# Homework Help: Mental math question

1. Mar 11, 2010

### PhysicsMark

1. The problem statement, all variables and given/known data
I am currently taking a mathematical methods in physics course. We were given a prerequisite inventory on the first day of class. There are 4 problems that we are assumed to be able to do in our head to 4 or 5 digit accuracy. I am not sure of how to compute these:

1/.9997
(.9997)^.5
sin(.025)
cos(.025)

2. Relevant equations
??? Perhaps linear approximation? (I can't recall how to use linear approximation)

3. The attempt at a solution

For the first one, using a calculator, I get 1.00030009 as an answer. Playing around with this and other combinations of numbers have led me to find a type of method ( I say type of method because I have no idea if it is valid for all numbers) to evaluate an expression of this kind.

.9997 = 100-99.0003

Bring the numerator into the position right of the decimal (i.e. numerator =1 then 1. or numerator = 2 then 2.). Then bring the .0003 from 99.0003 to the left of the decimal to get 1.0003. This also works for 1/.9996 = 1.0004. When the numerator is something other than 1, I think you have to multiply the 3 in .0003 by the numerator. I.e. 2/.9997 = 2.0006 or 2/.9996 = 2.0008.

Using my calculator and playing around has led me to believe there is a pattern that I am not fully describing here. I also see a relationship with the numbers past 5 digit accuracy. But I am not sure of precisely what it is.

For the other problems, I am lost. Does anyone know any tricks here? If possible I would like an explanation for a set of rules I can apply to the situation.

I have seen on the internet various "tricks" for evaluating large numbers. An example would be multiplying a 5 digit number like 34578*11 in your head. Often, these "tricks" do not fully explain what is going on. They usually give a set of conditions that need to be applied in order to use their method.

Sorry for totally butchering all terminology. Thanks for any help.

2. Mar 11, 2010

### Staff: Mentor

sin(.025) is about .025. If x is reasonably "small" sin(x) $\approx$ x.

I believe that the basis for the approximations you're being asked to do is linear approximations of functions. Each of the functions you're being asked to approximate has a Taylor series (actually a Maclaurin series for some of these problems) that can be truncated after the first degree term to give reasonable accuracy when x is near zero.

For sin(x), the Maclaurin series is sin(x) = x - x^3/3! + x^5/5! + ...

For cos(x), the Maclaurin series is cos(x) = 1 - x^2/2! + x^4/4! + ..., so for small x, cos(x) is about 1.

3. Mar 11, 2010

### willem2

taylor series: $$sin{x} \approx x$$

$$cos(x) \approx 1 - \frac{1}{2}x^2$$

4. Mar 11, 2010

### Gokul43201

Staff Emeritus
And we leave it to you to write out the Taylor expansion for (1+x)^n and choose a suitable approximation from it.

5. Mar 11, 2010

### PhysicsMark

Thank you for all of the responses. They have provided insight, and I'm thinking differently about these approximations now.

I'm assuming you're talking about the problem : (.9997)^.5

Hmm, I'm not seeing how a taylor expansion will help.

The taylor expansion for the form (1+x)^n is(I think):

$$f(x)=\sum_{n=0}^\infty\a_n(x-c)^n=a_o+a_1(x-c)^1+a_2(x-c)^2...$$

Would that make the problem I'm talking about be :

0+.1(1-.0003)^.1+.2(1-.0003)^.2+...etc?

If that is correct, then I am confused as to how that would help me evaluate (.9997)^.5 because I cannot evaluate (.9997)^.1 or .2 or .3.

I think I am missing something here.

6. Mar 11, 2010