always solve it with for instance Wolfram. (Thinking)

 

[mathmari]

I have done the following:

\begin{align*}&a_1=-\frac{\langle x\cdot P_0, P_0\rangle_\omega}{\langle P_0, P_0\rangle_\omega}
= -\frac{\langle x\cdot 1, 1\rangle_\omega}{\langle 1, 1\rangle_\omega}
= -\frac{\int_0^1 \omega(x)x\,dx}{\int_0^1 \omega(x)\,dx}
= -\frac{\int_0^1 x\,dx}{\int_0^1 dx}
= -\frac {\frac 12}1 \\
&P_1(x)=(a_1+x)P_0(x)+c_1P_{-1}(x)
= \left(-\frac 12+x\right)1+c_1\cdot 0 = x-\frac 12
\end{align*}

\begin{align*}&a_2=-\frac{\langle x\cdot P_{1}, P_{1}\rangle_w}{\langle P_{1}, P_{1}\rangle_w}=-\frac{\langle x\cdot \left (x-\frac 12\right ), x-\frac 12\rangle_w}{\langle x-\frac 12, x-\frac 12\rangle_w}=-\frac{\int_0^1x\cdot \left (x-\frac 12\right )\cdot w(x)\, dx}{\int_0^1\left (x-\frac 12\right )^2\cdot w(x)\, dx}=-\frac{\int_0^1\left (x^2-\frac x2\right )\, dx}{\int_0^1\left (x^2-x+\frac 14\right )\, dx}=-\frac{\frac{1}{12}}{\frac{1}{12}}=-1 \\ & c_2=-\frac{\langle P_{1}, P_{1}\rangle_w}{\langle P_{0}, P_{0}\rangle_w}=-\frac{\langle x-\frac 12, x-\frac 12\rangle_w}{\langle 1, 1\rangle_w}=-\frac{\int_0^1\left (x-\frac 12\right )^2\cdot w(x)\, dx}{\int_0^1 w(x)\, dx}=-\frac{\int_0^1\left (x-\frac 12\right )^2\, dx}{\int_0^1 1\, dx}=-\frac{\frac{1}{12}}{1}=-\frac{1}{12} \\ & P_2(x)=(a_2+x)P_{1}(x)+c_2P_{0}(x)=(-1+x)\cdot \left (x-\frac{1}{2}\right )-\frac{1}{12}\cdot 1=x^2-\frac{3x}{2}+\frac{5}{12}
\end{align*}

\begin{align*}&a_3=-\frac{\langle x\cdot P_{2}, P_{2}\rangle_w}{\langle P_{2}, P_{2}\rangle_w}=-\frac{\langle x\cdot \left (x^2-\frac{3x}{2}+\frac{5}{12} \right ), x^2-\frac{3x}{2}+\frac{5}{12}\rangle_w}{\langle x^2-\frac{3x}{2}+\frac{5}{12}, x^2-\frac{3x}{2}+\frac{5}{12}\rangle_w}=-\frac{\int_0^1 x\cdot \left (x^2-\frac{3x}{2}+\frac{5}{12} \right )^2\, dx}{\int_0^1 \left ( x^2-\frac{3x}{2}+\frac{5}{12}\right )^2\, dx}=-\frac{\frac{11}{1440}}{\frac{19}{720}}=-\frac{11}{38} \\ & c_3=-\frac{\langle P_{2}, P_{2}\rangle_w}{\langle P_{1}, P_{1}\rangle_w}=-\frac{\langle x^2-\frac{3x}{2}+\frac{5}{12}, x^2-\frac{3x}{2}+\frac{5}{12}\rangle_w}{\langle x-\frac 12, x-\frac 12\rangle_w}=-\frac{\int_0^1\left ( x^2-\frac{3x}{2}+\frac{5}{12}\right )^2\, dx}{\int_0^1\left ( x-\frac 12\right )^2\, dx}=-\frac{\frac{19}{720}}{\frac{1}{12}}=-\frac{19}{60} \\ & P_3(x)=(a_3+x)P_{2}(x)+c_3P_{1}(x)=\left (-\frac{11}{38}+x\right )\left (x^2-\frac{3x}{2}+\frac{5}{12}\right )-\frac{19}{60}\left (x-\frac 12\right ) \\ & =x^3-\frac{34}{19}x^2+\frac{203}{380}x+\frac{43}{1140}\end{align*}

\begin{align*}&a_4=-\frac{\langle x\cdot P_{3}, P_{3}\rangle_w}{\langle P_{3}, P_{3}\rangle_w}=-\frac{\int_0^1x\cdot \left (x^3-\frac{34}{19}x^2+\frac{203}{380}x+\frac{43}{1140}\right )^2\, dx}{\int_0^1\left (x^3-\frac{34}{19}x^2+\frac{203}{380}x+\frac{43}{1140}\right )^2\, dx}=-\frac{\frac{120583}{12129600}}{\frac{2087}{159600}}=-\frac{120583}{158612} \\ &c_4=-\frac{\langle P_{3}, P_{3}\rangle_w}{\langle P_{2}, P_{2}\rangle_w}=-\frac{\int_0^1 \left (x^3-\frac{34}{19}x^2+\frac{203}{380}x+\frac{43}{1140} \right )^2\, dx}{\int_0^1\left (x^2-\frac{3x}{2}+\frac{5}{12}\right )^2\, dx}=-\frac{\frac{2087}{159600}}{\frac{19}{720}}=-\frac{6261}{12635} \\ &P_4= (a_4+x)P_{3}(x)+c_4P_{2}(x)=\left (-\frac{120583}{158612}+x\right )\left (x^3-\frac{34}{19}x^2+\frac{203}{380}x+\frac{43}{1140}\right )-\frac{6261}{12635}\left (x^2-\frac{3x}{2}+\frac{5}{12}\right )\\ & =x^4-\frac{21285}{8348}x^3+\frac{1553409}{1110284}x^2+\frac{24973673}{66617040}x-\frac{15664717}{66617040}\end{align*}

But I must have done something wrong because the roots (as we see in Wolfram) are not the same as the ones that are in Wiki. Do you have an idea what I have done wrong? (Wondering)

 

[I like Serena]

I have done the following:

\begin{align*}&a_1=-\frac{\langle x\cdot P_0, P_0\rangle_\omega}{\langle P_0, P_0\rangle_\omega}
= -\frac{\langle x\cdot 1, 1\rangle_\omega}{\langle 1, 1\rangle_\omega}
= -\frac{\int_0^1 \omega(x)x\,dx}{\int_0^1 \omega(x)\,dx}
= -\frac{\int_0^1 x\,dx}{\int_0^1 dx}
= -\frac {\frac 12}1 \\
&P_1(x)=(a_1+x)P_0(x)+c_1P_{-1}(x)
= \left(-\frac 12+x\right)1+c_1\cdot 0 = x-\frac 12
\end{align*}

\begin{align*}&a_2=-\frac{\langle x\cdot P_{1}, P_{1}\rangle_w}{\langle P_{1}, P_{1}\rangle_w}
=-\frac{\langle x\cdot \left (x-\frac 12\right ), x-\frac 12\rangle_w}{\langle x-\frac 12, x-\frac 12\rangle_w}
=-\frac{\int_0^1x\cdot \left (x-\frac 12\right )\cdot w(x)\, dx}{\int_0^1\left (x-\frac 12\right )^2\cdot w(x)\, dx}=-\frac{\int_0^1\left (x^2-\frac x2\right )\, dx}{\int_0^1\left (x^2-x+\frac 14\right )\, dx}
=-\frac{\frac{1}{12}}{\frac{1}{12}}=-1
\end{align*}

Don't we have:
$$a_2=-\frac{\langle x\cdot P_{1}, P_{1}\rangle_w}{\langle P_{1}, P_{1}\rangle_w}=-\frac{\langle x\cdot \left (x-\frac 12\right ), x-\frac 12\rangle_w}{\langle x-\frac 12, x-\frac 12\rangle_w}=-\frac{\int_0^1x\cdot \left (x-\frac 12\right )^2\cdot w(x)\, dx}{\int_0^1\left (x-\frac 12\right )^2\cdot w(x)\, dx}$$
That is, with a square in the numerator? (Thinking)

But I must have done something wrong because the roots (as we see in Wolfram) are not the same as the ones that are in Wiki. Do you have an idea what I have done wrong? (Wondering)

We wouldn't get exactly the same roots, since we have scaled [-1,1] to [0,1].
But we should get more or less the same distribution yes.
And I would also expect the roots to be in the interval [0,1], because otherwise we cannot use them to calculate $f(x_i)$. (Worried)

 

[mathmari]

Don't we have:
$$a_2=-\frac{\langle x\cdot P_{1}, P_{1}\rangle_w}{\langle P_{1}, P_{1}\rangle_w}=-\frac{\langle x\cdot \left (x-\frac 12\right ), x-\frac 12\rangle_w}{\langle x-\frac 12, x-\frac 12\rangle_w}=-\frac{\int_0^1x\cdot \left (x-\frac 12\right )^2\cdot w(x)\, dx}{\int_0^1\left (x-\frac 12\right )^2\cdot w(x)\, dx}$$
That is, with a square in the numerator? (Thinking)

Ahh yes. I did the calculations again and now I get the following:
\begin{align*}&a_1=-\frac{\langle x\cdot P_0, P_0\rangle_\omega}{\langle P_0, P_0\rangle_\omega}
= -\frac{\langle x\cdot 1, 1\rangle_\omega}{\langle 1, 1\rangle_\omega}
= -\frac{\int_0^1 \omega(x)x\,dx}{\int_0^1 \omega(x)\,dx}
= -\frac{\int_0^1 x\,dx}{\int_0^1 dx}
= -\frac {\frac 12}1 \\
&P_1(x)=(a_1+x)P_0(x)+c_1P_{-1}(x)
= \left(-\frac 12+x\right)1+c_1\cdot 0 = x-\frac 12
\end{align*}

\begin{align*}&a_2=-\frac{\langle x\cdot P_{1}, P_{1}\rangle_w}{\langle P_{1}, P_{1}\rangle_w}=-\frac{\langle x\cdot \left (x-\frac 12\right ), x-\frac 12\rangle_w}{\langle x-\frac 12, x-\frac 12\rangle_w}=-\frac{\int_0^1x\cdot \left (x-\frac 12\right )^2\cdot w(x)\, dx}{\int_0^1\left (x-\frac 12\right )^2\cdot w(x)\, dx}=-\frac{\int_0^1x\cdot \left (x-\frac 12\right )^2\, dx}{\int_0^1\left (x-\frac 12\right )^2\, dx}=-\frac{\frac{1}{24}}{\frac{1}{12}}=-\frac{1}{2} \\ & c_2=-\frac{\langle P_{1}, P_{1}\rangle_w}{\langle P_{0}, P_{0}\rangle_w}=-\frac{\langle x-\frac 12, x-\frac 12\rangle_w}{\langle 1, 1\rangle_w}=-\frac{\int_0^1\left (x-\frac 12\right )^2\cdot w(x)\, dx}{\int_0^1 w(x)\, dx}=-\frac{\int_0^1\left (x-\frac 12\right )^2\, dx}{\int_0^1 1\, dx}=-\frac{\frac{1}{12}}{1}=-\frac{1}{12} \\ & P_2(x)=(a_2+x)P_{1}(x)+c_2P_{0}(x)=\left (-\frac{1}{2}+x\right )\cdot \left (x-\frac{1}{2}\right )-\frac{1}{12}\cdot 1=x^2-x+\frac{1}{6}
\end{align*}

\begin{align*}&a_3=-\frac{\langle x\cdot P_{2}, P_{2}\rangle_w}{\langle P_{2}, P_{2}\rangle_w}=-\frac{\langle x\cdot \left (x^2-x+\frac{1}{6} \right ), x^2-x+\frac{1}{6}\rangle_w}{\langle x^2-x+\frac{1}{6}, x^2-x+\frac{1}{6}\rangle_w}=-\frac{\int_0^1 x\cdot \left (x^2-x+\frac{1}{6} \right )^2\, dx}{\int_0^1 \left ( x^2-x+\frac{1}{6}\right )^2\, dx}=-\frac{\frac{1}{360}}{\frac{1}{180}}=-\frac{1}{2} \\ & c_3=-\frac{\langle P_{2}, P_{2}\rangle_w}{\langle P_{1}, P_{1}\rangle_w}=-\frac{\langle x^2-x+\frac{1}{6}, x^2-x+\frac{1}{6}\rangle_w}{\langle x-\frac 12, x-\frac 12\rangle_w}=-\frac{\int_0^1\left ( x^2-x+\frac{1}{6}\right )^2\, dx}{\int_0^1\left ( x-\frac 12\right )^2\, dx}=-\frac{\frac{1}{180}}{\frac{1}{12}}=-\frac{1}{15} \\ & P_3(x)=(a_3+x)P_{2}(x)+c_3P_{1}(x)=\left (-\frac{1}{2}+x\right )\left (x^2-x+\frac{1}{6}\right )-\frac{1}{15}\left (x-\frac 12\right )=x^3-\frac{3}{2}x^2+\frac{3}{5}x-\frac{1}{20}\end{align*}

\begin{align*}&a_4=-\frac{\langle x\cdot P_{3}, P_{3}\rangle_w}{\langle P_{3}, P_{3}\rangle_w}=-\frac{\int_0^1x\cdot \left (x^3-\frac{3}{2}x^2+\frac{3}{5}x-\frac{1}{20}\right )^2\, dx}{\int_0^1\left (x^3-\frac{3}{2}x^2+\frac{3}{5}x-\frac{1}{20}\right )^2\, dx}=-\frac{\frac{1}{5600}}{\frac{1}{2800}}=-\frac{1}{2} \\ &c_4=-\frac{\langle P_{3}, P_{3}\rangle_w}{\langle P_{2}, P_{2}\rangle_w}=-\frac{\int_0^1 \left (x^3-\frac{3}{2}x^2+\frac{3}{5}x-\frac{1}{20} \right )^2\, dx}{\int_0^1\left (x^2-x+\frac{1}{6}\right )^2\, dx}=-\frac{\frac{1}{2800}}{\frac{1}{180}}=-\frac{9}{140} \\ &P_4= (a_4+x)P_{3}(x)+c_4P_{2}(x)=\left (-\frac{1}{2}+x\right )\left (x^3-\frac{3}{2}x^2+\frac{3}{5}x-\frac{1}{20}\right )-\frac{9}{140}\left (x^2-x+\frac{1}{6}\right )=x^4-2x^3+\frac{9}{7}x^2-\frac{2}{7}x+\frac{1}{70}\end{align*}

Is there a way s that we can verify if we got the correct polynomial? (Wondering)

We wouldn't get exactly the same roots, since we have scaled [-1,1] to [0,1].
But we should get more or less the same distribution yes.
And I would also expect the roots to be in the interval [0,1], because otherwise we cannot use them to calculate $f(x_i)$. (Worried)

Now we get these roots: Wolfram, which are in the interval $[0,1]$.

(Wondering)

 

[I like Serena]

Is there a way that we can verify if we got the correct polynomial?

Yes. We can verify if $\langle P_i,P_j\rangle_w=\delta_{ij}$. (Thinking)

Now we get these roots: Wolfram, which are in the interval $[0,1]$.

That looks about right. Nice! (Happy)

 

[mathmari]

Yes. We can verify if $\langle P_i,P_j\rangle_w=\delta_{ij}$. (Thinking)

But it doesn't hold that $\langle P_i,P_i\rangle_w=1$, does it? (Wondering)

If we took the other weight function, $w(x)=\frac{1}{x+3}$ instead of $w(x)=1$, would we get other polynomials? (Wondering)

 

[I like Serena]

But it doesn't hold that $\langle P_i,P_i\rangle_w=1$, does it?

True. The polynomials are orthogonal, but they have not been normalized to $1$. (Blush)
So we should have for each $i\ne j$ that $\langle P_i,P_j\rangle_w=0$.

If we took the other weight function, $w(x)=\frac{1}{x+3}$ instead of $w(x)=1$, would we get other polynomials?

I think so yes. (Thinking)

 

[mathmari]

True. The polynomials are orthogonal, but they have not been normalized to $1$. (Blush)
So we should have for each $i\ne j$ that $\langle P_i,P_j\rangle_w=0$.

Ah ok! (Malthe)

I think so yes. (Thinking)

But both are corrects, aren't they? (Wondering)

 

[I like Serena]

Ah ok!

But both are corrects, aren't they?

Yes. (Nod)