Question

1. The continuous function f(x) is defined by \[f(x)=\frac{ e ^{(x-1)^{2}} }{ (x-1)^{2} }\] for \[x \neq 1 \] and f(1) = 1. the function f(x) has derivatives of all orders at x=1.

Answer 1

1. The continuous function f(x) is defined by \[f(x)=\frac{ e ^{(x-1)^{2}} }{ (x-1)^{2} }\] for \[x \neq 1 \] and f(1) = 1. the function f(x) has derivatives of all orders at x=1.

a) Write the first four nonzero terms and the general term of the Taylor series for \[e^{(x-1)^2}\] about x=1

b) Use the Taylor series found in part (a) to write about the first four nonzero terms and the general term of the Taylor series for f(x) about x=1.

c) Use the ratio test to find the interval of convergence for the Taylor series found in part (b).

d)Use the Taylor series for f(x) about x=1 to determine whether the graph of f(x) has any points of inflection.

Answer 2

we are also told that the maclaurin series for e^x is \[e^x=1+x+\frac{ x^2 }{ 2! }+...\frac{ x^n }{ n! } + ...\]

Answer 3

why would we do e^(1-x^2) when its e^(x-1)^2

Answer 4

Latex typo *

Answer 5

$$

\large
e^{{(x-1)}^2}=1+{(x-1)}^2+\frac{ {(({x-1})^2})^2 }{ 2! }+...\frac{ {(({x-1})^2})^n }{ n! } + ...


$$

Answer 6

@perl i get that this would be for a, but how would u use this for b, i dont realy understand stuff about general terms.

Answer 7

$$ \therefore \text{ means therefore} $$

Answer 8

General term is
$$
\large
{e^{{(x-1)}^2}=1+{(x-1)}^2+\frac{ {(({x-1})^2})^2 }{ 2! }+...\frac{ {(({x-1})^2})^n }{ n! } + ...
\\ \iff
\\{e^{{(x-1)}^2}={(x-1)}^{2\cdot0}+{(x-1)}^{2\cdot1}+\frac{ {({x-1}})^{2\cdot 2} }{ 2! }+...\frac{ {({x-1})^{2n}} }{ n! } + ...}


\\ \therefore

\\ e^{{(x-1)}^2}= \sum_{k=0}^{\infty}\frac{(x-1)^{2k}}{k!}

}

$$

Answer 9

@perl what is the difference between a and b?

Answer 10

i mean, how do u incorporate the change in equation? do u make the bottom (x-1)^2 k! ?

Answer 11

b is the general term. a) is the first 4

Answer 12

err, the first 4 are

Answer 13

@perl isnt b asking about the first 4 terms of f(x)

and \[f(x)=\frac{ e ^{(x-1)^{2}} }{ (x-1)^{2} }\]
so how would u go about to solving this?

Answer 14

$$
a) \large \text {The first four terms are } \\
\large
{e^{{(x-1)}^2}=1+{(x-1)}^2+\frac{ {(({x-1})^2})^2 }{ 2! }+\frac{ {(({x-1})^2})^3 }{ 3! }\\
=1+{(x-1)}^2+\frac{ {({x-1})^4} }{ 2 }+\frac{ ( x-1)^ 6 }{ 6 } \\

\large \text { The general term is }\\

\\{e^{{(x-1)}^2}={(x-1)}^{2\cdot0}+{(x-1)}^{2\cdot1}+\frac{ {({x-1}})^{2\cdot 2} }{ 2! }+...\frac{ {({x-1})^{2n}} }{ n! } + ...}


\\ \therefore \\
\\ e^{{(x-1)}^2}= \sum_{k=0}^{\infty}\frac{(x-1)^{2k}}{k!}

}


$$

Answer 15

b) $$
\large { e^{{(x-1)}^2}= \sum_{k=0}^{\infty}\frac{(x-1)^{2k}}{k!}
\\ \therefore \\
\Large \frac{e^{(x-1)^2}}{(x-1)^2} = \frac{\sum_{k=0}^{\infty}\frac{(x-1)^{2k} }{k!} }{(x-1)^2} = \sum_{k=0}^{\infty}\frac{(x-1)^{2k} }{k!(x-1)^2}
\\\text{ }\\
\Large =\sum_{k=0}^{\infty}\frac{(x-1)^{2k-2} }{k!}


} $$

Answer 16

and what steps do u take for c? i have a general idea of how to do the ratio test, but im not too sure

Answer 17

so i would get (x-1)^2 / (n+1)
i would make the abs value of that = 1? or would imake it equal to 0, if 0, than y?

Answer 18

the limit of that is zero

Answer 19

why would the limit be 0?

Answer 20

$$
\\\Large { \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| \\[0.2 in] \\

\\
\lim_{n \to \infty} |\frac{ \frac{ (x-1)^{2(n+1)-2} }{(n+1)!}} {\frac{(x-1)^{2n-2} }{n!}}| \\
= \lim_{n \to \infty} |{ \frac{ (x-1)^{2n} }{(n+1)!} \cdot \frac{ n! }{(x-1)^{2n-2} } } |
\\
= \lim_{n \to \infty} |\frac{(x-1)^2}{n+1}|\\
= (x-1)^2 \cdot \lim_{n \to \infty} | \frac{1}{n+1}|
= (x-1)^2 \cdot 0
= 0

}
$$

Answer 21

Since the limit is always less than 1, the series converges for all x.

Answer 22

so the interval of convergance would be 0?

Answer 23

the interval of convergence is all real numbers

Answer 24

and how would u solve d?

Answer 25

you have a mistake

, the maclaurin series for e^x is about x = 0, not x = 1

Answer 26

no, he gives us that maclaurin series at the start, than the questions are about the taylor series where x = 1

Answer 27

Write the first four nonzero terms and the general term of the Taylor series for
$$ e^{(x-1)^2}$$
about x=1.
We plugged (x-1)^2 into the maclaurin series for e^x, which is centered at x = 0.
we cant do that

Answer 28

but if u center it at 1, x-1 is 0, which means that the graph would only shift, right?

Answer 29

hmm, we might be able to do that

Answer 30

so how would we do d?

Answer 31

i know that inflection is second derivative
so do i find the second derivative and plug it in?