>>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 theanswer> >>> >>> 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 >> >> >>Tim, >> >>Yes you may do this without using calculus! Just use Newton'sgeneralized>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 >> > >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 >A 'zeroth' order approximation, as it were. _____________________________ Posted through www.DSPRelated.com
[OT] Calculus Problem
Started by ●August 19, 2013
Reply by ●August 20, 20132013-08-20
Reply by ●August 20, 20132013-08-20
On Tuesday, August 20, 2013 11:35:40 AM UTC-4, jacobfenton wrote:> >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.com > > > > > > > > >Tim, > > > > > >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 > > > > > > > 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.comI showed that much work to illustrate a technique I think many haven't seen. Glen knows of it, but then he and I both have physics backgrounds and the binomial expansion for nonintegral powers is a main stay of that discipline. It is easy to see f(0)=1 by simple substitution. And of course a McLauren series (Taylor's where it is expanded about 0) yields the same result. Clay
Reply by ●August 20, 20132013-08-20
>On Tuesday, August 20, 2013 11:35:40 AM UTC-4, jacobfenton wrote: >> >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 theanswer>> >> >> >> >> >> >> >> 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 >> >> > >> >> > >> >> >Tim, >> >> > >> >> >Yes you may do this without using calculus! Just use Newton'sgeneralized>> >> 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 >> >> > >> >> >> >> Do you even need to do that much work. If x << 1, squaring a numberless>> >> 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 > >I showed that much work to illustrate a technique I think many haven'tseen. Glen knows of it, but then he and I both have physics backgrounds and the binomial expansion for nonintegral powers is a main stay of that discipline.> >It is easy to see f(0)=1 by simple substitution. And of course a McLaurenseries (Taylor's where it is expanded about 0) yields the same result.> >Clay > > > >I wasn't knocking your work, just stating it could be done by inspection, and now I will read about binomial expansion, as your right, I didn't know about that. _____________________________ Posted through www.DSPRelated.com
Reply by ●August 20, 20132013-08-20
On Tuesday, August 20, 2013 2:51:55 PM UTC-4, jacobfenton wrote:> >On Tuesday, August 20, 2013 11:35:40 AM UTC-4, jacobfenton wrote: > > >> >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.com > > >> > > >> > > > >> > > >> > > > >> > > >> >Tim, > > >> > > >> > > > >> > > >> >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 > > >> > > >> > > > >> > > >> > > >> > > >> 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 > > > > > >I showed that much work to illustrate a technique I think many haven't > > seen. Glen knows of it, but then he and I both have physics backgrounds and > > the binomial expansion for nonintegral powers is a main stay of that > > discipline. > > > > > >It is easy to see f(0)=1 by simple substitution. And of course a McLauren > > series (Taylor's where it is expanded about 0) yields the same result. > > > > > >Clay > > > > > > > > > > > > > > I wasn't knocking your work, just stating it could be done by inspection, > > and now I will read about binomial expansion, as your right, I didn't know > > about that. > > > > _____________________________ > > Posted through www.DSPRelated.comNo worries! Clay
Reply by ●August 21, 20132013-08-21
On Tuesday, August 20, 2013 10:17:26 AM UTC-4, glen herrmannsfeldt wrote:> > 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. > > > > -- glenHi Glen, That works in this case. In the case where y(x) is more complicated - it may introduce terms into the Taylor series coefficient. You'll need to evaluate the Taylor series coefficients of the form: dg(y)/dx = dg(y)/dy x dy/dx. Cheers, Dave
Reply by ●August 21, 20132013-08-21
Dave <dspguy2@netscape.net> wrote: (snip, I wrote)>> 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.(snip)> That works in this case. In the case where y(x) is more > complicated - it may introduce terms into the Taylor > series coefficient. You'll need to evaluate the Taylor > series coefficients of the form:> dg(y)/dx = dg(y)/dy x dy/dx.Yes. But for integer powers of x, you can easily rewrite a known expansion. For example, if you know the Taylor series for exp(x), and you want exp(x^2) or exp(x^3) you can easily write down the series expansion with little work. Or, even the series for cos(sqrt(x)), which would otherwise be a lot of work. -- glen
