Why does it work?Tangent line to polynomial curve without calculus

BigBeachBanana

BigBeachBanana

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I've recently stumbled on an interesting method of finding the tangent line to a polynomial through long division. If we want to find the tangent line at [imath]x=a[/imath] for some polynomial function [imath]f(x)[/imath], then divide [imath]f(x)[/imath] by [imath](x-a)^2[/imath] using long division. The remainder of the division is the tangent line at [imath]x=a[/imath] and has the form [imath]\alpha x+\beta[/imath]. I have verified this technique with different polynomials, and it all seemed to work. I'm looking for some theory behind this method (no luck on the internet).

Example: Find the tangent line at [imath]x=2[/imath] of [imath]f(x)=5x^3-4x+7[/imath].
Divide [imath]f(x)=5x^3-4x+7[/imath] by [imath](x-2)^2=x^2-4x+4[/imath]. We get a remainder of [imath]56x+73[/imath], which is, in fact, the tangent line.

Why does it work? Does it work for all polynomials?
 
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BigBeachBanana said:
If we want to find the tangent line at [imath]x=a[/imath] for some polynomial function [imath]f(x)[/imath], then divide [imath]f(x)[/imath] by [imath](x-a)^2[/imath] using long division. The remainder of the division is the tangent line at [imath]x=a[/imath] and has the form [imath]\alpha x+\beta[/imath]. I have verified this technique with different polynomials, and it all seemed to work. I'm looking for some theory behind this method (no luck on the internet).

Why does it work? Does it work for all polynomials?
Click to expand...

The division with remainder tells us that [math]f(x) = (x-a)^2+bx+c[/math] where I'm calling the remainder [imath]g(x) = bx+c[/imath].

Let [imath]x=a[/imath], and we find that [math]f(a) = (a-a)^2+ba+c=ba+c = g(a),[/math]
so the line [imath]g(x)[/imath] passes through the point [imath](a, f(a)[/imath].

Now take the derivative: [math]f'(x) = 2(x-a)+b[/math]
Let [imath]x=a[/imath], and we find that [math]f'(a) = 2(a-a)+b = b,[/math]
so the line [imath]g(x)[/imath] has the same slope at the point [imath](a, f(a)[/imath] as [imath]f(x)[/imath].

So [imath]g(x)[/imath] is the tangent line to [imath]f(x)[/imath] at [imath]x=a[/imath].

Interesting! I don't think I ever heard of this, but it's clearly true.

There are probably other ways to see what is happening, perhaps something with the graph.
 
BigBeachBanana said:
I've recently stumbled on an interesting method of finding the tangent line to a polynomial through long division. If we want to find the tangent line at [imath]x=a[/imath] for some polynomial function [imath]f(x)[/imath], then divide [imath]f(x)[/imath] by [imath](x-a)^2[/imath] using long division. The remainder of the division is the tangent line at [imath]x=a[/imath] and has the form [imath]\alpha x+\beta[/imath]. I have verified this technique with different polynomials, and it all seemed to work. I'm looking for some theory behind this method (no luck on the internet).

Example: Find the tangent line at [imath]x=2[/imath] of [imath]f(x)=5x^3-4x+7[/imath].
Divide [imath]f(x)=5x^3-4x+7[/imath] by [imath](x-2)^2=x^2-4x+4[/imath]. We get a remainder of [imath]56x+73[/imath], which is, in fact, the tangent line.

Why does it work? Does it work for all polynomials?
Click to expand...
Never heard of this before, but this is very neat!
The remainder represents the linear approximation at x=a, and the rest is the second order error. Real cool!
 
Dr.Peterson said:
The division with remainder tells us that [math]f(x) = (x-a)^2+bx+c[/math] where I'm calling the remainder [imath]g(x) = bx+c[/imath].

Let [imath]x=a[/imath], and we find that [math]f(a) = (a-a)^2+ba+c=ba+c = g(a),[/math]
so the line [imath]g(x)[/imath] passes through the point [imath](a, f(a)[/imath].

Now take the derivative: [math]f'(x) = 2(x-a)+b[/math]
Let [imath]x=a[/imath], and we find that [math]f'(a) = 2(a-a)+b = b,[/math]
so the line [imath]g(x)[/imath] has the same slope at the point [imath](a, f(a)[/imath] as [imath]f(x)[/imath].

So [imath]g(x)[/imath] is the tangent line to [imath]f(x)[/imath] at [imath]x=a[/imath].

Interesting! I don't think I ever heard of this, but it's clearly true.

There are probably other ways to see what is happening, perhaps something with the graph.
Click to expand...
There's a major typo here, because I saw this just before getting off for the night, and was rushing. Where I wrote
[math]f(x) = (x-a)^2+bx+c,[/math]I clearly meant
[math]f(x) = (x-a)^2q(x)+bx+c,[/math]and so on.

So here's a redo:

The division with remainder tells us that [math]f(x) = (x-a)^2q(x)+bx+c[/math] where I'm calling the remainder [imath]g(x) = bx+c[/imath].

Let [imath]x=a[/imath], and we find that [math]f(a) = (a-a)^2q(a)+ba+c=ba+c = g(a),[/math]
so the line [imath]g(x)[/imath] passes through the point [imath](a, f(a)[/imath].

Now take the derivative: [math]f'(x) = 2(x-a)q(x)+(x-a)^2q'(x)+b[/math]
Let [imath]x=a[/imath], and we find that [math]f'(a) = 2(a-a)q(a)+(a-a)^2q'(a)+b = b,[/math]
so the line [imath]g(x)[/imath] has the same slope at the point [imath](a, f(a)[/imath] as [imath]f(x)[/imath].

So [imath]g(x)[/imath] is the tangent line to [imath]f(x)[/imath] at [imath]x=a[/imath].

Interesting! I don't think I ever heard of this, but it's clearly true.
 
This reminds me of something and i think it is related...

Suppose you want the equation of the tangent at the point (2,6) to the circle
[math]x^2+y^2+4x-6y=12[/math]
We can do this:

[math](x-2)^2+(y-6)^2+8x+6y-52=0[/math]
The linear term

[math]8x+6y-52=0[/math]
is the equation of the tangent!

Is this the same sort of thing?
 
And something else, suppose you want the equation of the tangent to [math]y=x^2[/math] at the point x=3
You can rewrite [math]x^2= (x-3)^2+ 6x-9[/math]
So [math]y=6x-9[/math] is the equation of the tangent

This clearly is the same as above but i think there is something going on graphically which we might be able to see better...

Havent thought this through but there might be something in this?

1647278439349.png
 
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Let's see if this works.
Suppose f(x) = x^2 and you want the equation of the tangent line at (a,f(a))
The slope of the tangent line is m = 2a
So the equation of the line is y = 2a(x-a) + f(a)

Now x^2 = (x-a)^2 +2ax - a^2.
Now 2ax-a^2 = 2ax - 2a^2 + a^2 = 2a(x-a) + f(a).

This is amazing!
 
Does it work for y=x^3.
The equation of the tangent line at (a,f(a)) if y= 3a^2(x-a) + f(a)

y = (x-a)^3 + 3ax^2 - 3a^2x + a^3
Now 3ax^2 - 3a^2x + a^3 = x(3ax - 3a^2) + a^3 = x[3a(x-a)] +f(a)

Almost, but the equation of the tangent line is easily obtained from this method.

You can even do this if f(x) = (x-b)^3
 
apple2357 said:
This reminds me of something and i think it is related...

Suppose you want the equation of the tangent at the point (2,6) to the circle
[math]x^2+y^2+4x-6y=12[/math]
We can do this:

[math](x-2)^2+(y-6)^2+8x+6y-52=0[/math]
The linear term

[math]8x+6y-52=0[/math]
is the equation of the tangent!

Is this the same sort of thing?
Click to expand...
I need to see this for myself.
Suppose x^2 + y^2 + rx+sy = c>0.
Find the equation of the tangent line at (a, b).
2x+2yy' + r + sy' = 0
y' = -(2x+r)/(2y+s)
y'(a,b) = -(2a+r)/(2b+s)
The equation of the tangent line at (a,b) is y-b = -(2a+r)/(2b+s)(x-a)
or (y-b)(2b+s) + (2a+r)(x-a)=0
2by-2b^2+sy-bs + 2ax -2a^2+rx-ar=0
(r+2a)x +(s+2b)y -(a^2+b^2) - (a^2 + b^2 + bs+ar)=0
(r+2a)x +(s+2b)y -(a^2+b^2 + c) =0


(x-a)^2 + (y-b)^2 + rx+sy = c -2ax + a^2 -2by+b^2
(x-a)^2 + (y-b)^2 + rx+sy - c +2ax -a^2 +2by -b^2=0
(x-a)^2 + (y-b)^2 + (r+2a)x + (s+2b)y - (a^2+b^2+c)=0

So this is really true! Wow!
 
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apple2357 said:
And something else, suppose you want the equation of the tangent to [math]y=x^2[/math] at the point x=3
You can rewrite [math]x^2= (x-3)^2+ 6x-9[/math]
So [math]y=6x-9[/math] is the equation of the tangent

This clearly is the same as above but i think there is something going on graphically which we might be able to see better...

Havent thought this through but there might be something in this?

View attachment 31671
Click to expand...
For the record, you graphed y=(x+3)^2, not y=(x-3)^2
 
Steven G said:
For the record, you graphed y=(x+3)^2, not y=(x-3)^2
Click to expand...

I think i meant (x+3)^2

The equation of the tangent of y=(x+3)^2 at x=0 is 6x+9
because (x+3)^2 = x^2 +6x+9 and the 6x+9 is the linear approx to the curve at x=0

we can then replace x by x-3

(x-3+3)^2 =(x-3)^2+6(x-3)+9

x^2= (x-3)^2+6x-18+9
= (x-3)^2+6x-9

Does this make sense?
 
Dr.Peterson said:
There's a major typo here, because I saw this just before getting off for the night, and was rushing. Where I wrote
[math]f(x) = (x-a)^2+bx+c,[/math]I clearly meant
[math]f(x) = (x-a)^2q(x)+bx+c,[/math]and so on.

So here's a redo:

The division with remainder tells us that [math]f(x) = (x-a)^2q(x)+bx+c[/math] where I'm calling the remainder [imath]g(x) = bx+c[/imath].

Let [imath]x=a[/imath], and we find that [math]f(a) = (a-a)^2q(a)+ba+c=ba+c = g(a),[/math]
so the line [imath]g(x)[/imath] passes through the point [imath](a, f(a)[/imath].

Now take the derivative: [math]f'(x) = 2(x-a)q(x)+(x-a)^2q'(x)+b[/math]
Let [imath]x=a[/imath], and we find that [math]f'(a) = 2(a-a)q(a)+(a-a)^2q'(a)+b = b,[/math]
so the line [imath]g(x)[/imath] has the same slope at the point [imath](a, f(a)[/imath] as [imath]f(x)[/imath].

So [imath]g(x)[/imath] is the tangent line to [imath]f(x)[/imath] at [imath]x=a[/imath].

Interesting! I don't think I ever heard of this, but it's clearly true.
Click to expand...
Dr P,
There is a minor flaw in your proof. You failed to say that deg(f(x))> 2.
Maybe you are thinking that q(x) can equal 0 and your proof is valid?
Personally, I would have broken this up into three cases (the 2nd and 3rd case would be when the degree of p(x) is 0 and 1).
 
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Steven G said:
Dr P,
There is a minor flaw in your proof. You failed to say that deg(f(x))> 2.
Maybe you are thinking that q(x) can equal 0 and your proof is valid?
Personally, I would have broken this up into three cases (the 2nd and 3rd case would be when the degree of p(x) is 0 and 1).
Click to expand...
I neither called it a proof nor stated exactly what was being proved. You may write it up formally if you wish; I was just interested in the question asked: Why would it work?
 
I'm a bit late to this party, but here's my take on it.
If [imath]p(x)[/imath] is a polynomial of degree [imath]\ge 2[/imath], then its Taylor series about [imath]x=a[/imath] is[math]p(x) = p(a) + p'(a)(x-a) + O(x-a)^2[/math]The division algorithm gives [imath]p(x) = (x-a)^2Q(x) +\alpha x + \beta[/imath]
Comparing these two make it clear the remainder is the same as the tangent line.
 
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