} else {
      // This is a generalized version of the calculation in add(double, double) above. Note that
      // non-finite inputs will have sumOfProductsOfDeltas = NaN, so non-finite values will result
      // in NaN naturally.
      sumOfProductsOfDeltas +=
          values.sumOfProductsOfDeltas()
              + (values.xStats().mean() - xStats.mean())
                  * (values.yStats().mean() - yStats.mean())

      assertWithMessage("" + value)
          .that(Floats.indexOf(new float[] {5f, value, value, 5f}, new float[] {value, value}))
          .isEqualTo(1);
    }
    assertThat(Floats.indexOf(new float[] {5f, NaN, NaN, 5f}, new float[] {NaN, NaN}))
        .isEqualTo(-1);
  }

  public void testLastIndexOf() {
    assertThat(Floats.lastIndexOf(EMPTY, (float) 1)).isEqualTo(-1);

    }
    return false;
  }

  /**
   * Returns the base 2 logarithm of a double value.
   *
   * <p>Special cases:
   *
   * <ul>
   *   <li>If {@code x} is NaN or less than zero, the result is NaN.
   *   <li>If {@code x} is positive infinity, the result is positive infinity.
   *   <li>If {@code x} is positive or negative zero, the result is negative infinity.
   * </ul>
   *

      fail();
    } catch (IllegalArgumentException expected) {
    }
  }

  public void testCheckNonNegative_nanFloat() {
    try {
      MathPreconditions.checkNonNegative("float", Float.NaN);
      fail();
    } catch (IllegalArgumentException expected) {
    }
  }

  public void testCheckNonNegative_zeroDouble() {
    MathPreconditions.checkNonNegative("double", 0d);
  }

    assertEquals(0.0, DoubleUtils.ensureNonNegative(Double.NEGATIVE_INFINITY));
    assertThrows(IllegalArgumentException.class, () -> DoubleUtils.ensureNonNegative(Double.NaN));
  }

  public void testOneBits() {
    assertEquals(DoubleUtils.ONE_BITS, Double.doubleToRawLongBits(1.0));
  }

  private static void failFormat(String template, Object... args) {

  }

  public void testMapping_nanY1() {
    assertThrows(
        IllegalArgumentException.class, () -> LinearTransformation.mapping(1.2, Double.NaN));
  }

  public void testMappingAnd_nanX2() {
    assertThrows(
        IllegalArgumentException.class,
        () -> LinearTransformation.mapping(1.2, 3.4).and(Double.NaN, 7.8));
  }

  }

  public void testMapping_nanY1() {
    assertThrows(
        IllegalArgumentException.class, () -> LinearTransformation.mapping(1.2, Double.NaN));
  }

  public void testMappingAnd_nanX2() {
    assertThrows(
        IllegalArgumentException.class,
        () -> LinearTransformation.mapping(1.2, 3.4).and(Double.NaN, 7.8));
  }

     *    value (whether it is NaN or infinity).
     * 2. If the new value is finite and the previous mean is non-finite then the mean is unchanged
     *    (whether it is NaN or infinity).
     * 3. If both the previous mean and the new value are non-finite and...
     * 3a. ...either or both is NaN (so mean != value) then the new mean is NaN.

      assertWithMessage("" + value)
          .that(Floats.indexOf(new float[] {5f, value, value, 5f}, new float[] {value, value}))
          .isEqualTo(1);
    }
    assertThat(Floats.indexOf(new float[] {5f, NaN, NaN, 5f}, new float[] {NaN, NaN}))
        .isEqualTo(-1);
  }

  public void testLastIndexOf() {
    assertThat(Floats.lastIndexOf(EMPTY, (float) 1)).isEqualTo(-1);

 * uniform distribution on [0,1]."
 *
 * <h3>Handling of non-finite values</h3>
 *
 * <p>If any values in the input are {@link Double#NaN NaN} then all values returned are {@link
 * Double#NaN NaN}. (This is the one occasion when the behaviour is not the same as you'd get from
 * sorting with {@link java.util.Arrays#sort(double[]) Arrays.sort(double[])} or {@link