Question

Calculus early transcendental functions fourth edition. f(x)= ln ((x^2)+1), c=0 find the Taylor Series. Can anyone please help me on this?

Answer 1

sure you want to do it step by step?

Answer 2

yes that would help me understand it so much!

Answer 3

ok there is probably a snappy way to do it, but we can work it out slowly. you are expanding about 0 so first term is
\[f(0)=\ln(1)=0\]

Answer 4

isn't it 0?

Answer 5

i am lost already what happened to the (x^2 +1)

Answer 6

ok back up

Answer 7

first term is
\[f(0)\]
\[f(x)=\ln(x^2+1)\] right?

Answer 8

yes

Answer 9

so
\[f(0)=\ln(0^2+1)=\ln(1)=0\] is that ok?

Answer 10

ok

Answer 11

now we need
\[f'(0)\] because the next term will be
\[f'(0)x\] and to find
\[f'(0)\] we first find
\[f'(x)\] then replace x by 0

Answer 12

so our next job is to find the derivative of
\[\ln(x^2+1)\] and this requires the chain rule. we get
\[f'(x)=\frac{2x}{x^2+1}\]

Answer 13

\[f'(x) = \frac{1}{x^2 + 1}* 2x = \frac{2x}{x^2+1}\]

Answer 14

and therefore
\[f'(0)=\frac{2\times 0}{0^2+1}=\frac{0}{1}=0\]

Answer 15

so first two terms of the taylor series are 0

Answer 16

\[f'(0) = 0\ => f'(0)*x = 0\]

Answer 17

the next term will be
\[\frac{f''(0)}{2}x^2\]

Answer 18

and before we go any further, i give you a hint.
\[\ln(x^2+1)\] is an even function, because if you replace x by -x you get the same thing back. so all terms will have only even powers. the odd powers will all have coefficient zero

Answer 19

ok i understand the first derivative, n ot to sure about getting the second derivative. I get the even function so no odd exponents

Answer 20

doctor agdgdgdgwngo has found the second derivative. but i think there is a mistake there

Answer 21

lets check hold on

Answer 22

There is: it should be \frac{-2(x^2-1)}{(x^2+1)^2}

Answer 23

how did you get the second derivative?

Answer 24

\[f''(x) = \frac{-2(x^2-1)}{(x^2+1)^2}\]

Answer 25

you take the derivative of the first derivative

Answer 26

\[f''(0) = 2\]
\[\frac{f''(0)}{2}x^2 = x^2\]

Answer 27

now
\[f''(0)=2\] so
\[\frac{f''(0)}{2}x^2=x^2\] that is the first term of the series

Answer 28

next term will be 0 because you have no odd powers here

Answer 29

yes I understand that I was just wondering what rule you used to get the second derivative

Answer 30

quotient probably right?

Answer 31

there is a much easier way to do this, because these derivatives are really going to suck in a minute, but the answer is very simple

Answer 32

i dont get that the equation, (second derivative(0)/2 )=x^2

Answer 33

third derivative = \[f'''(x)=\frac{4x(x^2-3)}{(x^2+1)^3}\]
fourth derivative=\[f''''(x) = \frac{-12(x^4-6x^2+1)}{(x^2+1)^4}\]

Answer 34

yes it is
\[\frac{f''(0)}{2}x^2\] and a general terms looks like
\[\frac{f^{(n)}(0)}{n!}x^n\]

Answer 35

lets please do it the easy way ok?

Answer 36

fifth derivative\[f'''''(x) = \frac{48x(x^4-10x^2+5)}{(x^2+1)^5}\]

Answer 37

tell me if I am getting this, you keep on taking the derivatives and plugging in 0 for x then because you were given c=0?

Answer 38

because these derivatives really get annoying fast

Answer 39

ok satellite I get that formula

Answer 40

yes

Answer 41

can we do it the easy way now please?

Answer 42

yes go ahead

Answer 43

instead of finding the expansion for
\[f(x)=\ln(x^2+1)\] instead find it for
\[\ln(x+1)\] which is both well known and much much easier to find. ready?

Answer 44

yes

Answer 45

the derivatives are really simple, you do not need the quotient rule

Answer 46

\[f'(x)=\frac{1}{x+1}=(x+1)^{-1}\]
\[f''(x)=-1(x+1)^{-2}\]
\[f'''(x)=2(x+1)^{-3}\]
\[f^{(4)}(x)=-6(x+1)^{-4}\] etc

Answer 47

I smell factorial

Answer 48

ok i get how you got those derivatives

Answer 49

and in general from the pattern you see
\[f^{(n)}(x)=(-1)^{n-1}(n-1)!(x+1)^{-n}\]

Answer 50

now we only want these at x = 0 so really we are only interested in the coefficients right?

Answer 51

for example
\[f'(0)=1\]
\[f''(0)=-1\]
\[f'''(0)=2\]
\[f^{(4)}(0)=-6\]
\[f^{(5)}(0)=4!\] and so on

Answer 52

whew. and now since we are dividing each of these by n! we have each term looks like
\[(-1)^{n-1}\frac{x^n}{n}\]

Answer 53

this because
\[\frac{(n-1)!}{n!}=\frac{1}{n}\]

Answer 54

let me know if i have completely lost you and we can back up

Answer 55

i got -1 when I plugged in n=1 and x=0 into f prime but understand what u are doing, i just dont know where you got that formula, and not getting we are dividing by n! now part

Answer 56

ok lets to back to the formula and worry about the n! later

Answer 57

the derivatives look like
\[(x+1)^{-1}\]
\[-(x+1)^{-2}\]
\[2(x+1)^{-3}\]
\[-6(x+1)^{-4}\]
\[24(x+1)^{-5}\] what would the next one be?

Answer 58

120(x+1)^-6

Answer 59

sorry -120(x+1)^-6

Answer 60

right.

Answer 61

and since we want these at x = 0 we are really only interested in the numbers out front right? they are
1
-1
2
-6
24
-120
etc

Answer 62

ok

Answer 63

and where did the -120 come from? it came from -5*24
and the 24 came from 4*6 and the 6 came from 3*2
so we found 120 via
\[120=6\times 5\times 4\times 3\times 2=6!\]

Answer 64

so each time you take a derivative, you are multiplying a successive number by what you had previously

Answer 65

ok

Answer 66

that is why the nth derivative has coefficient (n-1)!

Answer 67

ok formula i will have to know thank you for explaining that to me

Answer 68

well not exactly, we have to adjust for the fact that it is alternating. so the coefficient of the nth derivative is actually
\[(-1)^{n-1}(n-1)!\]

Answer 69

ok with the -1 it alternates from positive t negative right

Answer 70

right

Answer 71

now we can finish more easily. each term looks like
\[\frac{f^{(n)}(0)}{n!}x^n\]

Answer 72

and we know
\[f^{(n)}(0)=(-1)^{n-1}(n-1)!\]

Answer 73

and we also know that
\[\frac{(n-1)!}{n!}=\frac{1}{n}\]

Answer 74

so each term of the expansion of
\[\ln(x+1)\] is
\[(-1)^{n-1}\frac{x^n}{n}\]

Answer 75

in other words
\[\ln(x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} +...\] it is a simple one

Answer 76

also a well known one. you see the pattern immediately. and finally (finally) we are ready to solve your problem because you did not have
\[\ln(x+1)\] you had
\[\ln(x^2+1)\]

Answer 77

ok how do we get from there to (x^2+1)

Answer 78

ok well all the hard work was done, even though the answer is fairly simple. now since you know the expansion for
\[\ln(x+1)\] if you want the expansion for
\[\ln(x^2+1)\] you ...
replace x by x^2

Answer 79

and get
\[x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\frac{x^{10}}{5}-...\]

Answer 80

if you want fancy notation each term looks like
\[(-1)^{k-1}\frac{x^{2k}}{k}\]

Answer 81

ok and if I had lnx and knew that answer and was asked lnx^2 i would do the same thing?

Answer 82

so you can write
\[\ln(x^2+1)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}x^{2k}}{k}\]

Answer 83

well you cannot expand
\[\ln(x)\] about 0 because
\[\ln(0)\] is undefined
that is why you generally see it as the expansion of
\[\ln(x+1)\] about 0

Answer 84

you can expand
\[\ln(x)\] about 1 if you like

Answer 85

you get the same thing we got with
\[\ln(x+1)\] just with
\[(x-1)\] in place of x

Answer 86

yes thats the formula i have, thank you so much for explaining this to me,taking online calc 2 course and I spend so much time looking elsewhere trying to anwer hw cause i can't get the answers out of this book thank you so much

Answer 87

\[\ln(x)=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-...\]

Answer 88

yw and good luck.
btw if it is homework on line cheat when you can

Answer 89

i try, sometimes its the only way that I can understand how to do the problems, I am self teaching myself calc 2 whith a bad book instead of reading and studying I waste an unbelievable amount of time trying to find answers and explainations to this course