# Download A Survey of Free Math Fonts for TEX and LATEX by Stephen G.Hartke PDF By Stephen G.Hartke

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65. , If S = ∑ ax i i ∑ a x , then the is replaced by or truncated to S = i i i=1 1 error developed is a truncation error. A truncation error is a type of algorithm error. Also, if ex = 1 + x + 1+x+ x2 x3 x 4 + + + ...... ∞ = X (say) is truncated to 2! 3! 4! x2 x3 + = X′ (say), then truncation error = X – X′ 2! 3! 1 if 5 (i) The first three terms are retained in expansion. (ii) The first four terms are retained in expansion. Sol. (i) Error = True value – Approximate value Example. Find the truncation error for ex at x = F GH = 1+ x + I F JK GH x2 x 3 x2 + + ......

Here, ∴ tan A = a c A = tan–1 δA = = FG aIJ H cK ∂A ∂A δa + δc ∂a ∂c c 2 a +c 2 δa − a 2 a + c2 δc ERRORS or |δA|≤ = c a2 + c2 . δa + a a2 + c 2 51 . δc 15 6 . 1) + . 0103 radians. ∴ Example 12. In a ΔABC, a = 30 cm, b = 80 cm, ∠B = 90°, find the maximum 1 1 error in the computed value of A if possible errors in a and b are % and % , 3 4 respectively. Sol. 00235. 0909, respectively. 0909. Sol. 0909 are correct to four places of decimal. The Example 13. 00005. 00043 . 0001. 0667 respectively. 0667.

The roots of the equation x2 – 32x + 1 = 0 are 32 − (32) 2 − 4 2 32 + (32) 2 − 4 2 and 32 − 1020 = 16 − 255 2 I Algorithm. 03 The smaller root is II Algorithm. Smaller root = (16 − 255 ) . 97 = 16 + 255 The second algorithm is evidently a better one, as gives the result correct to 4 figures. Example 10. If X = x + e, prove that Sol. FG H X − x = X − X − e = X − X 1− = X– X + e 2 X ≈ e X − x≈ e e X 2 X IJ K 1/2 = . FG H X − X 1− e 2X IJ K . 2 X Example 11. In a ΔABC, a = 6 cm, c = 15 cm, ∠B = 90°.