Taylor's Theorem for Sin(a+x) and Proving Convergence | Homework Solution

[Lucy Yeats]

Homework Statement

Taylor's theorem can be stated f(a+x)=f(a)+xf'(a)+(1/2!)(x^2)f''(a)+...+(1/n!)(x^n)Rn
where Rn=fn(a+y), 0≤y≤x
Use this form of Taylor's theorem to find an expansion of sin(a+x) in powers of x, and show that in this case, mod($\frac{x^n Rn}{n!}$)$\rightarrow$0 as n$\rightarrow$$\infty$ for all x.

Homework Equations

The Attempt at a Solution

sin(a+x)=sin(a)+xcos(a)-$\frac{1}{2!}$x^2sin(a)-$\frac{1}{3!}$x^3cos(a)...

I don't know how to prove the next bit. Also, I don't understand why Rn=fn(a+y) rather than Rn=fn(A). Any help would be great.

 

[Lucy Yeats]

I'm still stuck on this. Any help would be brilliant!

 

[I like Serena]

Hi Lucy Yeats!

$R_n=f^{(n)}(a+y)$ is part of the remainder term.
Taylor's theorem states that there is such an y so that the remainder term is equal to the sum of the remaining terms in the series.

You already have the expansion for sin(a+x), although perhaps you should find a generic formula for the terms.

Furthermore you would need to find the limit of the remainder term.
Can you think of an upper and a lower bound for Rn?

 

[Lucy Yeats]

Hello again!

The question says 0≤y≤x, so maybe fn(a)≤Rn≤fn(a+x)?

 

[I like Serena]

Hello again!

The question says 0≤y≤x, so maybe fn(a)≤Rn≤fn(a+x)?

That would only be true if fn would be a monotonically increasing function.
But sines and cosines are notorious for it that they are not.

What do you think $f^{(n)}(u)$ looks like, knowing that f(u)=sin(u)?

 

[Lucy Yeats]

fn(u) is always between 1 and -1?

 

[I like Serena]

How did you come to that idea?

 

[Lucy Yeats]

Because the gradient is always sin, cos, -sin, or -cos, which have ranges between -1 and 1.

 

[I like Serena]

Right!

So...

 

[Lucy Yeats]

Can I have another hint? I really can't see the next step.

 

[I like Serena]

You are supposed to find the limit of $|\frac{x^n R_n}{n!}|$.
What do you know and what can you say about this limit?

 

[Lucy Yeats]

So mod(Rn) is between -1 and 1, so I thought about the x^n/n! part. The limit will only be zero if x is less than one. But x could be greater than 1, so I'm confused.

 

[I like Serena]

So suppose x>1, say x=100.
What will x^n/n! be for large(r) values of n?

 

[Lucy Yeats]

So if x=100, 100^n>n!

 

[I like Serena]

So if x=100, 100^n>n!

How do you know that 100^n>n! ?
Is that true for every n?

 

[Lucy Yeats]

Is that ony true if n<x?

 

[I like Serena]

Let's pick a smaller value for x, say x=3.

Then x^3 = 3x3x3 > 1x2x3 = n!
x^4 = 3x3x3x3 > 1x2x3x4 = n!
What do you get for n=5, 6, 7, 10, 100?

 

[I like Serena]

Hey Lucy!

Did you give up on this thread?
That would be a pity!