Taylor expansions confusion f(x+a) and f(x) at a... Watch

xfootiecrazeesarax
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Okay so my understanding of

f(x)≈ f(a) + f'(a)(x-a) + f''(a)(x-a)^{2}/2+ ... [1]

at a point a, is that we are approximating the function at this point so it can be a good approximation for points close to a, and so the point is we know f(x) at other regions in the domain but not around a?

Now my main question is, taylor series is also given as:

f(x+Δx) ≈ f(x) + f'(x)Δx+ f''(x) (Δx)^{2}/2 +...

- how is this derived from [1]?
- what is the idea behind this expansion, like for above I have a reason for it
- does Δx need to be small compared to x to do this? and why? how does this tie with the reason of the expansion and deriving it from [1], if so? ( Δx here is a bad choice for this question since the symbol generally denotes this, but in the general case being small is required?)

Many thanks in advance
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DylanJ42
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(Original post by xfootiecrazeesarax)
Okay so my understanding of

f(x)≈ f(a) + f'(a)(x-a) + f''(a)(x-a)^{2}/2+ ... [1]

at a point a, is that we are approximating the function at this point so it can be a good approximation for points close to a, and so the point is we know f(x) at other regions in the domain but not around a?

Now my main question is, taylor series is also given as:

f(x+Δx) ≈ f(x) + f'(x)Δx+ f''(x) (Δx)^{2}/2 +...

- how is this derived from [1]?
- what is the idea behind this expansion, like for above I have a reason for it
- does Δx need to be small compared to x to do this? and why? how does this tie with the reason of the expansion and deriving it from [1], if so? ( Δx here is a bad choice for this question since the symbol generally denotes this, but in the general case being small is required?)

Many thanks in advance
it looks like they are assuming x is a value close to a and letting  \displaystyle \delta x = x - a to get

 \displaystyle f(x) = f(x- \delta x) + f'(x - \delta x) \delta x + f''(x - \delta x) \frac{(\delta x)^2}{2} + ...

 \therefore \displaystyle f(x + \delta x) = f(x) + f'(x) \delta x + f''(x) \frac{(\delta x)^2}{2} + ...

1. I think its derived using the above method
2. The idea behind it, im not sure
3. pretty sure x and a must be close in value, and deltax must be small yea
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xfootiecrazeesarax
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(Original post by DylanJ42)
it looks like they are assuming x is a value close to a and letting  \displaystyle \delta x = x - a to get

 \displaystyle f(x) = f(x- \delta x) + f'(x - \delta x) \delta x + f''(x - \delta x) \frac{(\delta x)^2}{2} + ...

 \therefore \displaystyle f(x + \delta x) = f(x) + f'(x) \delta x + f''(x) \frac{(\delta x)^2}{2} + ...

1. I think its derived using the above method
2. The idea behind it, im not sure
3. pretty sure x and a must be close in value, and deltax must be small yea
ahh, many thanks,

delta x small < => x and a are close, so they're equivalent criteria,
in the initial expansion i gave as [1] it's valid, as in a good approx, only if x is close to a, so i guess that explains why
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B_9710
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(Original post by xfootiecrazeesarax)
Okay so my understanding of

f(x)≈ f(a) + f'(a)(x-a) + f''(a)(x-a)^{2}/2+ ... [1]

at a point a, is that we are approximating the function at this point so it can be a good approximation for points close to a, and so the point is we know f(x) at other regions in the domain but not around a?

Now my main question is, taylor series is also given as:

f(x+Δx) ≈ f(x) + f'(x)Δx+ f''(x) (Δx)^{2}/2 +...

- how is this derived from [1]?
- what is the idea behind this expansion, like for above I have a reason for it
- does Δx need to be small compared to x to do this? and why? how does this tie with the reason of the expansion and deriving it from [1], if so? ( Δx here is a bad choice for this question since the symbol generally denotes this, but in the general case being small is required?)

Many thanks in advance
From what you've put it seems that we are assuming that  \Delta x is small and that  x is close to a.
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