- Thread starter
- #1

#### DeusAbscondus

##### Active member

- Jun 30, 2012

- 176

In an introductory calculus course I am doing I have just come across the following problem:

"Given that $\sin(x)=e^{-x}$ has a solution near x=1, use Newton's method to find the solution to 4 decimal places."

My question will strike you as very basic, however, I *am* a beginner and I *have* been away from the books for 3 whole weeks, and things which were looking obvious to me last time I looked at them have grown strange since:

to put the above into "functional format", ie: to express the above equation in terms of a function of x, I would do this:

$0=e^{-x}-\sin(x) \Rightarrow f(x)=e^{-x}-\sin(x)$

1. Would this be the correct first step

2. What justifies the simple setting of the equation to equal zero as a means of rendering it a function? It looks and feels like "cheating"

thx for any help in the form of light from above (me!)

DeusAbs

"Given that $\sin(x)=e^{-x}$ has a solution near x=1, use Newton's method to find the solution to 4 decimal places."

My question will strike you as very basic, however, I *am* a beginner and I *have* been away from the books for 3 whole weeks, and things which were looking obvious to me last time I looked at them have grown strange since:

to put the above into "functional format", ie: to express the above equation in terms of a function of x, I would do this:

$0=e^{-x}-\sin(x) \Rightarrow f(x)=e^{-x}-\sin(x)$

1. Would this be the correct first step

2. What justifies the simple setting of the equation to equal zero as a means of rendering it a function? It looks and feels like "cheating"

thx for any help in the form of light from above (me!)

DeusAbs

Last edited by a moderator: