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Question-219678

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Question Number 219678 by universe last updated on 30/Apr/25
Commented by mr W last updated on 30/Apr/25
what do you mean with 1∙2? 1∙2=1×2? etc.
$${what}\:{do}\:{you}\:{mean}\:{with}\:\mathrm{1}\centerdot\mathrm{2}?\: \\ $$$$\mathrm{1}\centerdot\mathrm{2}=\mathrm{1}×\mathrm{2}?\:{etc}. \\ $$
Commented by universe last updated on 30/Apr/25
yes sir 1∙2 =1×2=2
$${yes}\:{sir}\:\mathrm{1}\centerdot\mathrm{2}\:=\mathrm{1}×\mathrm{2}=\mathrm{2} \\ $$
Answered by Ghisom last updated on 01/May/25
f(90)−(1/(90^2 ))=−((2431)/(50)) f(x)=−(x^7 /(16 460 236 800))+((13x^6 )/(895 795 200))−((3 791x^5 )/(2 743 372 800))+((221 983x^4 )/(3 292 047 360))−((82 499x^3 )/(45 722 880))+((23 987 239x^2 )/(914 457 600))−((605 341x)/(3 175 200))+((3 427 741)/(63 350 400))
$${f}\left(\mathrm{90}\right)−\frac{\mathrm{1}}{\mathrm{90}^{\mathrm{2}} }=−\frac{\mathrm{2431}}{\mathrm{50}} \\ $$$${f}\left({x}\right)=−\frac{{x}^{\mathrm{7}} }{\mathrm{16}\:\mathrm{460}\:\mathrm{236}\:\mathrm{800}}+\frac{\mathrm{13}{x}^{\mathrm{6}} }{\mathrm{895}\:\mathrm{795}\:\mathrm{200}}−\frac{\mathrm{3}\:\mathrm{791}{x}^{\mathrm{5}} }{\mathrm{2}\:\mathrm{743}\:\mathrm{372}\:\mathrm{800}}+\frac{\mathrm{221}\:\mathrm{983}{x}^{\mathrm{4}} }{\mathrm{3}\:\mathrm{292}\:\mathrm{047}\:\mathrm{360}}−\frac{\mathrm{82}\:\mathrm{499}{x}^{\mathrm{3}} }{\mathrm{45}\:\mathrm{722}\:\mathrm{880}}+\frac{\mathrm{23}\:\mathrm{987}\:\mathrm{239}{x}^{\mathrm{2}} }{\mathrm{914}\:\mathrm{457}\:\mathrm{600}}−\frac{\mathrm{605}\:\mathrm{341}{x}}{\mathrm{3}\:\mathrm{175}\:\mathrm{200}}+\frac{\mathrm{3}\:\mathrm{427}\:\mathrm{741}}{\mathrm{63}\:\mathrm{350}\:\mathrm{400}} \\ $$
Commented by mr W last updated on 01/May/25
but how?
$${but}\:{how}? \\ $$
Commented by Ghisom last updated on 01/May/25
simply f(x)=Σ_(j=0) ^7 c_j x^j we have 8 pairs of (x, f(x)) given, this leads to a system of 8 linear equations for 8 c_j which I just solved
$$\mathrm{simply} \\ $$$${f}\left({x}\right)=\underset{{j}=\mathrm{0}} {\overset{\mathrm{7}} {\sum}}{c}_{{j}} {x}^{{j}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{8}\:\mathrm{pairs}\:\mathrm{of}\:\left({x},\:{f}\left({x}\right)\right)\:\mathrm{given},\:\mathrm{this}\:\mathrm{leads} \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{system}\:\mathrm{of}\:\mathrm{8}\:\mathrm{linear}\:\mathrm{equations}\:\mathrm{for}\:\mathrm{8}\:{c}_{{j}} \\ $$$$\mathrm{which}\:\mathrm{I}\:\mathrm{just}\:\mathrm{solved} \\ $$
Commented by mr W last updated on 01/May/25
thanks!
$${thanks}! \\ $$
Answered by mr W last updated on 01/May/25
since f(x) is a degree 7 polynomial, then g(x)=x^2 f(x) is a degree 9 polynomial without x^0 and x^1 terms. for k_i =i(i+1) with i=1,2,...,8 we have f(k_i )=(1/k_i ^2 ), i.e. k_i ^2 f(k_i )=1 ⇒g(k_i )=1 ⇒g(x)=(ax+b)(x−k_1 )(x−k_2 )...(x−k_8 )+1 with a, b being contants. since g(x) has no x^0 and x^1 terms, bk_1 k_2 ...k_8 +1=0 ak_1 k_2 ...k_8 −bk_1 k_2 ...k_8 ((1/k_1 )+(1/k_2 )+...+(1/k_8 ))=0 ⇒b=−(1/(k_1 k_2 ...k_8 )) ⇒a=−(1/(k_1 k_2 ...k_8 ))((1/k_1 )+(1/k_2 )+...+(1/k_8 )) k_1 k_2 ...k_8 =(1×2)(2×3)...(8×9) =9×(8!)^2 (1/k_1 )+(1/k_2 )+...+(1/k_8 )=(1/(1×2))+(1/(2×3))+...+(1/(8×9)) =((1/1)−(1/2))+((1/2)−(1/3))+...+((1/8)−(1/9)) =1−(1/9)=(8/9) ⇒a=−(1/((8!)^2 ×9))×(8/9)=−(8/((9!)^2 )) ⇒b=−(1/((8!)^2 ×9))=−(9/((9!)^2 )) f(x)=((g(x))/x^2 ) ⇒f(x)=(1/x^2 )[1−(((8x+9))/((9!)^2 ))(x−1×2)(x−2×3)...(x−8×9)] f(90)−(1/(90^2 ))=−(((8×90+9))/((90×9!)^2 ))(90−1×2)(90−2×3)...(90−8×9) =−((2431)/(50)) ✓
$${since}\:{f}\left({x}\right)\:{is}\:{a}\:{degree}\:\mathrm{7}\:{polynomial}, \\ $$$${then}\:{g}\left({x}\right)={x}^{\mathrm{2}} {f}\left({x}\right)\:{is}\:{a}\:{degree}\:\mathrm{9}\: \\ $$$${polynomial}\:\underline{{without}}\:{x}^{\mathrm{0}} \:{and}\:{x}^{\mathrm{1}} \:{terms}. \\ $$$$ \\ $$$${for}\:{k}_{{i}} ={i}\left({i}+\mathrm{1}\right)\:{with}\:{i}=\mathrm{1},\mathrm{2},…,\mathrm{8} \\ $$$${we}\:{have}\:{f}\left({k}_{{i}} \right)=\frac{\mathrm{1}}{{k}_{{i}} ^{\mathrm{2}} },\:{i}.{e}.\:{k}_{{i}} ^{\mathrm{2}} {f}\left({k}_{{i}} \right)=\mathrm{1} \\ $$$$\Rightarrow{g}\left({k}_{{i}} \right)=\mathrm{1} \\ $$$$\Rightarrow{g}\left({x}\right)=\left({ax}+{b}\right)\left({x}−{k}_{\mathrm{1}} \right)\left({x}−{k}_{\mathrm{2}} \right)…\left({x}−{k}_{\mathrm{8}} \right)+\mathrm{1} \\ $$$${with}\:{a},\:{b}\:{being}\:{contants}. \\ $$$${since}\:{g}\left({x}\right)\:{has}\:{no}\:{x}^{\mathrm{0}} \:{and}\:{x}^{\mathrm{1}} \:{terms}, \\ $$$${bk}_{\mathrm{1}} {k}_{\mathrm{2}} …{k}_{\mathrm{8}} +\mathrm{1}=\mathrm{0} \\ $$$${ak}_{\mathrm{1}} {k}_{\mathrm{2}} …{k}_{\mathrm{8}} −{bk}_{\mathrm{1}} {k}_{\mathrm{2}} …{k}_{\mathrm{8}} \left(\frac{\mathrm{1}}{{k}_{\mathrm{1}} }+\frac{\mathrm{1}}{{k}_{\mathrm{2}} }+…+\frac{\mathrm{1}}{{k}_{\mathrm{8}} }\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=−\frac{\mathrm{1}}{{k}_{\mathrm{1}} {k}_{\mathrm{2}} …{k}_{\mathrm{8}} } \\ $$$$\Rightarrow{a}=−\frac{\mathrm{1}}{{k}_{\mathrm{1}} {k}_{\mathrm{2}} …{k}_{\mathrm{8}} }\left(\frac{\mathrm{1}}{{k}_{\mathrm{1}} }+\frac{\mathrm{1}}{{k}_{\mathrm{2}} }+…+\frac{\mathrm{1}}{{k}_{\mathrm{8}} }\right) \\ $$$${k}_{\mathrm{1}} {k}_{\mathrm{2}} …{k}_{\mathrm{8}} =\left(\mathrm{1}×\mathrm{2}\right)\left(\mathrm{2}×\mathrm{3}\right)…\left(\mathrm{8}×\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:=\mathrm{9}×\left(\mathrm{8}!\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{k}_{\mathrm{1}} }+\frac{\mathrm{1}}{{k}_{\mathrm{2}} }+…+\frac{\mathrm{1}}{{k}_{\mathrm{8}} }=\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{8}×\mathrm{9}} \\ $$$$\:\:\:\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{9}}\right) \\ $$$$\:\:\:\:\:\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{8}}{\mathrm{9}} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{1}}{\left(\mathrm{8}!\right)^{\mathrm{2}} ×\mathrm{9}}×\frac{\mathrm{8}}{\mathrm{9}}=−\frac{\mathrm{8}}{\left(\mathrm{9}!\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{b}=−\frac{\mathrm{1}}{\left(\mathrm{8}!\right)^{\mathrm{2}} ×\mathrm{9}}=−\frac{\mathrm{9}}{\left(\mathrm{9}!\right)^{\mathrm{2}} } \\ $$$${f}\left({x}\right)=\frac{{g}\left({x}\right)}{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left[\mathrm{1}−\frac{\left(\mathrm{8}{x}+\mathrm{9}\right)}{\left(\mathrm{9}!\right)^{\mathrm{2}} }\left({x}−\mathrm{1}×\mathrm{2}\right)\left({x}−\mathrm{2}×\mathrm{3}\right)…\left({x}−\mathrm{8}×\mathrm{9}\right)\right] \\ $$$${f}\left(\mathrm{90}\right)−\frac{\mathrm{1}}{\mathrm{90}^{\mathrm{2}} }=−\frac{\left(\mathrm{8}×\mathrm{90}+\mathrm{9}\right)}{\left(\mathrm{90}×\mathrm{9}!\right)^{\mathrm{2}} }\left(\mathrm{90}−\mathrm{1}×\mathrm{2}\right)\left(\mathrm{90}−\mathrm{2}×\mathrm{3}\right)…\left(\mathrm{90}−\mathrm{8}×\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{2431}}{\mathrm{50}}\:\checkmark \\ $$

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