How can I correctly write and display df and f' in a recent post?

[Math Amateur]

In a recent post i tried to write/state the following ... ... ...

I am most interested in how/why we know that

$$\text{df} (h) = ( ​

\text{df}_1 (h), / ... / ... / ... / ​

\text{df}_m (h) )​

$$

... and also that ...$$f' (p) = \begin(bmatrix) f'_1 (p) \\ f'_2 (p) \\ . \\ . \\ . \\ f'_n (p) \end(bmatrix) $$

​

... ... ...
BUT ...... the above did not post and display correctly ...
... ... can someone inform me of the errors in the above ...Peter
​

 

[MountEvariste]

First one didn't work because you inserted colour at several places.

$$\text{df} (h) =  ( [FONT=Tahoma]\text{df}_1 (h), / ... / ... / ... /  [FONT=Tahoma][FONT=Tahoma]\text{df}_m (h) )$$[LEFT][FONT=Tahoma][LEFT][FONT=Tahoma]

You wanted $df(h) = (df_1(h), \ldots, df_m(h))$ which is given by

df(h) = (df_1(h), \ldots, df_m(h))

I used \ldots for the dotting since that's more natural-looking.

The second one didn't render because you used ( instead of {. This

\displaystyle f' (p) = \begin{bmatrix} f'_1 (p) \\ f'_2 (p) \\ . \\ . \\ . \\ f'_n (p) \end{bmatrix}

$\displaystyle f' (p) = \begin{bmatrix} f'_1 (p) \\ f'_2 (p) \\ . \\ . \\ . \\ f'_m (p) \end{bmatrix}$

$\displaystyle\mathbf{f}' (\mathbf{p}) = \begin{bmatrix} f'_1 (p) \\ f'_2 (p) \\  \vdots \\ f'_n (p) \end{bmatrix}$

$\displaystyle\mathbf{f}' (\mathbf{p}) = \begin{bmatrix} f'_1 (\mathbf{p}) \\ f'_2 (\mathbf{p}) \\ \vdots \\ f'_m (\mathbf{p}) \end{bmatrix}$

Using \vdots for the $\vdots$ and \mathbf{p} to write $\mathbf{p}$

 

[Math Amateur]

First one didn't work because you inserted colour at several places.

$$\text{df} (h) =  ( [FONT=Tahoma]\text{df}_1 (h), / ... / ... / ... /  [FONT=Tahoma][FONT=Tahoma]\text{df}_m (h) )$$[LEFT][FONT=Tahoma][LEFT][FONT=Tahoma]

You wanted $df(h) = (df_1(h), \ldots, df_m(h))$ which is given by
I used \ldots for the dotting since that's more natural-looking.

The second one didn't render because you used ( instead of {. This

\displaystyle f' (p) = \begin{bmatrix} f'_1 (p) \\ f'_2 (p) \\ . \\ . \\ . \\ f'_n (p) \end{bmatrix}

$\displaystyle f' (p) = \begin{bmatrix} f'_1 (p) \\ f'_2 (p) \\ . \\ . \\ . \\ f'_m (p) \end{bmatrix}$

$\displaystyle\mathbf{f}' (\mathbf{p}) = \begin{bmatrix} f'_1 (p) \\ f'_2 (p) \\  \vdots \\ f'_n (p) \end{bmatrix}$

$\displaystyle\mathbf{f}' (\mathbf{p}) = \begin{bmatrix} f'_1 (\mathbf{p}) \\ f'_2 (\mathbf{p}) \\ \vdots \\ f'_m (\mathbf{p}) \end{bmatrix}$

Using \vdots for the $\vdots$ and \mathbf{p} to write $\mathbf{p}$

Thanks for the help MountEvariste ...

Really appreciate your help ...

No idea how the colour codes got there ...

Peter