Approximating sin13 with Taylor Series using TI84 (n=150)

[kakashi1027]

Homework Statement

approximate sin13 by using the Taylor series using the TI84.
Add for n=150

Homework Equations

the infinite sums for sinx is ((-1)^n)(x^(2n+1)/(2n+1)!)

The Attempt at a Solution

I'm new to programming so i don't have any idea on where to start.
I was thinking about using the For(n,0) command, but I have no idea on how to bypass the overflow error since at n=150, it is 301!, which I'm pretty sure the calc won't do.

 

[SammyS]

I assume that is 13 radians, since 13° should only take a few terms of the Taylor series for sin(13°).

The Taylor Polynomial you need to use is a 301 degree polynomial with 150 non-zero terms.

Use Horner's scheme (see http://en.wikipedia.org/wiki/Horner_scheme" ) to evaluate the polynomial starting at the highest degree term in a "nested" fashion. There will be no need to use an exponential (except for (-1)ⁿ) and no need to use the factorial function.

As an example of this, consider the Polynomial: $$P(x)=4+3x-7x^2+7x^3-5x^4+9x^5-2x^6$$

Factor x⁵out of the two highest degree terms: $$P(x)=4+3x-7x^2+7x^3-5x^4+x^5(9-2x)$$

Now factor x⁴ out of the last two "trems": $$P(x)=4+3x-7x^2+7x^3+x^4(-5+x(9-2x))$$

Continuing: $$P(x)=4+3x-7x^2+x^3(7+x(-5+x(9-2x)))$$

etc, until you get: $$P(x)=4+x(3+x(-7+x(7+x(-5+x(9-2x)))))$$

Now, to see how this works: Try to calculate P(5) without a calculator. Not too easy with the standard form of P(x)

$$P(5)=4+3(5)-7(5)^2+7(5)^3-5(5)^4+9(5)^5-2(5)^6$$

Now try this (from the inside out): $$P(5)=4+5(3+5(-7+5(7+5(-5+5(9-2\cdot5)))))$$

9-2(5)=-1, then -5+5(-1)=-10, then 7+5(-10)=-43, then -7+5(-43)=-7-215=-222, (well that is a bit much), then 3+5(-222)=3-1110=-1107, then 4+5(-1107)=4-5535=-5531

Look at the 4 highest degree terms of the Taylor Polynomial you want to evaluate:

$$\sin(x)\approx \dots\,+\,\frac{(-1)^{147}\,x^{295}}{295!}\,+\,\frac{(-1)^{148}\,x^{297}}{297!}\,+\,\frac{(-1)^{149}\,x^{299}}{299!}\,+\,\frac{(-1)^{150}\,x^{301}}{301!}$$

$$\sin(x)\approx \dots\,+\,\frac{(-1)^{147}\,x^2}{294\cdot295}\left(1\,+\,\frac{(-1)^{148}\,x^2}{296\cdot297}\left(1\,+\,\frac{(-1)^{149}\,x^2}{298\cdot299}\left(1\,+\,\frac{(-1)^{150}\,x^2}{300\cdot301}\right)\right)\right)$$

Of course x=13.

Programming my not be all that easy if you haven't done any.

 

[SammyS]

..
Look at the 4 highest degree terms of the Taylor Polynomial you want to evaluate:

$$\sin(x)\approx \dots\,+\,\frac{(-1)^{147}\,x^{295}}{295!}\,+\,\frac{(-1)^{148}\,x^{297}}{297!}\,+\,\frac{(-1)^{149}\,x^{299}}{299!}\,+\,\frac{(-1)^{150}\,x^{301}}{301!}$$

There is a error in the next line:

$$\sin(x)\approx \dots\,+\,\frac{(-1)^{147}\,x^2}{294\cdot295}\left(1\,+\,\frac{(-1)^{148}\,x^2}{296\cdot297}\left(1\,+\,\frac{(-1)^{149}\,x^2}{298\cdot299}\left(1\,+\,\frac{(-1)^{150}\,x^2}{300\cdot301}\right)\right)\right)$$

Starting with: $$\sin(x)\approx \dots\,+\,\frac{(-1)^{147}\,x^{295}}{295!}\,+\,\frac{(-1)^{148}\,x^{297}}{297!}\,+\,\frac{(-1)^{149}\,x^{299}}{299!}\,+\,\frac{(-1)^{150}\,x^{301}}{301!}$$

$$\sin(x)\approx \dots\,+\,\frac{(-1)^{147}\,x^{295}}{295!}\,+\,\frac{(-1)^{148}\,x^{297}}{297!}\,+\,\frac{(-1)^{149}\,x^{299}}{299!}\left(1\,+\,\frac{(-1)\,x^2}{(300)(301)}\right)$$

$$\sin(x)\approx \dots\,+\,\frac{(-1)^{147}\,x^{295}}{295!}\left(1\,+\,\frac{(-1)\,x^2}{296\cdot297}\left(1\,+\,\frac{(-1)\,x^2}{298\cdot299}\left(1\,+\,\frac{(-1)\,x^2}{300\cdot301}\right)\right)\right)$$

Getting the degree 1 term is a little tricky.

$$\sin(x)\approx x\,\left(1+\,\frac{(-1)\,x^{2}}{2\cdot3}\left(\dots\,\left(1\,+\,\frac{(-1)\,x^2}{296\cdot297}\left(1\,+\,\frac{(-1)\,x^2}{298\cdot299}\left(1\,+\,\frac{(-1)\,x^2}{300\cdot301}\right)\right)\right)\dots\right)\right)$$

But I have been having trouble getting calculator program to give good results.

Note added in edit:

I got it to work!

The result agrees with the calculator's value of sin(13) to the 8th decimal place.

 

[kakashi1027]

hi, thanks for the responce!
what was your code that you used on the calculator? did you use the for loop?

 

[SammyS]

Yes, I used a "For loop".

I have a TI-86, which is an older model than a TI-84. The manual's description of the "For loop" made me skeptical of its ability to count down, so used For(I,0,N-1) , then computed N-I → J, and used J as my index.