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Question Number 215893 by hardmath last updated on 20/Jan/25

Find: 1) lim_(x→0) ((1 − cos2x)/x^2 ) = ? 2) Σ_(n=1) ^(n=∞) (n^2 /3^n ) = ?

$$\mathrm{Find}: \\ $$$$\left.\mathrm{1}\right)\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}\:−\:\mathrm{cos2x}}{\mathrm{x}^{\mathrm{2}} }\:=\:? \\ $$$$\left.\mathrm{2}\right)\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}=\infty} {\sum}}\:\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}} }\:=\:? \\ $$

Answered by mathmax last updated on 23/Jan/25

cos(2x)∼1−((4x^2 )/2) =1−2x^2 ⇒((1−cos(2x))/x^2 )∼((2x^2 )/x^2 ) ⇒ lim_(x→0) ((1−cos(2x))/x^2 ) =2 2)let S(x)=Σ_(n=0) ^∞ x^n and ∣x∣<1 ⇒S^′ (x)=Σ_(n=1) ^∞ nx^(n−1) and xS^′ (x)=Σ_(n=1) ^∞ nx^n by detivation S^′ (x) +xS^((2)) (x) =Σn^2 x^(n−1) ⇒ xS^′ (x)+x^2 S^((2)) (x) =Σ_(n≥1) n^2 x^n x=(1/3) ⇒Σ_(n=1) ^∞ (n^2 /3^n )=(1/3)S^′ ((1/3))+(1/9)S^((2)) ((1/3)) S(x)=(1/(1−x)) ⇒S^′ (x)=(1/((1−x)^2 )) et S^((2)) (x)=−2(((−1)(1−x))/((1−x)^4 ))=(2/((1−x)^3 )) ⇒S^′ ((1/3))=(1/(((2/3))^2 ))=(9/4) S^((2)) ((1/3))=(2/(((2/3))^3 ))=2×((27)/8) =((27)/4) ⇒ S=(1/3).(9/4)+(1/9).((27)/4)=(3/4)+(3/4)=(6/4)=(3/2)

$${cos}\left(\mathrm{2}{x}\right)\sim\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:\Rightarrow\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} }\sim\frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} }\:=\mathrm{2} \\ $$$$\left.\mathrm{2}\right){let}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{and}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{S}^{'} \left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \\ $$$${and}\:{xS}^{'} \left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}} \:\:{by}\:{detivation} \\ $$$${S}^{'} \left({x}\right)\:+{xS}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\Sigma{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} \:\Rightarrow \\ $$$${xS}^{'} \left({x}\right)+{x}^{\mathrm{2}} {S}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\sum_{{n}\geqslant\mathrm{1}} \:{n}^{\mathrm{2}} {x}^{{n}} \:\:\: \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{n}^{\mathrm{2}} }{\mathrm{3}^{{n}} }=\frac{\mathrm{1}}{\mathrm{3}}{S}^{'} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{9}}{S}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${S}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{S}^{'} \left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${et}\:{S}^{\left(\mathrm{2}\right)} \left({x}\right)=−\mathrm{2}\frac{\left(−\mathrm{1}\right)\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }=\frac{\mathrm{2}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow{S}^{'} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${S}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }=\mathrm{2}×\frac{\mathrm{27}}{\mathrm{8}}\:=\frac{\mathrm{27}}{\mathrm{4}}\:\Rightarrow \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}.\frac{\mathrm{27}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{6}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by AntonCWX last updated on 21/Jan/25

1) By differentiating numerator and denominator two times we get lim_(x→0) (((4cos(2x))/2))=lim_(x→0) (2cos(2x))=2

$$\left.\mathrm{1}\right) \\ $$$${By}\:{differentiating}\:{numerator}\:{and}\:{denominator}\:{two}\:{times}\:{we}\:{get} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{4}{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\right)=\mathrm{2} \\ $$$$ \\ $$

Commented by hardmath last updated on 21/Jan/25

1. Answer: 1

$$\mathrm{1}.\:\mathrm{Answer}:\:\mathrm{1} \\ $$

Commented by AntonCWX last updated on 21/Jan/25

Since you have the answer. I won′t bother to solve the second one.

$${Since}\:{you}\:{have}\:{the}\:{answer}. \\ $$$${I}\:{won}'{t}\:{bother}\:{to}\:{solve}\:{the}\:{second}\:{one}. \\ $$

Commented by mr W last updated on 21/Jan/25

to hardmath: you should show why the answer for 1) is 1. i think the solution and result from AntonCWX sir is correct.

$${to}\:{hardmath}: \\ $$$${you}\:{should}\:{show}\:{why}\:{the}\:{answer}\:{for} \\ $$$$\left.\mathrm{1}\right)\:{is}\:\mathrm{1}. \\ $$$${i}\:{think}\:{the}\:{solution}\:{and}\:{result}\:{from} \\ $$$${AntonCWX}\:{sir}\:{is}\:{correct}. \\ $$

Answered by mr W last updated on 21/Jan/25

1) lim_(x→0) ((1−cos 2x)/x^2 ) =lim_(x→0) ((2 sin^2 x)/x^2 ) =2 lim_(x→0) (((sin x)/x))^2 =2×1^2 =2 ✓ 2) Σ_(n=1) ^∞ x^n =(1/(1−x))−1 Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) Σ_(n=1) ^∞ nx^n =(x/((1−x)^2 )) Σ_(n=1) ^∞ n^2 x^(n−1) =(1/((1−x)^2 ))+((2x)/((1−x)^3 )) Σ_(n=1) ^∞ n^2 x^n =(x/((1−x)^2 ))+((2x^2 )/((1−x)^3 ))=((x(1+x))/((1−x)^3 )) set x=(1/3) Σ_(n=1) ^∞ (n^2 /3^n )=(((1/3)×(4/3))/(((2/3))^3 ))=(3/2) ✓

$$\left.\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{{x}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}×\mathrm{1}^{\mathrm{2}} =\mathrm{2}\:\checkmark \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}}−\mathrm{1} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{nx}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{nx}^{{n}} =\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{\mathrm{2}} {x}^{{n}} =\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\frac{{x}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$${set}\:{x}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{\mathrm{3}^{{n}} }=\frac{\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{3}}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }=\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$

Answered by MathematicalUser2357 last updated on 21/Jan/25

the first one: 1 the second one: (3/2)

$${the}\:{first}\:{one}:\:\mathrm{1} \\ $$$${the}\:{second}\:{one}:\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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