A Better Approximation for ln(2.1) using Small Increment Method

[Appleton]

Homework Statement

If f(x) = x ln (1+x), find an approximation for the increase in f(x) when x increases by δx.
Hence estimate the value of ln (2.1), given that ln 2 = 0.6931.

Homework Equations

δy ≈ (dy/dx)δx

The Attempt at a Solution

δy ≈ [ln (1+x) + x/(1+x)] δx

When x = 1,

x ln(1+x) = ln 2 ≈ 0.6931

I would like to find a value for x such that

x ln(1+x) = 2.1

However, I am unable to solve this equation. Also when I try to derive the value of δx from the answer given in my textbook , 0.72, I get 0.040931534, which is not a solution to

x ln(1+x) = 2.1

So I'm a bit stuck, and would appreciate any help someone might be able to offer.

 

[Stephen Tashi]

The estimation procedure is easiest to describe in terms of using $f'(x)$ to denote the derivative of $f(x)$.
You need a "relevant equation" that says $f(x+h) \approx f(x) + h f'(x)$.

In this problem, you are to estimate $f(1.1) = ln(2.1)$ by using $x = 1, \ h = (1.1 - 1.0) , \ f(x)= x\ln(1+x)$.

In the $\delta$ notation, it is hard to express where the function $y$ is evaluated.
Use $\delta x = (1.1 - 1.0)$ and substitute that value and the value $x = 1$ in the equation $\delta y =( ln(1+x) + \frac{1}{1+x})\delta x$ to find $\delta y$. Then find the function y evaluated at x = 1.1 by the sum of y evaluated at x =1 plus $\delta y$ .

 

[Appleton]

Thanks for your reply. I'm still a little puzzled by 2 things.

When I use the method you kindly outlined for me I get 0.8124, which differs significantly from the answer in my textbook: 0.72.

Also, why choose 0.1 for δx? Are there not better numbers to use, for instance 0.05 which when added to 1 would be closer to the solution of x ln(1+x) = 2.1 and give a more accurate estimate?

 

[Stephen Tashi]

When I use the method you kindly outlined for me I get 0.8124, which differs significantly from the answer in my textbook: 0.72.

I agree with your answer. I conclude your text wants you to use a more sophisticated estimation procedure. Have you studied Taylor series? With a truncated Taylor series you can use terms involving higher powers of $\delta x$.

Also, why choose 0.1 for δx? Are there not better numbers to use, for instance 0.05 which when added to 1 would be closer to the solution of x ln(1+x) = 2.1 and give a more accurate estimate?

The problem asks us to pretend that we only know $ln(2) = f(1)$. It asks for $ln(2.1) = f(1 + 0.1)$. If you used $\delta x = 0.05$ you could use it in a Taylor series for $f(1 + 0.5)= ln(2.05)$ but it would be less straighforward to find a formula using $\delta x = 0.05$ that estimated $f( 1+ 0.1)$.

 

[Ray Vickson]

Homework Statement

If f(x) = x ln (1+x), find an approximation for the increase in f(x) when x increases by δx.
Hence estimate the value of ln (2.1), given that ln 2 = 0.6931.

Homework Equations

δy ≈ (dy/dx)δx

The Attempt at a Solution

δy ≈ [ln (1+x) + x/(1+x)] δx

When x = 1,

x ln(1+x) = ln 2 ≈ 0.6931

I would like to find a value for x such that

x ln(1+x) = 2.1

However, I am unable to solve this equation. Also when I try to derive the value of δx from the answer given in my textbook , 0.72, I get 0.040931534, which is not a solution to

x ln(1+x) = 2.1

So I'm a bit stuck, and would appreciate any help someone might be able to offer.

Why are you trying to find x that solve the equation x ln(1+x) = 2.1? Where in the question were you asked to do that?

The way you copied the question, it seems to be asking for an estimate of ln(2.1), rather than an estimate of f(1.1) = (1.1) ln(2.1); but whatever the case, it is not asking for a solution of x ln(x+1) = 2.1.

 

[Stephen Tashi]

ln(2.1), rather than an estimate of f(1.1) = (1.1) ln(2.1).

Ray has a good point. The estimate we made is for $f(2.1) = (1.1) ln(2.1) \approx .8124$, so the estimate for $ln(2.1)$ should be $.8124/1.1$.

 

[Appleton]

So I can assume that x ln(1+x) is a bit of a red herring and using f(x) = ln(1+x), x = 1 and δx = 0.1

ln 2.1 ≈ 0.1[1/(1+x)] + ln 2
ln 2.1 ≈ 0.5(0.1) + 0.6931 = 0.7431

Which is a more accurate answer than the textbook answer 0.72.

The textbook did note that this method is equivalent to the linear approximation obtained from a Taylor series. But I can't see how or why you would use the Taylor series to obtain a less accurate 0.72. I think I shall just put this curious number down to a misprint.

Thanks for your help.

 

[Stephen Tashi]

So I can assume that x ln(1+x) is a bit of a red herring and using f(x) = ln(1+x), x = 1 and δx = 0.1

If the objective is only to approximate ln(2.1) you can do that. But if the objective is to solve the problem as stated in the book, you should use x ln(1+x) to do the estimate - i.e. estimate (1.1) ln(2.1) and then divide that estimate by 1.1.

 

[BvU]

So I can assume that x ln(1+x) is a bit of a red herring and using f(x) = ln(1+x), x = 1 and δx = 0.1

ln 2.1 ≈ 0.1[1/(1+x)] + ln 2
ln 2.1 ≈ 0.5(0.1) + 0.6931 = 0.7431

Which is a more accurate answer than the textbook answer 0.72.

The textbook did note that this method is equivalent to the linear approximation obtained from a Taylor series. But I can't see how or why you would use the Taylor series to obtain a less accurate 0.72. I think I shall just put this curious number down to a misprint.

Thanks for your help.

No red herring: you found the derivative. $$(x+dx) \; \ln(1+(x+dx)) = x\;ln(1+x) + \left ( \ln(1+x) + {x\over 1+x} \right ) \; dx \ \Rightarrow\\ \qquad 1.1\; \ln(2.1) = \ln 2 + \left (\ln 2 + {1\over 2} \right ) \; 0.1 = \ln 2 + 0.11931 = 0.8124\ \Rightarrow\\ \qquad\qquad \ln 2.1 = 0.8124/1.1 = 0.7385 $$