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Question Number 58267 by salahahmed last updated on 20/Apr/19

lim_(x→0) [((1−cos (x)cos (2x)cos (3x)cos (4x))/x^2 )]

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}−\mathrm{cos}\:\left({x}\right)\mathrm{cos}\:\left(\mathrm{2}{x}\right)\mathrm{cos}\:\left(\mathrm{3}{x}\right)\mathrm{cos}\:\left(\mathrm{4}{x}\right)}{{x}^{\mathrm{2}} }\right] \\ $$

Commented by Smail last updated on 20/Apr/19

Commented by Smail last updated on 20/Apr/19

When n=4 lim_(x→0) ((1−cosxcos(2x)cos(3x)cos(4x))/x^2 )=((4(4+1)(4×2+1))/(12)) =((180)/(12))=15

$${When}\:\:{n}=\mathrm{4} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−{cosxcos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{3}{x}\right){cos}\left(\mathrm{4}{x}\right)}{{x}^{\mathrm{2}} }=\frac{\mathrm{4}\left(\mathrm{4}+\mathrm{1}\right)\left(\mathrm{4}×\mathrm{2}+\mathrm{1}\right)}{\mathrm{12}} \\ $$$$=\frac{\mathrm{180}}{\mathrm{12}}=\mathrm{15} \\ $$

Commented by maxmathsup by imad last updated on 20/Apr/19

sir smail why (1−(x^2 /2))(1−(((2x)^2 )/2) )...(1−(((nx)^2 )/2))+o(x^2 ) =1−((x^2 /2) +(((2x)^2 )/2) +.....+(((nx)^2 )/2)) +o(x^2 )....

$${sir}\:{smail}\:\:{why}\:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:\right)...\left(\mathrm{1}−\frac{\left({nx}\right)^{\mathrm{2}} }{\mathrm{2}}\right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{1}−\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:+.....+\frac{\left({nx}\right)^{\mathrm{2}} }{\mathrm{2}}\right)\:+{o}\left({x}^{\mathrm{2}} \right).... \\ $$

Commented by Smail last updated on 21/Apr/19

1 coms from 1 times 1 times 1...n times and ((x^2 /2)+(((2x)^2 )/2)+...(((nx)^2 )/2)) comes from 1 times each term whose degree is equal 2. However,I am not interested in the other terms whose degree are greater than 2 because when I divide everything by x^2 , the terms whose degree are greater than 2 will equal 0 since x→0

$$\mathrm{1}\:{coms}\:{from}\:\:\mathrm{1}\:{times}\:\mathrm{1}\:{times}\:\mathrm{1}...{n}\:{times} \\ $$$${and}\:\:\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+...\frac{\left({nx}\right)^{\mathrm{2}} }{\mathrm{2}}\right)\:\:{comes}\:{from}\:\mathrm{1}\:{times}\:\:{each}\:{term}\:{whose}\:{degree}\:{is}\:{equal}\:\mathrm{2}. \\ $$$${However},{I}\:{am}\:{not}\:{interested}\:{in}\:{the}\:{other}\:{terms}\:{whose}\:{degree} \\ $$$${are}\:{greater}\:{than}\:\mathrm{2}\:{because}\:{when}\:{I}\:{divide} \\ $$$${everything}\:{by}\:\:{x}^{\mathrm{2}} ,\:\:{the}\:{terms}\:{whose}\:{degree} \\ $$$${are}\:{greater}\:{than}\:\mathrm{2}\:\:{will}\:{equal}\:\mathrm{0}\:{since}\:{x}\rightarrow\mathrm{0} \\ $$

Commented by Smail last updated on 21/Apr/19

In other words P(x)=(1−(x^2 /2))(1−(((2x)^2 )/2))...(1−(((nx)^2 )/2)) =Σ_(i=0) ^n a_i x^(2i) =1+Σ_(i=1) ^n b_i x^(2i) =1−((x^2 /2)+(((2x)^2 )/2)+...+(((nx)^2 )/2))+Σ_(i=2) ^n c_i x^(2i) ((P(x))/x^2 )=(1/x^2 )−(1/2)(1+2^2 +3^2 +...+n^2 )+Σ_(i=2) ^n c_i (x^(2i) /x^2 ) =(1/x^2 )−((n(n+1)(2n+1))/(2×6))+Σ_(i=2) ^n c_i x^(2(i−1)) ((1−P(x))/x^2 )=((n(n+1)(2n+1))/(12))−Σ_(i=1) ^(n−1) c_i x^(2i) as x→0 Σ_(i=1) ^(n−1) c_i x^(2i) →0

$${In}\:{other}\:{words}\: \\ $$$${P}\left({x}\right)=\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\right)...\left(\mathrm{1}−\frac{\left({nx}\right)^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{\mathrm{2}{i}} =\mathrm{1}+\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{b}_{{i}} {x}^{\mathrm{2}{i}} \\ $$$$=\mathrm{1}−\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+...+\frac{\left({nx}\right)^{\mathrm{2}} }{\mathrm{2}}\right)+\underset{{i}=\mathrm{2}} {\overset{{n}} {\sum}}{c}_{{i}} {x}^{\mathrm{2}{i}} \\ $$$$\frac{{P}\left({x}\right)}{{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+{n}^{\mathrm{2}} \right)+\underset{{i}=\mathrm{2}} {\overset{{n}} {\sum}}{c}_{{i}} \frac{{x}^{\mathrm{2}{i}} }{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}×\mathrm{6}}+\underset{{i}=\mathrm{2}} {\overset{{n}} {\sum}}{c}_{{i}} {x}^{\mathrm{2}\left({i}−\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}−{P}\left({x}\right)}{{x}^{\mathrm{2}} }=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}}−\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{c}_{{i}} {x}^{\mathrm{2}{i}} \\ $$$${as}\:{x}\rightarrow\mathrm{0}\:\:\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{c}_{{i}} {x}^{\mathrm{2}{i}} \rightarrow\mathrm{0} \\ $$

Commented by Tawa1 last updated on 21/Apr/19

What is O sir

$$\mathrm{What}\:\mathrm{is}\:\mathrm{O}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 21/Apr/19

I mean, the + Ox^2

$$\mathrm{I}\:\mathrm{mean},\:\:\:\mathrm{the}\:\:+\:\mathrm{Ox}^{\mathrm{2}} \\ $$

Commented by Tawa1 last updated on 21/Apr/19

And why do you chose to stop at power 2 sir

$$\mathrm{And}\:\mathrm{why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{chose}\:\mathrm{to}\:\mathrm{stop}\:\mathrm{at}\:\:\mathrm{power}\:\:\mathrm{2}\:\mathrm{sir} \\ $$

Commented by Smail last updated on 21/Apr/19

I used Tylor series to find this limite and the Tylor series of cosx near 0 is cosx=Σ_(n=0) ^∞ (((−1)^n x^(2n) )/((2n)!))=1−(x^2 /2)+(x^4 /(4!))+... and as more terms you add the more accurate you get I choose to stop at degree to because the degree of the denominator is equal to 2 and O(x^2 ) this symbol means the degree I am stopping at. I can stop at any degree that is greater than 2, but when I divide by x^2 , those terms will equal 0

$${I}\:{used}\:{Tylor}\:{series}\:{to}\:{find}\:{this}\:{limite} \\ $$$${and}\:{the}\:{Tylor}\:{series}\:{of}\:{cosx}\:{near}\:\mathrm{0}\:{is} \\ $$$${cosx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+... \\ $$$${and}\:{as}\:{more}\:{terms}\:{you}\:{add}\:{the}\:{more}\:{accurate}\:{you}\:{get} \\ $$$${I}\:{choose}\:{to}\:{stop}\:{at}\:{degree}\:{to}\:{because}\:{the}\:{degree} \\ $$$${of}\:{the}\:{denominator}\:{is}\:{equal}\:{to}\:\mathrm{2} \\ $$$${and}\:\:{O}\left({x}^{\mathrm{2}} \right)\:\:{this}\:{symbol}\:{means}\:{the}\:{degree}\: \\ $$$${I}\:{am}\:{stopping}\:{at}. \\ $$$${I}\:{can}\:{stop}\:{at}\:{any}\:{degree}\:{that}\:{is}\:{greater}\:{than}\:\mathrm{2}, \\ $$$${but}\:{when}\:{I}\:{divide}\:{by}\:{x}^{\mathrm{2}} \:,\:{those}\:{terms}\:{will}\:{equal}\:\mathrm{0} \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 21/Apr/19

thanks sir for this clarification.

$${thanks}\:{sir}\:{for}\:{this}\:{clarification}. \\ $$

Commented by Smail last updated on 21/Apr/19

You are welcome and I hope my comments were helpful.

$${You}\:{are}\:{welcome}\:{and}\:{I}\:{hope}\:{my}\:{comments}\:{were}\:{helpful}. \\ $$

Commented by Tawa1 last updated on 21/Apr/19

Now, i understand sir. more knowledge.

$$\mathrm{Now},\:\mathrm{i}\:\mathrm{understand}\:\mathrm{sir}.\:\mathrm{more}\:\mathrm{knowledge}. \\ $$

Commented by maxmathsup by imad last updated on 21/Apr/19

we have cos(x)cos(2x)=(1/2){cos(3x) +cos(x)} ⇒ cis(x)cos(2x)cos(3x) =(1/2){cos(3x) cos(3x)+cos(x)cos(3x)} =(1/4){ cos(6x) +1 + cos(4x) +cos(2x)} ⇒ cos(x)cos(2x)cos(3x)cos(4x) =(1/4){cos(6x)cos(4x) +cos(4x)+cos(4x) cos(4x) +cos(2x)cos(4x)} =(1/8){cos(10x) +cos(4x) +cos(8x) +1 +cos(6x) +cos(2x)} let f(x) =1−(1/8){1+cos(2x) +cos(4x) +cos(6x) +cos(8x) +cos(10x)} and g(x) =x^2 let use hospital theorem we have f^′ (x) =(1/8){2sin(2x)+4sin(4x)+6sin(6x) +8sin(8x) +10 sin(10x)} f^(′′) (x) =(1/8){4 cos(2x) +16 cos(4x) +36 cos(6x) +64 cos(8x) +100cos(10x)} ⇒lim_(x→0) f^(′′) (x) =(1/8){ 4+16 +36 +64 +100} =(1/8){20 +200} =((220)/8) =((110)/2) =55 we have g(x)=x^2 ⇒ g^′ (x)=2x and g^(′′) (x) =2 ⇒ lim_(x→0) ((1−cos(x)cos(2x)cos(3x)cos(4x))/x^2 ) =((55)/2) . =

$${we}\:{have}\:{cos}\left({x}\right){cos}\left(\mathrm{2}{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\mathrm{3}{x}\right)\:+{cos}\left({x}\right)\right\}\:\Rightarrow \\ $$$${cis}\left({x}\right){cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{3}{x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\mathrm{3}{x}\right)\:{cos}\left(\mathrm{3}{x}\right)+{cos}\left({x}\right){cos}\left(\mathrm{3}{x}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\:{cos}\left(\mathrm{6}{x}\right)\:+\mathrm{1}\:+\:\:{cos}\left(\mathrm{4}{x}\right)\:+{cos}\left(\mathrm{2}{x}\right)\right\}\:\Rightarrow \\ $$$${cos}\left({x}\right){cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{3}{x}\right){cos}\left(\mathrm{4}{x}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\left\{{cos}\left(\mathrm{6}{x}\right){cos}\left(\mathrm{4}{x}\right)\:+{cos}\left(\mathrm{4}{x}\right)+{cos}\left(\mathrm{4}{x}\right)\:{cos}\left(\mathrm{4}{x}\right)\right. \\ $$$$\left.+{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left\{{cos}\left(\mathrm{10}{x}\right)\:+{cos}\left(\mathrm{4}{x}\right)\:+{cos}\left(\mathrm{8}{x}\right)\:+\mathrm{1}\:+{cos}\left(\mathrm{6}{x}\right)\:+{cos}\left(\mathrm{2}{x}\right)\right\} \\ $$$${let}\:{f}\left({x}\right)\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\left\{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\:+{cos}\left(\mathrm{4}{x}\right)\:+{cos}\left(\mathrm{6}{x}\right)\:+{cos}\left(\mathrm{8}{x}\right)\:+{cos}\left(\mathrm{10}{x}\right)\right\} \\ $$$${and}\:\:{g}\left({x}\right)\:={x}^{\mathrm{2}} \:\:\:{let}\:{use}\:{hospital}\:{theorem}\:{we}\:{have} \\ $$$${f}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}\left\{\mathrm{2}{sin}\left(\mathrm{2}{x}\right)+\mathrm{4}{sin}\left(\mathrm{4}{x}\right)+\mathrm{6}{sin}\left(\mathrm{6}{x}\right)\:+\mathrm{8}{sin}\left(\mathrm{8}{x}\right)\:+\mathrm{10}\:{sin}\left(\mathrm{10}{x}\right)\right\} \\ $$$${f}^{''} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}\left\{\mathrm{4}\:{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{16}\:{cos}\left(\mathrm{4}{x}\right)\:+\mathrm{36}\:{cos}\left(\mathrm{6}{x}\right)\:+\mathrm{64}\:{cos}\left(\mathrm{8}{x}\right)\:+\mathrm{100}{cos}\left(\mathrm{10}{x}\right)\right\} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{f}^{''} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}\left\{\:\mathrm{4}+\mathrm{16}\:+\mathrm{36}\:+\mathrm{64}\:+\mathrm{100}\right\}\:=\frac{\mathrm{1}}{\mathrm{8}}\left\{\mathrm{20}\:+\mathrm{200}\right\}\:=\frac{\mathrm{220}}{\mathrm{8}}\:=\frac{\mathrm{110}}{\mathrm{2}}\:=\mathrm{55} \\ $$$${we}\:{have}\:{g}\left({x}\right)={x}^{\mathrm{2}} \:\Rightarrow\:{g}^{'} \left({x}\right)=\mathrm{2}{x}\:\:{and}\:{g}^{''} \left({x}\right)\:=\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{cos}\left({x}\right){cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{3}{x}\right){cos}\left(\mathrm{4}{x}\right)}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{55}}{\mathrm{2}}\:. \\ $$$$= \\ $$

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