1.असमान अन्तराल के लिए अन्तर्वेशन (Interpolation for Unequal Intervals),लग्रांज का असमान अन्तराल के लिए अन्तर्वेशन सूत्र (Lagrange Interpolation Formula for Unequal Intervals):

Interpolation for Unequal Intervals

असमान अन्तराल के लिए अन्तर्वेशन (Interpolation for Unequal Intervals) के इस आर्टिकल में असमान अन्तराल पर आधारित अन्तर्वेशन के सवालों को लग्रांज के सूत्र से हल करेंगे।
आपको यह जानकारी रोचक व ज्ञानवर्धक लगे तो अपने मित्रों के साथ इस गणित के आर्टिकल को शेयर करें।यदि आप इस वेबसाइट पर पहली बार आए हैं तो वेबसाइट को फॉलो करें और ईमेल सब्सक्रिप्शन को भी फॉलो करें।जिससे नए आर्टिकल का नोटिफिकेशन आपको मिल सके।यदि आर्टिकल पसन्द आए तो अपने मित्रों के साथ शेयर और लाईक करें जिससे वे भी लाभ उठाए।आपकी कोई समस्या हो या कोई सुझाव देना चाहते हैं तो कमेंट करके बताएं।इस आर्टिकल को पूरा पढ़ें।

Also Read This Article:- Newton Backward Interpolation Formula

2.असमान अन्तराल के लिए अन्तर्वेशन के उदाहरण (Interpolation for Unequal Intervals Illustrations):

Illustration:6(b).दिया हुआ है (Given that)
f(1)+f(2)+f(3)=25,f(4)=29 तथा (and) f(5)+f(6)=113 ;Estimate the value of f(7)
Solution:माना u_6=f(1)+f(2)+f(3)+f(4)+f(5)+f(6) \\ \Rightarrow u_6=25+29+113=167 \\ u_3=f(4)+f(5)+f(6) \\ \Rightarrow u_3=29+113=142 \\ \Rightarrow u_2=f(5)+f(6)=113 \\ \Rightarrow u_1=f(4)=29

\begin{array}{cc} \hline x & u_n \\ \hline 1 & 29 \\ 2 & 113 \\ 3 & 142 \\ 6 & 167 \\ \hline \end{array}
लग्रांज सूत्र से:

f(x)=\frac{\left(x-x_1\right) \cdot\left(x-x_2\right) \ldots \left(x-x_n\right)}{\left(x_0-x_1\right)\left(x_0-x_2 \right) \ldots \left(x_0-x_n\right)} f\left(x_0\right) +\frac{\left(x-x_0\right)\left(x-x_2\right) \ldots\left(x-x_n\right)}{\left(x_1-x_0\right)\left(x_1-x_2\right) \ldots\left(x_1-x_n\right)}f\left(x_1\right) +\cdots+ \frac{\left(x-x_0\right) \left(x-x_1\right) \ldots \left(x-x_n-1\right)}{\left(x_n-x_0\right)\left(x_n-x_1\right) \ldots \left(x_n-x_{n-1}\right)} f\left(x_n\right) \\ u_7=\frac{(7-2)(7-3)(7-6)}{(1-2)(1-3)(1-6)} \times 29 +\frac{(7-1)(7-3)(7-6)}{(2-1)(2-3)(2-6)} \times 113+\frac{(7-1)(7-2)(7-6)}{(3-1)(3-2)(3-6)} \times 142+\frac{(7-1)(7-2)(7-3)}{(6-1)(6-2)(6-3)} \times 167 \\ = \frac{5 \times 4 \times 1}{(-1) \times(-2) \times(-5)} \times 29+\frac{6 \times 4 \times 1}{1 \times(-1) \times(-4)} \times 113 +\frac{6 \times 5 \times 1}{2 \times 1 \times(-3)} \times 142+\frac{6 \times 5 \times 4}{5 \times 4 \times 3} \times 167 \\=-58+678-710+334 \\ \Rightarrow u_7=244 \\ \Rightarrow f(7)=u_7-46 \\ \Rightarrow f(7)=244-167 \\ \Rightarrow f(7)=77
Illustration:7.स्वतन्त्र चर की चार स्थितियों 3,7,9 तथा 10 पर फलन के मान क्रमशः 168,120, 72 तथा 63 पाये गए,स्वतन्त्र चर की स्थिति 6 के लिए फलन का सर्वोत्तम मान क्या दिया जा सकता है?
(The observed values of a function are respectively 168,120,72 and 63 at the four position 3,7,9 and 10 of the independent variable.What is the best estimate you can give for the values of the function at the positition of 6 of the independent variable?)
Solution:दिया है:f(3)=168,f(7)=120,f(9)=72,f(10)=63,f(6)=?
लग्रांज सूत्र से:

f(x)=\frac{\left(x-x_1\right) \left(x-x_2\right) \ldots \left(x-x_n\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right) \ldots \left(x_0-x_n\right)} \times f\left(x_0\right) +\frac{\left(x-x_0\right)\left(x-x_2\right) \ldots\left(x-x_n \right)}{\left(x_1-x_0\right)\left(x_1-x_2\right) \ldots \left(x_1-x_n\right)} f\left(x_1\right) + \ldots \ldots +\frac{\left(x-x_0\right)\left(x-x_1\right) \ldots \left(x-x_{n-1}\right)}{\left(x_n-x_0\right) \left(x_n-x_1\right) \ldots \left(x_n-x_{n-1}\right)} f\left(x_n\right) \\ f(6)=\frac{(6-7)(6-9)(6-10)}{(3-7)(3-9)(3-10)} \times 168 +\frac{(6-3)(6-9)(6- 10)}{(7-3)(7-9)(7-10)} \times 120 +\frac{(6-3)(6-7)(6-10)}{(9-3)(9-7)(9-10)} \times 72 +\frac{(6-3)(6-7)(6-9)}{(10-3)(10-7)(10-9)} \times 63 \\ = \frac{168}{14}+\frac{3}{2} \times 120-72+\frac{3}{7} \times 63 \\ =12+180-72+27 \\ \Rightarrow f(6)=147
Illustration:9.न्यूनतम घात वाला बहुपद ज्ञात कीजिए जो मान 3,12,15,-21 ग्रहण करता है,जबकि x के मान क्रमशः 3,2,1,-1 हैं।
(Find the polynomial of the lowest possible degree which assumes the values 3,12,15,-21,when x has the values 3,2,1,-1 respectively.)
Solution:माना f(-1)=-21,f(1)=15,f(2)=12,f(3)=3 \\ x_0=-1, x_1=1, x_2=2, x_3=3 \\ f\left(x_0\right)=-21, f\left(x_1\right)=15, f\left(x_2\right)=12, f\left(x_3\right)=3
लग्रांज सूत्र से:

Interpolation for Unequal Intervals

Illustration:10.एक बहुपद ज्ञात कीजिए जो निम्न से सन्तुष्ट होता है:
(Find the polynomial satisfied by)
(-4,1245),(-1,33),(0,5),(2,9),(5,1335)
Solution: x_0=-4, x_1=-1, x_2=0 ,x_3=2, x_4=5 \\ f\left(x_0\right)=1245, f\left(x_1\right)=33, f\left(x_2\right)=5, f\left(x_3\right)=9, f\left(x_4\right)=1335
लग्रांज सूत्र से:

f(x)=\frac{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right)\left(x_0-x_3\right)\left(x_0-x_4\right)} f(x_0) +\frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right) }{\left(x_1-x_0\right)\left(x_1-x_2\right)\left(x_1-x_3\right) \left(x_1-x_4\right)} f\left(x_1\right) +\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_3\right)\left(x-x_4\right)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_2\right)\left(x_2-x_4\right)} f\left(x_2\right) +\frac{\left(x-x_0\right) \left(x-x_1\right)\left(x-x_2\right)\left(x-x_4\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3-x_2\right) \left(x_3-x_4\right)} f\left(x_4\right)+\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right) \left(x-x_3 \right)}{\left(x_4-x_0\right)\left(x_4-x_1\right)\left(x_4-x_2\right) \left(x_4 x_3\right)} f\left(x_4\right) \\ =\frac{(x+1)(x-0)(x-2)(x-5)}{(-4+1)(-4-0)(-4-2)(-4-5)} \times 1245 +\frac{(x+4)(x-0)(x-2)(x-5)}{(-1+4)(-1-0)(-1-2)(-1-5)} \times 33 +\frac{(x+4)(x+1)(x-2)(x-5) }{(0+4)(0+1)(0-2)(0-5)} \times 5 +\frac{(x+4)(x+1)(x-0)(x-5)}{(2+4)(2+1)(2-0)(2-5)} \times 9 +\frac{(x+4)(x+1)(x-0)(x-2)}{(5+4)(5+1)(5-0)(5-2)} \times 1335 \\ =\frac{\left(x^4-6 x^3+3 x^2+10 x\right)}{216} \times 415+ \frac{\left(x^4-3 x^3-18 x^2+40 x\right) }{(-18)} \times 11 +\frac{\left(x^4-2 x^3-21 x^2+22 x+40\right)}{8} +\frac{\left(x^4-21 x^2-20 x\right)}{(-12)}+\frac{\left(x^4+3 x^3-6 x^2-8 x\right)}{54} \times 89 \\ =\frac{\begin{array}{cc} 415 x^4-2490 x^3+1245 x^2+4150 x -132 x^4+396 x^3 \\ +2376 x^2-5280 x +27 x^4-54 x^2-567 x^2+594 x+1080 \\ -18 x^4+318 x^2+360 x+356 x^4+1068 x^3-2136 x^2-2848 x\end{array}  }{216}\\ =\frac{648 x^4-1080 x^3+1296 x^2-3024 x+1080 }{216}\\ \Rightarrow f(x)=3 x^4-5 x^3+6 x^2-14 x+5
Illustration:11.निम्न सारणी से क्रमशः फलन ज्ञात कीजिए:
(Find the form of the function respectively given by the following table):
Illustration:11(i).
Solution: x_0=0, x_1=1, x_2=2, x_3=5 , f\left(x_0\right)=2, f\left(x_1\right)=3, f\left(x_2\right)=12, f\left(x_3\right)=147
लग्रांज सूत्र से:

f(x)=\frac{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right) \left(x_0-x_3\right)} \times f\left(x_0\right)+ \frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_1-x_0 \right) \left(x_1-x_2\right)\left(x_1-x_3\right)} f\left(x_1\right)+ \frac{\left(x-x_0\right)\left(x-x_1\right) \left(x-x_3\right)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_3\right)} f\left(x_2\right)+ \frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3-x_2\right)} \times f\left(x_3\right) \\ =\frac{(x-1)(x-2)(x-5)}{(0-1)(0-2)(0-5)} \times 2+\frac{(x-0)(x-2)(x-5)}{(1-0)(1-2)(1-5)} \times 3 +\frac{(x-0)(x-1)(x-5)}{(2-0)(2-1)(2-5)} \times 12+\frac{(x-0)(x-1)(x-2)}{(5-0)(5-1)(5-2)} \times 147 \\ =\frac{\left(x^3-8 x^2+17 x-10\right)}{(-5)}+\frac{\left(x^3-7 x^2+10 x\right) }{4} \times 3+ \frac{\left(x^3-6 x^2+5 x\right)}{(-1)} \times 2 + \frac{\left(x^3-3 x^2+2 x\right)}{20} \times 49 \\ =\frac{\begin{array}{cc} -4 x^3+32 x^2-68 x+40+15 x^3-105x^2+150 x \\ -40 x^3+240 x^2-200 x+49 x^3-147 x^2+98 x \end{array}}{20} \\ =\frac{20 x^3+20 x^2-20 x+40}{20} \\ \Rightarrow f(x)=x^3+x^2-x+2
Illustration:11(ii).
Solution: x_0=0, x_1=1, x_2=2, x_3=4 \\ f\left(x_0\right)=1, f\left(x_1\right)=1, f\left(x_2\right)=2, f\left(x_3\right)=5
लग्रांज सूत्र से:

f(x)=\frac{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right)\left(x_0-x_3\right)} \times f (x_0)+ \frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_1-x_0\right) \left(x_1-x_2\right)\left(x_1-x_3\right)} f\left(x_1\right)+ \frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_3\right)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_3\right)} f\left(x_2\right)+ \frac{\left(x-x_0\right) \left(x-x_1\right)\left(x-x_2\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3-x_2\right)} \times f\left(x_3\right) \\=\frac{(x-1)(x-2)(x-4)}{(0-1)(0-2)(0-4)} \times 1+\frac{(x-0)(x-2)(x-4)}{(1-0)(1-2)(1-4)} \times 1 +\frac{(x-0)(x-1)(x-4)}{(2-0)(2-1)(2-4)} \times 2+\frac{(x-0)(x-1)(x-2)}{(4-0)(4-1)(4-2)} \times 5 \\=\frac{\left(x^3-7 x^2+14 x-8\right)}{(-8)}+\frac{\left(x^3-6 x^2+8 x\right)}{3} +\frac{\left(x^3-5 x^2+4 x\right)}{(-2)}+\frac{\left(x^3-3 x^2+2 x\right) }{24} \times 5 \\= \frac{\begin{array}{cc} -3 x^3+21 x^2 -42 x+24 +8 x^3-48 x^2+64 x \\-12 x^3+60 x^2-48 x-15 x^3-15 x^2+10 x \end{array}}{24} \\ =\frac{-2 x^3+18 x^2-16 x+24}{24} \\ \Rightarrow f(x)=\frac{1}{12}\left(-x^3+9 x^2-8 x+12\right)
Illustration:11(iii).  
Solution: x_0=0, x_1=1, x_2=3, x_3=5, x_4=6,x_5=9 \\ f\left(x_0\right)=-18, f\left(x_1\right)=0, f\left(x_2\right)=0, f\left(x_2\right)=-248, f\left(x_4\right)=0, f(45)=13104
लग्रांज सूत्र से:

y=f(x)=\frac{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right)\left(x_0-x_3\right)\left(x_0-x_4\right)\left(x_0-x_5\right)} \times f\left(x_0\right) +\frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)}{\left(x_1-x_0\right)\left(x_1-x_2\right)\left(x_1-x_1\right)\left(x_1-x_3\right)\left(x_1-x_4\right)\left(x_1-x_5\right)} \times f\left(x_1\right) +\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_3\right)\left(x-x_4\right)(x-x_5)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_3\right)\left(x_2-x_4\right)\left(x_2-x_5\right)} \times f\left(x_2\right) +\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)\left(x-x_4\right)\left(x-x_5\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3-x_2\right)\left(x_3-x_4\right)\left(x_3-x_5\right)} \times f\left(x_3\right) +\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_5\right)}{\left(x_4-x_0\right)\left(x_4-x_1\right)\left(x_4-x_2\right)\left(x_4-x_3\right)\left(x_4-x_5\right)} \times f\left(x_4\right) +\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right) }{\left(x_5-x_0\right)\left(x_5-x_1\right)\left(x_5-x_2\right)\left(x_5-x_3\right)\left(x_5-x_4\right)} \times f\left(x_5\right)\\ =\frac{(x-1)(x-3)(x-5)(x-6)(x-9)}{(0-1)(0-3)(0-5)(0-6)(0-9)} \times(-18) +\frac{(x-0)(x-3)(x-5)(x-6)(x-9)}{(1-0)(1-3)(1-5)(1-6)(1-9)} \times 0 +\frac{(x-0)(x-1)(x-5)(x-6)(x-9)}{(3-0)(3-1)(3-5)(3-6)(3-9)} \times 0 +\frac{(x-0)(x-1)(x-3)(x-6)(x-9)}{(5-0)(5-1)(5-3)(5-6)(5-9)} \times(-248) + \frac{(x-0)(x-1)(x-3)(x-5)(x-9)}{(6-0)(6-1)(6-3)(6-5)(6-9)} \times 0 +\frac{(x-0)(x-1)(x-3)(x-5)(x-6)}{(9-0)(9-1)(9-3)(9-5)(9-6)} \times 13104 \\ =(x-1)(x-3)(x-6)\left[\frac{(x-5)(x-9)}{45}+\frac{x(x-9)(-31)}{20}+\frac{x(x-5)(91)}{36}\right] \\ =(x-1)(x-3)(x-6)\left[\frac{4 x^2-56 x+180-279 x^2+2511 x+455 x^2-2275 x}{180}\right] \\ =(x-1)(x-3)(x-6) \left[\frac{180 x^2+180 x+180}{180}\right] \\ \Rightarrow y=(x-1)(x-3)(x-6)\left(x^2+x+1\right)
Illustration:11(iv).
Solution: x_0=0, x_1=1, x_2=4, x_3=6 \\ y_0=1, y_1=-1, y_2=1, y_3=-1
लग्रांज सूत्र से:

y=\frac{\left(x-x_1\right)\left(x-x_2\right)\left(x_1-x_3\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right)\left(x_0-x_3\right)} \times y_0+ \frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right) }{\left(x_1-x_0\right) \left(x_1-x_2\right)\left(x-x_3\right)} \times y_1 + \frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_3\right)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_3\right)} \times y_2 +\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3 -x_2\right)} \times y_3 \\ =\frac{(x-1)(x-4)(x-6)}{(0-1)(0-4)(0-6)} \times 1+\frac{(x-0)(x-4)(x-6)}{(1-0)(1-4)(1-6)} \times(-1) +\frac{(x-0)(x-1)(x-6)}{(4-0)(4-1)(4-6)} \times 1+\frac{(x-0)(x-1)(x-4)}{(6-0)(6-1)(6-4)} \times(-1) \\ =-\frac{(x-1)(x-4)(x-6)}{24}-\frac{x(x-4)(x-6)}{15}-\frac{x(x-1)(x-6)}{24}-\frac{x(x-1)(x-4)}{60}
Illustration:12.यदि y_1, y_2, \ldots y_9 एक श्रेणी के क्रमागत पद हों,तो सिद्ध कीजिए कि
(If y_1, y_2, \ldots y_9 are the consecutive terms of a series,then prove that):

y_5=\frac{4}{5}\left(y_4+y_6\right)-\frac{2}{5}\left(y_3+y_7\right)+\frac{4}{35} \left(y_2+y_8\right)-\frac{1}{70}\left(y_1+y_9\right)
Solution: 
लग्रांज सूत्र से:

f(x)=\frac{\left(x_1-x_1\right)\left(x-x_2\right) \cdots\left(x-x_n\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right) \cdots\left(x_0-x_n\right)} \times f(x_0) +\frac{\left(x-x_0\right) \left(x-x_2\right) \ldots \left(x-x_n\right) }{\left(x_1-x_0\right)\left(x_1-x_2\right) \ldots \left(x_1-x_n\right)} f\left(x_1\right) + \cdots+\frac{\left(x-x_0\right)\left(x-x_1\right) \ldots \left(x-x_{n-1}\right) }{\left(x_n-x_0\right)\left(x_n-x_1\right)\ldots \left(x_n-x_{n-1}\right)} f\left(x_n\right) \\ \Rightarrow \frac{f(x)}{\left(x-x_0\right)\left(x-x_1\right) \ldots \left(x-x_n\right)}=\frac{f\left(x_0\right)}{\left(x-x_0\right)\left(x_0-x_1\right)\left(x_0-x_2\right) \ldots \left(x_0-x_n\right)} +\frac{f\left(x_1\right)}{\left(x-x_1\right)\left(x_1-x_0\right)\left(x_1-x_2\right) \ldots \left(x_1-x_n\right)} +\cdots+\frac{f\left(x_n\right)}{\left(x-x_n\right) \left(x_n-x_0\right)\left(x_n-x_1\right) \ldots \ldots \left(x_n-x_{n-1}\right)} \\ \text{put } x_0=1, x_1=2, x_2=3, x_3=4, x_4=6, x_5=7, x_6=8, x_7=9, x=5 \\ f(1)=y_1, f(2)=y_2, f(3)=y_3, f(4)=y_4, f(5)=y_5, f(6)=y_6, f(7)=y_7, f(8)=y_8 ,f(9)=y_9 \\ \Rightarrow \frac{y_5}{(5-1)(5-2)(5-3) (5-4)(5-6)(5-7)(5-8)(5-9)}= \frac{y_1}{(5-1)(1-2)(1-3)(1-4)(1-6)(1-7)(1-8)(1-9)} +\frac{y_2}{(5-2)(2-1)(2-3)(2-4)(2-6)(2-7)(2-8)(2-9)}+\frac{y_3}{(5-3)(3-1)(3-2)(3-4)(3-6)(3-7)(3-8)(3-30} +\frac{y_4}{(5-4)(4-1)(4-2)(4-3)(4-6)(4-7)(4-8)(4-9)}+\frac{y_6}{(5-6)(6-1)(6-2)(6-3)(6-4)(6-7)(6-8)(6-9)} +\frac{y_7}{(5-7)(7-1)(7-2)(7-3)(7-4)(7-6)(7-8)(7-9)} +\frac{y_8}{(5-8)(8-1)(8-2)(8-3)(8-4)(8-6)(8-1)(8-1)} +\frac{y_9}{(5-9)(9-1)(9-2)(9-3)(9-4)(9-6)(9-8)} \\ \Rightarrow \frac{y_5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 1 \cdot 2 \cdot 3 \cdot 4}=\frac{-y_1}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}+\frac{y_2}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}-\frac{y_3}{2 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}+\frac{y_4}{6 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}+\frac{y_6}{6 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} -\frac{y_7}{2 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}+\frac{y_8}{1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 \cdot 7 }-\frac{y_9}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} \\ \Rightarrow \frac{y_5}{24} =-\frac{y_1}{1680}+\frac{y_2}{210}-\frac{y_3}{60}+\frac{y_4}{30}+\frac{y_6}{30}+\frac{y_7}{60}+\frac{y_8}{20}-\frac{y_9}{1880} \\ =\frac{1}{30}\left(y_4+y_6\right)-\frac{1}{60} \left(y_3+y_7\right)+\frac{1}{210}\left(y_2+y_8\right)-\frac{1}{1680}\left(y_1+y_9\right) \\ \Rightarrow y_5 =\frac{4}{5}\left(y_4+y_6\right)-\frac{2}{5}\left(y_3+y_7\right)+\frac{4}{35}\left(y_2+y_8\right)-\frac{1}{70}\left(y_1+y_9\right)
Illustration:13.सिद्ध कीजिए कि लग्रांज सूत्र निम्न प्रकार से व्यक्त किया जा सकता है:
(Prove that Lagrange’s formula can be expressed as the following):
जहाँ P_n(x)=f(x)
Solution:माना (n+1) बिन्दुओं से गुजरने वाला बहुपद है:
बिन्दु: \left[x_0, f\left(x_0\right)\right] ;\left[x_1, f\left(x_1\right)\right], \cdots \cdots \left[x_n, f\left(x_n\right)\right]
बहुपद: P_n(x)=A_0+A_1 x+A_2 x^2+\cdots+A_n x^n \\ f\left(x_0\right)=A_0+A_1 x_0+A_2 x_0^2+\cdots+A_n x_0^n \\ f\left(x_1\right)=A_0+A_1 x_1+A_2 x_1^2+\cdots+A_n x_1^n \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ f\left(x_n\right)=A_0+A_1 x_n+A_2 x_n^2 \cdots+A_n x_n^n \\ A_0, A_1, \cdots A_n का विलोपन करने पर : 

उपर्युक्त उदाहरणों के द्वारा असमान अन्तराल के लिए अन्तर्वेशन (Interpolation for Unequal Intervals),लग्रांज का असमान अन्तराल के लिए अन्तर्वेशन सूत्र (Lagrange Interpolation Formula for Unequal Intervals) को समझ सकते हैं।

Interpolation for Unequal Intervals

Also Read This Article:- Forward Difference Interpolation

3.असमान अन्तराल के लिए अन्तर्वेशन (Frequently Asked Questions Related to Interpolation for Unequal Intervals),लग्रांज का असमान अन्तराल के लिए अन्तर्वेशन सूत्र (Lagrange Interpolation Formula for Unequal Intervals) से सम्बन्धित अक्सर पूछे जाने वाले प्रश्न:

असमान अन्तराल के लिए अन्तर्वेशन (Interpolation for Unequal Intervals) के इस आर्टिकल
में असमान अन्तराल पर आधारित अन्तर्वेशन के सवालों को लग्रांज के सूत्र से हल करेंगे।