Finite-difference interpolation formulas (Newton-Gregory, Gauss, Bessel)

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Using the table

xxln⁡x\ln x1010Δln⁡x10^{10} \Delta \ln x1010Δ2ln⁡x10^{10} \Delta ^{2} \ln x
1, 0053{,}0021660594, 319, 184{,}7−4 291,3-4\, 291{,}3
1, 0063{,}0025979784, 314, 893{,}4−4 282,8-4\, 282{,}8

compute ln⁡1005,237\ln 1005{,}237, determining the precision of the computation. The numbers in the table are given to within 12\frac{1}{2} of the last digit.

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Using the table of problem 5208, find xx given ln⁡x=3,002268651\ln x = 3{,}002268651.

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Using the table

uuln⁡sh⁡u\ln \operatorname {sh} u105Δln⁡sh⁡u10^{5} \Delta \ln \operatorname {sh} uln⁡ch⁡u\ln \operatorname {ch} u105Δln⁡ch⁡u10^{5} \Delta \ln \operatorname {ch} u
0{,}5009{,}7169294{,}00{,}0521720{,}1
0{,}5019{,}7178693{,}80{,}0523720{,}1
0{,}5029{,}7187993{,}70{,}0525720{,}1
0{,}5039{,}7197393{,}50{,}0527720{,}2
0{,}5049{,}7206693{,}30{,}0529720{,}2

find sh⁡0,50235\operatorname {sh} 0{,}50235.

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Find the root of the equation ch⁡u=1,1287\operatorname {ch} u = 1{,}1287. See the table of problem 5210.

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Using the data of the table

xxsin⁡x\sin x
41∘41^{\circ }0{,}6560590290
42∘42^{\circ }0{,}6691306064
43∘43^{\circ }0{,}6819983601
44∘44^{\circ }0{,}6946583705

compute, with the greatest possible precision, sin⁡41∘20′\sin 41^{\circ }20'.

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Using the data of the table

φ\varphiF(φ)=∫0φdφ1−12sin⁡2φF(\varphi ) = \int _{0}^{\varphi } \dfrac {d\varphi }{\sqrt{1 - \frac{1}{2}\sin ^{2}\varphi }}
21,0∘21{,}0^{\circ }0{,}3706344
21,5∘21{,}5^{\circ }0{,}3796626
22,0∘22{,}0^{\circ }0{,}3887052
22,5∘22{,}5^{\circ }0{,}3977625

(table 13.1) compute F(φ)F(\varphi ) at φ=21∘15′\varphi = 21^{\circ }15'.

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Using the data of the table

φ\varphiF(φ)=∫0φdφ1−12sin⁡2φF(\varphi ) = \int _{0}^{\varphi } \dfrac {d\varphi }{\sqrt{1 - \frac{1}{2}\sin ^{2}\varphi }}
28,0∘28{,}0^{\circ }0{,}49846361
28,5∘28{,}5^{\circ }0{,}50772435
29,0∘29{,}0^{\circ }0{,}51700422
29,5∘29{,}5^{\circ }0{,}52630357

(table 13.2) compute the root of the equation F(φ)=0,5F(\varphi ) = 0{,}5.

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Using the data of the table

xxψ(x)=dln⁡Γ(1+x)dx\psi (x) = \dfrac {d \ln \Gamma (1+x)}{dx}
0{,}710{,}2164501462
0{,}720{,}2242928871
0{,}730{,}2320769593
0{,}740{,}2398032061

(table 13.3) compute ψ(0,715)\psi (0{,}715). To estimate the error one should take into account that ψ′(x)=1(x+1)2+1(x+2)2+…\psi '(x) = \dfrac {1}{(x+1)^{2}} + \dfrac {1}{(x+2)^{2}} + \ldots

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Using the data of the table

xxψ(x)=dln⁡Γ(1+x)dx\psi (x) = \dfrac {d \ln \Gamma (1+x)}{dx}
0{,}830{,}3068612
0{,}840{,}3140589
0{,}850{,}3212000
0{,}860{,}3282922

(table 13.4) compute the root of the equation ψ(x)=0,31\psi (x) = 0{,}31.

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Using a table of values of reciprocals of integers and an interpolation formula taken up through the fourth differences inclusive, find the value of the fraction 112,5\dfrac {1}{12{,}5} by means of the formulas:

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(a)

Newton--Gregory, with y0=112y_{0} = \dfrac {1}{12}, y1=113y_{1} = \dfrac {1}{13}, y2=114,…y_{2} = \dfrac {1}{14}, \ldots;

(b)

Newton--Gauss, and

(c)

Newton--Bessel, comparing the results obtained with the exact value.

Remark.
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Newton--Gregory formula: yx=y0+xΔy0+x(x−1)2!Δ2y0+x(x−1)(x−2)3!Δ3y0+…y_{x} = y_{0} + x \Delta y_{0} + \dfrac {x(x-1)}{2!}\Delta^{2} y_{0} + \dfrac {x(x-1)(x-2)}{3!}\Delta^{3} y_{0} + \ldots

Newton--Gauss formula: yx=y0+xΔy0+x(x−1)2!Δ2y−1+x(x2−1)3!Δ3y−1+…y_{x} = y_{0} + x \Delta y_{0} + \dfrac {x(x-1)}{2!}\Delta^{2} y_{-1} + \dfrac {x(x^{2}-1)}{3!}\Delta^{3} y_{-1} + \ldots

Newton--Bessel formula: yx=12(y0+y1)+(x−12)Δy0+x(x−1)2!⋅Δ2y−1+Δ2y02+…y_{x} = \dfrac {1}{2}(y_{0}+y_{1}) + \left(x - \dfrac {1}{2}\right)\Delta y_{0} + \dfrac {x(x-1)}{2!} \cdot \dfrac {\Delta^{2} y_{-1} + \Delta^{2} y_{0}}{2} + \ldots

Here y0,y1,…,yny_{0}, y_{1}, \ldots , y_{n} are the values of yy at x=a,a+h,…,a+nhx=a, a+h, \ldots , a+nh; the quantities Δy0,Δ2y0,…\Delta y_{0}, \Delta^{2} y_{0}, \ldots are the successive differences.

A similar question for the value 12,25\sqrt{12{,}25}; y0=12y_{0} = \sqrt{12}, y1=13y_{1} = \sqrt{13}, y2=14,…y_{2} = \sqrt{14}, \ldots

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Of 1000 born, the number surviving to age 20, 30, 40, 50, 60 years is respectively 560, 530, 480, 400, and 300 people. How many people out of 1000 survive to age 35 and 45?

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In a certain locality, the numbers of people whose ages fall into three consecutive five-year periods are as follows:

37 237,40 118,42 652. 37\, 237, \quad 40\, 118, \quad 42\, 652.

How many people are at the age of the middle year of the second five-year period?

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The numbers of people in the ages 25--30, 30--35, 35--40, 40--45 years are respectively as follows:

458 572,441 424,423 123,402 918. 458\, 572, \quad 441\, 424, \quad 423\, 123, \quad 402\, 918.

How many people are in their 38th year?

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A table contains, in one column, 20 consecutive values of a function. The sums of three consecutive columns are as follows: 630 611630\, 611, 1 946 9351\, 946\, 935, 3 246 3193\, 246\, 319. With what number does the fourth column begin?

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References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.