Display powers of Phi with Fibonacci precision Write some code that takes a single non-negative integer \$n\$ and outputs the \$n\$th power of phi (\$\phi\$, the Golden Ratio, approximately 1.61803398874989) with the same number of decimal digits as the \$n\$th Fibonacci number.
Your code must produce the correct sequence of digits for all inputs up to at least 10 (55 decimal digits). The output must be human-readable decimal. You may choose whether to round the last digit to the closest value, or truncate the value. Please specify which one your code uses.
\$n\$ and output, up to 10, rounding down:
 0   1
 1   1.6
 2   2.6
 3   4.23
 4   6.854
 5  11.09016
 6  17.94427190
 7  29.0344418537486
 8  46.978713763747791812296
 9  76.0131556174964248389559523684316960
10 122.9918693812442166512522758901100964746170048893169574174

\$n\$ and output, up to 10, rounding to the closest value:
 0   1
 1   1.6
 2   2.6
 3   4.24
 4   6.854
 5  11.09017
 6  17.94427191
 7  29.0344418537486
 8  46.978713763747791812296
 9  76.0131556174964248389559523684316960
10 122.9918693812442166512522758901100964746170048893169574174

The 7th Fibonacci number is 13, so the output for \$n=7\$, \$\phi^7\$, has 13 decimal places. You must not truncate trailing zeros that would display too few digits; see output for 6 in the first table, which ends in a single zero to keep the decimal precision at 8 digits.
Maybe as a bonus, say what the highest number your program can correctly output is.
dc, 26 bytes

99k5v1+2/?^d5v/.5+0k1/k1/p


Due to the initial precision of 99 digits after the comma, this will work up yo input 11. A dynamic (or higher static) precision is possible, but would elevate the byte count.

Test cases

$ for ((i = 0; i &lt; 11; i++)) { dc -e '99k5v1+2/?^d5v/.5+0k1/k1/p' &lt;&lt;&lt; $i; }
1
1.6
2.6
4.23
6.854
11.09016
17.94427190
29.0344418537486
46.978713763747791812296
76.0131556174964248389559523684316960
122.9918693812442166512522758901100964746170048893169574174


How it works

Since the desired output is &phi;n, we can calculate the the Fibonacci number F(n) as &lfloor;&phi;n ÷ &radic;5 + 0.5&rfloor; with little additional effort.

99k                         Set the precision to 99.
   5v                       Compute the square root of 5.
     1+                     Add 1.
       2/                   Divide by 2.
                            This pushes the golden ratio.
         ?                  Read the input from STDIN.
          ^                 Elevate the golden ratio to that power.
           d                Push a copy.
            5v/             Divide it by the square root of 5.
               .5+          Add 0.5.
                  0k        Set the precision to 0.
                    1/      Divide by 1, truncating to the desired precision.
                            This pushes F(n).
                      k     Set the precision to F(n).
                       1/   Divide by 1, truncating to the desired precision.
                         p  Print.