SQRTPI in Collabora Office: Find the Square Root of Pi Times a Number

• Introduced: original Collabora release and launch date are not established by the sources reviewed.
• Availability: documented in Collabora Office 26.04 help, scoped to the desktop-help release.
• Platform scope: support in every online or mobile deployment is not inferred from desktop documentation.
• Checked: October 14, 2026. Function-specific documentation and independently derived controls; no native Collabora test was performed.

=SQRTPI(number)

The official Collabora reference explicitly equates the function to SQRT(PI()*number). Pi is inside the square root. This differs from multiplying the argument by the square root of pi.

For real arithmetic, the argument must be nonnegative because pi is positive. The zero result is zero. The guide’s domain statement is mathematical; an exact native negative-input error display has not been tested. Examples use English function names and decimal-point numbers.

Enter 0, 1, 2, 4 in A2:A5. In B2 enter =SQRTPI(A2) and fill down.

The output is the principal nonnegative square root. It does not return both positive and negative roots as two cells.

=SQRTPI(4)
=SQRT(PI()*4)
=4*SQRT(PI())

The first two expected values are approximately 3.544907702. The third is approximately 7.089815404. Moving the factor four outside the root without taking its own square root changes the calculation.

For these positive inputs, SQRTPI(4) equals 2*SQRTPI(1). It does not equal four times SQRTPI(1). The function’s response scales with the square root of the argument.

Consider a constructed circle with radius 3 units. Its area is pi×3². A square with the same area has side length equal to the square root of that area:

=SQRTPI(3^2)

Expected side length: approximately 5.317361553 units. Squaring that unrounded result reconstructs the area, approximately 28.274333882 square units.

This result is a side length, not the circle’s circumference or diameter. The model explicitly squares the radius before calling SQRTPI. Using SQRTPI(3) would instead calculate the square root of 3×pi, a different quantity.

Keep the source radius and its unit visible. A negative radius is not meaningful in this geometry, even though squaring a negative number would hide its sign before the function receives it.

For a direct argument of 2, calculate SQRTPI(2) and keep its full stored result in D2. Then:

=D2^2/PI()

Expected recovered argument: approximately 2. The inverse relationship is number=result^2/pi on the nonnegative branch.

Retyping a shortened displayed value such as 2.5066 would introduce rounding before the inverse check. Keep the unrounded formula result and use an appropriate tolerance when comparing the reconstruction.

Compare inputs 1, 4 and 9. Expected outputs are approximately 1.772453851, 3.544907702 and 5.317361553. Their ratio is 1:2:3, while the inputs’ ratio is 1:4:9.

If an application requires a response proportional to the input, SQRTPI does not supply that relationship. SQRT explains the underlying root operation, and PI supplies the constant.

Check the placement of parentheses, whether the intended operation includes pi inside the root, and the meaning and units of the argument. A valid square-root result does not establish that the geometric or numerical model was appropriate.

The direct source establishes the function identity but not every native error, imported-text or extreme-value policy. The examples use ordinary finite nonnegative inputs and independently calculated expectations, without inferring a universal precision guarantee from displayed decimals.

• Collabora Office 26.04: SQRTPI.
• Editorial testing standards.

The direct product reference establishes the function definition and syntax. Worked examples were independently calculated; no native Collabora execution or broader client certification is claimed.