Menu Close

Question-74863

Tinku Tara 0 Comments
Pin




Question Number 74863 by mrS last updated on 02/Dec/19
Commented by abdomathmax last updated on 03/Dec/19
we have S=Σ_(n=1) ^(45) (1/n^2 ) =Σ_(p=1) ^([((45)/2)]) (1/((2p)^2 )) +Σ_(p=0) ^([((45−1)/2)]) (1/((2p+1)^2 )) =(1/4)Σ_(p=1) ^(22) (1/p^2 ) +Σ_(p=0) ^(22) (1/((2p+1)^2 )) Σ_(p=1) ^(22) (1/p^2 ) =Σ_(k=1) ^(11) (1/((2k)^2 )) +Σ_(k=0) ^([((22−1)/2)]) (1/((2k+1)^2 )) =(1/4) Σ_(p=1) ^(11) (1/p^2 ) +Σ_(p=0) ^(10) (1/((2p+1)^2 )) ⇒ S =(1/4)( (1/4) Σ_(p=1) ^(11) (1/p^2 ) +Σ_(p=0) ^(10) (1/((2p+1)^2 )))+Σ_(p=0) ^(22) (1/((2p+1)^2 )) =(1/4^2 )Σ_(p=1) ^(11) (1/p^2 ) +(1/4)Σ_(p=0) ^(10) (1/((2p+1)^2 )) +Σ_0 ^(22) (1/((2p+1)^2 )) also Σ_(p=1) ^(11) (1/p^2 ) =Σ_(p=1) ^(10) (1/(4p^2 )) +Σ_(p=0) ^5 (1/((2p+1)^2 )) ⇒ S =(1/4^2 )((1/4)Σ_(p=1) ^(10) (1/p^2 ) +Σ_(p=0) ^5 (1/((2p+1)^2 )))+(1/4)Σ_(p=0) ^(10) (1/((2p+1)^2 )) +Σ_(p=0) ^(22) (1/((2p+1)^2 )) S=(1/4^3 ) Σ_(p=1) ^(10) (1/p^2 ) +(1/4^2 ) Σ_(p=0) ^5 (1/((2p+1)^2 )) +(1/4)Σ_(p=0) ^(10) (1/((2p+1)^2 )) +Σ_(p=0) ^(22) (1/((2p+1)^2 )) S=(1/4^3 )Σ_(p=1) ^(10) (1/p^2 ) +((1/4^2 )+(1/4))Σ_(p=0) ^5 (1/((2p+1)^2 ))+(1/4)Σ_(p=6) ^(10) (1/((2p+1)^2 )) +((1/4) +1)Σ_(p=0) ^(10) (1/((2p+1)^2 )) +Σ_(p=11) ^(22) (1/((2p+1)^2 )) Σ_(p=6) ^(10) (1/((2p+1)^2 )) =_(p−6=k) Σ_(k=0) ^4 (1/((2(6+k)+1)^2 )) =Σ_(k=0) ^4 (1/((2k+13)^2 )) Σ_(p=11) ^(22) (1/((2p+1)^2 )) =_(p−11=k) Σ_(k=0) ^(11) (1/((2k+23)^2 )) and Σ_(p=1) ^(10) (1/p^2 ) can also be simplified so the value of S is p.p known....
$${we}\:{have}\:{S}=\sum_{{n}=\mathrm{1}} ^{\mathrm{45}} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\left[\frac{\mathrm{45}}{\mathrm{2}}\right]} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\left[\frac{\mathrm{45}−\mathrm{1}}{\mathrm{2}}\right]} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{p}=\mathrm{1}} ^{\mathrm{22}} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\mathrm{22}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{p}=\mathrm{1}} ^{\mathrm{22}} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{11}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)^{\mathrm{2}} }\:+\sum_{{k}=\mathrm{0}} ^{\left[\frac{\mathrm{22}−\mathrm{1}}{\mathrm{2}}\right]} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{p}=\mathrm{1}} ^{\mathrm{11}} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\mathrm{10}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\:\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{p}=\mathrm{1}} ^{\mathrm{11}} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\right)+\sum_{{p}=\mathrm{0}} ^{\mathrm{22}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }\sum_{{p}=\mathrm{1}} ^{\mathrm{11}} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{p}=\mathrm{0}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:+\sum_{\mathrm{0}} ^{\mathrm{22}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${also}\:\:\sum_{{p}=\mathrm{1}} ^{\mathrm{11}} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{\mathrm{4}{p}^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{4}}\sum_{{p}=\mathrm{1}} ^{\mathrm{10}} \:\:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{p}=\mathrm{0}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$+\sum_{{p}=\mathrm{0}} ^{\mathrm{22}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }\:\sum_{{p}=\mathrm{1}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }\:\sum_{{p}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{p}=\mathrm{0}} ^{\mathrm{10}} \frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$+\sum_{{p}=\mathrm{0}} ^{\mathrm{22}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }\sum_{{p}=\mathrm{1}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\right)\sum_{{p}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{p}=\mathrm{6}} ^{\mathrm{10}} \frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$+\left(\frac{\mathrm{1}}{\mathrm{4}}\:+\mathrm{1}\right)\sum_{{p}=\mathrm{0}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{11}} ^{\mathrm{22}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{p}=\mathrm{6}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:=_{{p}−\mathrm{6}={k}} \:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\frac{\mathrm{1}}{\left(\mathrm{2}\left(\mathrm{6}+{k}\right)+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{13}\right)^{\mathrm{2}} } \\ $$$$\sum_{{p}=\mathrm{11}} ^{\mathrm{22}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:=_{{p}−\mathrm{11}={k}} \:\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{11}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{23}\right)^{\mathrm{2}} } \\ $$$${and}\:\sum_{{p}=\mathrm{1}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:\:{can}\:{also}\:{be}\:{simplified}\:\:{so}\:{the}\: \\ $$$${value}\:{of}\:{S}\:{is}\:{p}.{p}\:{known}…. \\ $$
Answered by MJS last updated on 02/Dec/19
all you can do is calculate it
$$\mathrm{all}\:\mathrm{you}\:\mathrm{can}\:\mathrm{do}\:\mathrm{is}\:\mathrm{calculate}\:\mathrm{it} \\ $$
Commented by tw000001 last updated on 03/Dec/19
I know Σ_(n=1) ^∞ (1/n^2 )=(π^2 /6)≈1.644, then Σ_(n=1) ^(45) (1/n^2 )≈1.622.
$$\mathrm{I}\:\mathrm{know}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\approx\mathrm{1}.\mathrm{644}, \\ $$$$\mathrm{then}\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{45}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\approx\mathrm{1}.\mathrm{622}. \\ $$
Answered by tw000001 last updated on 03/Dec/19
Σ_(n=1) ^(45) (1/n^2 )=(π^2 /6)−(1/(45))≈1.622
$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{45}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{45}}\approx\mathrm{1}.\mathrm{622} \\ $$
Commented by MJS last updated on 03/Dec/19
Σ_(n=1) ^(45) (1/n^2 )=Σ_(n=1) ^∞ (1/n^2 )−(1/(45)) (1/(45))+Σ_(n=1) ^(45) (1/n^2 )=Σ_(n=1) ^∞ (1/n^2 ) (k∈N^★ ⇒ Σ_(n=1) ^k (1/n^2 )∈Q)∧(1/(45))∈Q ⇒ ⇒ ((1/(45))+Σ_(n=1) ^(45) (1/n^2 ))∈Q ⇒ (π^2 /6)∈Q which is wrong ⇒ Σ_(n=1) ^(45) (1/n^2 )≠Σ_(n=1) ^∞ (1/n^2 )−(1/(45))
$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{45}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{45}} \\ $$$$\frac{\mathrm{1}}{\mathrm{45}}+\underset{{n}=\mathrm{1}} {\overset{\mathrm{45}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\left({k}\in\mathbb{N}^{\bigstar} \:\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\in\mathbb{Q}\right)\wedge\frac{\mathrm{1}}{\mathrm{45}}\in\mathbb{Q}\:\Rightarrow \\ $$$$\Rightarrow\:\left(\frac{\mathrm{1}}{\mathrm{45}}+\underset{{n}=\mathrm{1}} {\overset{\mathrm{45}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\in\mathbb{Q}\:\Rightarrow\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\in\mathbb{Q}\:\mathrm{which}\:\mathrm{is}\:\mathrm{wrong} \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{45}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\neq\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{45}} \\ $$
Commented by tw000001 last updated on 03/Dec/19
OK, I know the answer is 1.622. But is there any possible solution that is approximately on 1.622?
$$\mathrm{OK},\:\mathrm{I}\:\mathrm{know}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{1}.\mathrm{622}. \\ $$$$\mathrm{But}\:\mathrm{is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{possible}\:\mathrm{solution}\:\mathrm{that}\:\mathrm{is} \\ $$$$\mathrm{approximately}\:\mathrm{on}\:\mathrm{1}.\mathrm{622}? \\ $$
Commented by MJS last updated on 03/Dec/19
the exact value is (m/n) with m, n > 10^(37) you can use a calculator to get an approximation but there′s no unique approximation I get Σ_(n=1) ^(45) (1/n^2 )≈1.6229569293973 the exact value is ((2i0 571 823 268 450 072 862 674 893 950 786 869 803)/(12 675 520 154 492 970 709 544 386 574 878 080 000))
$$\mathrm{the}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{is}\:\frac{{m}}{{n}}\:\mathrm{with}\:{m},\:{n}\:>\:\mathrm{10}^{\mathrm{37}} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{a}\:\mathrm{calculator}\:\mathrm{to}\:\mathrm{get}\:\mathrm{an}\:\mathrm{approximation} \\ $$$$\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{unique}\:\mathrm{approximation} \\ $$$$\mathrm{I}\:\mathrm{get} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{45}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\approx\mathrm{1}.\mathrm{6229569293973} \\ $$$$\mathrm{the}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{is} \\ $$$$\frac{\mathrm{2}{i}\mathrm{0}\:\mathrm{571}\:\mathrm{823}\:\mathrm{268}\:\mathrm{450}\:\mathrm{072}\:\mathrm{862}\:\mathrm{674}\:\mathrm{893}\:\mathrm{950}\:\mathrm{786}\:\mathrm{869}\:\mathrm{803}}{\mathrm{12}\:\mathrm{675}\:\mathrm{520}\:\mathrm{154}\:\mathrm{492}\:\mathrm{970}\:\mathrm{709}\:\mathrm{544}\:\mathrm{386}\:\mathrm{574}\:\mathrm{878}\:\mathrm{080}\:\mathrm{000}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *