It's true that zero-padding in the time domain corresponds to interpolation in the frequency domain. If you have a length $N$ signal $x[n]$, its discrete Fourier transform (DFT) is given by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{1}$$

The signal $x[n]$ can be expressed in terms of its DFT coefficients $X[k]$ by the inverse DFT

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{2}$$

Since the coefficients $X[k]$ contain the same information as the signal $x[n]$, anything that can be computed from $x[n]$ can also be computed from $X[k]$.

Let $\tilde{X}(\omega)$ denote the discrete-time Fourier transform (DTFT) of $x[n]$:

$$\tilde{X}(\omega)=\sum_{n=0}^{N-1}x[n]e^{-jn\omega}\tag{3}$$

Note that $\omega$ is a continuous variable. Comparing (1) and (3) shows that the DFT is simply a sampled version of the DTFT:

$$X[k]=\tilde{X}(2\pi k/N),\quad k=0,\ldots,N-1\tag{4}$$

Furthermore, a length $L$ DFT ($L>N$) of $x[n]$ simply corresponds to a more densely sampled version of $\tilde{X}(\omega)$:

$$X_L[l]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nl/L}=\tilde{X}(2\pi l/L),\quad l=0,\ldots,L-1\tag{5}$$

Now let's express $\tilde{X}(\omega)$ in terms of the length $N$ DFT of $x[n]$. Rewriting (3) using (2) gives

$$\begin{align}\tilde{X}(\omega)&=\sum_{n=0}^{N-1}\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}e^{-jn\omega}\\ &=\sum_{k=0}^{N-1}X[k]\underbrace{\frac{1}{N}\sum_{n=0}^{N-1}e^{-jn(\omega-2\pi k/N)}}_{G(\omega-2\pi k/N)} \end{align}\tag{6}$$

where $G(\omega)$ is an interpolation function which can be expressed by

$$G(\omega)=\frac{1}{N}\sum_{n=0}e^{-jn\omega}=\frac{e^{-j\omega(N-1)/2}}{N}\frac{\sin(N\omega/2)}{\sin(\omega/2)}\tag{7}$$

By setting $\omega=2\pi l/L$ in (6) and using (5) you can see that the length $L$ zero-padded DFT of $x[n]$ can be computed from the length $N$ DFT using the interpolation function given by (7):

$$X_L(l)=\tilde{X}(2\pi l/L)=\sum_{k=0}^{N-1}X[k]G(2\pi l/L-2\pi k/N)\tag{8}$$

It's true that zero-padding in the time domain corresponds to interpolation in the frequency domain. If you have a length $N$ signal $x[n]$, its discrete Fourier transform (DFT) is given by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{1}$$

The signal $x[n]$ can be expressed in terms of its DFT coefficients $X[k]$ by the inverse DFT

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{2}$$

Since the coefficients $X[k]$ contain the same information as the signal $x[n]$, anything that can be computed from $x[n]$ can also be computed from $X[k]$.

Let $\tilde{X}(\omega)$ denote the discrete-time Fourier transform (DTFT) of $x[n]$:

$$\tilde{X}(\omega)=\sum_{n=0}^{N-1}x[n]e^{-jn\omega}\tag{3}$$

Note that $\omega$ is a continuous variable. Comparing (1) and (3) shows that the DFT is simply a sampled version of the DTFT:

$$X[k]=\tilde{X}(2\pi k/N),\quad k=0,\ldots,N-1\tag{4}$$

Furthermore, a length $L$ DFT ($L>N$) of $x[n]$ simply corresponds to a more densely sampled version of $\tilde{X}(\omega)$:

$$X_L[l]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nl/L}=\tilde{X}(2\pi l/L),\quad l=0,\ldots,L-1\tag{5}$$

Now let's express $\tilde{X}(\omega)$ in terms of the length $N$ DFT of $x[n]$. Rewriting (3) using (2) gives

$$\begin{align}\tilde{X}(\omega)&=\sum_{n=0}^{N-1}\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}e^{-jn\omega}\\ &=\sum_{k=0}^{N-1}X[k]\underbrace{\frac{1}{N}\sum_{n=0}^{N-1}e^{-jn(\omega-2\pi k/N)}}_{G(\omega-2\pi k/N)} \end{align}\tag{6}$$

where $G(\omega)$ is an interpolation function which can be expressed by

$$G(\omega)=\frac{1}{N}\sum_{n=0}^{N-1}e^{-jn\omega}=\frac{e^{-j\omega(N-1)/2}}{N}\frac{\sin(N\omega/2)}{\sin(\omega/2)}\tag{7}$$

By setting $\omega=2\pi l/L$ in (6) and using (5) you can see that the length $L$ zero-padded DFT of $x[n]$ can be computed from the length $N$ DFT using the interpolation function given by (7):

$$X_L(l)=\tilde{X}(2\pi l/L)=\sum_{k=0}^{N-1}X[k]G(2\pi l/L-2\pi k/N)\tag{8}$$

It's true that zero-padding in the time domain corresponds to interpolation in the frequency domain. If you have a length $N$ signal $x[n]$ its discrete Fourier transform (DFT) is given by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{1}$$

It can be expressed in terms of its DFT coefficients $X[k]$ by the inverse DFT

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{2}$$

Since the coefficients $X[k]$ contain the same information as the signal $x[n]$, anything that can be computed from $x[n]$ can also be computed from $X[k]$.

Let $\tilde{X}(\omega)$ denote the discrete-time Fourier transform (DTFT) of $x[n]$:

$$\tilde{X}(\omega)=\sum_{n=0}^{N-1}x[n]e^{-jn\omega}\tag{3}$$

Note that $\omega$ is a continuous variable. Comparing (1) and (3) shows that the DFT is simply a sampled version of the DTFT:

$$X[k]=\tilde{X}(2\pi k/N),\quad k=0,\ldots,N-1\tag{4}$$

Furthermore, a length $L$ DFT ($L>N$) of $x[n]$ simply corresponds to a more densely sampled version of $\tilde{X}(\omega)$:

$$X_L[l]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nl/L}=\tilde{X}(2\pi l/L),\quad l=0,\ldots,L-1\tag{5}$$

Now let's express $\tilde{X}(\omega)$ in terms of the length $N$ DFT of x[n]. Rewriting (3) using (2) gives

$$\begin{align}\tilde{X}(\omega)&=\sum_{n=0}^{N-1}\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}e^{-jn\omega}\\ &=\sum_{k=0}^{N-1}X[k]\underbrace{\frac{1}{N}\sum_{n=0}e^{-jn(\omega-2\pi k/N)}}_{G(\omega-2\pi k/N)} \end{align}\tag{6}$$

where $G(\omega)$ is an interpolation function which can be expressed by

$$G(\omega)=\frac{1}{N}\sum_{n=0}e^{-jn(\omega-2\pi k/N)}=\frac{e^{-j\omega(N-1)/2}}{N}\frac{\sin(N\omega/2)}{\sin(\omega/2)}\tag{7}$$

So by setting $\omega=2\pi l/L$ in (6) and using (5) you can see that the length $L$ zero-padded DFT of $x[n]$ can be computed from the length $N$ DFT using the interpolation function given by (7):

$$X_L(l)=\tilde{X}(2\pi l/L)=\sum_{k=0}^{N-1}X[k]G(2\pi l/L-2\pi k/N)\tag{8}$$

It's true that zero-padding in the time domain corresponds to interpolation in the frequency domain. If you have a length $N$ signal $x[n]$, its discrete Fourier transform (DFT) is given by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{1}$$

The signal $x[n]$ can be expressed in terms of its DFT coefficients $X[k]$ by the inverse DFT

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{2}$$

Since the coefficients $X[k]$ contain the same information as the signal $x[n]$, anything that can be computed from $x[n]$ can also be computed from $X[k]$.

Let $\tilde{X}(\omega)$ denote the discrete-time Fourier transform (DTFT) of $x[n]$:

$$\tilde{X}(\omega)=\sum_{n=0}^{N-1}x[n]e^{-jn\omega}\tag{3}$$

Note that $\omega$ is a continuous variable. Comparing (1) and (3) shows that the DFT is simply a sampled version of the DTFT:

$$X[k]=\tilde{X}(2\pi k/N),\quad k=0,\ldots,N-1\tag{4}$$

Furthermore, a length $L$ DFT ($L>N$) of $x[n]$ simply corresponds to a more densely sampled version of $\tilde{X}(\omega)$:

$$X_L[l]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nl/L}=\tilde{X}(2\pi l/L),\quad l=0,\ldots,L-1\tag{5}$$

Now let's express $\tilde{X}(\omega)$ in terms of the length $N$ DFT of $x[n]$. Rewriting (3) using (2) gives

$$\begin{align}\tilde{X}(\omega)&=\sum_{n=0}^{N-1}\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}e^{-jn\omega}\\ &=\sum_{k=0}^{N-1}X[k]\underbrace{\frac{1}{N}\sum_{n=0}^{N-1}e^{-jn(\omega-2\pi k/N)}}_{G(\omega-2\pi k/N)} \end{align}\tag{6}$$

where $G(\omega)$ is an interpolation function which can be expressed by

$$G(\omega)=\frac{1}{N}\sum_{n=0}e^{-jn\omega}=\frac{e^{-j\omega(N-1)/2}}{N}\frac{\sin(N\omega/2)}{\sin(\omega/2)}\tag{7}$$

By setting $\omega=2\pi l/L$ in (6) and using (5) you can see that the length $L$ zero-padded DFT of $x[n]$ can be computed from the length $N$ DFT using the interpolation function given by (7):

$$X_L(l)=\tilde{X}(2\pi l/L)=\sum_{k=0}^{N-1}X[k]G(2\pi l/L-2\pi k/N)\tag{8}$$