5. Find the Taylor series for f(x) centered at the given...

5. Find the Taylor series for $f(x)$ centered at the given value of $a$, and find the radius of convergence. [Assume $f$ has a series expansion at $a$. Don't show $R_n(x)\rightarrow 0$.] a) $f(x) = x^5 + 2x^3 + x$, $a = 2$ b) $f(x) = 1/x$, $a = -3$ c) $f(x) = \frac{1}{\sqrt{x}}$, $a = 1$ Hint: $1 \cdot 3 \cdot 5 \cdots (2n + 1) = \frac{1 \cdot 2 \cdot 3 \cdots (2n + 1)}{2 \cdot 4 \cdot 6 \cdots (2n)} = \frac{(2n + 1)!}{2^n n!}$

Transcript

00:01 And my students, welcome.

00:02 We have a question here.

00:03 And in that question, we have to solve that tamr series for f centered for f, which is centered itself.

00:27 And it is given that if this is n plus 4 n is equal to z.

00:43 And now we need to find the series.

00:48 Of a 90 series, the pen 7 is equal to minus 1 to the bar of n, pectonian n divided by pi to the part of n plus full.

01:04 Now we have to find the value of the summation of n is n is equal to 0 to infinite and the radius of the convergence.

01:14 So let's solve this question.

01:16 So here, as you can see, let's stay this equation as the first equation.

01:28 So we know that, you know that, know that, the parallel series for, of fx x is equal to p.

01:50 So this will be as fx is equal to summation of n is equal to 0 to infinite.

02:00 This will be f n a divided by factorial n x minus a to the bar of n.

02:15 So using now 1 at is equal to 7, which is given to a question.

02:23 So the failure series will be series will be as summation of n is equal to 0 to infinite.

02:38 This is equal to minus 1 to the bar of n, factorial n divided by 5 to the power of n, to n plus 4.

02:52 And here we will have x minus 8, which is x 7 now.

02:56 So x minus 7 to the whole are n divided by factorial n.

03:04 So now factorial an in factorial and cancelor.

03:08 And here we have submission of n is equal to 0 to infinite and be equal to minus 1 to the part of n.

03:19 X minus 7 this is x minus 7 upon after 5 to the part of n x plus.

03:27 So now using ratio test now by using ratio test if you find that for the values of x values of x for which the series converses.

04:06 So now this will be as a n plus 1 divided by a .n is equal to, let's put down the value of a n plus 1 series.

04:21 This is equal to minus 1 to the power of n plus 1.

04:26 X minus 7 to the bar of n plus 1 divided by 5 to the power of n plus 1 multiplied by n plus 1 multiplied by n plus n plus 4.

04:42 This will be for a n plus 1.

04:46 And now for the other part of the ratio, let's come down value of a .n here.

04:56 This will be minus 1 to the part of n, x minus 7 to the power of n divided by 5 to the bar of n, end plus 4.

05:11 So here you can see this is the value of ratio and the absolute value we need to find.

05:17 So this will be equal to minus 1 to the bar of n minus 1.

05:24 This will be n plus 1, the whole bar by the rule of exponent.

05:28 And now x minus 7 to the power of n, again x minus 7, divided by similarly 5 to the bar of n into 5.

05:43 And then n plus 5 1 plus 4 is 5.

05:48 So 4 divide and again into the denominator will be in the same minus 1 to the power of l, x minus 7 to the power of n divided by 5 to the power of n x plus 4...

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