Understanding Small Changes in Calculus

[roger]

Hi

I have a basic question on calculus ...please can someone explain to me :


Let f(x) = x^2

y + delta y = f ( x + delta x )

The book says that a small increase in x will cause a small increase in y.

But if I put let's say 3 and 2 for x and dx y is 5^2 which is 25.

But this isn't the same as f(3) + f(2) which equals 13 ?

What have I done wrong ?




Also, can f ( a+b+c ) be treated as f(a) + f(b) + f(c) ?


Thanks a lot for any help


Roger

 

[Nylex]

As I've always been told, dx is an infinitesimal change in x, so really really really small. Also, f(a + b + c) can't be treated as f(a) + f(b) + f(c), I don't think. f(a + b + c) can't be treated like that in your example anyway:

f(x) = x^2
f(a + b + c) = (a + b + c)^2
f(a) + f(b) + f(c) = a^2 + b^2 + c^2

 

[Mellow^Guy]

f(a+b+c) doesn't wqual to f(a)+f(b)+f(c)

now suppose that f(x)=X+1

and let a=1 & b=2 & c=3

so a+b+c=6

f(6)=6+1=7

but

f(1)=1+1=2
f(2)=2+1=3
f(3)=3+1=4

f(1)+f(2)+f(3)=2+3+4=9

that is f(a+b+c) does not equal to f(a)+f(b)+f(c)

 

[HallsofIvy]

In general f(x+ y) is NOT the same as f(x)+ f(y).

Except for some extremely weird function, the only functions for which
f(x+ y)+ f(x)+ f(y) are f(x)= Cx- that is, a constant time x.

 

[roger]

In general f(x+ y) is NOT the same as f(x)+ f(y).

Except for some extremely weird function, the only functions for which
f(x+ y)+ f(x)+ f(y) are f(x)= Cx- that is, a constant time x.

Sorry I don't quite understand the last bit ?
please can you explain further ?




But my main question is the first section I originally wrote :

Let f(x) = x^2

y + delta y = f ( x + delta x )

The book says that a small increase in x will cause a small increase in y.

But if I put let's say 3 and 2 for x and dx y is 5^2 which is 25.

But this isn't the same as f(3) + f(2) which equals 13 ?



I understand dx is infinitesimall but I just plugged in a larger value for ease of calculation.

In the example above, why doesn't change in x give a change in y ?

The book gave the statement I outlined in red above...

But isn't it strictly speaking inaccurate because a small change in x gives a small change in y but the value of that small change in y is still bigger than x
because the function is applied to x ( in this case f(x)=x^2 ) ?


please can someone explain ?
thanks

roger

 

[matt grime]

Do the math, as they say:

(x+dx)^2 = x^2+2xdx+(dx)^2

where on Earth do you even get that f(2)+f(3) should be 25? What are you doing there?

Are you saying you think dy = f(dx)? Cos that's how it appears.

dy = f'(x)dx +o((dx)^2) is what's going on really,.

 

[roger]

Do the math, as they say:

(x+dx)^2 = x^2+2xdx+(dx)^2

where on Earth do you even get that f(2)+f(3) should be 25? What are you doing there?

My function is x^2 Therefore f(2)+f(3)=13

So I let x be 2 and dx be 3
but the answer for f(x+dx) = 25

So I've got 2 different answers and the thing I don't quite understand is the book states a change in x gives a change in y but in the above example, I put 3 to be dx so I added the function applied to 3 to the function applied to x and obviously, the two results are different ?

This is what I am trying to find out.

Are you saying you think dy = f(dx)? Cos that's how it appears.

Why is that wrong ?

dy = f'(x)dx +o((dx)^2) is what's going on really,.

Thanks

Roger

 

[matt grime]

Obviously f(x+dx) doesn't equal f(x)+f(dx), and no one says it does.

A small change of dx changes y by approximately dx.f'(x) at x. that is what the derivative does.

 

[roger]

Obviously f(x+dx) doesn't equal f(x)+f(dx), and no one says it does.

A small change of dx changes y by approximately dx.f'(x) at x. that is what the derivative does.

Is it a small change in x or dx ?

Why is it only approximate ?

 

[arildno]

roger:
You should ALWAYS think of \bigtriangleup{y} as (when y=f(x)):
\bigtriangleup{y}=f(x+\bigtriangleup{x})-f(x)
Hence, in your example, we have:
\bigtriangleup{y}=2x\bigtriangleup{x}+(\bigtriangleup{x})^{2}
When \bigtriangleup{x} is TINY, the second term on the right-hand side is much less in magnitude than the first term (provided x is non-zero).
Hence, we may then write:
\bigtriangleup{y}\approx{2x}\bigtriangleup{x}
which you should recognize as:
\bigtriangleup{y}\approx{f}'(x)\bigtriangleup{x}

 

[matt grime]

Is it a small change in x or dx ?

Why is it only approximate ?

To repeat myself:

(x+dx)^2 = ...

 

[roger]

roger:
You should ALWAYS think of \bigtriangleup{y} as (when y=f(x)):
\bigtriangleup{y}=f(x+\bigtriangleup{x})-f(x)
Hence, in your example, we have:
\bigtriangleup{y}=2x\bigtriangleup{x}+(\bigtriangleup{x})^{2}
When \bigtriangleup{x} is TINY, the second term on the right-hand side is much less in magnitude than the first term (provided x is non-zero).is this because the last term dx is squared ?
Hence, we may then write:
\bigtriangleup{y}\approx{2x}\bigtriangleup{x}
which you should recognize as:
\bigtriangleup{y}\approx{f}'(x)\bigtriangleup{x}

thanks.

roger.