# Show (1+x)^k=1+kx near x=0

1. Aug 5, 2007

1. The problem statement, all variables and given/known data
Using Local linear approx show that $$(1+x)^k\approx1+kx$$ at x=0

2. Relevant equations $$L(x)=f(a)+f'(a)(x-a)$$

3. The attempt at a solution
I believe this is probably easier than I am making it...I may be confusing what to use for$$a$$

This is what I have...

$$f(0)=1^k$$and $$f'(0)=k^{k-1}$$
$$\Rightarrow L(x)=1^k+kx^{k-1}$$

Where do we go now? :does Axl Rose dance:

Last edited: Aug 5, 2007
2. Aug 5, 2007

### morphism

Are you sure f'(0)=k^(k-1)?

3. Aug 5, 2007

No...oops. it should be $$kx(1+x)^{k-1}$$...right? Let me get crackin again..

4. Aug 5, 2007

### meopemuk

Perhaps,

$$f'(x) = k(1+x)^{k-1}$$?

Eugene.

5. Aug 5, 2007

Sorry, I got ahead of myself,
so $$L(x)=1^k+kx(1+x)^{k-1}$$
...is this right do far?...
My goal is to make it look like 1+kx right?

Thanks

Last edited: Aug 5, 2007
6. Aug 5, 2007

$$\Rightarrow L(x)=1+kx^{k-1}$$ now what is happening to k-1 that must make it negligible...

7. Aug 5, 2007

### morphism

How are you getting that?

L(x) = f(a) + f'(a)(x - a)

We want a=0. And meopemuk told you what f'(x) looks like.

8. Aug 5, 2007

f(a)=(1+0)^K
f'(a)=k(1+0)^k-1
(x-o)=x
so
L(0)=1^k+[k(1)^k-1]*x

oh...
I attached the k-1 to x not 1......in revising i see that 1^k-1 =1
therefore L(0)=1+k*x
Thanks people.

9. Aug 5, 2007

### morphism

I don't understand what you're doing. How is 1^k - 1= 1?

f(x) = (1 + x)^k
f'(x) = k (1 + x)^(k-1)

So for a=0,
L(x) = f(0) + (x - 0)f'(0) = 1 + kx

10. Aug 6, 2007

### VietDao29

What do you mean? How did you differentiate f(x) with respect to x? It does not seem right at all. =.=" Can you show us your steps?

What is 1k? Is it 1, or 2? >"<

Well, I think you should read the book again, from the very first chapter of differentiating. You shouldn't settle down solving problem unless you understand the core concept.

Well, it can be pretty time consuming, but it's extremely good in this case. Go ahead, and re-read the book. It may help. :) Then, let's try the problem once more, and see if you can get it.

11. Aug 6, 2007

### huyen_vyvy

binomial expansion is the easiest way to answer this question though,
(1+x)^k=1+kx+k(k-1)/2*x^2+... and we can ignore those higher degrees of x.

12. Aug 6, 2007

If my function is $$(1+x)^k$$ where a=0 then i get (1+0)^k right.
f'(x) using the power rule would say that d/dx[$$(1+x)^k$$] =$$k(1+x)^{k-1}$$ right. Nowf'(a) at a=o = k(1+0)^k-1 right?

so L(x)=f(a)+f'(a)(x-a)
$$\Rightarrow$$L(x)=$$(1+0)^k+k(1+0)^{k-1}(x-o)$$

I really don't see why this does not work. Belive it or not I have taken Calculus I only this past semester; so if it is do to some oversight please explain.

These problems I am doing are for "fun" and not for Homework. I just want to understand this better, so feel free to just give me the proper procedure and allow me to see where I am erring.

Casey

Oh; and as far as $$1^k$$ and $$1^{k-1}$$ equaling 1... I am just going on the assumption that 1 to the any power is 1....is this incorrect within the context of Calculus?

Last edited: Aug 6, 2007
13. Aug 6, 2007

I don't understand what it is that you don't understand? Please be more specific; I thought I had included all of my steps. If you could maybe point to one that does not make sense...

Edit:perhaps it is my laziness 1^k-1 was supposed to be $$1^{k-1}$$ I do not see how this would not =1. what could k be that does not cause that expression to equal 1?

Last edited: Aug 6, 2007
14. Aug 6, 2007

### nrqed

Yes, it's simply that 1 raised to any power is 1, as you pointed out. That's all there is to it, really!

15. Aug 6, 2007

This is why I am confused as to what the others are confused about?
VietDao29...what do you mean is 1^k=1 or is=2? That staement does not make sense to me Please elaborate.

16. Aug 6, 2007

### nrqed

It's just that you made a few mistakes in a few simple calculations and then in the notation which is what confused everybody. First there was some confusion about 1^k-1 instead of 1^(k-1). Then there was some confusion about x versus "a" in the expansion (it's misleading to write$k x (1+x)^{k-1}$ instead of $k x (1+a)^{k-1}$because the first expression does not distinguish the variable x and the point x=a around which the expansion is made) . Then VietDao29 got confused because you again wrote 1^k-1 =1 instead of writing 1^(k-1) = 1.

It's little details but in maths it's important to get the little details of notation right!

17. Aug 7, 2007

### Gib Z

In this case I agree with huyen_vyvy: Binomial expansion is the easiest way to go. We are asked for a linear approximation, so we take the linear terms and constants, and ignore the higher degrees.

18. Aug 8, 2007

### VietDao29

Yup, this is correct.

It always seems much better to write mathematical terms in LaTeX. :) Well, your notation in the post previous to this one is a little bit confusing. And I didn't read it carefully enough to grasp what you mean. Sorry. My bad.

Btw, remember to use parentheses in the future. It's very important. One missing parenthesis can cause a lot of confusion.