1a find the following maclaurin series lnx1sumn0infty...

1.a. Find the following Maclaurin series ln(x+1)=sum_(n=0)^(infty ) ((-1)^(n+1))/(n+1)x^(n+1)=sum_(n=1)^(infty ) ((-1)^(n))/(n)x^(n) using the definition of a Taylor series centered at x=0. 1.a.Find the following Maclaurin series 1n(x+1 (-1)n+1 xn+1 n+1 n=0 n=1 using the definition of a Taylor series centered atx=O

Transcript

00:01 To start this problem, we're going to calculate the first three derivatives of 1 over x squared, and then we're going to figure out what those values are at a, and plug them into the formula for the taylor series and find a general form.

00:16 So if we start with the function 1 over x squared, 1 over x squared is equivalent to x to the negative 2 power.

00:26 If we take the first derivative of x to the negative 2, we can use the power rule, bring the exponent to the front times x, subtract 1 from the exponent, negative 2, x to negative 3.

00:38 Second derivative, following the same manner, it would be negative 2 times negative 3, 6x to the negative 4, and then f triple prime of 3 will be negative 4 times 6, which is negative 24, x to the negative 5.

00:57 Now if we plug in a, which is 1, to each of these, we get f of 1 is 1 to the negative 2, which is just 1, since 1 raise to any exponent is just 1.

01:08 The first derivative at 1, it would be negative 2 times 1 to the negative 3, which is just negative 2.

01:16 Second derivative at 1 will be 6 times 1 to the 4th, which is just 6.

01:23 And then the third derivative evaluated at 1, be negative 24.

01:28 Times 1 to the negative 5th, which is negative 24.

01:33 Now we're going to plug these into the taylor series definition, which starts with f of a, which is 1, plus f prime of a, which is negative 2.

01:45 This will be 1 minus 2 times x minus a, which is 1, plus f double prime at a, which is 6 over 2 factorial times x minus 1, squared.

02:00 Minus 24 over 3 factorial times x minus 1 cubed and so on.

02:10 And here we can see that we have somewhat of a general pattern.

02:15 Since we can write this in some notation using k equals 0 to infinity, the coefficients are 1 to 6 over 2 factorial.

02:27 2 factorial is 2 times 1 is just 2.

02:30 6 divided by 2 is 3, 24 over 3 factorial is 24 over 3 times 2 times 1.

02:43 Cancelling out a factor of 2 and the top and bottom, leave us 12 in the numerator.

02:49 It will be 12 divided by 3, which is 4.

02:52 So this coefficient is 4, and this is 3.

02:57 So the coefficient is just increasing 1 and vary negative sign...

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