what's the reason for getting an output at the point itself though it is not in range

Question

dareen 2024-7-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2133626-what-s-the-reason-for-getting-an-output-at-the-point-itself-though-it-is-not-in-range

编辑： dareen 2024-7-4

i have used taylor series to find the first derivative of the following function cos(x) and got 2 diffrent approximations ,

i wanted to check which one has more precision,

i started by understanding how to graph the function and its derviative wrote the 2 approximations for it, then tried to graph each graphs error , i have not yet added axis names and so on , but this graph

plot(x+h,approx_1(f,x,h),'*',x-h,approx_1(f,x,-h),'*') is for the first approximation of the value of the point

and the second plot(x+h,approx_2(f,x,h),'*',x-h,approx_2(f,x,-h),'*')

yet i get a number at the point it self though the range of h does not include 0 what may be the cause

,thanks for the guidness in advance

%testing a bit too much

f = @cos;

x = 1;

h=0:0.1:8*pi;

plot(x+h,f(x+h),'*',x-h,f(x-h),'*')

d_f =@(x) -sin(x);

exact = d_f(x);

h=0:0.1:8*pi;

hold on

plot(x+h,d_f(x+h),'*',x-h,d_f(x-h),'*')

exact = d_f(x);

hold off

d_f(1)

ans = -0.8415

f(1)

ans = 0.5403

%testing near the point of interest but closer

h2=0:0.001:10^-2

h2 = 1x11

0 0.0010 0.0020 0.0030 0.0040 0.0050 0.0060 0.0070 0.0080 0.0090 0.0100

plot(x+h2,f(x+h2),'*',x-h2,f(x-h2),'*')

plot(x+h2,d_f(x+h2),'*',x-h2,d_f(x-h2),'*')

%identifiy

approx_1=@(f, x, h) (f(x + h) - f(x))./ h;

approx_2=@(f, x, h) (f(x + h) - f(x-h)) ./ (2*h);

d_f(1)

ans = -0.8415

h = linspace(10^-15,10^-1,20)

h = 1x20

0.0000 0.0053 0.0105 0.0158 0.0211 0.0263 0.0316 0.0368 0.0421 0.0474 0.0526 0.0579 0.0632 0.0684 0.0737 0.0789 0.0842 0.0895 0.0947 0.1000

<mw-icon class=""></mw-icon>

plot(x+h,approx_1(f,x,h),'*',x-h,approx_1(f,x,-h),'*')

plot(x+h,approx_2(f,x,h),'*',x-h,approx_2(f,x,-h),'*')

v_1_up=approx_1(f,x,h)-d_f(1)

v_1_up = 1x20

-0.1577 -0.0014 -0.0028 -0.0042 -0.0056 -0.0070 -0.0084 -0.0098 -0.0111 -0.0125 -0.0138 -0.0152 -0.0165 -0.0178 -0.0191 -0.0204 -0.0217 -0.0230 -0.0243 -0.0256

<mw-icon class=""></mw-icon>

v_2_up=approx_2(f,x,h)-d_f(1)

v_2_up = 1x20

-0.0467 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0010 0.0011 0.0013 0.0014

<mw-icon class=""></mw-icon>

v_1_down=approx_1(f,x,-h)-d_f(1)

v_1_down = 1x20

0.0643 0.0014 0.0029 0.0043 0.0057 0.0072 0.0087 0.0101 0.0116 0.0131 0.0146 0.0161 0.0176 0.0191 0.0207 0.0222 0.0237 0.0253 0.0268 0.0284

<mw-icon class=""></mw-icon>

v_2_down=approx_2(f,x,-h)-d_f(1)

v_2_down = 1x20

-0.0467 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0010 0.0011 0.0013 0.0014

<mw-icon class=""></mw-icon>

%plot(h,v_1_up,'*',h,v_1_down,'*')

%plot(h,v_2_up,'*',h,v_2_down,'*')

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Answer 1

Walter Roberson 2024-7-1

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2133626-what-s-the-reason-for-getting-an-output-at-the-point-itself-though-it-is-not-in-range#answer_1479476

在 MATLAB Online 中打开

h=0:0.1:8*pi;

approx_1=@(f, x, h) (f(x + h) - f(x)) / h;

Your h is a vector. You need to use ./ instead of /

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dareen 2024-7-2

it really slipped over my mind thank you

dareen 2024-7-2

编辑：dareen 2024-7-4

there are some weird stuff at the point itself in the approximaiton i should not be getting a value there as for the rest i have written a code that does whats asked from me log log ploting where the x axis is h i used x+h before so i got some really weird results

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