Solving Mental Math Questions for Mathematical Methods in Physics

[PhysicsMark]

Homework Statement

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)

Homework Equations

? Perhaps linear approximation? (I can't recall how to use linear approximation)

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.

 

[Mark44]

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.

 

[willem2]

taylor series: sin{x} \approx x

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

 

[Gokul43201]

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

 

[PhysicsMark]

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

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

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.

 

[Mark44]

The first couple of terms of the binomial series for (1 + x)^n are 1 + nx + ...
x can be negative.

For more information, see http://en.wikipedia.org/wiki/Binomial_series.