The Fourier series expansion of x3in the interval −1 ≤ x < 1 with periodic continuation has

Correct Option: "only sine terms"

Step-by-Step Explanation:

To determine the nature of the terms in the Fourier series expansion of the function
$f\left(x\right)=x^3$
in the symmetric interval
$\&\#x2212;1\&\#x2264;x\&lt;1$,
we can analyze the symmetry (even or odd properties) of the function.

1. Define the Fourier Series and Coefficients:

As shown in the formula visible in the reference image, the general Fourier series expansion of a periodic function $f\left(x\right)$ over an interval $\&\#x2212;L\&\#x2264;x\&lt;L$ (where $L=1$ in this case) is given by:

$$f(x)=\frac{a_0}{2}+\underset{n=1}{\overset{\mathrm{\&\#x221E;}}{\&\#x2211;}}[a_n\cos \left(\frac{n\mathrm{\&\#x3C0;}x}{L}\right)+b_n\sin \left(\frac{n\mathrm{\&\#x3C0;}x}{L}\right)]$$

Where the coefficients are defined as:

$$a_0=\frac{1}{L}{\&\#x222B;}_{\&\#x2212;L}^{L}f(x)dx$$

$$a_n=\frac{1}{L}{\&\#x222B;}_{\&\#x2212;L}^{L}f(x)\cos \left(\frac{n\mathrm{\&\#x3C0;}x}{L}\right)dx$$

$$b_n=\frac{1}{L}{\&\#x222B;}_{\&\#x2212;L}^{L}f(x)\sin \left(\frac{n\mathrm{\&\#x3C0;}x}{L}\right)dx$$

2. Analyze Function Symmetry:

To determine if the function is even, odd, or neither, we evaluate $f(\&\#x2212;x)$:

$$f(\&\#x2212;x)=(\&\#x2212;x{)}^3=\&\#x2212;x^3=\&\#x2212;f(x)$$

Since $f(\&\#x2212;x)=\&\#x2212;f\left(x\right)$, the function
$f\left(x\right)=x^3$
is a strictly odd function.

3. Evaluate Fourier Coefficients:

For an odd function integrated over a symmetric interval $\&\#x2212;L\&\#x2264;x\&lt;L$:

• The integral of any odd function over a symmetric interval is zero, so:
  $a_0=0$
• The product of an odd function $f\left(x\right)$ and an even function $\cos (n\mathrm{\&\#x3C0;}x/L)$ is an odd function. Integrating this product over a symmetric interval yields:
  $a_n=0$
• The product of an odd function $f\left(x\right)$ and an odd function $\sin (n\mathrm{\&\#x3C0;}x/L)$ is an even function. Therefore, the sine coefficients $b_n$ are generally non-zero:
  $b_n=\frac{2}{L}{\&\#x222B;}_0^L{x}^3\sin \left(\frac{n\mathrm{\&\#x3C0;}x}{L}\right)dx\&\#x2260;0$

Conclusion:

Since both the constant term coefficient $a_0$ and the cosine coefficients $a_n$ are identically zero, the Fourier series expansion of $x^3$ consists only of sine terms.