Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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mathsyperson
2005-12-12 09:39:21

True. That method wouldn't work if the range had a number of the form (n+0.5)π in it, because the sign changes at every point with that form, but they are not roots.

Ricky
2005-12-11 11:23:54

Just don't forget to also state the function is continuous from [1.1, 1.2].

mathsyperson
2005-12-10 23:18:36

To show that there is a root between 1.1 and 1.2, just find the values of the functions at each of those points and show that there is a change of sign.

f(1.1) ≈ 0.23
f(1.2) ≈ -0.17

There is a change of sign, so a root must exist between the two.

Flowers4Carlos
2005-12-10 16:29:54

i was working on problem one in degree mode and my calculator kept spitting out positiive results even for very large values of x.  then i switched over to radian mode and it worked!  sometimes a lil trial and error will get you on the right track!

2)

f(x) = x² + sin(x/2)                f(0) = 0
f'(x) = 2x + (1/2)cos(x/2)       f'(0) = 1/2
f''(x) = 2 - (1/4)sin(x/2)          f''(0) = 0
f'''(x) = -(1/8)cos(x/2)            f'''(0) = -1/8
f''''(x) = (1/16)sin(x/2)            f''''(0) = 0
f'''''(x) = (1/32)cos(x/2)          f'''''(0) = 1/32

0 + [(1/2)x]/1! + 0x²/2! - [(1/8)x³]/3! + [0x^(4)]/4! + [(1/32)x^5]/5!

irspow
2005-12-10 12:33:52

The first one is 1.165561185 and then my little solar calculator couldn't go any further.  It took six iterations using Newton's Method.

x - f(x)/f '(x)

Oh, and be sure to use radians because degrees will not give a solution.

You can use degrees but then your equation will be:  y = 2x - tan(180x/pi)

Mike
2005-12-09 23:55:16

I have this mathematics question that is really giving me a headache.......

Show that the equation 2x-tanx=0 has a root in the interval [1.1, 1.2]. hence find this root correct to 2 decimal places.

b) Find the Macluarin series expansion for f(x) = x2 + sin (x/2) up to and including the term in x5.