### <Success>Infinite series lecture 3</Success> <div style='float: left; text-align:left width:100%;'> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/014-0027.6.jpg"> <img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/19188ca9-c374-4a57-21da-1e3184b3af00/public"></p> </div>

### <Success>Recall</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$(1-x)^{-n} \rarr$ integer</p> <p>$(1-x)^{p/q} \rarr$ rational number</p> <p>In particular we were looking at $(1-x)^{-\frac{1}{2}}$.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/024-0099.6.jpg"></p> </div>

### <Success>Example</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>What is the series expansion for $(1-x)^{\frac{1}{2}}$?</p> <p><strong>Solution:</strong></p> <p>We have $(1-x)^{\frac{1}{2}} \times (1-x)^{\frac{1}{2}} = (1 - x)$</p> <p>$\therefore$ If we write</p> <p>$(1-x)^{\frac{1}{2}}=a_0+a_1{x}+a_2 x^2$</p> <p>then by multiply it with itself we should get $(1 - x).$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/9ec8430c-462b-40af-3996-3ff424644000/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Let $(1-x)^{\frac{1}{2}}=a_0+a_1 x +a_2 x^2+\cdots$</p> <p>$\therefore (a_0+a_1{x}+a_2 x^2+…) \times (a_0+a_1{x}+a_2 x^2+.. .)$</p> <p>$= 1 - x$</p> <p>$\therefore$ coeff of $x^0=a_0^2=1$</p> <p>Taking the $(+)$ve root $a_0=1$</p> <p>coeff of $x=2 a_0 a_1=-1$</p> <p>$\therefore 2 a_1=-1$</p> <p>$\therefore a_1=-\frac{1}{2}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/f864541b-cb1e-445e-82e9-427eb0340100/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Coeff of $x^2, \ $ $\ a_0 a_2+a_1^2+a_2 a_0 = 0$</p> <p>$\because \nexists x^2$ in $(1 - x)$</p> <p>$\therefore 2 a_2+a_1^2=0$</p> <p>or $2 a_2+\frac{1}{4}=0$</p> <p>$\therefore a_2=-\frac{1}{8}$</p> <p>$\therefore$ We can write $(1-x)^{\frac{1}{2}} =\sqrt{1-x}$ as $1-\frac{1}{2} x-\frac{1}{8} x^2+\cdots$</p> <p>If $x$ is small we aften ignore higher powers.</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/75fffd4d-db52-4d1b-8706-d8e5c9e08b00/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>In a similar way we can find $\sqrt{1+x}$.</p> <p>say $\sqrt{1+x}=a_0+a_1 x+a_2 x^2+\cdots$</p> <p>$\therefore(a_0+a_1 x+a_2 x^2+\cdots) \times (a_0+a_1 x+a_2 x^2)\cdots = 1 + x$</p> <p>$a_0^2=1 \text { or } a_0=1 $ (Taking +ve value)</p> <p>$a_0 a_1+a_1 a_0=1$</p> <p>or $2 a_0 a_1=1 \quad \therefore a_1=\frac{1}{2}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/080-0550.8.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ a_0 a_2+a_1^2+a_0 a_2=0$</p> <p>or $2 a_2+a_1^2=0$</p> <p>$\therefore 2 a_2=-a_1^2=-\frac{1}{4}$</p> <p>$\therefore a_2=-\frac{1}{8}$</p> <p>$\therefore$ We get $\sqrt{1+x}$</p> <p>$=1+\frac{1}{2} x-\frac{1}{8} x^2 \ldots .$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/093-0618.0.jpg"></p> </div>

### <Success>Question 1</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Find $\sqrt{17}$.</p> <p><strong>Solution:</strong></p> <p>We can write it as</p> <p>$(1+16)^{\frac{1}{2}}$ $\larr$ we are making a mistake</p> <p>In $(1 + x) \larr$ $|x| < 1$</p> <p>We write in a different way</p> <p>$ 16^{\frac{1}{2}}(1+\frac{1}{16})^{1/2}$</p> <p>$|{\frac{1}{16}}| < 1$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d7814c0b-c437-4f8c-bb47-b232c97a6400/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>To obtain $\sqrt{17}$</p> <p>we write it as $16^{\frac{1}{2}}\left(1+\frac{1}{16}\right)^{\frac{1}{2}}$</p> <p>$=4 \times (1+\frac{1}{16})^{\frac{1}{2}}$</p> <p>In the expansion of $(1+x)^{\frac{1}{2}}$</p> <p>$=1+\frac{1}{2} x-\frac{1}{8} x^2$</p> <p>Putting $x=\frac{1}{16}$ we get</p> <p>$ =1+\frac{1}{32}-\frac{1}{8 \times 16^2}$</p> <p>$\therefore \sqrt{17}=4{\times}\left(1+\frac{1}{32}-\frac{1}{8 \times 16^2}\right)$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/aa9ee9db-15cf-4f78-94f7-80199aeb6200/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$4 \times 1=4$</p> <p>$4 \times \frac{1}{32}=\frac{1}{8}=0.125$</p> <p>$4 \times \biggl( -\frac{1}{8 \times 16^2}\biggl )=-\frac{1}{2 \times 16^2}$</p> <p>$ =-\frac{1}{2 \times 256} =-\frac{1}{512}$</p> <p>$ =-0.0019$</p> <p>$\therefore \sqrt{17} =4.125-0.0019$</p> <p>$= 4.1231$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/000001.png"></p> </div>

### <Success>Notes</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Therefere the final expansion is $(1+x)^{p/q}$</p> <p>$=1+\frac{p}{q} x+\frac{\frac{p}{q}\left(\frac{p}{q}-1\right)}{2 !} x^2 +\frac{\frac{p}{q}\left(\frac{p}{q}-1\right)\left(\frac{p}{q}-2\right)}{3 !} x^3 +…..$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/166-1131.6.jpg"></p> </div>

### <Success>Notes</Success> <div style='float: left; text-align:left width:99%;font-size:30px;'> <p>What we have done so far is not a proof, we have verified certain results.</p> <p>The basic assumption was:</p> <p>Weierstrass Approximation Theorem.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/177-1208.4.jpg"></p> </div>

### <Success>Weierstrass Approximation Theorem</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Every continous function in a closed interval can be approximated as closely as possible by a polynomial function.</p> <p>$(1+x)^k \leftarrow$ negative integral or rational.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/187-1302.0.jpg"></p> </div>

### <Success>Exapansion of the series</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>How to obtain the coefficients of the given function $f(x)$?</p> <p>it is possible to have such a polynomial expansion about a point $a.$</p> <p>$f(x)=a_ 0 +a_ 1(x-a) +a_ 2(x-a)^2 +a_3(x-a)^3 +……$</p> <p>Advantage of a polynomial function is that</p> <p>If it is of degree $n$, it can be differentiatal $(x + 1)$ times.</p> <p>& If we take an infinite polynomial, we can differentiate infinite number of times.</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/53d23789-d01f-4449-d6bf-3a5ac1bddd00/public"></p> </div>

### <Success>Exapansion of the series</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$f(x)=a_0+ a_1(x-a)+a_2(x-a)^2 +a_3(x-a)^3+..$</p> <p>$\therefore f(a)=a_0 \therefore a_0=f(a)$</p> <p>$f^{(1)}(a)=a_ 1+2 a_ 2(x-a) +3 a_ 3(x-a)^2 +4 a_4(x-a)^3+…..$</p> <p>$f^{(2)}=2 a_2+3 \cdot 2 a_3(x-a) +4 \cdot 3 \cdot a_4(x-a)^2+…..$</p> <p>$\therefore f^{(2)}(a)= 2 \cdot a_2 $</p> <p>or $a_2=f^{(2)}(a) / 2$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d9f05e28-c9d6-4cfa-4774-6d14c5571800/public"></p> </div>

### <Success>Exapansion of the series</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$f^{(3)}(x)= 3 \cdot 2 \cdot 1 \cdot a_3$</p> <p>$ \quad +4 \cdot 3 \cdot 2(x-a)+…..$</p> <p>terms with higher powers of $(x - a)$</p> <p>$ \therefore f^{(3)}(a)=3 ! a_3$</p> <p>$ \therefore a_3=\frac{f^{(3)}(a)}{3 !}$</p> <p>$f^{(4)}(x)=4 \cdot 3 \cdot 2 \cdot 1 a_4$ + power of $(x - a)$</p> <p>$\therefore a_4=\frac{f(4)(a)}{4 !}$</p> <p>This expansion is called <strong>Taylor Series expansion</strong> of $f(x).$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/c8bdf7aa-19f7-43c6-e1dc-033be854cd00/public"> <img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/6c9b0848-17ad-4cdc-2762-9d577f1dd400/public"></p> </div>

### <Success>Exapansion of the series</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore$ $f(x)$ can be written as:</p> <p>$f(a) + f^{(1)}(a)(x-a) +\frac{f^{(2)}(a)}{2 !} (x-a)^2$</p> <p>$+\frac{f^{(3)}(a)}{3 !} (x-a)^3$</p> <p>$+\frac{f^{(4)}(a)}{4 !} (x-a)^4 +…….$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/265-1837.2.jpg"></p> </div>

### <Success>$\text{Exapansion of} \ (1-x)^{-2}$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Let us consider $(1-x)^{-2} \ \ \therefore f(x) =(1-x)^{-2}$</p> <p>$f^{(1)}(x) =-2(1-x)^{-3}(-1)=2(1-x)^{-3}$</p> <p>$f^{(2)}(x) =-3 \cdot 2 \cdot(1-x)^{-4}(-1) =3 !(1-x)^{-4}$</p> <p>$f^{(3)}(x) =4 !(1-x)^{-5}$</p> <p>$\therefore f(0) = 1$</p> <p>$f^{(1)}(0)=2$</p> <p>$f^{(2)}(0)=3!$</p> <p>$f^{(4)}(0)=4!$</p> </div>

### <Success>$\text{Exapansion of} \ (1-x)^{-2}$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore$ we get</p> <p>$f(x) = f(0) + 2(x - 0) +3 ! \frac{(x-0)^2}{2 !} +4 ! \frac{(x-0)^3}{3 !}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/310cde2f-1cb0-4fa0-17f1-a9774c248400/public"></p> </div>

### <Success>$\text{Exapansion of} \ (1-x)^{-2}$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore$ We can write</p> <p>$f(x)= f(a)+f^{(1)}(a) \cdot(x-a) +f^{(2)}(a)\frac{(x-a)^2}{2 !} +f^{(3)}(a)\frac{(x-a)^3}{3 !}$</p> <p>$+f^{(4)}(a)\frac{(x-a)^4}{4 !} +……$</p> <p>Putting the values:</p> <p>$f(x)=(1-x)^{-2}$</p> <p>$=f(0)+f^{(1)}(0) x+f^{(2)}(0) \frac{x^2}{2 !} +f^{(3)}(0)\frac{x^3}{3 !} + f^{(4)}(0)\frac{x^4}{4 !}$</p> </div>

### <Success>$\text{Exapansion of} \ (1-x)^{-2}$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$f(0)+2 \cdot x+3 ! \frac{(x-0)^2}{2 !}+4 ! \frac{(x-0)^3}{3 !}+…..$</p> <p>$=1+2 x+3 x^2+4 x^3+\cdots$</p> <p>These are the terms that we have already seen.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/000003.png"></p> </div>

### <Success>Trigonometric functions</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Looking Beyond Polynomials.</p> <p>Consider trigonometric Functions:</p> <p>$\sin x$</p> <p>$\cos x$</p> <p>$\tan x$</p> <p>$ 0^{\circ}, \pi / 6, \pi / 4, \pi / 3, \pi / 2, \pi $</p> <p>$ 15^{\circ}, 18^{\circ} \ \text{etc.}$</p> <p>$ \sin 1^{\circ}, \sin 5^{\circ}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/e2f9a6df-a104-4ea9-bc05-2781363e7d00/public"></p> </div>

### <Success>$\text{Expansion of} \sin x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Consider $\sin x.$</p> <p>$f(x) = \sin x \ \ \therefore f(0) = 0$</p> <p>$f^{(1)}(x)=\cos x \ \ \therefore f^{(1)}(0)=1$</p> <p>$f^{(2)}(x)=-\sin x \ \ \therefore f^{(2)}(0)=0$</p> <p>$f^{(3)}(x)=-\cos x \ \ \therefore f^{(3)}(0)=-1$</p> <p>$f^{(4)}(x)=\sin x \ \ \therefore f^{(4)}(0)=0$</p> <p>$f^{(5)}(x)=\cos x \ \ \therefore f^{(5)}(0)=1$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/000004.png"></p> </div>

### <Success>$\text{Expansion of} \sin x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>From Taylor’s series</p> <p>$ f(x)= f(0)+f’(0)(x-0)+f^{(3)}(0) \frac{(x-a)^2}{2 !}$</p> <p>$+f^{(3)}(0) \frac{(x-a)^3}{3 !}+f^{(4)}(0) \frac{(x-a)^4}{4 !} +f^{(5)}(0) \frac{(x-a)^5}{5 !}+\ldots$</p> <p>$f(0) = 0$</p> <p>$f^{(1)}(0) = 1$</p> <p>$f^{(2)}(0)=0$</p> <p>$f^{(3)}(0)=-1$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/000005.png"></p> </div>

### <Success>$\text{Expansion of} \sin x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$f^{(4)}(0)=0$</p> <p>$f^{(5)}(0)=+1$</p> <p>Putting these values we get,</p> <p>$\sin (x) = \sin (0) +1 \cdot(x-0)+0 \cdot \frac{(x-0)^2}{2 !} -1 \frac{(x-0)^3}{3 !}+0 \cdot \frac{(x-0)^4}{4 !} +1 \cdot \frac{(x-0)^5}{5 !}+\cdots$</p> <p>$= 1 \cdot x-\frac{x^3}{3 !}+\frac{x^5}{5 !}$</p> <p>$= x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\frac{x^7}{7 !}+\cdots$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/383-2739.6.jpg"></p> </div>

### <Success>$\text{Expansion of} \cos x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Similarly we can work out for $\cos x$.</p> <p>$\cos x=1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !} \cdots$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/397-2811.6.jpg"></p> </div>

### <Success>$\text{Expansion of} \tan x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Find the Taylor series expansion for $\tan (x).$</p> <p>Again we expand about $0.$</p> <p>$f(0) = \tan (0) = 0$</p> <p>$f^{(1)}(x)=\frac{d}{d x} \tan x=\sec ^2 x$</p> <p>$=1+\tan ^2 x$</p> <p>$\therefore f^{(1)}(0)= 1.$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/000006.png"></p> </div>

### <Success>$\text{Expansion of} \tan x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$f^{(2)}(x)=\frac{d}{d x}(1+\tan^2 x)$</p> <p>$=2 \tan x(\sec^2 x)$</p> <p>$=2 \tan x(1+\tan^2 x)$</p> <p>$=2 \tan x+2 \tan^3 x$</p> <p>$\therefore f^{(2)}(0)=0.$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/430-2960.4.jpg"></p> </div>

### <Success>$\text{Expansion of} \tan x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$f^{(3)}(x)=2(1+\tan^2 x)+6 \tan^2 x(1+\tan^2 x)$</p> <p>$=2+2 \tan^2 x+6 \tan^2(x) +6 \tan^{(4)} x$</p> <p>$\therefore f^{(3)}(0)=2$</p> <p>$f^{(4)}(x)=\frac{d}{d x}(8 \tan^2 x+6 \tan^4 x)$</p> <p>$=16 \tan x+16 \tan^3 x + 24 \tan^3 x (1+\tan^2 x)$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/444-3087.6.jpg"></p> </div>

### <Success>$\text{Expansion of} \tan x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore f^{(n)}(0)=0$</p> <p>$f^{(5)}(x)=16(1+\tan^2 x)+$ terms with $\tan x$</p> <p>$\therefore f^{(5)}(0)=16$</p> <p>$\tan x=0+\frac{1}{1 !} x+0 \frac{x^2}{2 !}$</p> <p>$+2 \cdot \frac{x^3}{3 !}+0 \cdot \frac{x^4}{4 !} +16 \frac{x^5}{5 !}+\cdots $</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/3c3f35ab-fd09-47f7-fb00-58e1e61bda00/public"></p> </div>

### <Success>$\text{Expansion of} \tan x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\tan x= 0+x + 0 \cdot x^2 +2 \cdot \frac{x^3}{3 !}+0 x^4 +16 \frac{x^5}{5 !}$</p> <p>$=x+\frac{x^3}{3}+\frac{2}{15} x^5 + \text{higher powers of} \ x$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-15-infinite-series-l-3_5-0np2qebyqma/img/000007.png"></p> </div>

<div> <h2><Success>Thank you</Success></h2> </div>