How To solve this equation?

[Matt J]

I see no equal sign, nor any unknowns.

[Alif]

idk my professor just gave me this equation and I don't know how to solve it.

[Torsten]

Do you see a " = " sign somewhere ? I don't . So it's not an equation. And if it's not an equation, there is nothing to solve for.

[Sam Chak]

@Alif, I think your professor wants you to sum up the expression from to . It should be relatively easy to calculate that in Excel if your professor didn't specifically instruct you to use MATLAB.

[Alif]

there instruct to use a mathlab @Sam Chak

this the instruction "Calculate the result of the following equation using MATLAB"

Answer 1

[Sam Chak]

Hi @Alif

Here is the numerical approach.

format long g
x = 1:1000;
y = (nthroot(5*x + 2, 5))./x;
S = sum(y)

S = 
          21.5330354759271

[Sam Chak]

@Alif, This is the for-loop method, but I'm unsure why the last digit (2) differs from the last digit (1) computed using the sum() command.

format long g
y = 0;       % Initialize y to store the sum later
for x = 1:1000
    y = y + (nthroot(5*x + 2, 5))/x;
end
disp(y)

          21.5330354759272

[Bruno Luong]

Summing in reverse seems to be more accurate as it is sum the smallest elemenst first

format long g
y = 0;       % Initialize y to store the sum later
for x = 1000:-1:1
    y = y + (nthroot(5*x + 2, 5))/x;
end
disp(y)

          21.5330354759271

[Sam Chak]

Thank you, @Bruno Luong. I tried this on Octave.

Forward summation:

Backward summation:

[Bruno Luong]

I guess one can obtain more digits with MATLAB too

For reference Star Strider's vpa result is 21.533035475927133

x = 1:1000;
y = (nthroot(5*x + 2, 5))./x;
S = sum(y);
fprintf('sum command = %2.15f\n', S)

sum command = 21.533035475927122

y = 0;       % Initialize y to store the sum later
for x = 1:1000
    y = y + (nthroot(5*x + 2, 5))/x;
end
fprintf('sum forward = %2.15f\n', y)

sum forward = 21.533035475927161

yb = 0;       % Initialize y to store the sum later
for x = 1000:-1:1
    yb = yb + (nthroot(5*x + 2, 5))/x;
end
fprintf('sum backward = %2.15f\n', yb)

sum backward = 21.533035475927125

yvpa = 21.533035475927133;

% LSB differences
(y-yvpa)/eps(yvpa)  % 8 LSB, not so good

ans = 8

(yb-yvpa)/eps(yvpa)  % -2 LSB, not bad

ans = -2

(S-yvpa)/eps(yvpa) % -3 LSB, OK

ans = -3

Answer 2

[Star Strider]

One approach —

syms x 
f = symsum(((5*x+2)^(1/5))/x, x, 1, 1E+3)

f = 

fvpa = vpa(f)

fvpa = 

21.53303547592713329236358151811

.

[Dyuman Joshi]

@Star Strider, the expression you have written is incorrect -

syms x 
((5*x+2)^1/5)/x

ans = 

You need to put the power inside parenthesis.

[Star Strider]

Fixed. I thought that the order of precedence would take care of that. (Too early here.)

[Dyuman Joshi]

^ takes precedence over /, so it's logical that would be executed first.