Heh. This actually had me stumped, to the extent where much of the following text was a post where I was actually asking for help (well, screaming for help). Fortunately for my pride, I figured out the answer before I hit "send": Consider f(x) = sqrt(1 + x^2). Now take its derivative. I get d/dx f(x) = (2 * x) / (2 * sqrt(1 + x^2)) = x / sqrt(1 + x^2). So, for x << 1, how does one approximate f(x)? (Hint: answers that do not use the first derivative of f(x) are perfectly acceptable, as long as you explain _why_ you can't use the first derivative of f(x)). -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
[OT] Calculus Problem
Started by ●August 19, 2013
Reply by ●August 19, 20132013-08-19
>Heh. This actually had me stumped, to the extent where much of the >following text was a post where I was actually asking for help (well, >screaming for help). Fortunately for my pride, I figured out the answer >before I hit "send": > >Consider f(x) = sqrt(1 + x^2). > >Now take its derivative. I get d/dx f(x) = (2 * x) / (2 * sqrt(1 + x^2))>= x / sqrt(1 + x^2). > >So, for x << 1, how does one approximate f(x)? > >(Hint: answers that do not use the first derivative of f(x) are perfectly>acceptable, as long as you explain _why_ you can't use the first >derivative of f(x)). > >-- >Tim Wescott >Control system and signal processing consulting >www.wescottdesign.com >f(x) approximately = 1 + x for x approximately = 0. Is that what you got? _____________________________ Posted through www.DSPRelated.com
Reply by ●August 19, 20132013-08-19
On 8/19/13 1:29 PM, Tim Wescott wrote:> Heh. This actually had me stumped, to the extent where much of the > following text was a post where I was actually asking for help (well, > screaming for help). Fortunately for my pride, I figured out the answer > before I hit "send": > > Consider f(x) = sqrt(1 + x^2). > > Now take its derivative. I get d/dx f(x) = (2 * x) / (2 * sqrt(1 + x^2)) > = x / sqrt(1 + x^2). > > So, for |x| << 1, how does one approximate f(x)? > > (Hint: answers that do not use the first derivative of f(x) are perfectly > acceptable, as long as you explain _why_ you can't use the first > derivative of f(x)).lessee, so d/dx f(x) = x / sqrt(1 + x^2). might it be that for |x| << 1, that d/dx f(x) is virtually 0? we know that for real x such that |x| << 1, that (1 + x)^p =approx 1 + p*x i would just use that fact. f(x) =approx 1 + 0.5*x^2 and for big |x|, you can say that f(x) =approx |x| . connecting 1 + 0.5*x^2 to |x| would be the next problem. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●August 19, 20132013-08-19
On Monday, August 19, 2013 4:29:23 PM UTC-4, Tim Wescott wrote:> Heh. This actually had me stumped, to the extent where much of the > > following text was a post where I was actually asking for help (well, > > screaming for help). Fortunately for my pride, I figured out the answer > > before I hit "send": > > > > Consider f(x) = sqrt(1 + x^2). > > > > Now take its derivative. I get d/dx f(x) = (2 * x) / (2 * sqrt(1 + x^2)) > > = x / sqrt(1 + x^2). > > > > So, for x << 1, how does one approximate f(x)? > > > > (Hint: answers that do not use the first derivative of f(x) are perfectly > > acceptable, as long as you explain _why_ you can't use the first > > derivative of f(x)). > > > > -- > > Tim Wescott > > Control system and signal processing consulting > > www.wescottdesign.comTim, Yes you may do this without using calculus! Just use Newton's generalized binomial theorem. (x+y)^r =(x^r)(y^0)+r*(x^r-1)(y^1)+(r)(r-1)(x^r-2)(y^2)/2! + ... Let y=1; r=0.5; x=x^2 (1+x^2)^0.5 = 1+(1/2)x^2 -(1/8)(x^4)+(1/16)(x^6) ... So for very small x, the answer is essentially 1. Use as many terms as necessary. Clay
Reply by ●August 19, 20132013-08-19
Tim Wescott <tim@seemywebsite.please> wrote:> Heh. This actually had me stumped, to the extent where much of the > following text was a post where I was actually asking for help (well, > screaming for help). Fortunately for my pride, I figured out the answer > before I hit "send":> Consider f(x) = sqrt(1 + x^2).> Now take its derivative. I get d/dx f(x) = (2 * x) / (2 * sqrt(1 + x^2)) > = x / sqrt(1 + x^2).> So, for x << 1, how does one approximate f(x)?> (Hint: answers that do not use the first derivative of f(x) are perfectly > acceptable, as long as you explain _why_ you can't use the first > derivative of f(x)).Well, the easy answer is binomial expansion of (1+x^2)^(1/2), and so 1+x^2/2, If you do Taylor series around 0, it should give the same result. As you note, the first derivative is zero at x==0, so the first order approximation is f(x)=1. For the second order, you need the second derivative. -- glen
Reply by ●August 19, 20132013-08-19
>>Heh. This actually had me stumped, to the extent where much of the >>following text was a post where I was actually asking for help (well, >>screaming for help). Fortunately for my pride, I figured out the answer>>before I hit "send": >> >>Consider f(x) = sqrt(1 + x^2). >> >>Now take its derivative. I get d/dx f(x) = (2 * x) / (2 * sqrt(1 +x^2))> >>= x / sqrt(1 + x^2). >> >>So, for x << 1, how does one approximate f(x)? >> >>(Hint: answers that do not use the first derivative of f(x) areperfectly> >>acceptable, as long as you explain _why_ you can't use the first >>derivative of f(x)). >> >>-- >>Tim Wescott >>Control system and signal processing consulting >>www.wescottdesign.com >> > >f(x) approximately = 1 + x > >for x approximately = 0. > >Is that what you got? > >_____________________________ >Posted through www.DSPRelated.com >Whoops, goofed it. Bleh. _____________________________ Posted through www.DSPRelated.com
Reply by ●August 20, 20132013-08-20
On Tuesday, August 20, 2013 8:29:23 AM UTC+12, Tim Wescott wrote:> Heh. This actually had me stumped, to the extent where much of the > > following text was a post where I was actually asking for help (well, > > screaming for help). Fortunately for my pride, I figured out the answer > > before I hit "send": > > > > Consider f(x) = sqrt(1 + x^2). > > > > Now take its derivative. I get d/dx f(x) = (2 * x) / (2 * sqrt(1 + x^2)) > > = x / sqrt(1 + x^2). > > > > So, for x << 1, how does one approximate f(x)? > > > > (Hint: answers that do not use the first derivative of f(x) are perfectly > > acceptable, as long as you explain _why_ you can't use the first > > derivative of f(x)). > > > > -- > > Tim Wescott > > Control system and signal processing consulting > > www.wescottdesign.comTaylor series
Reply by ●August 20, 20132013-08-20
On Tuesday, August 20, 2013 12:42:50 AM UTC-4, gyans...@gmail.com wrote:> On Tuesday, August 20, 2013 8:29:23 AM UTC+12, Tim Wescott wrote: > > > Heh. This actually had me stumped, to the extent where much of the > > > > > > following text was a post where I was actually asking for help (well, > > > > > > screaming for help). Fortunately for my pride, I figured out the answer > > > > > > before I hit "send": > > > > > > > > > > > > Consider f(x) = sqrt(1 + x^2). > > > > > > > > > > > > Now take its derivative. I get d/dx f(x) = (2 * x) / (2 * sqrt(1 + x^2)) > > > > > > = x / sqrt(1 + x^2). > > > > > > > > > > > > So, for x << 1, how does one approximate f(x)? > > > > > > > > > > > > (Hint: answers that do not use the first derivative of f(x) are perfectly >> Taylor seriesYup - the first term in the Taylor series (expanding about the point x=0) is equal to 0 i.e. f'(x)|x=0 = 0. This makes sense because you are essentially looking at terms involving x^2 in the function f(x)=sqrt(1+x^2). f''(x) has 2 terms but evaluating them at x=0 gives a coefficient of 1. The resulting Taylor series is f(x) =1 +0.5 *x^2 + ... Cheers, Dave
Reply by ●August 20, 20132013-08-20
Dave <dspguy2@netscape.net> wrote: (snip)>> > Consider f(x) = sqrt(1 + x^2).(snip)>> Taylor series> Yup - the first term in the Taylor series (expanding about the > point x=0) is equal to 0 i.e. f'(x)|x=0 = 0. > This makes sense because you are essentially looking at terms > involving x^2 in the function f(x)=sqrt(1+x^2).Or, use g(y)=sqrt(1+y) and y=x^2. g(y) expands to first order as 1+y/2, and substitute x^2 for y. -- glen
Reply by ●August 20, 20132013-08-20
>On Monday, August 19, 2013 4:29:23 PM UTC-4, Tim Wescott wrote: >> Heh. This actually had me stumped, to the extent where much of the >> >> following text was a post where I was actually asking for help (well, >> >> screaming for help). Fortunately for my pride, I figured out the answer>> >> before I hit "send": >> >> >> >> Consider f(x) = sqrt(1 + x^2). >> >> >> >> Now take its derivative. I get d/dx f(x) = (2 * x) / (2 * sqrt(1 +x^2))>> >> = x / sqrt(1 + x^2). >> >> >> >> So, for x << 1, how does one approximate f(x)? >> >> >> >> (Hint: answers that do not use the first derivative of f(x) areperfectly>> >> acceptable, as long as you explain _why_ you can't use the first >> >> derivative of f(x)). >> >> >> >> -- >> >> Tim Wescott >> >> Control system and signal processing consulting >> >> www.wescottdesign.com > > >Tim, > >Yes you may do this without using calculus! Just use Newton's generalizedbinomial theorem.> >(x+y)^r =(x^r)(y^0)+r*(x^r-1)(y^1)+(r)(r-1)(x^r-2)(y^2)/2! + ... > >Let y=1; r=0.5; x=x^2 > >(1+x^2)^0.5 = 1+(1/2)x^2 -(1/8)(x^4)+(1/16)(x^6) ... > >So for very small x, the answer is essentially 1. Use as many terms asnecessary.> >Clay >Do you even need to do that much work. If x << 1, squaring a number less then one, makes an even smaller number, thus 1+x^2 will approach 1 as x approaches 0. sqrt(1) is 1. _____________________________ Posted through www.DSPRelated.com
